Sharp total variation bounds from spherical rearrangement
Henry Zweiman
September 16, 2026
Abstract
We study the sharp control of total variation by a higher derivative under homogeneous endpoint conditions. For every derivative order , we prove that the variation norm on is twice the point-evaluation norm, with coincident midpoint-symmetric extremizers. A Jacobi-polynomial representation identifies the constrained Green form with a positive mixture of spherical Poisson interactions, permitting rearrangement to a cap. We also prove the complete measure-endpoint identity and equality classification in every order, using extreme moment measures, unimodal splines and contact motion. A weighted zero-count argument establishes strict midpoint height maximization and unique symmetric height extremizers at every finite exponent.
For each fixed order, compactness and persistence of a single peak give the full variation statement on intervals around two and immediately above one. We then prove finite nondegenerate switching and simple residual zeros for every finite-exponent variation maximizer, derive its switching Hessian, and exclude all two-switch and all midpoint-symmetric three-switch maximizers. An exact rational counterexample rules out an unrestricted extension to positive derivative weights.
Finally, we settle derivative order three for every finite exponent : the variation and height extremizers coincide, are symmetric about the midpoint, and have exactly one interior critical point. A conserved residual quantity parametrizes the nodal intervals by their quadratic vertices. A positive action is strictly concave in the vertex depths and strictly convex after optimization in the mean parameter. Comparing its values for different nodal counts constructs a better competitor whenever there are at least four intervals; the earlier two-switch exclusion completes the proof. At the uniform endpoint, a duality and regularity argument extends the positive action and gives the complete equality classification with an explicit extremizer. Thus the order-three conjecture holds at every exponent, with the measure interpretation at one. In every order, an endpoint conserved quantity also proves saturation of the highest derivative, finitely many simple critical points and simple residual zeros. For each fixed order, the critical-point count is uniformly bounded over all exponents above one, including the uniform endpoint. The endpoint switching objective has an explicit Hessian supported on the simple residual zeros. A semialgebraic description of the resulting finite splines proves that the uniform-endpoint sharp variation constant is real algebraic in every order, with an algebraic maximizing spline. An atomic affine response, positive-weight approximation and separated-zero count extend the two-switch and midpoint-symmetric three-switch exclusions to the uniform endpoint, including a boundary sign count for the flat case in order four. The necessary response-location condition for nonsymmetric three-critical-point candidates extends to that endpoint as well. The full all-order variation conjecture remains open for the remaining cases in orders at least four.
1. The sharp inequality
All functions and measures in this paper are real. For , we use the homogeneous norm on . The endpoint conditions are
Let and denote the sharp constants in
The inequality follows from the variation of a function that vanishes at both endpoints. Nazarov and Shcheglova [NS, Conjecture 4.14] conjecture equality for all and , together with coincidence and midpoint symmetry of the extremizers. Their Remark 4.15 records the previously known case .
Theorem 1. For every integer ,
The nonzero extremizers are precisely the scalar multiples of the Riesz representer of evaluation at in . In particular, the maximizing sets in (2) coincide, and their elements are even about .
The point-evaluation formula used here is prior work. Kalyabin [K, Theorem 2] gives, after affine rescaling,
Our contribution is the upper bound for the variation norm and its equality classification, obtained through the rearrangement principle in Section 3. Best polynomial approximation already occurs in this embedding problem [GS]. The sharp to embedding of the function itself was proved by Hindov, Nitzan, Olsen and Rydhe [HNOR]; their theorem does not include the additional zero-mean constraint on a derivative that appears below.
2. A Jacobi representation of the clamped Green form
Fix , and set
Let be the orthonormal polynomials for , with positive leading coefficients. These are normalized shifted Jacobi polynomials with parameters ; in particular . Put
Lemma 2. The functions form an orthonormal basis of in its energy inner product. Moreover,
Proof. The left side of (7) has degree . If is a polynomial of degree less than , integration by parts times yields
Every boundary term vanishes because one of its two factors has fewer than derivatives and retains an endpoint zero. Orthogonality therefore shows that the left side of (7) is proportional to . Comparing leading coefficients proves (7). A further integration by parts gives
If is orthogonal to every , weak integration by parts gives for every . The span the polynomials, which are dense in , so . This proves completeness.
The mean functional has Riesz representer . Therefore
is obtained by removing the zeroth basis vector. For a bounded real load , Parseval gives the squared dual norm of on :
Equivalently, if is the clamped Green kernel and , then
Formula (11) is sufficient for the proof and avoids any pointwise convergence issue for a Green series.
3. Spherical rearrangement of the load
Set and let be normalized surface measure on . The coordinate has law . Thus
identifies with the zonal subspace of . Under this identification, is the normalized zonal harmonic of degree . This follows from the orthogonal decomposition of polynomials into spherical harmonics: a zonal polynomial of degree orthogonal to all lower zonal degrees has only its degree- component.
The spherical Poisson operator , , has eigenvalue on degree- harmonics and kernel
We use normalized surface measure, as in [ABR, (1.15)]. The eigenvalue follows by harmonic extension of a homogeneous harmonic polynomial. The beta integral gives
Tonelli's theorem applied to the nonnegative spectral terms in (11) therefore gives
The deletion of the constant harmonic is exactly the mean subtraction needed in (12).
Theorem 3. If is measurable and satisfies , then
If , equality holds only when agrees, up to null sets, with an interval abutting one endpoint and having that -measure.
Proof. Lift to the zonal subset using (13). For fixed , is positive, bounded and strictly decreasing in geodesic distance. Spherical rearrangement [BH, Theorem 2] maximizes the self-interaction of at fixed surface measure by a cap. Choose that cap about the negative coordinate axis. It corresponds to . The mean term in (16) is unchanged, and integration proves (17).
For clarity, the equality hypotheses in [BH] are satisfied with two functions and integrand . Its mixed finite difference is strictly positive; the kernel is positive and strictly decreasing; and the nontrivial indicators are nonconstant with finite interaction. Hence equality at fixed forces to be a cap up to rotation. If equality holds in (17), the nonnegative rearrangement deficit has zero integral in (16), so equality holds for almost every , and in particular for one . Thus is a cap.
Its axis must be the original coordinate axis up to reversal. Indeed, zonal invariance makes the centroid of parallel to that axis. A nontrivial cap has a nonzero centroid parallel to its own axis. Consequently is an endpoint interval, as claimed.
The same argument proves a broader load rearrangement statement. For a bounded real , let be its nonincreasing equimeasurable rearrangement relative to . Adding a constant makes nonnegative without changing . Applying [BH] in (16) gives
This is rearrangement of a load in a Green quadratic form. It does not assert that ordinary rearrangement preserves a higher derivative norm.
4. Proof of Theorem 1
Put . The map is an isometric bijection from to : integrating recovers , and the zero-mean condition supplies .
Duality gives
To justify the second equality, write a bounded with as the average of over uniform . Convexity of the squared Hilbert dual norm bounds its value by the supremum over sign functions. Constants are annihilated in (11), giving .
For an endpoint interval,
Theorem 3, (4), and (19) now prove (3). In particular, the maximizing interval is a half-interval.
Suppose that attains equality, and take , choosing either sign where . Both signs occur on sets of positive measure. Equality in the Hilbert dual bound and in Theorem 3 forces to be a half-interval sign function, up to its global sign and null sets. Its functional on is . Equality in Cauchy--Schwarz therefore makes a scalar multiple of .
Conversely, this representer maximizes point evaluation. Its variation is at least twice its maximum and at most the bound just proved, so it also maximizes variation. Formula (4) has its unique maximum at , and uniqueness of a Hilbert-space Riesz representer gives precisely the same classification for the point-norm extremizers. Finally, reflection preserves the energy inner product and fixes evaluation at , so uniqueness makes reflection invariant.
5. The measure endpoint
Let consist of real finite signed measures on such that
Their compactly supported potentials are
The moments make vanish off , and . For , is continuous and absolutely continuous, with integrable first derivative. Endpoint atoms are allowed in this relaxed formulation.
Theorem 4. For every ,
The maximizing measures for the two quantities coincide. The relaxed constants equal and , respectively. No symmetry assertion at is included.
Proof of the constant identity. Every extreme point of has norm one and exactly support points. To prove the support bound, suppose has norm one and at least disjoint Borel sets of positive -measure. A nonzero function constant on those sets can satisfy the equations and the extra equation . Then are distinct norm-one admissible measures for small . Such a is not extreme. A nonzero measure of norm less than one is excluded by scalar perturbation. Zero is the midpoint of opposite nonzero moment measures, which exist on any distinct nodes, and is not extreme either. A nonzero measure on at most points is excluded by the Vandermonde determinant.
On nodes , the moment nullspace is one-dimensional, with weights
Thus every extreme measure is a normalized signed multiple of . Its potential is a B-spline, up to scale and sign. More explicitly, the Genocchi--Hermite identity [dB, (52)] identifies
with the density of , where is uniform on the standard -simplex. After parallel sections of the simplex are identified by translation, their Minkowski interpolation is contained in the section at the interpolated coordinate. The classical Brunn--Minkowski inequality [GHW, (3), ] therefore makes their volumes to the power concave. The coarea factor relating section volume to is constant, so is concave and is unimodal. It is positive between its end knots, continuous, and zero at those knots. Every extreme potential therefore satisfies
Write . The set is weak-star compact. Moreover, is weak-star lower semicontinuous: it is the supremum of the continuous linear functionals , where and
Its sublevel set at is closed and convex and contains all extreme points by (26). Krein--Milman therefore gives the upper bound in (23). The elementary variation lower bound gives the reverse inequality. The height supremum is attained: on the bounded measure set, joint continuity of the kernels in (22) makes the potential map weak-star to uniform continuous. Consequently the variation supremum is attained as well.
Coincidence. Let maximize variation, choose , and let be the compact exposed face on which . Its extreme points are extreme in . For each such point ,
Thus agrees with the sign of wherever the latter is nonzero, and every maximizes height. Its end knots must be and : a potential supported on a shorter interval of length can be stretched to , keeping its derivative-measure norm one and multiplying its height by .
All have the same global sign, since opposite signs give opposite nonzero derivatives on a common interval near zero, contradicting their agreement with . Change that common sign to positive. Concavity of makes its maximum set a closed interval, with strictly positive derivative almost everywhere before it and strictly negative derivative almost everywhere after it. Two disjoint maximum intervals would again force contradictory signs for . The maximum intervals therefore intersect pairwise, and hence have a common point . All . Evaluation is weak-star continuous, so Krein--Milman applied to gives . Conversely, a height maximizer is a variation maximizer by the already proved constant identity.
Recovery of the ordinary constants. For an ordinary admissible , the measure satisfies (21). Conversely, mollify the zero extension of with a nonnegative mollifier supported in , obtaining . With , set
It is smooth, supported on , satisfies all endpoint conditions, and has . Its maximum and variation converge to those of , by uniform convergence of the potential mollification and convergence of its first derivative. This proves equality of the relaxed and ordinary sharp constants. The proof classifies coincidence of relaxed maximizers and does not assert attainment in the ordinary class.
6. An open interval of exponents
The Hilbert theorem also yields a neighborhood of exponents for each fixed derivative order. The interval is not asserted to be uniform in the order. In this section write for the clamped space defined by (1), with norm .
Theorem 5 (local continuation in the exponent). For every fixed integer , there exists such that, if , then
All nonzero extremizers in the two inequalities are the scalar multiples of the unique positive norm-one extremizer for evaluation at . They are symmetric about the midpoint and, after a choice of global sign, have strictly positive derivative on and strictly negative derivative on .
The radius is allowed to depend on and is not estimated here. The theorem does not reach all , give an interval uniform in , or settle the full all-exponent variation conjecture.
6.1. Dual representation and compactness of variation maximizers
Let denote the polynomials of degree at most . Integration from the left endpoint identifies isometrically with
Indeed ; the right endpoint conditions are equivalent to the displayed moments. For bounded , put
Fubini gives . If , the norm of this functional on equals
Here is a direct justification without a quotient-duality assumption. A minimizing polynomial exists by finite-dimensional coercivity and is unique for . Its residual satisfies
If , the function has norm one and belongs to (6.2). It realizes the bound in Hölder's inequality. The zero-residual case is immediate. This proves (6.4).
For every fixed , the variation supremum is attained. A norm-bounded sequence is bounded in by successive integration, and its derivatives through order are uniformly bounded and equicontinuous, using Hölder's inequality for the highest integral. A subsequence converges uniformly through order , and the highest derivatives have a weakly convergent subsequence in . The limit satisfies all endpoint conditions, has norm at most one, and retains the limiting variation. The supremum is positive, so a maximizer must have norm exactly one by scaling.
For a norm-one variation maximizer , choose , with any value in on the zero set. The functional in (6.4) is at most , since . On it equals . Thus its residual satisfies , and equality in Hölder gives
This formula gives more compactness than the original Sobolev bound.
Lemma 6 (compactness of maximizers). Suppose and is any sequence of norm-one variation maximizers in . Then, after a subsequence and a choice of global signs, converges in to the positive norm-one Hilbert maximizer .
Proof. Restrict to , so . The functions , with , are uniformly bounded and Lipschitz. For , this follows directly from ; for , differentiate (6.3) once and bound its remaining integral.
The minimizing polynomials satisfy
.
Their coefficients are uniformly bounded. To verify the needed uniform norm equivalence, a contrary sequence of polynomials can be normalized to have coefficient norm one and norm tending to zero. A coefficient-convergent subsequence would give a nonzero polynomial vanishing almost everywhere, a contradiction. Consequently both and their derivatives are uniformly bounded, and the residuals are uniformly bounded and Lipschitz.
The quantities are bounded above by (6.4) with the zero polynomial. They are bounded away from zero by testing the one fixed nonzero function , whose highest derivative is bounded. Formula (6.6) now makes uniformly bounded and equicontinuous. For example, signed power maps with exponent in on a fixed bounded interval have a common Hölder modulus with exponent , after enlarging the constant. Arzelà--Ascoli supplies a uniformly convergent subsequence of the highest derivatives. Integration and the left endpoint conditions give convergence in to some .
Uniform boundedness and imply
.
All right endpoint conditions pass to the limit. Also
.
Testing gives
, while the Hilbert inequality applied to the limit gives the reverse bound. Thus is a Hilbert maximizer. Theorem 1 identifies it with or , proving the lemma.
6.2. Persistence of a single peak
We give the details that weak or uniform convergence alone would miss. The Hilbert midpoint representer, up to a positive factor, can be written
To check this representation, expand the integrand. The resulting expression for is an even polynomial of degree at most , plus a multiple of . Thus (6.7) is polynomial of degree at most on each half-interval and has matching derivatives through order at the midpoint. Near , writing shows that is a positive constant times , to leading order. Therefore all endpoint derivatives below order vanish. The jump of the derivative of order is nonzero. The weak derivative is a scalar multiple of the midpoint delta. Testing against shows this scalar is positive because and . Uniqueness of the Riesz representer proves the claim up to positive normalization.
For , differentiation gives
Also . Hence
on , on , and . The endpoint expansion gives
and .
These properties imply that every sufficiently close perturbation satisfying the same clamped endpoint conditions has exactly one interior critical point, a strict maximum. Indeed, near zero its highest derivative remains positive and
The reflected argument gives negativity near one. On compact intervals away from the endpoints and midpoint the strict derivative signs persist. On a small interval about the midpoint, the second derivative stays negative, so there is precisely one zero of the first derivative, bracketed by its signs on either side.
Lemma 6 therefore implies that for all sufficiently close to two, every norm-one variation maximizer is, up to sign, positive and strictly unimodal. Such a maximizer has variation exactly twice its height. Since every admissible function has variation at least twice its height, this proves (6.1). It also proves coincidence of maximizing sets: a variation maximizer has height , and every height maximizer has variation between and .
The only conclusion still needed for the local theorem is midpoint symmetry. Single-peak geometry by itself does not supply it.
6.3. Smoothness of the point-evaluation profile near two
Let , and define
By the same moment argument as (6.4), the evaluation norm is . We need uniform continuity on the whole interval in , and twice continuous differentiability in near .
First, minimizing polynomials have coefficients bounded uniformly for in a compact subset of and , by the argument used above. On this bounded coefficient set the integrands in (6.10) vary continuously with , uniformly in . Minimizing over that compact set proves continuity of , including uniform convergence as . Strict convexity gives a unique minimizing polynomial; compactness and uniqueness prove continuous dependence of its coefficients.
The following elementary sublevel estimate handles exponents just below two. If is a polynomial of degree at most on an interval of length bounded above and below, and , then
The constant is uniform for intervals in the stated length range. To prove this, let the sublevel set have length . Divide its mass into sufficiently many equal portions and choose points from alternate portions. They can be chosen in the sublevel set with consecutive separations bounded below by a fixed multiple of : coordinate separation is at least the measure of the intervening portion. Lagrange interpolation at these points gives , with the interval-length factor absorbed into the constant. Rearrangement gives (6.11); approximation of the chosen points handles essential rather than attained sublevel membership.
Consequently, for , integration of distribution functions gives
For example, split the region into dyadic bands , apply (6.11) to each, and sum the convergent geometric series. The estimates also show uniform integrability of these inverse powers.
At , the minimizing residual in (6.10) is a nonzero polynomial on each of the two half-intervals. If it were identically zero on one half, its orthogonality to constants would fail on the other: the remaining residual would be a nonzero constant multiple of the nonnegative truncated power. Continuity of the coefficients therefore supplies a uniform positive lower bound for the sup norm of each polynomial piece on its own interval when and the polynomial coefficients are sufficiently close to this point. The intervals' lengths also stay bounded away from zero. Thus (6.11)--(6.12) apply uniformly to both pieces.
Write and . The normal equations are
Take a small open neighborhood with . The equations are continuously differentiable in there. Their derivatives in are obtained by multiplying the variation of by , and the derivative in has integrand , with value zero at . These assertions follow first away from a small sublevel set of , by ordinary differentiation. The uniform bound (6.12) makes the omitted derivatives uniformly integrable and arbitrarily small as the sublevel threshold tends to zero. The logarithmic derivative is bounded by a constant times a small positive power of near zero. This also proves continuity of the derivatives by splitting into the same two regions. The moving interface causes no additional term, since and its first -derivative are continuous there for .
At the base point, the coefficient derivative of (6.13) is the negative definite Gram matrix . The finite-dimensional implicit function theorem yields a continuously differentiable coefficient map . It is the minimizing polynomial map by strict convexity and continuity already established.
The envelope derivative and its second derivative are
where
and
.
For , the last expression is the indicator of , with its value at immaterial. The coefficient term in the first derivative vanishes by (6.13). The second derivative exists and is continuous by (6.12), boundedness of these kernels, and dominated convergence away from the moving interface. In particular no claim that all residual roots are simple is needed.
At , the credited point-evaluation formula (4) gives
It has a unique maximum at , with strictly negative second derivative there. By continuity of (6.14), it follows that is strictly concave on one fixed neighborhood of , for all close enough to two. Uniform convergence on forces every global maximum into that neighborhood. Reflection of the original evaluation problem gives . Strict concavity and this symmetry show that is the unique global maximizing point.
6.4. Completion of the proof
Shrink so that both the variation-maximizer argument and the evaluation-profile argument apply. We have established the constant identity and coincidence of maximizing sets. Every height maximizer must realize evaluation at its absolute maximum, and Section 6.3 forces that point to be . The norm-one positive evaluation extremizer is unique by equality in Hölder in (6.4), or by strict convexity of the unit ball of . Reflection preserves the norm and evaluation at , so uniqueness makes this function midpoint symmetric. Section 6.2 already gives its strict derivative signs and its unique peak. This proves Theorem 5.
6.5. The unresolved global step
Compactness away from the endpoints of the exponent range. The argument in Lemma 6 gives the following more general statement. If and are norm-one variation maximizers, a subsequence converges in to a norm-one variation maximizer at , and . To see this, keep in any fixed compact subinterval of ; the exponents then have a positive lower bound, so (6.6) still gives a common Hölder modulus. The limit has norm one at . For the lower bound on the limiting optimal value, test a maximizer at ; (6.6) makes its highest derivative bounded, so it is admissible at all nearby exponents and its varying norms converge. The upper bound follows by applying the inequality to the limit. This proves the statement without assuming uniqueness or symmetry at .
Thus a local continuation criterion at another exponent is available: if all its normalized variation maximizers are scalar-sign copies of one single-peak function with nonzero endpoint highest derivatives and a nondegenerate peak, and its point-evaluation profile has a unique midpoint maximum with a continuously persisting negative second derivative, the same argument proves the conjecture in a neighborhood of that exponent. These hypotheses must be established there; they cannot be imported from the case .
The strongest compactness assertion here is convergence of variation maximizers, established from their dual optimality equation. It is not asserted for arbitrary normalized Sobolev sequences. This distinction is necessary to preserve the derivative signs near the clamped endpoints.
The point-profile calculation uses finite-dimensional best approximation only after proving uniform control of possible multiple zeros in its polynomial residual. Omitting (6.11)--(6.12) would leave the two-sided neighborhood in unjustified, since creates negative powers at residual zeros.
This argument does not prove that the set of valid exponents is closed. At a hypothetical first exponent where the result ceases to hold, compactness of the old single-peak branch does not exclude a different multiple-peak global maximizer with the same value. Theorem 10 below resolves the separate point-evaluation symmetry question at every finite exponent. The possible coexistence of multiple-peak variation maximizers remains a substantive global obstacle.
The remaining task is to exclude such competing branches or to find a counterexample. Theorem 5 supplies a local result, not the full conjecture. No numerical experiment is used in its proof.
7. Complete symmetry at the fourth-order measure endpoint
We first settle the endpoint symmetry problem explicitly in order four by a finite alternation argument. The theorem below concerns the measure class (21); its constants also apply to the ordinary clamped Sobolev space by Theorem 4.
Theorem 8 (fourth-order endpoint). One has
The norm-one relaxed extremizers for height and variation are exactly and , where
Their potentials are even about . No nonzero extremizer exists in the ordinary class. Thus the full constant, coincidence and symmetry assertions in [NS, Conjecture 4.14] hold for , , with the endpoint understood in the measure sense.
The all-order symmetry assertion is proved separately in Section 8. The point estimates for orders two and three are already included in [NS, Theorem 4.5]; those cases are not new here.
7.1. Extreme measures and a dual cubic
Let be the maximal relaxed height. It is positive and attained by Section 5. Choose a maximizing point and sign. The corresponding evaluation functional exposes a nonempty compact face of . An extreme point of that face is extreme in and still maximizes height. By the moment-circuit argument in Theorem 4, after choosing its positive potential it has nodes
The end nodes are and because stretching a shorter support, with derivative-measure norm fixed, multiplies height by the inverse cube of its length. The signs of the weights in (7.3) are .
Write for this positive potential and let maximize it. Before it is a positive multiple of ; after it is a positive multiple of . Its derivatives at and have strict opposite signs. Hence . Reflecting if necessary, assume
Put . There is a polynomial of degree at most three such that
Here are the duality details. The dual of is the space of finite measures annihilating , with total-variation norm. Hahn-Banach and the Riesz representation theorem therefore identify the distance of to with the norm of its evaluation functional on . That norm equals , because the chosen measure attains the global height. A best polynomial exists by finite-dimensional coercivity. Equality in forces the five values in (7.5). Since is continuously differentiable, its derivative vanishes at .
Let , so that on . The cubic has a maximum at and a minimum at . Consequently, for some ,
Equality of its values at and gives, on putting ,
Indeed .
7.2. The remaining algebraic constraints
On the left of , . The equations and yield
Set . Dividing the two equations, and using , gives
For clarity, . Its positivity gives , and its upper bound gives .
The potential itself is stationary at . Only the nodes contribute to its first derivative under (7.4), so reads
Combining this with (7.8) gives . Also implies
The integral equals by (7.7). All divided quantities are positive.
Substitute (7.7) and (7.9) into (7.10)-(7.11). The resulting two quadratic equations in are
where
One can check the second equation without elimination software: after substitution, the difference between the two sides of (7.10), multiplied by , is
The first equation is . These identities justify (7.12) with no loss from a vanishing factor.
Two quadratics in (7.12) with a common root must satisfy
where direct expansion gives
To verify the first equality directly, put , and . Subtracting multiples of the two equations gives and . Since , these imply . The factorization in the second equality is an ordinary polynomial identity.
There is a short sign check for its last factor. For ,
Indeed the four positive terms are at most . Since , equation (7.15) forces . Both equations in (7.12) then reduce to . The root makes and is inadmissible. The other gives
Thus every globally maximizing extreme measure has precisely these nodes. This conclusion concerns maximizers directly; it does not require a knot-reflection inequality for every configuration.
7.3. Constants and all equality cases
The symmetric dual cubic, after normalization, is the centered cubic Chebyshev spline already described in [dB03, p. 74]. The new issue in this section is unrestricted global optimality and its equality classification. The weights in (7.3), normalized to total variation one at (7.18), are exactly (7.2). Their potential is
It is symmetric and unimodal by Section 5. Substitution at gives . The extreme maximizing measure exists, and (7.18) identifies it, so this value is the unrestricted sharp height, not merely the maximum within symmetric configurations. Theorem 4 then gives the variation constant .
For an arbitrary norm-one height maximizer, expose a face by evaluation at one of its maximum-height points, with the appropriate sign. Every extreme point of that face is a height-maximizing extreme measure and therefore equals or . Only one of these two signs can belong to the chosen face, because the exposed value is positive. Krein-Milman makes the face a singleton. This proves the stated equality classification, and Theorem 4 transfers it to variation. An ordinary Sobolev extremizer would have an absolutely continuous fourth derivative, whereas (7.2) is atomic. Hence ordinary attainment is impossible. The approximation argument in Theorem 4 still recovers both sharp constants.
8. Symmetry in every order at the measure endpoint
The fourth-order calculation can be replaced by an all-order comparison. The key is to follow the contacts of the best uniform polynomial approximation as the evaluation point moves. Their velocities are strictly less than one. The monotonicity of B-spline peaks then compares the evaluation point with the peak of its own optimizing potential.
Theorem 9 (all-order measure endpoint). For every , the point-evaluation norm
is strictly increasing on and strictly decreasing on . There is a unique measure of total variation one attaining positive evaluation at . Its potential is positive in , symmetric about , and has exactly one peak. The normalized relaxed height and variation extremizers are precisely and . In particular, all assertions of [NS, Conjecture 4.14] hold at for every order, with the endpoint interpreted in the measure sense. The sharp constants in the ordinary clamped Sobolev class agree with the relaxed constants, but no nonzero ordinary extremizer exists.
The full conjecture at the other exponents is not asserted here. The B-spline peak monotonicity used below is credited to Foucart [F]; the endpoint dependence needed here is supplied explicitly.
8.1. A zero count and regular contacts
We first record an elementary zero-count fact. Suppose is polynomial of degree at most on each side of a single knot and belongs to , with neither polynomial piece identically zero. Then the number of isolated zeros, counted with polynomial multiplicity away from and the order of the first nonzero common derivative at , is at most . The only case where no common derivative through order is nonzero at a knot zero is
with ; this case has just one distinct zero.
Here is a proof of the count, including missing coefficients. Write the common Taylor coefficients through degree as and the two top coefficients as . If the first nonzero common coefficient has index , factor . Descartes' rule of signs bounds the zeros on the right and left by the sign variations in the remaining coefficient list and the list with signs alternating according to degree. For successive nonzero common coefficients whose degrees differ by , the sum of the two variation contributions is at most : it is one for odd , and zero or two for even . Their total is at most . The two independently chosen top coefficients add at most two. Adding the knot multiplicity gives . If all common coefficients vanish, the displayed monomial case applies. Endpoint zeros of the interval under consideration are among the roots counted by this argument.
Fix , put , and write
where is a best uniform polynomial of degree at most . The moment duality in Section 7 applies in every order. A maximizing evaluation face has an extreme measure supported on distinct nodes, and the signed weights force alternating contacts of .
Neither piece of can be constant: smooth matching at would make the other a constant plus a multiple of , yielding at most two alternating contacts. Thus the contact set is finite. Every interior contact is a strict local extremum. Applying the zero count to with shows that it has at most zeros counting multiplicity. Since there are already at least interior contacts, the contact set is exactly
All interior critical points are simple, and the endpoint derivatives are nonzero. For , a contact cannot equal : the two linear pieces of would otherwise have their common zero at , and a second critical point would force one piece to vanish identically. For , a multiple critical point at also violates the same zero count. Consequently, in every case,
Every measure maximizing positive evaluation at is supported on this contact set, by equality in the dual norm inequality. Vandermonde rank gives a one-dimensional nullspace for the moment conditions on nodes. Positive evaluation and total variation one fix its sign and magnitude. Denote this unique measure by . Its potential is the positive normalized B-spline on these nodes.
The contacts, polynomial coefficients and are functions of . Indeed, differentiate the contact-value equations and critical-point equations. The contact-value block in the polynomial coefficients and is invertible: a nonzero polynomial of degree at most cannot take nonzero alternating values at ordered points. The remaining diagonal block consists of the nonzero numbers . The implicit function theorem applies. Its equations are even when equals a contact for ; the case was excluded above. To identify this local branch with the actual optimizer, use uniqueness and compactness: best polynomial coefficients are locally bounded and their subsequential limits are best approximants; the norming measures converge to the unique norming measure, whose nonzero atoms prevent contact coalescence.
8.2. Every contact moves more slowly than evaluation
A dot denotes differentiation with respect to ; a prime denotes differentiation with respect to . Put and define
The identity shows that is again a degree-at-most- spline with one simple knot: it is a polynomial minus . Differentiating (8.2) and using at interior contacts gives
Neither piece of vanishes identically, since its endpoint values are nonzero. The zero count gives at most zeros counting multiplicity. The prescribed interior zeros must therefore all be simple and there can be no additional zero. To see the parity step explicitly, the endpoint signs in (8.5) require the parity of the number of crossings to be . An additional simple zero or one double prescribed zero would reverse this parity; restoring it costs another zero and exceeds the bound . A prescribed zero of multiplicity at least three already exceeds that bound. At a knot zero where the first common coefficients vanish, both pieces would be monomials and there would be only one distinct zero, contrary to the prescribed zeros. Thus this exception does not interfere with the parity argument.
It follows that has sign , the same as . Differentiating the critical-point equation yields
Hence
No lower bound on these velocities is needed.
8.3. The dependence of a B-spline peak on all its knots
Let be the conventional positive B-spline of degree on simple knots and let be its unique peak. Foucart [F, Proposition 6] proves
The peak is nondegenerate and depends differentiably on the knots. We need the same inequality for the two support endpoints. We prove this extension using the same divided-difference identities, keeping its scope separate from the cited statement.
By affine scaling take the support to be , and let be the degree- B-spline obtained by repeating the endpoint . The knot-insertion identity in [F, Lemma 3 and Formulae 5] gives
At , differentiating once and eliminating yields
For the left endpoint , integrating (8.8) with gives
For the right endpoint, the reflected formula gives the same strict inequality. Thus . Differentiation of the divided difference defining with respect to an endpoint gives
The derivative of the support-length normalization contributes a multiple of , so it contributes nothing to (8.11). Since , implicit differentiation of proves (8.7) for the endpoints too.
Translation equivariance now gives
8.4. Strict midpoint comparison and equality
Let be the peak of . The normalizing factor of this potential does not affect its peak. Equations (8.6) and (8.12), together with , show that
Reflection and uniqueness of the positive-evaluation norming measure at give . Therefore is strictly decreasing and
Finally, uniqueness and weak-star continuity of , together with joint continuity of for , give the envelope identity
It follows directly by testing the optimizer at each of two neighboring points in the other evaluation problem and taking the two difference quotients. The B-spline potential is strictly increasing before its unique peak and strictly decreasing after it. Equation (8.14) therefore proves the asserted strict monotonicity of .
Every global height maximizer must consequently attain its absolute height at . The unique norming measure there, up to the sign of evaluation, is . Its potential is symmetric by reflection and positive by the moment-circuit formula. Theorem 4 transfers the complete classification to variation. The ordinary constants agree by the recovery argument there; an ordinary extremizer would have an absolutely continuous highest derivative, whereas the unique relaxed one is a nonzero atomic circuit. This rules out ordinary attainment.
For , the direct formula and its unique three-point norming measure give the same conclusions. This completes the proof.
9. Point symmetry at every finite exponent
The preceding cancellation also has a finite-exponent form. Instead of following finitely many contacts, we integrate an auxiliary spline against a positive weight and count its zeros. Throughout this section is the dual exponent; it is not the target exponent in [NS, Conjecture 4.3].
Theorem 10 (finite-exponent point symmetry). Let and . The point-evaluation norm
is strictly increasing on and strictly decreasing on . The nonzero extremizers for the height inequality in (2) are precisely the scalar multiples of the unique positive norm-one midpoint normer. That normer is symmetric, positive in the interior, and has a unique nondegenerate maximum.
The order-two statement and order-three midpoint symmetry are prior results [W, Theorem 1.1 and Lemma 1.3], recorded also in [NS, Theorem 4.5]. Watanabe et al. explicitly note in [W, Remark 1.4] that their symmetrization proof does not extend to orders at least four. The theorem above gives the all-order comparison. Together with Theorem 9 and the credited theorem [NS, Theorem 4.12], it establishes the , target- case of [NS, Conjecture 4.3] for every and . It does not establish the remaining variation identity in Conjecture 4.14.
9.1. Simple residual zeros and a single peak
We prove the theorem for and write . As in Section 6, let be the unique best polynomial approximant of degree at most to . Set
The normal equations (6.5) make clamped at both endpoints and give
Here is the unique positive-evaluation normer. This follows from the dual formula and equality in Hölder's inequality, as in Section 6; the best-approximation formula itself is credited to [GS, Theorem 2 and Corollary 1].
The residual has at least sign changes. Otherwise the polynomial having its sign-change points as roots, with a suitable sign, has degree at most and gives a strictly positive integral against , contradicting (6.5). Neither polynomial piece of vanishes identically: smooth matching would make the other a nonzero multiple of , of constant sign on its side, contradicting the constant moment. The one-knot zero count of Section 8.1 bounds the total number of zeros, with multiplicity, by . Its exceptional two-monomial case has only one distinct zero and is excluded. Thus has exactly simple interior zeros, and its endpoint values are nonzero.
Repeated Rolle counting shows that has no interior zero and has exactly one simple interior zero. Indeed, has endpoint zeros of multiplicity . An additional interior zero would force at least distinct zeros of after differentiations. Applied to , whose endpoint multiplicities are , the same count excludes either two distinct interior zeros or a degenerate zero. Consequently in and its unique peak satisfies
The last two signs follow from the endpoint Taylor expansions.
9.2. Differentiation through the residual zeros
The preceding simple-root result removes the restriction to exponents near two in the regularity argument. Fix and . The residual is in , since . Its roots are transverse even if one equals the kernel knot. In a small neighborhood of the minimizing polynomial coefficients and evaluation point, there remain exactly simple roots with inverse slopes uniformly bounded. For , this gives the locally uniform estimate
It follows by changing variables from to in disjoint monotone neighborhoods of the roots; outside them is bounded below. For the weight has no singularity. Differentiating the normal equations in and the polynomial coefficients is therefore justified by the same sublevel splitting as in Section 6.3. The coefficient Jacobian is
a negative definite Gram matrix. The implicit function theorem and uniqueness give a minimizing coefficient map for every fixed .
The same estimate shows that is continuously differentiable as a map into . Its derivative is . As a function of , is absolutely continuous, with derivative . Integration gives continuous differentiability of into , as well as . The nondegenerate peak in (9.4) therefore depends continuously differentiably on . In what follows, a dot denotes the derivative and a prime the derivative.
9.3. A weighted comparison for peak motion
Fix and abbreviate , , , and . Define
The identity makes a polynomial of degree at most . Almost everywhere,
The function is a degree-at-most- spline with one simple knot. If both pieces are nonzero, Section 8.1 bounds its isolated zeros by , with the two-monomial exception having only one distinct zero. The weight in (9.8) is positive almost everywhere and integrable.
The function satisfies
In particular, is positive near zero and negative near one. We claim
Here is a zero-count proof that includes the possible singularities of the weight. If , the opposite endpoint signs force another root on each side of . If and , there is a double root and another root. If , there is a root of multiplicity at least three. These derivatives exist because . Thus failure of (9.10) supplies interior zeros of total multiplicity at least three, besides the two endpoint zeros of multiplicity .
Apply ordinary Rolle's theorem times. It yields at least distinct zeros of : the initial count is at least , and all retained multiplicities are exhausted by this stage. On each of the intervening disjoint intervals, the integral of is zero. If both pieces of are nonzero, it is not identically zero on any interval. Positivity of the weight then forces a sign change of on each interval, contradicting its zero bound.
If one piece of vanishes identically, smooth matching makes the other a constant multiple of . Then is the nonzero constant or on the zero side and is strictly monotone on the other side. It has at most one zero, again a contradiction. If both pieces vanish, it is constant nonzero and has none. This proves (9.10).
Differentiate . Equations (9.7) and (9.10) give
Since , we obtain for every .
9.4. Strict midpoint maximization and the remaining gap
Reflection and uniqueness give . Thus is strictly decreasing, positive before the midpoint and negative after it. The envelope identity is
The strict single-peak signs of prove the stated monotonicity. Any height extremizer must attain its absolute maximum at ; uniqueness of the midpoint normer gives all equality cases, and reflection gives its symmetry. This proves Theorem 10.
Two useful boundary identities follow from the same cancellation. With , integration by parts and the normal equations give
For the second identity use plus a polynomial of degree at most . The theorem therefore also proves the strict ordering of the endpoint residual magnitudes on either side of the midpoint.
This argument classifies height extremizers for every finite exponent. It does not imply that every variation maximizer has a single peak. Its dual residual in (6.6) comes from , which may have many changes of polynomial piece when . The one-knot bound in (9.8) is then unavailable. This is the remaining obstruction to the full variation conjecture.
10. Variation continuation from the measure endpoint
The unique endpoint measure also controls arbitrary variation maximizers as the input exponent decreases to one. Two issues require care: ordinary highest derivatives develop concentration, and in order three weak measure convergence does not directly give uniform second-derivative convergence.
Theorem 11 (continuation above one). For every there exists such that, for ,
The nonzero extremizers in the two inequalities are exactly the scalar multiples of the unique positive norm-one midpoint normer. They are symmetric and strictly unimodal up to sign. The interval width is not estimated, and overlap with the interval in Theorem 5 is not asserted. The case is already known for all by [NS, Remark 4.15].
10.1. Compactness of arbitrary endpoint sequences
Fix . Let tend to one and let be any norm-one variation maximizers, whose existence was proved in Section 6. Write and . Since the interval has length one,
A subsequence converges weak-star to some . For the potential kernel and its first derivative are jointly continuous on the square. Their integrals over the unit measure ball are uniformly equicontinuous. Pointwise weak-star convergence and this equicontinuity therefore give in .
The constants satisfy . Conversely, given , the recovery argument in Theorem 4 supplies one nonzero smooth clamped function with
Its highest-derivative norm tends to its norm. Testing this fixed function and then letting decrease to zero proves . Thus maximizes endpoint variation. Theorems 4 and 9 identify with or . After choosing the common sign along the subsequence, we have
The last limit follows from lower semicontinuity and the upper bound one.
We also need convergence of the absolute measures:
Indeed, any weak-star limit of a subsequence of dominates . To see this, for nonnegative continuous and continuous with , pass to the limit in
and use the dual description of variation. Both and have mass one by (10.2), so they agree. Every subsequence has this same limit, proving (10.3).
10.2. The limiting dual residual fixes the endpoint signs
Put and let , with any choice in where . The strict signs of imply pointwise convergence except at to the midpoint sign function , which is positive before and negative after it. Dominated convergence gives in . Hence converges uniformly to
Let minimize over polynomials of degree at most , where . The bound
and finite-dimensional norm equivalence give bounded polynomial coefficients. Take any subsequential coefficient limit . For each fixed finite , each polynomial in the same space, and all sufficiently large ,
First let tend to infinity, using uniform convergence, and then let tend to infinity. Thus is a best uniform approximant to . Uniqueness of best uniform polynomial approximation and (10.4) identify
Here is the truncated-power residual in Section 8; every subsequence has the same limit. The second identity is the dual optimality equation (6.6).
The endpoint residual values in (8.2) are nonzero, positive at zero and of sign at one. Uniform convergence therefore fixes these signs for on small endpoint neighborhoods, independently of once it is large. Equation (10.6) fixes the same strict signs for , although its magnitude need not remain bounded. Integration from the clamped endpoints gives near zero and near one. On compact sets separated from the endpoints and midpoint, uniform convergence of to preserves the required signs as well.
10.3. Curvature at the midpoint and completion
For , the twice-differentiated potential kernel is jointly continuous. The compactness argument above then gives uniformly. The B-spline peak is nondegenerate, as used in Section 8 and [F, Proposition 6]. Thus on a fixed central interval for all sufficiently large .
For , the unique symmetric measure has four distinct knots , with . The midpoint lies in a knot-free interval, where is a strictly negative constant. Choose a closed central subinterval entirely inside this gap. Equation (10.3) and imply . Also : continuous approximation of the interval indicator, together with (10.3), justifies this passage because its relative boundary has zero limiting variation mass. The endpoint atom at zero is included on both sides. Since , we obtain
Hence the middle curvature is strictly negative in order three too.
Combine this curvature with the derivative signs already proved near the endpoints and on the intervening compact intervals. Each has exactly one critical point for large , with positive derivative before it and negative derivative after it. Because the original sequence consisted of arbitrary variation maximizers, a contradiction-sequence argument supplies for which every normalized maximizer has this geometry, up to sign, whenever .
Such a maximizer has variation twice its height, so . The reverse bound always holds. A variation maximizer is therefore a height maximizer, and a height maximizer is a variation maximizer. Theorem 10 gives the midpoint symmetry and unique equality class. This proves Theorem 11.
11. Finite switches and the stability matrix
The remaining variation problem can be reduced to finite, regular switching configurations without assuming this regularity in advance.
Theorem 12 (regularity of variation maximizers). Fix and , and put . Every norm-one variation maximizer has finitely many interior critical points, all nondegenerate. If their number is , then and its dual residual has exactly simple interior zeros. Moreover,
The number is uniformly bounded when ranges over a compact subinterval of , with fixed. No explicit bound or assertion is made.
11.1. Endpoint identities without a switching assumption
Write , , and . By Section 6,
In particular , is continuous and . For some constant , differentiation of the load gives
The chain rule for is valid regardless of its zero set. It gives
Integrate by parts times. All boundary terms vanish by the clamped conditions. In the second identity use
and . The result is
The absolutely continuous function has derivative . Therefore
Equation (11.1) follows from (11.2), since . In particular is bounded away from zero on endpoint neighborhoods. No derivative of the sign function was taken.
11.2. A degenerate critical point is unstable
Suppose at an interior point. By (11.1), choose open intervals to its left and to its right on which has a positive lower bound. There exists such that
Indeed, if the last functional vanished on the common kernel of the moment functionals, finite-dimensional linear algebra would express it as their linear combination. Testing smooth functions supported in the two intervals would make its kernel a polynomial of degree at most there. That polynomial vanishes on and hence identically, but the kernel is nonzero on , a contradiction.
Let . The moment conditions make it a smooth clamped function, and . Since is supported where is bounded below, the map is twice continuously differentiable near zero, including when . Applying the sharp variation inequality to and and averaging gives
The scalar identity
identifies the left side with for . But and is bounded below near . For every fixed , integration over gives a lower bound for all sufficiently small , where is independent of . Choosing arbitrarily large contradicts (11.7). Thus at every interior zero of .
All these zeros are isolated. The clamped Taylor expansions and nonzero endpoint values of fix a nonzero sign of near each endpoint. An infinite set of zeros in the remaining compact interval would have an accumulation point, necessarily with . Hence there are finitely many zeros. Each is a sign switch, and at least one occurs because and is nonzero.
11.3. The residual zeros are also simple
Let the critical points be . The moment equations force at least sign changes of . Repeated mean-value arguments then force to take both positive and negative values on sets of positive measure. Equation (11.3), with taking only the values almost everywhere, implies .
Consequently has degree exactly on each of the knot intervals. Its derivative is continuous piecewise affine, with a nonzero slope on each interval, and has at most distinct zeros. The interior zeros of , together with its endpoint zeros of exact multiplicity , force at least distinct zeros of after applications of Rolle's theorem. These are also zeros of .
Conversely, applying Rolle times to bounds its distinct zeros by . A multiple root would add another zero in this count and force at least distinct zeros of , which is impossible. This argument uses only a double root and continuity of , including at a switching knot when . Thus the residual has exactly simple interior zeros.
For the uniform bound, suppose stays in a compact subinterval of and the corresponding maximizing functions have unbounded critical-point counts. After a subsequence , the compactness argument in Section 6.5 gives convergence in to a normalized maximizer at . Its endpoint highest derivatives are nonzero and all its interior critical points are nondegenerate. Hence the endpoint signs persist, each interior critical point has exactly one nearby critical point for large , and the intervening compact sets contain none. The critical-point counts are eventually equal to that of the limit, a contradiction. This completes the proof of Theorem 12.
11.4. A justified second-variation condition
Let be the sign of just before , so the load jumps there by . Nearby ordered switching points define nearby sign loads. Modulo a polynomial of degree at most , their dual load is
Put . At the maximizing configuration set
where is orthogonal projection onto in the weighted inner product . The weight is positive almost everywhere and integrable. By Theorem 12 all residual roots are simple, so the transverse-root estimates of Section 9 justify continuously differentiating the normal equations in the switching parameters. Their coefficient Gram matrix is positive definite. The same sublevel argument and the envelope formula make twice continuously differentiable near this configuration.
With the unnormalized potential , differentiation gives
For , the second derivative of a moving kernel is an indicator. Its contribution is still justified by dominated convergence against the bounded signed residual power; no boundary term is introduced. The weighted term is controlled by the integrable transverse-root estimates.
Define and . Every nearby sign load has dual norm at most , so the Hessian in (11.9) must be negative semidefinite. Equivalently,
where denotes comparison of quadratic forms. This is a necessary condition for every finite-exponent variation maximizer. Theorem 12 justifies its finite, nondegenerate setting; Theorem 13 below excludes . It remains to exclude every configuration with satisfying it, or prove that such a configuration has a smaller value than the midpoint load. No such general exclusion is asserted here.
12. Two switches are unstable
The translation direction in the Hessian has a stronger consequence than the finite Hilbert diagnostic: it excludes a two-switch maximizer at every finite exponent, without a symmetry assumption.
Theorem 13 (two-switch exclusion). Let and . A nonzero global variation maximizer cannot have exactly two interior critical points. Consequently its critical-point count is either one or at least three.
12.1. A polynomial cancellation under translation
Suppose otherwise, and write the critical points as . Choose the sign so that near zero. Put , , and , so is a positive multiple of . By Theorem 12, the sign load is on the three successive intervals, all residual roots are simple, and
The signs follow from the residual roots, and equality of the endpoint magnitudes follows from (11.5).
Translate both switches to . Modulo , the load is . Let be its best residual and . A dot denotes differentiation in at zero. Since differentiation in cancels differentiation in the integration variable for each translated kernel,
Define
The simple-root estimates used in Section 11 justify differentiation of the signed residual power in , even when . They also give . Thus , and (12.3) is an almost-everywhere identity with a positive almost-everywhere integrable weight.
The moment equations make every clamped. Their differentiated equations make clamped as well. Therefore
12.2. The auxiliary function has a single peak
Put . Its endpoint zeros have exact multiplicity . Equations (12.1) and (12.4) show that just after zero and just before one. We claim that has exactly one interior zero.
Here is the required weighted Rolle count. If selected interior zeros of have total multiplicity , counting a zero with twice, repeated Rolle arguments through derivatives give at least distinct zeros of . Between consecutive zeros the integral of vanishes. Positivity of the weight forces a root of the polynomial in every such open interval. The polynomial is nonzero: otherwise would be a polynomial of degree at most with endpoint zeros of multiplicity at both ends, contradicting . Hence
This reasoning uses an integral between zeros of the absolutely continuous function ; it does not differentiate the weight or assign a finite value to it at a residual zero.
There must be a sign-changing zero of because its endpoint signs are opposite. Two distinct sign-changing zeros would require a third to give these endpoint signs. Any additional tangency contributes multiplicity at least two and, together with the sign-changing zero, also violates (12.5). An infinite zero set is excluded by selecting three distinct zeros. Thus has exactly one interior zero , and
In particular is strictly unimodal and positive in the interior.
12.3. Orthogonality gives a positive second variation
Since , the moment equations give , and (12.3) gives . Integrating by parts times, the only possible boundary term is
by (11.5). Using (11.3) and , we obtain
Strict unimodality and now imply . Let . The envelope identity near zero and its derivative at zero give
Every nearby translated sign load has dual norm at most the sharp variation norm attained at . The strict positivity in (12.8) contradicts this local maximum, proving Theorem 13.
For three or more switches, the same translation cancellation gives an alternating sum in (12.7). It no longer equates the values at just two points, so the sign conclusion in (12.8) does not follow. No exclusion of all such configurations is asserted.
13. Affine motion and the three-switch obstruction
The dilation direction complements common translation. It excludes a midpoint-symmetric three-switch maximizer and imposes a further necessary geometric condition on every remaining three-switch candidate.
Theorem 14 (three-switch dilation test). Let and . A global variation maximizer that is symmetric about the midpoint cannot have exactly three interior critical points. More generally, if a global maximizer has exactly three critical points , the unique minimum of the auxiliary function for dilation about , defined below, must lie outside the closed interval .
13.1. The general affine cancellation
Use the notation of Section 11, put , , and set , . Fix an affine velocity and move the switches along . A dot denotes differentiation at zero. The kernel identity
shows that
Define
As in Section 12, simple residual roots justify parameter and spatial differentiation of the signed residual power in . Thus , its endpoint derivatives through order vanish, and
Polynomial orthogonality gives . After integrations by parts, the surviving boundary term is . Hence
Let be the Hessian in (11.9), and let . Since , differentiating the envelope formula gives
Indeed , and the last term vanishes at the switches. The switch path is affine in , so there is no acceleration term.
13.2. Dilation about the middle switch
Suppose there are exactly three switches , and choose the sign so that . Their signs are . The simple residual zeros and equal endpoint magnitudes give . Take . Equation (13.3) implies that near zero and near one, with endpoint zeros of exact multiplicity .
The weighted Rolle argument in (12.5) applies verbatim to (13.2): has at most two interior zeros counted with the same multiplicity convention. Its opposite endpoint signs exclude a second zero or tangency. Consequently there is a unique interior point such that
The auxiliary function is negative in the interior and has its unique minimum at . If , the middle switch has zero velocity and (13.5) becomes
Both summands are positive when . At either endpoint of , one vanishes and the other remains positive. This contradicts maximality. Any remaining three-switch global maximizer therefore requires or .
13.3. Midpoint symmetry excludes three switches
If , its three simple critical points are , with . Dilation about preserves this symmetry of the switch set. Every moved sign load is antisymmetric. Reflection and uniqueness of best polynomial approximation give ; the clamped potentials are consequently symmetric. Thus , and are symmetric as well.
The unique minimum in (13.6) is therefore at , between the outer switches. In particular,
contradicting maximality and completing the proof of Theorem 14.
The theorem does not establish symmetry of arbitrary variation maximizers. For a nonsymmetric three-switch candidate, it remains to exclude the response-minimum locations and , or find another positive Hessian direction. Configurations with at least four switches also remain open.
14. A limitation of weighted Hilbert reduction
The constant derivative weight in Theorem 1 cannot be replaced by an arbitrary positive weight. This matters for a possible reduction of the remaining finite-exponent problem. For a positive bounded weight bounded away from zero, put
where the suprema run over .
Theorem 15 (failure of the arbitrary-weight extension). There exists a positive bounded piecewise-constant weight , bounded away from zero, such that
Such weights also exist in . In particular, the arbitrary-weight version of and the assertion that every weighted variation maximizer is single-peaked are false. This is a weighted counterexample; it does not disprove [NS, Conjecture 4.14].
14.1. Dual formulas and an explicit weight
Write and . Let be the orthogonal projection onto in this inner product. The endpoint conditions on are equivalent to the three moments , . Weighted Hilbert duality therefore gives
For a sign load with switches and successive signs , integration of the load against the derivative kernel gives modulo . Hence
These formulas have no numerical approximation. For example, if and with zero initial derivatives, the moment equations give the right endpoint conditions, , and . Thus .
Take , and
with
In particular , so satisfies all the hypotheses in (14.1).
14.2. A rational certificate on the entire evaluation interval
Here the explicit counterexample is proved by finite exact polynomial arithmetic, not by testing finitely many evaluation points. We give the formulas and a positive Bernstein certificate covering every .
Let , , and let . Then
The matrix is positive definite. All its entries and the required polynomial coefficients are rational. For full reproducibility, define
On a constant-weight interval lying to the left of , its contribution to is . On the interval containing , replace by . Intervals to the right contribute zero. Use in (14.7). On each of the thirteen constant-weight intervals, is thus a polynomial of degree at most ten.
For , write on a subinterval
A polynomial of lower degree is padded by zero coefficients. The following table gives rational lower bounds: on each row every is at least the displayed integer divided by . The only added subdivision of a constant-weight interval is at .
Lower bound numerator
0
29
899999
29
31
899999
31
59
899579
59
61
899452
61
129
863685
129
131
860777
131
490
96932
490
849
88748
849
851
339729
851
869
337888
869
871
480234
871
969
511776
969
971
899999
971
1000
899999
The Bernstein basis is nonnegative and sums to one on . Every entry is positive, so the table proves on the entire closed interval. Exact integration and the same moment projection give
Equations (14.3)--(14.4) now prove (14.2). The accompanying certificate generator computes the piecewise polynomials and their exact Bernstein coefficients. A separate interpolation checker evaluates the integrals at eleven rational points per interval, reconstructs the degree-at-most-ten polynomial, and verifies every printed bound and the exact fraction (14.10). Interpolation is exact here because the degree bound is already proved. Both programs use rational arithmetic for all decisive comparisons. The initial floating-point search is not part of this proof.
14.3. Smooth weights and the limit of freezing
Extend by outside and convolve with nonnegative smooth approximate identities. The resulting are smooth, obey the same positive lower and upper bounds, and converge to in . The moment matrices converge to , so their inverses converge. The kernels are uniformly bounded for ; therefore all moments in (14.7), and hence , converge uniformly in . The fixed two-switch energy also converges to . The strict gaps in (14.2) persist for sufficiently small . Taking proves the smooth assertion. Weighted variation maximizers exist by the compact embedding of the equivalent Hilbert norm into . A single-peaked function vanishing at both endpoints has variation twice its height, so none can maximize when .
The counterexample closes one tempting route to the original problem. If and is a norm-one global maximizer with , define away from its finite zero set. For every clamped with finite weighted energy, Holder's inequality gives
Thus also maximizes variation in this weighted Hilbert problem. However, a general weighted single-peak theorem is false by Theorem 15. Any successful use of (14.11) must exploit the special relation between the weight and the original maximizer. Our example is not shown to have this relation. The corresponding transfer for is not established by (14.11).
The recent weighted result of Hindov and Lokharu [HL, Theorem 1] places a weight on the function's norm and leaves the derivative energy unweighted. It has no additional mean constraint of the kind obtained by setting . It therefore supplies neither the arbitrary derivative-weight assertion disproved here nor the missing constrained variation theorem.
15. Nodal-length concentration in order three
In derivative order three, independent changes to the lengths of the nodal intervals of give another necessary condition. The changes preserve at the joins, so they remain admissible even though they need not preserve higher derivatives.
Theorem 16 (two dominant nodal intervals). Let and let be a global variation maximizer, normalized by . Let be the finitely many intervals of constant nonzero sign of , and put
At most two intervals satisfy
If , there is a set of at most two intervals such that
Thus the total length outside two intervals tends to zero at an explicit rate as . Additional short intervals are not excluded by this theorem.
15.1. An admissible finite-dimensional family
Put , so and . Theorem 12 supplies finitely many nondegenerate interior zeros of . Let be its sign on , and write
where . Every is positive: if on , the zero endpoint values would make identically zero there.
Choose with , and concatenate intervals of lengths in their original order. If is the new left endpoint, set on the th interval
At every join, and the one-sided values of agree with the same original value of . Hence and at the outer endpoints. The second derivative on this interval is . If also , integration gives an admissible .
The functionals are exactly
The point is a local maximum of on , . These two constraints have independent gradients: the vector is strictly positive, while has both signs. Their tangent space is therefore
Only the explicit smooth finite-dimensional functionals in (15.6) are differentiated. Differentiability of in a Sobolev norm is not needed.
15.2. The constrained second variation
Put and . Lagrange multipliers give
The value of follows by summing and using (15.4). On the tangent space (15.7), the same equation yields
The Hessian of , restricted to that tangent space, is
Before applying (15.9), the two contributions to the rank-one term are
Their sum becomes . The diagonal part is
Substituting (15.8) gives (15.10).
If three indices had , a nonzero vector supported on those three indices could satisfy the two equations (15.7). Its diagonal contribution in (15.10) would be strictly positive, and the rank-one contribution is nonnegative. This contradicts and proves (15.2), since .
For , let consist of the indices in (15.2). It is nonempty, because
It contains at most two indices. On its complement, , so
The strict inequality uses the positive energy on the nonempty set . Since , this proves (15.3).
Theorem 16 gives a quantified restriction on the geometry of any remaining maximizer in order three. For example, the two selected intervals cover more than of the domain at , and more than at . It does not bound the number of additional short intervals. The direct rescaling argument does not extend to higher derivative orders, because preserving at the joins does not preserve all the additional derivatives then required.
16. The complete finite-exponent theorem in order three
We now settle the variation identity and the full equality classification in derivative order three at every finite exponent.
Theorem 17 (order three at every finite exponent). For every ,
The nonzero variation extremizers are exactly the scalar multiples of the unique positive norm-one midpoint height extremizer. They are symmetric about the midpoint and have exactly one interior critical point. In particular, [NS, Conjecture 4.14] holds for and every finite .
The proof retains the residual equation, which contains more information than the interval-energy conditions in Section 15. It reconstructs every possible global maximizer from its residual vertices, then compares a positive action across different nodal counts. The comparison is a hand proof for every count and exponent.
16.1. Normalization and the conserved quantity
Let . Use the unnormalized dual potential and set , so
The endpoint conditions are at both endpoints and . For a global maximizer, Theorem 12 supplies finitely many simple interior zeros of , and . Its dual residual is continuously differentiable and quadratic between those zeros. Put .
On any interval of constant sign of , direct differentiation gives
This remains valid for : the derivative of is the continuous function . No derivative of is needed. At a switch, jumps but , so has no jump. The endpoint conditions imply
Integrating (16.2) on an interval of total length , and integrating by parts using , gives
When , this recovers the published endpoint normalization. The identity (16.4) uses the two endpoint conditions on ; the extra mean condition is not needed for this particular identity.
16.2. Strict interior bound and one vertex in every nodal interval
Lemma. Every interior point satisfies . In every nodal interval of there is exactly one zero of , and the sign of there is opposite to the sign of .
At an interior maximum of , one has . If this point is not a switch, then . Equations (16.2)-(16.3) imply . If it is a switch, they imply . Thus the maximum over the closed interval is at most .
An interior equality point must therefore be a switch with . The two nonzero constant curvatures on its sides have opposite signs. Since vanishes at the switch, it has the same strict sign immediately on both sides. Hence passes strictly through its value there, which contradicts the global bound . This proves strictness.
At an interior switch, (16.2)-(16.3) now give
Let be the sign of on a nodal interval and . The derivative is affine with slope . At an interior left endpoint it has sign , and at an interior right endpoint it has sign , by (16.5) and the simple crossing signs of .
At the outer left endpoint, : this follows from and . Strict decrease of into the interval gives ; equality would contradict the positive curvature of . The corresponding right endpoint argument gives . Consequently every nodal interval has exactly one interior vertex of .
Finally, on a nodal interval ,
Thus is negative somewhere. Its unique minimum is therefore for some . This proves the lemma.
For the normalized original maximizer, the lemma also gives at every interior point. This is stronger than a bound only at the endpoints but does not by itself force the average energy-density threshold in Theorem 16.
16.3. Matching consecutive quadratic pieces
Let the nodal intervals have alternating signs after a global sign choice, and set . Write the residual on interval , in coordinates centered at its vertex, as
At the join between and , continuity of and gives
Indeed, , and substitution of the matching slopes into the matching values yields (16.7). The outer endpoint values give
Thus all half-lengths are determined by , , and the positive vertex depths .
16.4. The exact nodal equation
On interval , put . At its vertex, (16.2) gives
Since and is even, for some real tilt one has
Define
The two endpoint equations are equivalent to
A useful derivative identity is
To check it, use to integrate . The terms cancel against . In particular, is strictly decreasing wherever .
16.5. Sufficiency and the remaining mean condition
Conversely, take , , , and , with half-lengths defined by (16.7)-(16.8). Suppose all equations (16.10) hold. At every internal join the residual value is
which lies strictly between and . The quadratic pieces themselves remain in this range away from the outer endpoints. Hence (16.11) proves between its two zeros: on the right,
and the left half is identical with in place of .
The pieces (16.9) glue to a continuously differentiable . In fact, at either internal join their one-sided derivatives both satisfy (16.5), with the same nonzero . Thus these derivatives agree. The outer derivatives vanish because there. Since is continuous, the resulting is and has the required simple interior zeros.
Set
Integrating the conserved quantity on a nodal interval gives its positive area
Consequently the last condition needed for to have all three zero endpoint traces is precisely
Under (16.13), annihilates every polynomial of degree at most two by integration by parts. Also , so differs from the integrated sign load by a quadratic. These are exactly the normal equations for its unique best approximation by quadratics. Thus the constructed profile is a genuine sign-compatible stationary dual configuration, not merely an abstract collection of lengths and energies. This converse does not assert local or global maximality.
16.6. Removing the overall scale and identifying the comparison target
Normalize the endpoint residual magnitude to , and solve (16.10) and (16.13) for the unknowns , with . Let
The unit-interval residual is obtained by multiplying lengths by and residual values by . Its energy and variation quotient are
Indeed (16.4) on the normalized interval gives , and the rescaling factor in its energy is . Also follows from (16.13), so Holder equality identifies the quotient.
Every global maximizer is covered by this finite system by Theorem 12. The next argument constructs a better admissible competitor whenever . Published Theorem 13 already excludes .
At , the single-switch solution has and . Its half-lengths are and in each interval, giving and , in agreement with Theorem 1. The all- single-switch profile is supplied by Theorem 10.
16.7. A positive action for matched intervals
Fix , , alternating signs , depths , and . Put and . Define the positive outer contribution
For consecutive depths and curvatures with , define
The maximizing join is unique and interior. Indeed the derivative in is
Its bracket is strictly decreasing from positive infinity to negative infinity. Since on , its unique zero is
This is precisely the matching join in (16.7). Consequently
Every contribution is positive.
16.8. The depths maximize a strictly concave function
For fixed , direct differentiation of the last expression in (16.18) gives
Here is the endpoint accounting. An internal join has slope magnitude . Varying changes the sum of its two half-lengths by . The two matching residual values have the same absolute value, so their endpoint contribution is . The integral contribution on the changed interval is . Since , these are exactly in the alternative identity
At an outer endpoint , the endpoint contribution is zero and this same formula applies.
For fixed and , differentiating the original definition in gives
The integral is finite for all : its possible zero is simple, because . At a zero located exactly at , the factor also removes any endpoint concern. The derivative follows by splitting at the moving simple zero, or by local power estimates. Thus the depth gradient is continuously differentiable in the open cube.
The derivative is negative at an internal half-length by (16.11). That half-length has equal positive derivatives with respect to its own depth and its neighboring depth. The outer half-length has positive own-depth derivative and there. Therefore the Hessian of is symmetric, has negative diagonal entries and nonpositive off-diagonal entries, and is strictly diagonally dominant: the two strictly negative fixed- terms (16.20) supply the strict excess on each diagonal. Hence it is negative definite. So is strictly concave.
A maximum exists in the interior. The function extends continuously to by its integral definitions; zero-length internal pieces contribute zero. If is sufficiently close to zero, uniformly over the other depths, both relevant terms are positive. One bound is
and the half-lengths have a fixed upper bound when is fixed. If is sufficiently close to one, both terms are negative uniformly over the other depths. At , , , and (16.11) is strictly negative up to the outer endpoint. All half-lengths have a uniform positive lower bound when is near one. Compactness and continuity give the asserted uniform negativity. Thus every maximizing depth lies in an interior cube. Strict concavity gives a unique maximizing depth vector , and (16.19) says it solves exactly the nodal equations. Conversely every solution of those equations is this maximizer.
For a second uniqueness argument not requiring (16.20), choose an index with maximal absolute difference between two proposed solutions. After exchanging the solutions, its own depth increases and the sum with either neighboring depth does not decrease. Hence both half-lengths do not decrease. Each decreases strictly with its own depth and decreases with the half-length throughout the admissible range. Its nodal equation cannot remain zero. This contradicts the existence of two solutions.
16.9. Strict convexity in the mean parameter
For fixed , each integral in (16.16) is nonnegative and independent of ; its coefficient is a positive multiple of , a strictly convex function of . Thus is convex in , being the maximum over a fixed interval of convex functions. The outer terms are strictly convex and positive. Therefore is strictly convex.
Set . This is convex as a supremum of convex functions, and is strictly convex because the positive outer terms supply strictness. More explicitly, for two distinct parameter values take the maximizing at their strict convex combination, apply strict convexity for that same , and then bound its endpoint values by the corresponding maxima.
Moreover as . For , the positive first outer interval suffices. For , an even has a negative last outer interval. For odd , choose fixed positive depths and, in an internal edge adjacent to a negative interval, a fixed interior join such as zero; its positive integral has coefficient . The definition as a maximum gives divergence. Thus has a unique minimum at some .
The nondegenerate depth Hessian gives differentiable dependence on by the implicit function theorem. Differentiating (16.16) at its optimizing join and then (16.18) gives
The moving-join term vanishes because (16.17) is its critical point; the moving-depth terms vanish by (16.19). Hence is exactly the unique parameter satisfying the zero-mean condition. The shooting reconstruction proves that this minimum supplies an actual admissible sign-compatible stationary profile on rescaling to the unit interval.
For every , reconstruction gives the two zero endpoint traces of , whether or not its mean is zero. Therefore (16.4) yields
A stationary mean-zero profile with intervals thus has normalized total length and variation quotient given by (16.14).
16.10. Removing a pair of intervals improves the quotient
For and a fixed , take the maximizing depth vector of . Append two copies of its final depth. The final sign is unchanged by adding two intervals, so the two outer contributions and every old edge contribution are unchanged. Exactly two additional edge contributions appear, and both are strictly positive. Thus
Evaluate this at and use the minimum defining :
By (16.22) and (16.14), the admissible stationary profile with intervals has a strictly larger variation quotient than the profile with intervals. Therefore no global variation maximizer can have nodal intervals.
A global maximizer cannot have because its nonzero first derivative has zero mean. The case is excluded by published Theorem 13 (two interior critical points). Thus for every global maximizer in derivative order three and every finite . Such a function has a single peak, so its variation is twice its height. Published Theorem 10 supplies the unique midpoint-symmetric height extremizer, up to scalar multiplication. This establishes
with coincidence and midpoint symmetry of the extremizers. This proves Theorem 17.
16.11. The constant at the uniform endpoint
Corollary (uniform norm constant). One also has
Proof. The height constants decrease as increases and are bounded below by . To identify their limit, take normalized height maximizers with . For every fixed , their third derivatives eventually have norm at most one. Successive integration and compactness give a subsequence converging through the second derivative. Weak compactness in every fixed , followed by a diagonal extraction, gives a limit whose third derivative has norm at most one for every finite . On an interval of length one this implies its norm is at most one: any set on which its magnitude exceeded would contradict the bound for sufficiently large . Thus the limiting height is at most , and .
For any clamped with , Theorem 17 gives for every finite . Passing to the limit proves the upper bound in (16.25); the reverse bound follows from variation being at least twice the height.
The corollary proves the endpoint constant identity. Section 17 establishes the complete variation equality classification at this endpoint in order three. The measure endpoint and its equality classification were already settled by Theorem 9. For derivative orders at least four, the remaining finite-exponent variation problem is still open.
17. Complete equality at the uniform endpoint
Theorem 18 (uniform endpoint in order three). One has
Every nonzero variation extremizer is a scalar multiple of the following function. Put , let take the successive values on the intervals cut out by , and define
The height and variation extremizers coincide and are symmetric about the midpoint. The height constant is credited prior work [NS, Theorems 4.5 and 4.12]; the new step here is the classification of every variation maximizer. Together with Theorems 9 and 17, this completes [NS, Conjecture 4.14] in derivative order three for all exponents, with the measure interpretation at .
17.1. Duality for an arbitrary endpoint maximizer
The unit ball of clamped is compact through the second derivative and weak-star closed in the third derivative, so the variation supremum is attained. Let be any maximizer with , and set , , and , with sign zero defined to be zero. The third derivative annihilates every quadratic polynomial.
The quotient-space Hahn-Banach formula identifies the norm of the linear load on this annihilator with the distance of its integrated load from the quadratics in . A best quadratic exists by finite-dimensional coercivity. For any such best quadratic , put . Since supports this variation maximizer, equality in the dual pairing gives
for a real constant . On the zero set of , only is initially known. No finite-switch or bang-bang assumption has been made. The pairing equality follows explicitly from and the dual norm formula.
17.2. The mean parameter is strictly between minus and plus one
Suppose . Then is concave. If it is positive somewhere, its positive set is an interval, and its zeros outside that set have measure zero. On the negative pieces reaching the outer endpoints, , hence on the left piece and on the right piece. On the positive interval, and is convex with nonpositive endpoint values. Thus everywhere, contradicting its zero mean and nontriviality. This argument also covers a positive interval touching an outer endpoint, where the corresponding endpoint value of is zero. A concave function positive somewhere cannot have a separate zero interval, since its derivative is nonincreasing.
If instead everywhere, its zero set is an interval, a point, or empty. If that set has measure zero, almost everywhere, contradicting the two zero endpoint values of . Otherwise write the zero interval as . There , so almost everywhere. Necessarily and almost everywhere on its interior. But continuity at the left end gives , forcing ; the right end similarly forces . This would make identically zero, contradicting . The case follows by changing the signs of . Therefore
17.3. The conserved quantity proves endpoint regularity
The absolutely continuous function
has derivative almost everywhere. On the zero set of , this uses the fact that an absolutely continuous function has derivative zero almost everywhere on a level set. Thus the argument does not differentiate the discontinuous sign function and does not assume that the zero set is finite.
The endpoint traces and an integration by parts give
Indeed , , and .
At an interior maximum of , , so (17.4)-(17.6) show . Thus everywhere. If an interior zero of also had , (17.5) would give , and hence there. In a small neighborhood, has one nonzero sign , so and . Then and
contradicting . Every interior zero of is therefore simple. The nonzero endpoint residuals also give a fixed quadratic sign of near each outer endpoint. Compactness now rules out infinitely many interior zeros: an accumulation point would have both .
Finally, equality at an interior point forces . At this simple zero, would combine with the two opposite nonzero curvatures to make pass monotonically through that value, again contradicting its bound. Consequently
Exactly as in Section 16, each of the finitely many nodal intervals of has one residual vertex, of sign opposite to and magnitude . The opposite sign also follows from . The alternating vertices and the two endpoint values imply exactly simple residual roots when there are nodal intervals. Thus almost everywhere on the whole domain, and is . These conclusions were derived for an arbitrary endpoint maximizer.
17.4. The positive action extends to q=1
Normalize . Keep the half-lengths, outer action, edge action and total action of Section 16, with the integrand in place of . On a nodal interval write , where , , and put . The exact scalar function is now
In particular, is continuously differentiable even when a residual root meets a nodal endpoint. Simply substituting into (16.20) would omit the second term in ; formula (17.8) is the needed endpoint calculation.
The action gradient is still the sum of the two values. Differentiation through is justified by its Lipschitz bound and its simple zeros; (17.8) then gives the continuous second derivatives. The Hessian is negative definite by the same strict diagonal-dominance argument. Its off-diagonal entries come only from the matched half-lengths, whose own-depth and neighboring-depth derivatives agree; the strictly negative fixed-half-length derivative supplies the excess on the diagonal.
The depth maximum is again interior. To check the zero-depth face, let be either half-length, bounded above for fixed . If , then
If , then
uniformly for sufficiently small . The upper-depth face follows from at , strict decrease in , and the uniform positive lower bound on the half-lengths. This proves existence and uniqueness of the interior depth maximizer for each , just as before.
The edge action remains a maximum over a fixed joining-value interval of positive integrals times . Therefore the optimized action is strictly convex and coercive for every . Its derivative is the signed area. Its unique minimum gives a reconstructed clamped profile with zero mean of its first derivative and third derivative equal to the sign of its residual almost everywhere. The reconstruction matches the value and first derivative of at every join; bounded second derivatives suffice here, and continuity of those second derivatives is not claimed.
The normalized total length and quotient are
All ingredients of the two-interval insertion comparison remain valid. Thus for , minimizing after appending two copies of the last depth yields
and an actual admissible profile with intervals has a strictly larger variation. No endpoint global maximizer has .
17.5. The three-interval profile is strictly worse
For , reflection preserves the alternating sign sequence and the same parameter . The unique depth vector at that parameter is therefore reflection symmetric. Its residual and its third derivative are symmetric about the midpoint. The four simple residual roots have the form , and has successive signs .
The zero moments of give
The first zero of is , which lies between and . On this interval,
The symmetry of and its zero mean make antisymmetric. Its variation is therefore
This is strictly below the value in (17.1) of the admissible two-interval profile verified below. An elementary rational comparison suffices: because , and because . Hence . The same inequalities show . Thus is excluded without assuming the finite-exponent two-switch argument continues to the endpoint.
17.6. The two-interval profile and complete equality
Only remains, since contradicts zero mean. Reflection exchanges the signs and sends to , so is even. Its unique minimum is at zero, and the unique depth vector has equal entries. The residual is antisymmetric and has three simple roots . The third derivative of the maximizing function has the four signs in (17.2).
The first moment condition, after using antisymmetry, is
All three moments vanish for this control. On the first half, is positive until its zero at ; the second half follows by reflection. Direct integration gives
Thus every normalized variation maximizer is or . Its height is half its variation. Conversely, any height maximizer has variation at least twice that height and at most , so it is a variation maximizer as well. This proves the complete equality classification and the endpoint constant formula. The scalar multiples include all nonzero extremizers; the zero function gives the trivial equality.
18. Uniform-endpoint regularity in every order
The conserved quantity extends to all derivative orders if its additional mixed derivative terms are retained. It proves saturation and finite simple switching for every endpoint maximizer, even though it does not supply the quadratic comparison used in order three.
Theorem 19 (all-order endpoint regularity). Let , and let be any maximizer of variation in the clamped class with . Put , , , and , with . For any best polynomial approximant to , put . Then
for a real constant . With , one has
Both and have finitely many simple interior zeros. If has such zeros, then and has exactly . In particular,
The highest derivative is constant almost everywhere on a neighborhood of either endpoint, and is a nonzero signed monomial on each such neighborhood. For each fixed , the critical-point count is uniformly bounded over all , as proved below.
18.1. Duality and the conserved quantity
Existence follows from weak-star compactness of the unit ball under its moment constraints, together with uniform convergence of the integrated first derivative. A nonzero smooth clamped function shows that , and homogeneity makes the highest-derivative norm of every maximizer equal to one.
The annihilator of in is exactly the set of clamped th derivatives. The quotient norm of modulo is the norm of the load on this annihilator. It is at most , while its pairing with the present is . A best polynomial exists by finite-dimensional coercivity. This proves the integral equalities in (18.2), and equality in gives off the residual zero set. On that zero set only is initially known.
Both and belong to , and differentiating the integrated load gives the first identity in (18.1). Define the absolutely continuous function
where the sum is empty for . Differentiation of the sum telescopes. Using and (18.1), the chain rules give
Off the residual zero set this is dual equality; on that set almost everywhere by the absolutely continuous level-set derivative property. For the same displayed derivative follows directly. No sign function is differentiated.
All endpoint derivatives of through order vanish, so at both endpoints. Each summand in (18.4) has integral
Indeed, integrate by parts times; the clamped traces remove every boundary term, and . The remaining two contributions in (18.4) also integrate to . Hence
proving the endpoint identities before any restriction on or either zero set.
18.2. The strict mean-parameter bound
We first record a zero lemma. If has of one sign almost everywhere, then distinct zeros force to vanish throughout their convex hull. To prove this, take zeros . Their divided difference of order is zero. The repeated fundamental theorem of calculus expresses this divided difference as the integral of over the simplex of convex combinations of these knots. Its induced density is positive almost everywhere throughout . The one-signed derivative must therefore vanish almost everywhere there. Thus is a polynomial of degree at most on that interval, and its zeros make it identically zero. The integral formula for follows by mollification from the smooth formula. In particular, a function of this kind with no zero interval has at most distinct zeros.
Suppose now that . Then has one sign. If has no zero interval, it has at most zeros and almost everywhere. Choose a polynomial with roots exactly at the sign-changing zeros of , and choose its overall sign to agree with . Then and almost everywhere. The positive integral of this product contradicts the moment constraints.
Otherwise take a maximal nontrivial zero interval of . Its derivative traces through order vanish at both ends. Equation (18.1) gives almost everywhere inside it, so and almost everywhere there. The integral Taylor formulas give
and
The signs are strict outside the maximal zero interval: equality at an outside point would force almost everywhere between it and the interval, extending the zero interval by the same Taylor formula. Consequently to the left and to the right. The clamped endpoint traces imply
Continuity against almost everywhere inside the zero interval forces and . This would make identically zero, contradicting . Thus . Applying the same argument to gives .
18.3. Null zero sets and saturation
For a function in , all derivatives through order vanish almost everywhere on a level set. This follows by successively applying the absolutely continuous level-set derivative property, discarding a null set at each of the finitely many steps.
Suppose that had positive measure. Equation (18.1) would give almost everywhere on . Since takes values only in and , this forces and almost everywhere on . Every term in (18.4) then vanishes almost everywhere on , contradicting the nonzero constant (18.5). Hence is null, proving (18.3).
If had positive measure, the same level-set property would give almost everywhere on it. This contradicts almost everywhere. Thus the zero set of is also null.
Finally, continuity and the nonzero endpoint values of give a fixed nonzero sign on each endpoint neighborhood. Equation (18.3) makes constant there. Integration of the clamped traces gives, on the respective neighborhoods,
This proves saturation and the endpoint assertions. We next prove finiteness and simplicity.
18.4. Finitely many simple zeros
An infinite set of residual zeros would have an accumulation point in the interior, because is nonzero near both endpoints. Repeated Rolle arguments and give
Equation (18.5) then gives , so . Its sign is fixed on a neighborhood. There (18.1) makes a polynomial of degree exactly , since . It cannot have infinitely many zeros on that neighborhood. Thus has finitely many zeros.
The same argument applies to . Its endpoint monomials exclude accumulation at the boundary. At an interior accumulation point, for , and (18.5) gives . Hence has one fixed sign nearby, making a polynomial of degree exactly , again a contradiction.
Both functions are therefore splines of degree : has a knot only where changes sign, while has a knot only where changes sign. All polynomial pieces have nonzero leading coefficient. Every zero has finite multiplicity at most . Indeed, if all derivatives through vanish at a zero of either function, the conserved quantity makes the other function nonzero there; the first function then agrees locally with a polynomial of degree exactly .
Let be the number of sign changes of , let be the number of sign changes of , and let be the sum of the multiplicities of all interior zeros of . A spline of degree with knots and nonzero leading coefficient on every piece has at most zeros counted with multiplicity. This follows by applying Rolle's theorem times: its st derivative is continuous and affine with nonzero slope on each of the pieces, so it has at most distinct zeros. Multiplicities are reduced by one at each differentiation, as in Section 8. Endpoint zeros are included when present.
Apply this bound first to , whose endpoint zeros each have multiplicity , and then to , whose endpoint values are nonzero. If is the sum of the multiplicities of the interior residual zeros, then
Thus equality holds throughout. In particular : every zero of changes sign and has multiplicity one. Also , so every residual zero is simple and changes sign. Writing , there are exactly residual zeros. Since and , one has . For the same count applies directly to continuous piecewise affine functions. This proves Theorem 19.
Corollary (uniform critical-point bound over all exponents). For each fixed , there is a finite bound on the number of critical points of any variation maximizer, uniform over . The bound is not made explicit.
Proof. Theorem 11 bounds the count by one when is sufficiently close to one, and Theorem 12 gives uniform bounds on compact subintervals of . It remains to treat sequences , allowing .
The constants decrease with and are bounded below by . For normalized maximizers at finite , the highest derivative has norm at most one once , for every fixed finite . Compactness of its successive integrals, weak compactness for each , and a diagonal extraction give a limit whose highest derivative has every finite norm at most one and hence norm at most one. The first derivatives converge uniformly. Their variations converge to the limit of the constants, proving and showing that every such limit maximizes endpoint variation. The same compactness holds for sequences already at the endpoint.
Choose best polynomial residuals for the sequence. Their coefficients are bounded: , using the best-approximation inequality and finite-dimensional norm equivalence. By the just-proved regularity of the endpoint limit, its first derivative has only finitely many zeros, so uniform convergence gives almost everywhere and in . Thus, after extracting polynomial coefficients, uniformly. For finite exponents, and comparison against every fixed polynomial shows that is a best residual for the endpoint load. The same assertion follows directly for endpoint residuals.
Theorem 19 gives nonzero endpoint values of . Uniform residual convergence therefore fixes the highest-derivative signs on two common endpoint neighborhoods, by (6.6) for finite exponents and (18.3) at the endpoint. Successive integration from the clamped traces excludes critical points there. On the interior, also holds in : the functions have uniform Holder control from any fixed bound on the highest derivative with , also when . The endpoint limit's critical points are all simple. On disjoint small neighborhoods of those points, the second derivative has one fixed nonzero sign, and on the remaining compact set the first derivative stays away from zero. The sequence therefore has exactly the same critical-point count for all sufficiently large indices. This excludes an unbounded sequence of counts as and completes the proof.
18.5. A local obstruction to the order-three residual bound
For , the mixed term in (18.4) is , which vanishes at a residual extremum. For , it is . At the first term survives and has no prescribed sign. The following exact local solutions show why the conserved quantity alone cannot supply the order-three bound.
On , put and
Here and , while , and . At zero, has a strict local maximum of value two, exceeding the formal level obtained from . This is a local stationary solution; it is not clamped and is not a global maximizer.
The same obstruction is present at finite exponent. For , keep this and take
Then , , and
. Again the formal level is while . Direct triangle bounds on the stated interval give and ; in (18.9), .
These examples disprove only an inference from the local equations and negative conserved level to the residual bound. They do not disprove a bound with global maximizing hypotheses, or the variation conjecture. The remaining problem is to control the additional derivative traces and construct a better admissible competitor for every multiple-peak maximizer.
19. Endpoint second variation and algebraic sharp constants
The finite simple residual zeros in Theorem 19 permit a second-variation calculation at the uniform endpoint. The second derivative is a finite weighted sum over those zeros. It does not vanish when the conjugate exponent tends to one.
Theorem 20 (endpoint switching Hessian). Let and let be a normalized variation maximizer in . Write for its critical points and for the sign of immediately to the left of . Let be its best polynomial residual, with simple zeros , where . Define
where is the orthogonal projection onto the restrictions of to the residual zeros. The nearby switching objective
is twice continuously differentiable, and at this configuration
Consequently, with and
, one has
19.1. Differentiating the moving residual zeros
The Gram matrix of for is positive definite: there are distinct nodes, and a nonzero polynomial of degree at most cannot vanish at all of them. For a perturbation of a residual with simple zeros, differentiation of the signed moments gives
Indeed, a residual zero moves with velocity , and the sign jump there is . Differentiating the endpoints of the finitely many sign intervals gives (19.5), including its positive sign and factor two. The same formula holds for a continuous test kernel in place of .
For , is jointly continuously differentiable in . Thus the simple-root implicit function theorem applies even when a residual zero equals a load switch. It provides continuously differentiable root locations as functions of the switch positions and polynomial coefficients. All nearby residuals retain exactly these simple zeros: their derivatives have fixed nonzero signs on small root neighborhoods, and the residual is bounded away from zero on the remaining compact set, including both endpoints.
The normal equations are for . Their derivative in the polynomial coefficients is the negative Gram matrix for . The implicit function theorem therefore gives a continuously differentiable polynomial branch satisfying these equations. The zero signed moments and convexity of the objective make it a branch of global best approximants.
Each best polynomial on this branch is unique. If two polynomials were best, equality in convexity would force their residuals to have the same sign almost everywhere. At every simple zero of the first residual, continuity forces the second to vanish too; otherwise their signs disagree on one side. Their difference is a polynomial of degree at most with at least distinct zeros and hence is zero.
Differentiating the normal equations and the objective now gives
At the maximizing configuration, Theorem 19 gives . Differentiate the integral defining using (19.5) and the moving evaluation point. Polynomial orthogonality replaces by in the residual contribution and yields (19.3). These finite-root formulas also show that the gradient is continuously differentiable. When , the derivative of the moving kernel is an indicator; integration against the bounded sign control justifies the calculation by dominated convergence. No ordinary derivative of the discontinuous sign function is asserted.
Every nearby sign load in (19.2) has dual norm at most , with equality at the given maximizer. Its Hessian is therefore negative semidefinite, giving the first assertion in (19.4). The orthogonal complement of polynomial data in the -dimensional node space has dimension , giving the rank bound.
The same local differentiation proof applies to any configuration with at least simple residual zeros and nonzero endpoint residuals, before a maximality assertion. At a maximizer the rank bound supplies a strictly negative Hessian direction; for the Hessian is exactly . A negative direction is compatible with a local maximum. Excluding higher counts still requires a positive direction or a global value comparison. In particular, the finite-exponent exclusion proof cannot be transferred by formally setting : the measure in (19.1) replaces the vanishing factor in that formal substitution.
19.2. Semialgebraic description at a fixed control count
Theorem 21 (algebraicity at the uniform endpoint). For every integer , the sharp constant is a real algebraic number. There exists a normalized variation maximizer whose control knots and polynomial coefficients are all real algebraic numbers.
Proof. Fix an integer . For ordered knots
, take the alternating control on . It satisfies the required polynomial moment conditions precisely when
Its integrated potential is
The left clamped traces vanish by construction, and (19.7) is equivalent to all right clamped traces. The highest derivative has norm one.
We claim that the graph of the total variation is semialgebraic over , meaning that it is described by a finite Boolean combination of polynomial equalities and inequalities with rational coefficients. On each control interval, (19.8) is a polynomial in with rational polynomial coefficients in the knots. Its derivative has degree with nonzero leading coefficient. List all its interior critical points in increasing order, using at most variables per interval. Polynomial conditions require that these points are zeros; a universal real quantifier requires that there are no other zeros in the open interval. There are finitely many choices of list lengths.
Adjoin the two interval endpoints to each list. The variation on that interval is the sum of the absolute differences between consecutive polynomial values. Absolute value is expressed by polynomial sign alternatives. Summing these expressions gives a first-order formula over for . Quantifier elimination over real closed fields [B, Theorem 2.1] makes this graph semialgebraic over . This argument uses the finite critical-point description, not a claim that arbitrary integrals of semialgebraic functions are semialgebraic.
The set of attained values under (19.7) and the strict knot inequalities is consequently a one-dimensional semialgebraic set over . Any attained finite maximum of such a set is real algebraic: in a quantifier-free description, signs of all nonzero rational polynomials are constant near any transcendental point, so such a point cannot be a finite endpoint of the set.
Theorem 19 supplies a global maximizer with a finite control count . Change its overall sign to make its first control positive. It then has representation (19.8), while every admissible configuration at this same has variation at most . The set of attained values at this therefore has attained maximum exactly , proving algebraicity.
Finally impose in the first-order formula. It has a solution over . Quantifier elimination transfers this statement, with its real algebraic coefficients, to the real closed field of real algebraic numbers; this field is included in the notation of [B]. The resulting knots and auxiliary critical points are algebraic. Formula (19.8) gives a maximizing spline with algebraic coefficients on all pieces.
This proof does not assert algebraicity of every maximizing knot configuration, an explicit algebraic degree bound, or an effective stopping count for searching over . The uniform count bound in Section 18 is implicit. Theorem 21 determines the arithmetic nature of the sharp constant without computing it or settling the higher-order variation identity.
20. Two-switch and symmetric three-switch exclusion at the endpoint
The endpoint Hessian now permits the two finite-exponent exclusion results to be completed at . An approximation of the linear response supplies weak unimodality. A separated-zero count then recovers the strict second variation, with a separate boundary sign count in order four.
Theorem 22 (endpoint switch exclusions). Let . A nonzero variation maximizer in cannot have exactly two interior critical points. If it is symmetric about the midpoint, it cannot have exactly three interior critical points either. For a hypothetical maximizer with three critical points , the affine response defined below for has its unique minimum strictly outside . Thus Theorems 13 and 14, including this necessary response-location condition, extend to every .
Nonsymmetric three-critical-point configurations with the response minimum outside the two outer critical points remain unresolved.
20.1. An atomic affine response
Suppose that a normalized endpoint maximizer has critical points and left signs . Choose its overall sign so that near zero. Let and be as in Theorem 20, put , and move the switches with the affine velocity . Dots denote derivatives along this motion. The truncated-power identity used in Section 13 gives
For the associated clamped potentials , define at the base point
These parameter derivatives exist by the finite-root formulas of Theorem 20. Distributionally, and . Differentiating (20.2), the term from cancels . Also . Therefore
and the one-sided endpoint traces on the fixed-control neighborhoods are
Here mean those constant endpoint controls. The function is a spline of degree at most , with knots only at the residual zeros. Integration by parts gives, for every ,
This moment system uniquely determines because its polynomial Gram matrix is positive definite.
Two further identities will be used. First, since vanishes on the atomic support. Integrating by parts times, using and , gives
All intermediate boundary terms vanish by (20.3). The last equality uses at both endpoints. This calculation is valid for the spline and its derivative measure: has bounded variation, while the remaining integrations involve absolutely continuous functions.
Second, differentiation of the envelope gradient in Theorem 20 gives, with ,
Indeed , and . At a global maximizer, the expression in (20.7) must be nonpositive.
20.2. Unimodality from positive approximations
We need a linear-response fact whose hypotheses are just (20.3)--(20.5). Write and . If
then is weakly increasing up to an interior point and weakly decreasing thereafter. Reversing both signs gives weak decrease followed by weak increase.
To prove this, approximate the finite positive measure by positive smooth densities on , for example by mollifying the atoms and adding a positive background tending to zero. The associated polynomial Gram matrices converge to the positive definite matrix for . Let solve
Its coefficients converge to those of . Define
The moments (20.9) give exactly the clamped traces through at both ends and the two highest traces . The positive-weight Rolle argument of Section 12 applies to these functions without needing a stationary profile. For completeness, let . Its endpoint zeros have multiplicity . If selected interior zeros have total multiplicity , counting a tangency twice, Rolle's theorem through derivatives gives at least distinct zeros of . Between consecutive zeros, the integral of is zero, forcing a root of in that open interval. The polynomial is nonzero, since otherwise the boundary conditions would make a zero polynomial despite . Thus .
The endpoint signs in (20.8) force a crossing of from positive to negative. Any additional zero, or an interval of zeros, contradicts this count and the endpoint signs. Thus is strictly unimodal. Weak convergence of the measures and coefficient convergence give uniform convergence of through derivative : the integrated kernels of these orders are continuous and form uniformly continuous families on the compact square. Passing to a subsequence of the peak locations gives the asserted weak unimodality of . Its nonzero endpoint monomials place the limiting peak in the interior. Negation handles the reversed signs. This approximates the linear boundary-value response; it does not assume continuation of the original stationary profile to finite exponents.
20.3. Separated zeros and flat intervals
Two zeros of a continuous function are called separated if the function is not identically zero between them. We use the following elementary clamped spline count, in the form of [C, Lemma 2.4 and Proposition 2.5]: a spline of degree at most with at most interior knots and endpoint zeros of order at least has at most separated interior zeros, provided the endpoint zeros are separated.
For , the affine pieces have at most separated zeros altogether, including the endpoints; a flat zero piece contributes at most one. For the induction, choose a nonzero extremum between each consecutive pair of separated zeros. These are derivative zeros. Any two consecutive chosen derivative zeros are separated, since a constant derivative there would be zero, contradicting the original function's intervening zero and nonzero extremal value. The endpoint derivative zeros are separated from the chosen ones by the same reasoning. Thus separated interior zeros give at least for the derivative, which remains clamped of order . Induction yields . This argument permits zero intervals and lower-degree derivative pieces.
One consequence will supply strictness. Suppose that a function vanishes at , with , is nonzero somewhere between each consecutive pair, and has derivative zero at at least two of . Choose four nonzero extrema, one in each intervening interval. Together with those two tangency points, these give six separated interior zeros of . Nonzero extremal values and the intervening zeros ensure separation of this six-point list by the preceding argument.
We also need a flat-interval bound for the response. Put and , and suppose that vanishes on an open interval. Choose a point in it that is not a residual knot. If residual knots lie to its left, the polynomial identity on that interval is
Since , this is a nonzero signed measure on and the left knots annihilating all polynomials of degree at most . Polynomial interpolation requires at least distinct support points, so . The analogous right-end representation and require at least right knots. A flat interval therefore requires at least residual knots. Zero coefficients only strengthen this necessary count.
20.4. Two critical points
For , Theorem 18 already excludes the configuration. Let and suppose that the two critical points are . There are residual knots. Set , so the highest response traces are and . Section 20.2 makes weakly unimodal. The flat-interval bound would require knots, so has no zero interval. It follows that is strictly increasing to its unique peak and strictly decreasing thereafter, even though its derivative may vanish at isolated points on either side.
Equation (20.6) gives . Thus the peak lies strictly between and , and
If equality held, would vanish at both and , as well as at its central sign-changing zero. The outer two zeros are tangencies, so ; here because . The six-zero consequence in Section 20.3 would give six separated interior zeros of . But is a spline of degree , with at most knots and endpoint zeros of order . The clamped count allows only
separated interior zeros. This contradiction makes (20.12) strictly positive, contrary to maximality.
20.5. Three critical points and the response minimum
Let and suppose that the critical points are . Use . There are residual knots. The traces in (20.4) are and , giving near zero and near one. Section 20.2 gives weak decrease followed by weak increase.
We first rule out every flat interval of . The knot count does this for . For , suppose a knot-free open interval has . There must be three residual knots on each side, written and . The left moment identity (20.11), with and coefficient at zero, gives a measure on annihilating . Its nullspace is spanned by the barycentric weights
which have signs , by interpolation of polynomials of degree at most two. The negative coefficient at zero fixes a positive multiple. Thus the three left atomic coefficients have signs .
The analogous right identity is the measure
annihilating . One may derive it by subtracting the left zero-moment identity from the endpoint moment equations (20.5) restricted to . Here . Applying (20.14) to the ordered nodes again gives a positive multiple, so the right also have signs . Since is positive, the values of on each of the two triples have those signs. This forces at least two polynomial roots within each triple, hence at least four distinct roots. But . This contradiction excludes a flat interval in order four without a symmetry assumption.
Therefore is strictly decreasing to a unique minimum and strictly increasing thereafter. Suppose . Formula (20.7) gives
At a maximizer is negative semidefinite. Thus this quadratic form is zero, and . The latter implication follows, for example, by diagonalizing the real symmetric Hessian. The component identity in (20.7) then forces
Put . Its three zeros in (20.17) are separated, since there is no flat interval of . At most one can be the sign-changing zero at its unique minimum; at least two are tangencies and also zeros of , because . The six-zero consequence of Section 20.3 gives six separated interior zeros of . Yet this clamped spline has degree and at most knots, allowing only . The contradiction shows that lies strictly outside , including exclusion of its boundary points.
If is midpoint symmetric, then , , and reflection preserves the load under this dilation. Uniqueness of the best polynomial residual gives for the nearby branch, so and are midpoint symmetric. The unique minimum of must then be , contradicting the proved restriction. This excludes every midpoint-symmetric three-critical-point maximizer. For , Theorem 18 already excludes every such higher-count configuration. Theorem 22 follows.
Combining the endpoint result with Theorems 13 and 14 leaves nonsymmetric three-critical-point maximizers, whose dilation-response minimum lies outside their outer critical points, and configurations with at least four critical points as the remaining higher-order possibilities outside the established exponent intervals. Their exclusion requires an additional argument; the full all-order conjecture is still open.
21. Remaining questions
The order-three case of [NS, Conjecture 4.14] is now complete at every exponent: Theorem 17 gives the finite-exponent identity and equality classification, Theorem 18 gives the complete uniform endpoint, and Theorem 9 supplies the measure endpoint. The full all-order conjecture remains unresolved. In every order, Theorems 5 and 11 establish exponent intervals around two and immediately above one; their widths and overlap are not known. Theorem 10 proves height symmetry and uniqueness at every finite exponent.
For orders , Theorem 12 supplies finite nondegenerate critical points and simple dual-residual zeros. Theorem 13 excludes every two-switch maximizer, and Theorem 14 excludes midpoint-symmetric three-switch maximizers and restricts every remaining three-switch candidate. Excluding nonsymmetric three-switch configurations and all higher counts remains the finite-exponent task in those orders. At the uniform endpoint, Theorem 19 proves saturation, finite simple critical points and exactly simple residual zeros in every order. The critical-point count is uniformly bounded over all for each fixed order, but no explicit bound or higher-order single-peak conclusion is proved. The positive action in Sections 16 and 17 uses the quadratic residual pieces specific to order three. Section 18 identifies the additional mixed derivative terms and gives exact local obstructions to reusing the same pointwise argument; no higher-order action comparison is proved here.
Theorem 20 provides the endpoint switching Hessian, including its discrete root weights and polynomial projection. Theorem 22 uses the affine response to exclude two-critical-point and midpoint-symmetric three-critical-point maximizers at the endpoint in every order, completing these exclusions over all exponents above one. A remaining nonsymmetric three-critical-point candidate must have its dilation-response minimum strictly outside the two outer critical points. These and higher-count candidates still require a positive direction or another comparison. Theorem 21 proves that every uniform-endpoint sharp variation constant is algebraic and admits an algebraic maximizing spline. It gives neither an explicit degree bound nor an effective stopping count for a search over control knots.
Theorem 15 rules out a reduction using the same identity for arbitrary derivative weights. The special weights generated by an actual maximizer remain a possible restricted route in higher orders. The stronger centroid and knot-reflection inequalities in the research notes remain unproved and are not needed for the results above.
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