Mathematical Appendix: Certificates, Spectral Models, and Limitations for Cosine Sums

Companion to the Bass Plus “Mathematics of Sound” research report.
Version: 18 October 2026.
Scope: complete statements and proofs for the results promoted in the accompanying articles.

This appendix studies a precise model of synchronized, equally weighted harmonics. It does not prove or disprove Chowla’s cosine conjecture. Its purpose is to make the report independently checkable: every bound states its domain, every computational observation identifies its finite search space, and every change of model is explicit.

The moment hierarchy below is a specialization of Lasserre’s univariate pushforward-measure hierarchy Lasserre2021. Polynomial concentration using Chebyshev or needle polynomials is established prior art, including the analyses of Slot and Laurent SlotLaurent2021; SlotLaurent2022. We therefore make no claim to have invented either mechanism. The maximum-frequency-independent degree estimate proved here is a specialized bound obtained during this investigation; its novelty relative to all existing literature has not been established. The proofs below stand independently of a novelty claim.

Contents

• A. Model, normalization, and the constant
• B. Attainment at a fixed number of frequencies
• C. Exact moments and the polynomial hierarchy
• D. A phase-mass lemma independent of the highest frequency
• E. Uniform approximation using a Chebyshev filter
• F. Consecutive frequencies and low-degree certificates
• G. The lattice operator and a finite open box
• H. A discrete grid can miss the minimum
• I. Why integer amplitudes do not preserve the conjecture
• J. A positive Riesz-product certificate
• K. Negative moments, quantiles, and a saturating detector
• L. Gibbs weights and an entropy budget
• M. Frequency scaling and finite observation time
• N. What has and has not been established

A. Model, normalization, and the constant

Let

$$A=\{a_1,\ldots,a_N\}\subset\mathbb Z_{>0},\qquad N\ge1,$$

where the frequencies are distinct. Throughout the original problem,

$$f_A(x)=\sum_{a\in A}\cos(ax),\qquad K(A)=-\min_{x\in\mathbb R}f_A(x). \tag{A.1}$$

The variable belongs to the circle $\mathbb T=\mathbb R/(2\pi\mathbb Z)$. There is one common variable, not an independently adjustable phase for each frequency. All amplitudes equal one, and all initial phases equal zero. Write

$$\mathbb E u=\int_{\mathbb T}u(x)\,d\mu(x), \qquad d\mu(x)=\frac{dx}{2\pi}.$$

The measure has total mass one. Consequently, different positive integer harmonics are orthogonal and

$$\mathbb E f_A=0,\qquad \mathbb E f_A^2=\frac N2,\qquad -N\le f_A(x)\le N,\qquad f_A(0)=N. \tag{A.2}$$

For completeness, writing each cosine as $(e^{iax}+e^{-iax})/2$ reduces these assertions to $\mathbb E e^{ikx}=0$ for every nonzero integer $k$, and to one when $k=0$. Continuity makes the minimum in (A.1) attainable. Its value is negative, since a nonconstant continuous function of mean zero cannot be nonnegative everywhere.

Proposition A.1 — An elementary uniform bound. For every admissible $A$,

$$1\le K(A)\le N. \tag{A.3}$$

Proof. Only the lower bound requires an argument. Let $D=\max A$, and take the $D$ points

$$x_j=\frac{(2j+1)\pi}{D},\qquad 0\le j<D.$$

For $1\le a<D$, the average of $e^{iax_j}$ equals

$$\frac{e^{ia\pi/D}}D \sum_{j=0}^{D-1}e^{2\pi iaj/D}=0.$$

For $a=D$, every $\cos(Dx_j)=-1$. The average of $f_A$ over these points is therefore exactly $-1$, so at least one point has value at most $-1$. The upper bound follows from $f_A\ge-N$. $\square$

The coefficient attached to an individual set is

$$c(A)=\frac{K(A)}{\sqrt N}.$$

A physically scaled pressure perturbation might be represented by

$$p(t)=P_0 f_A(\omega_0t),\qquad P_0>0,\quad\omega_0>0.$$

Over a full period, $p_{\mathrm{rms}}=P_0\sqrt{N/2}$, while its negative peak magnitude is $P_0K(A)$. Thus its negative peak-to-RMS ratio is exactly $\sqrt2\,c(A)$. The ratio is dimensionless and unchanged by either common scale. This interpretation concerns the shape of the waveform; it does not identify $c(A)$ with an energy quantum or a material constant.

To formulate the open question, define

$$\kappa_N=\inf_{|A|=N}K(A),\qquad c_N=\frac{\kappa_N}{\sqrt N}. \tag{A.4}$$

Chowla’s assertion is the existence of an absolute $c>0$ for which

$$K(A)\ge c\sqrt{|A|}\quad\text{for every admissible }A. \tag{A.5}$$

Equivalently, $\inf_{N\ge1}c_N>0$. It is also equivalent to $\liminf\limits_{N\to\infty}c_N>0$: a positive eventual lower bound can be combined with $c_N\ge1/\sqrt N$ for the finitely many remaining sizes. In contrast, positivity of each individual $c_N$ follows already from (A.3) and does not settle (A.5).

A disproof would require actual admissible sets $A_j$ with $|A_j|\to\infty$ and $K(A_j)/\sqrt{|A_j|}\to0$. A lower-bound formula that becomes small does not supply such a sequence: the formula may merely fail to detect a much deeper minimum.

B. Attainment at a fixed number of frequencies

The infimum in (A.4) is taken over infinitely many frequency sets. Nevertheless it is attained. This fact does not give an effective bound on the largest frequency of an exact minimizer.

Theorem B.1 — Fixed-size attainment. For each integer $N\ge1$, some admissible set $A$ of size $N$ satisfies $K(A)=\kappa_N$.

Proof. Choose admissible ordered tuples $a^{(j)}=(a_1^{(j)},\ldots,a_N^{(j)})$ with $K(A_j)\to\kappa_N$, and let

$$H_j=\{a^{(j)}x\bmod 2\pi:x\in\mathbb T\}\subset\mathbb T^N.$$

Each $H_j$ is a compact circle subgroup. The nonempty compact subsets of a compact metric space are compact for the Hausdorff metric; one proof chooses convergent membership patterns on successively finer finite nets and takes a diagonal subsequence. Pass to a subsequence with Hausdorff limit $H$. This limit is a subgroup: approximating any $u,v\in H$ by $u_j,v_j\in H_j$ shows that $u-v$, the limit of $u_j-v_j$, belongs to $H$.

Define the integer annihilator

$$L(H)=\{k\in\mathbb Z^N:e^{ik\cdot\theta}=1 \text{ for every }\theta\in H\}.$$

We use two elementary facts about this group. First, every subgroup of $\mathbb Z^N$ is finitely generated. An induction proves this: project onto the first coordinate, whose image is either zero or $d\mathbb Z$; choose one lift of $d$ in the latter case, and apply the induction to the kernel. Second,

$$H=\{\theta:e^{ik\cdot\theta}=1\text{ for every }k\in L(H)\}. \tag{B.1}$$

Here is a direct justification of the separation statement (B.1). With a translation-invariant metric on the torus, the continuous function $r(\theta)=\operatorname{dist}(\theta,H)$ is invariant under translations by $H$. Its Fourier coefficient at $k$ vanishes unless $k\in L(H)$: translate the defining integral by any $h\in H$. Product Fejér convolutions approximate $r$ uniformly, and contain only Fourier coefficients already present in $r$. The uniform approximation follows from positivity, integral one, and concentration at the origin of the kernels

$$F_m(u)=\frac1m\left|\sum_{j=0}^{m-1}e^{iju}\right|^2;$$

outside any fixed neighborhood of zero, $F_m(u)\le1/[m\sin^2(u/2)]$. If $\theta$ annihilates every character in $L(H)$, each of these approximating polynomials takes the same value at $\theta$ and zero. Passing to the uniform limit yields $r(\theta)=r(0)=0$, proving (B.1).

Choose finite generators $k_1,\ldots,k_r$ of $L(H)$. For each fixed generator, Hausdorff convergence implies that all values of $e^{ik_i\cdot\theta}$, with $\theta\in H_j$, eventually lie in a proper small arc around one. But its image on $H_j$ is either the singleton $\{1\}$ or the entire circle: it is parametrized by $e^{i(k_i\cdot a^{(j)})x}$. The latter possibility is excluded. Therefore every generator annihilates $H_j$ for all sufficiently large $j$, simultaneously. By (B.1), $H_j\subset H$ eventually.

Finally set $\Phi(\theta)=-\sum_{i=1}^N\cos\theta_i$. Uniform continuity and Hausdorff convergence give

$$\max_H\Phi=\lim_j\max_{H_j}\Phi=\kappa_N.$$

For every sufficiently large $j$, inclusion gives $K(A_j)=\max_{H_j}\Phi\le\kappa_N$. The definition of the infimum gives the reverse inequality. Hence these original admissible sets attain $\kappa_N$. $\square$

The limit subgroup is an auxiliary compactness device; it need not itself be a one-dimensional orbit with distinct positive coordinate frequencies. The proof avoids that difficulty by recovering equality on original members of the sequence. It provides neither a search cutoff nor a stopping rule for a frequency enumeration.

C. Exact moments and the polynomial hierarchy

Write $M_r(A)=\mathbb E f_A^r$, including $M_0=1$. With $S=A\cup(-A)$, expansion and Fourier orthogonality give

$$M_r=2^{-r}\#\{(s_1,\ldots,s_r)\in S^r: s_1+\cdots+s_r=0\},\qquad r\ge1. \tag{C.1}$$

The tuples are ordered and repetitions are allowed. Thus every moment is rational, and its numerator is an exact count of integer relations. Equality modulo a finite integer is not the condition in (C.1).

For $d\in\mathbb Z_{\ge0}$, with $p$ ranging over real polynomials, define

$$B_d(A)=-\min_{p\ne0,\,\deg p\le d} \frac{\mathbb E[f_Ap(f_A)^2]}{\mathbb E[p(f_A)^2]}. \tag{C.2}$$

Only a polynomial of the value of the original sum is introduced. This construction does not change any amplitude or phase in $f_A$.

Proposition C.1 — A finite, attained optimization problem. Let

$$H_d=(M_{i+j})_{i,j=0}^d, \qquad F_d=(M_{i+j+1})_{i,j=0}^d.$$

The matrix $H_d$ is positive definite, and $B_d$ is the negative of the smallest generalized eigenvalue of $(F_d,H_d)$. In particular,

$$0=B_0\le B_1\le B_2\le\cdots\le K(A). \tag{C.3}$$

Proof. If $u$ is the coefficient vector of $p$, the denominator and numerator in (C.2) are $u^TH_du$ and $u^TF_du$. The continuous nonconstant function $f_A$ takes every value in the nondegenerate interval $[-K(A),N]$. A nonzero polynomial cannot vanish on this interval. At some point $p(f_A)\ne0$, and by continuity its square is positive on a set of positive measure. Thus $u^TH_du>0$ for nonzero $u$.

The normalization $u^TH_du=1$ is a compact ellipsoid, so the minimum is attained. Standard diagonalization of $H_d^{-1/2}F_dH_d^{-1/2}$ gives the eigenvalue formulation. Constants give quotient zero, enlarging the degree enlarges the admissible class, and every quotient is at least $-K(A)$. These observations prove (C.3). $\square$

Let $\nu_A$ be the pushforward of $\mu$ under $x\mapsto f_A(x)$. Equation (C.2) is the univariate measure-based hierarchy applied to $\nu_A$. Allowing a sum of squares $\sum_jp_j(t)^2$ instead of one square does not improve its optimum: its quotient is a weighted average of the individual quotients, with nonnegative weights $\mathbb E p_j(f_A)^2$. This makes the connection to Lasserre2021 exact, rather than merely analogous.

Proposition C.2 — Pointwise convergence in the frequency set. For each fixed admissible $A$, $B_d(A)\to K(A)$.

Proof. Set $m=\min f_A$, and choose $p_d(t)=(N-t)^d$. Fix $0<\eta<N-m$. The set $E=\{f_A\le m+\eta/2\}$ has positive measure. The unnormalized squared weight is at least $(N-m-\eta/2)^{2d}$ on $E$, and at most $(N-m-\eta)^{2d}$ on $\{f_A\ge m+\eta\}$. Consequently the normalized weight of the latter set tends to zero. Since $0\le f_A-m\le2N$, the corresponding Rayleigh quotient has limit superior at most $m+\eta$. Let $\eta\downarrow0$, then use $B_d\le K$. $\square$

This limit does not assert a rate uniform in $A$. Sections D and E provide a particular uniform rate. Neither statement supplies a lower bound of order $\sqrt N$ for the limiting value.

C.1. The first nontrivial level and a useful special class

The two matrices at $d=1$ are

$$H_1=\begin{pmatrix}1&0\\0&M_2\end{pmatrix},\qquad F_1=\begin{pmatrix}0&M_2\\M_2&M_3\end{pmatrix}.$$

The least $k$ for which $F_1+kH_1$ is positive semidefinite is the positive root of $M_2k^2+M_3k-M_2^2=0$. Hence

$$B_1(A)=\frac{\sqrt{M_3^2+4M_2^3}-M_3}{2M_2}. \tag{C.4}$$

If $R(A)=\#\{(a,b)\in A^2:a+b\in A\}$, then $M_3=3R(A)/4$: there are $6R(A)$ ordered signed zero-sum triples in (C.1). Thus, whenever $A\cap(A+A)=\varnothing$,

$$K(A)\ge B_1(A)=\sqrt{N/2}. \tag{C.5}$$

This is a result for an explicitly restricted class. The condition forbids $a+a=b$ as well as relations involving three different frequencies.

D. A phase-mass lemma independent of the highest frequency

A bound obtained from $\|f_A'\|_\infty\le\sum a$ deteriorates when the frequencies are large. The following different argument depends on the number of harmonics and not their magnitudes. It retains their common phase parameter throughout.

Lemma D.1 — Uniform mass near a minimum. Let $m=\min f_A$, let $\delta>0$, and choose an integer $M\ge2\pi N/\delta$. Then

$$\mu\{x:f_A(x)\le m+\delta\}\ge M^{-N}. \tag{D.1}$$

Proof. Partition $[0,2\pi)^N$ into $M^N$ half-open cubes of side $2\pi/M$. Pull this partition back to $\mathbb T$ by

$$x\longmapsto (a_1x,\ldots,a_Nx)\pmod{2\pi}.$$

One preimage $E$ has measure at least $M^{-N}$. Choose $x_0\in E$, and a minimizer $x_*$ of $f_A$. For each $x\in E$, every coordinate of $a(x-x_0)$ modulo $2\pi$ has a representative of absolute value at most $2\pi/M$, because the two phase points lie in the same cube. Since cosine is 1-Lipschitz,

$$f_A(x+x_*-x_0)\le f_A(x_*)+\frac{2\pi N}M\le m+\delta.$$

Translation preserves $\mu$. The translate of $E$ lies in the asserted sublevel set and retains its measure. $\square$

A phase cube is used only to locate a set of common times with similar relative phases. No step replaces the one-dimensional orbit by a product distribution of independent phases. The mass bound can be extremely small; it is a guarantee, not a claim that the minimum is easy to encounter in a short experiment.

E. Uniform approximation using a Chebyshev filter

Theorem E.1 — A sufficient degree depending only on $N$ and accuracy. Suppose $N\ge1$ and $0<\varepsilon\le N$. Define

$$M=\left\lceil\frac{8\pi N}{\varepsilon}\right\rceil, \qquad d_*= \left\lceil \sqrt{\frac N\varepsilon} \left[N\log M+\log\left(\frac{16N}{\varepsilon}\right)\right] \right\rceil. \tag{E.1}$$

For every integer $d\ge d_*$ and every admissible $A$ with $|A|=N$,

$$0\le K(A)-B_d(A)\le\varepsilon. \tag{E.2}$$

Here and below $\log$ is natural logarithm. There is no restriction on $\max A$.

Proof. Set $m=\min f_A$ and

$$D=N-m-\varepsilon/2, \qquad u(t)=\frac{2t-(N+m+\varepsilon/2)}D.$$

Because $-N\le m<0$ and $\varepsilon\le N$, we have $0<N/2<D\le2N$. The affine map $\nu$ sends $[m+\varepsilon/2,N]$ to $[-1,1]$. Choose

$$p(t)=T_d(\nu(t)),$$

where $T_d$ is the Chebyshev polynomial of the first kind. The identities

$$T_d(\cos\theta)=\cos(d\theta),\qquad T_d(-\cosh v)=(-1)^d\cosh(dv) \tag{E.3}$$

follow, for example, from $T_d((z+z^{-1})/2)=(z^d+z^{-d})/2$ NISTDLMF. They imply $|p(f_A)|\le1$ wherever $f_A\ge m+\varepsilon/2$.

On the smaller sublevel set $G=\{f_A\le m+\varepsilon/4\}$,

$$\nu(f_A)\le-1-\frac{\varepsilon}{2D} \le-1-\frac{\varepsilon}{4N}.$$

Put $s=\varepsilon/(4N)$, so $0<s\le1/4$. Direct integration gives

$$\operatorname{arcosh}(1+s) =\int_0^s\frac{dv}{\sqrt{v(v+2)}} \ge\frac{2\sqrt s}{\sqrt{2+s}} \ge\sqrt s. \tag{E.4}$$

The last inequality holds for $s\le2$, hence in the stated domain. By (E.3), on $G$,

$$|p(f_A)|\ge\frac12 \exp\left(\frac d2\sqrt{\frac\varepsilon N}\right).$$

Lemma D.1 with $\delta=\varepsilon/4$ yields $\mu(G)\ge M^{-N}$. Thus the normalization satisfies

$$W:=\mathbb E p(f_A)^2 \ge\frac14M^{-N}\exp\left(d\sqrt{\frac\varepsilon N}\right). \tag{E.5}$$

On $\{f_A<m+\varepsilon/2\}$, the excess $f_A-m$ is at most $\varepsilon/2$. On its complement, $f_A-m\le2N$ and $p(f_A)^2\le1$. Splitting this expectation gives

$$\frac{\mathbb E[(f_A-m)p(f_A)^2]}W \le\frac\varepsilon2+\frac{2N}W.$$

The remaining term satisfies

$$\frac\varepsilon2+\frac{2N}W \le\frac\varepsilon2+ 8NM^N\exp\left(-d\sqrt{\frac\varepsilon N}\right) \le\varepsilon. \tag{E.6}$$

The final inequality is precisely the degree condition (E.1). The quotient for this admissible $p$ is at most $m+\varepsilon$, so the optimized value satisfies $B_d\ge K-\varepsilon$. Its opposite bound follows from Proposition C.1. $\square$

E.1. Why the proof is not a circular algorithm

The comparison polynomial used in the proof contains the unknown minimum $m$. It proves existence of a polynomial in the admissible space with the required quotient. Computing $B_d$ means optimizing over that whole space through the matrices $F_d,H_d$, whose entries are the exact counts (C.1); its definition does not require $m$. The proof does not claim an explicit list of optimal coefficients without performing that optimization.

The matrix dimension is $d+1$, but a small dimension alone does not imply a low total cost. Moment enumeration, large integer sizes, conditioning, and exact eigenvalue certification all remain relevant. No large matrix at the worst-case sufficient degree was evaluated merely by proving this theorem.

E.2. A completely rational sufficient certificate

For a rational $\varepsilon>0$, no rounded logarithm, square root, or value of $\pi$ is necessary to certify a sufficient degree. Choose

$$M_0,s_0\in\mathbb Z_{>0},\qquad q,r\in\mathbb Z_{\ge0},$$

such that

$$M_0\varepsilon\ge32N,\quad s_0^2\varepsilon\ge N,\quad 2^q\ge M_0,\quad 2^r\varepsilon\ge16N. \tag{E.7}$$

Then

$$d_0=s_0(Nq+r) \tag{E.8}$$

is sufficient for (E.2). Indeed, $\pi<4$ makes $M_0$ a permitted choice in the phase-mass estimate, and positivity of $s_0$ gives

$$d_0\sqrt{\varepsilon/N}\ge Nq+r.$$

Furthermore,

$$e^{Nq+r}\ge2^{Nq+r} \ge M_0^N\frac{16N}{\varepsilon}.$$

Substitution into (E.6) proves the claim. The requirement $s_0>0$ is essential; the squared inequality by itself would allow a negative integer and would not imply the needed inequality for the exponent.

E.3. Comparison with the elementary power filter

For $0<\varepsilon\le1$, the earlier choice $p(t)=(N-t)^d$ gives the sufficient degree

$$d\ge\left\lceil \frac{2N^2}{\varepsilon} \log\left\lceil\frac{4\pi N}{\varepsilon}\right\rceil \right\rceil. \tag{E.9}$$

Here is a short derivation. Let $g=N-f_A$, $G=\max g=N+K$, and $r_0=2d$. Lemma D.1 gives

$$\|g\|_{r_0}\ge(G-\varepsilon/2) \exp\left(-\frac{N\log M'}{r_0}\right), \quad M'=\left\lceil\frac{4\pi N}{\varepsilon}\right\rceil.$$

Using $G\le2N$ and $1-e^{-u}\le u$ yields

$$G-\|g\|_{r_0}\le\frac\varepsilon2+ \frac{2N^2\log M'}{r_0}\le\varepsilon.$$

Hölder’s inequality gives $\mathbb E g^{r_0+1}/\mathbb E g^{r_0}\ge\|g\|_{r_0}$. The quotient associated with $(N-t)^d$ therefore proves (E.9).

The sufficient order improves from $O(N^2\varepsilon^{-1}\log(2N/\varepsilon))$ to $O(N^{3/2}\varepsilon^{-1/2}\log(2N/\varepsilon))$. This is a comparison of the two bounds proved in this report, not a claim to improve the best known general polynomial-optimization theory. In particular, SlotLaurent2021 studies convergence of the same pushforward approach using polynomial concentration. Its constants and geometric hypotheses must be compared carefully before any claim of novelty for (E.1).

For $\varepsilon=\alpha\sqrt N$ with fixed $0<\alpha\le1$, (E.1) gives $d=O_\alpha(N^{5/4}\log(2N))$. Approximating an unknown minimum to that accuracy still does not prove that its magnitude is at least a positive multiple of $\sqrt N$.

F. Consecutive frequencies and low-degree certificates

The interval family $A_N=\{1,\ldots,N\}$ is especially useful for testing a proposed certificate. It has strong arithmetic relations and a directly visible deep negative excursion.

Put $L=2N+1$. Summing a geometric progression gives

$$D_N(x):=1+2f_{A_N}(x) =\sum_{a=-N}^Ne^{iax} =\frac{\sin(Lx/2)}{\sin(x/2)}, \tag{F.1}$$

with its continuous value $L$ at multiples of $2\pi$. At $x=3\pi/L$,

$$K(A_N)\ge\frac12+\frac1{2\sin(3\pi/(2L))} \ge\frac12+\frac L{3\pi}. \tag{F.2}$$

The final step uses $\sin u\le u$ with $0<u\le\pi/2$. Thus any certificate that stays bounded on this family is missing an actual minimum of at least linear depth.

F.1. Exact formulas for all required moments

For an integer $r\ge1$, inclusion–exclusion gives

$$C_r(N):=\mathbb E D_N^r =\sum_{j=0}^{\lfloor r/2\rfloor}(-1)^j\binom rj \binom{(r-2j)N+r-1-j}{r-1}, \qquad C_0(N)=1. \tag{F.3}$$

Set $\binom uv=0$ for integers $0\le u<v$. To prove (F.3), shift each exponent in a zero-sum tuple from $[-N,N]$ to $[0,2N]$. The sum of the shifted entries must be $rN$. Unrestricted nonnegative solutions are counted by stars and bars; choosing $j$ entries that exceed $2N$ and subtracting $2N+1$ from them gives the displayed sum. Terms beyond $\lfloor r/2\rfloor$ cannot contribute. Then

$$M_r(N)=2^{-r}\sum_{j=0}^r(-1)^{r-j}\binom rjC_j(N). \tag{F.4}$$

In particular,

$$M_2=\frac N2,\quad M_3=\frac{3N(N-1)}8,\quad M_4=\frac{N(8N^2-6N+7)}{24}. \tag{F.5}$$

For $r\ge2$, the leading coefficient is

$$\lim_{N\to\infty}\left(\frac{M_r(N)}{N^{r-1}}\right) =\frac1{2^r(r-1)!} \sum_{j=0}^{\lfloor r/2\rfloor} (-1)^j\binom rj(r-2j)^{r-1}. \tag{F.6}$$

For $r=2,3,4,5,6,7$, these values are respectively

$$\frac12,\quad\frac38,\quad\frac13,\quad \frac{115}{384},\quad\frac{11}{40},\quad\frac{5887}{23040}.$$

These are finite combinatorial identities, not a conclusion from simulated limiting distributions.

F.2. The first two levels can miss the scale

Equation (C.4) and (F.5) give $B_1(A_N)\to2/3$. The second level also stays bounded:

$$B_2(A_N)\longrightarrow\frac{24}{11}. \tag{F.7}$$

To check the latter assertion, use the basis $(1,f/N,(f/N)^2)$. For fixed $k$, the lower-right block of $F_2+kH_2$ tends to

$$Q=\begin{pmatrix}3/8&1/3\\1/3&115/384\end{pmatrix}, \qquad\det Q=\frac{11}{9216}>0,$$

and the off-diagonal vector linking it to the first coordinate tends to $h=(1/2,3/8)^T$. The first diagonal entry is $k$; all remaining contributions from $kH_2$ tend to zero. The limiting Schur complement is

$$k-h^TQ^{-1}h=k-\frac{24}{11}.$$

For every fixed $k>24/11$, the matrix is positive definite for all sufficiently large $N$; for every fixed $k<24/11$, it is not positive semidefinite for all sufficiently large $N$. The threshold characterization of $B_2$ proves (F.7).

F.3. A cubic polynomial detects a linear depth on this family

Choose

$$p_N(t)=t(1-t/N)(1-2t/N).$$

Direct substitution of (F.3)–(F.4) in its Rayleigh quotient gives

$$B_3(A_N)\ge R_N= \frac{56N^5+1005N^4+5750N^3+12495N^2+11474N+3780} {12(43N^4+465N^3+1175N^2+1125N+372)}. \tag{F.8}$$

To make this algebra independently reproducible, write

$$\begin{aligned} U_N&=M_3-6M_4/N+13M_5/N^2-12M_6/N^3+4M_7/N^4,\\ V_N&=M_2-6M_3/N+13M_4/N^2-12M_5/N^3+4M_6/N^4. \end{aligned}$$

The identities being used are exactly

$$-N^4U_N= \frac{56N^6+1005N^5+5750N^4+12495N^3+11474N^2+3780N}{5760},$$

$$N^4V_N= \frac{43N^5+465N^4+1175N^3+1125N^2+372N}{480}.$$

They follow by polynomial expansion of the finite binomial formulas above. Every coefficient of $V_N$ is positive. Subtracting $(14/129)N(N^4V_N)$ from $-N^4U_N$ leaves, in ascending powers of $N$, the coefficients

$$0,\quad\frac{21}{32},\quad\frac{47255}{24768},\quad \frac{31619}{16512},\quad\frac{18145}{24768},\quad \frac{1145}{16512},\quad0.$$

Consequently $R_N>(14/129)N$ for every $N\ge1$. The direct bound (F.2) has a larger linear coefficient. The cubic calculation is useful because it demonstrates a change of scale within the hierarchy, not because it improves the interval’s elementary bound. No analogous all-set lower bound for $B_3$ is established here.

G. The lattice operator and a finite open box

Let $T$ be the bilateral shift on $\ell^2(\mathbb Z)$,

$$(T\psi)(n)=\psi(n-1),\qquad \mathcal H_A=\frac12\sum_{a\in A}(T^a+T^{-a}). \tag{G.1}$$

This is a bounded self-adjoint operator with norm at most $N$. In the language of a tight-binding model, the admissible hopping distances are the elements of $A$. The factor $1/2$ ensures that the two opposing directions contribute one unit-amplitude cosine after Fourier transformation. This is a mathematical model; arbitrary long hopping distances are not asserted to occur in a particular material.

Proposition G.1 — Exact spectral identity.

$$\operatorname{spec}(\mathcal H_A)=[-K(A),N], \qquad -K(A)=\inf_{\|\psi\|_2=1} \langle\psi,\mathcal H_A\psi\rangle. \tag{G.2}$$

Proof. The unitary Fourier map $U\psi(x)=\sum_n\psi(n)e^{inx}$ from $\ell^2(\mathbb Z)$ to $L^2(\mathbb T,\mu)$ sends $T$ to multiplication by $e^{ix}$. It sends $\mathcal H_A$ to multiplication by $f_A(x)$. The spectrum of a multiplication operator is its essential range. For this continuous function, every neighborhood of an attained value has a preimage of positive measure, so essential range and range agree. The range is the whole interval between its minimum and maximum by continuity and connectedness of $\mathbb T$. The Rayleigh infimum is at least the minimum; normalized indicators of small neighborhoods of a minimizing point approach it from above. $\square$

For the localized state $\delta_0$,

$$\langle\delta_0,\mathcal H_A\delta_0\rangle=0, \qquad \langle\delta_0,\mathcal H_A^2\delta_0\rangle=N/2.$$

Moreover, $U p(\mathcal H_A)\delta_0=p(f_A)$. The quotient in (C.2) is exactly the energy expectation of this filtered state after normalization. This gives a useful physical interpretation of the polynomial filter without changing the original waveform.

Discreteness of the lattice does not imply a discrete spectrum here. A nonconstant trigonometric polynomial takes any particular value on only finitely many points of the circle. A nonzero eigenfunction of multiplication by $f_A$ would have to be supported on such a set of measure zero, which is impossible in $L^2$. The spectral edge is approached by states; it is not a normalizable isolated ground state of this infinite translation-invariant operator.

G.1. Open boundaries and the exact loss at an edge

Restrict the lattice to $0,\ldots,L-1$, for an integer $L\ge1$, and let $\mathcal H_{A,L}$ have entry $1/2$ between indices $i,j$ when $|i-j|\in A$, and zero otherwise. Put

$$B_L^{\mathrm{box}}(A)=-\lambda_{\min}(\mathcal H_{A,L}).$$

Proposition G.2 — Finite-box bound.

$$0\le B_L^{\mathrm{box}}\le B_{L+1}^{\mathrm{box}}\le K(A), \qquad 0\le K(A)-B_L^{\mathrm{box}}(A) \le\sum_{a\in A}\min(a/L,1). \tag{G.3}$$

Proof. The matrix has trace zero, so its smallest eigenvalue is nonpositive. Extending a vector by zero embeds each box into the next and into the full lattice; the variational principle proves the first chain. For the normalized vector $\psi_x(n)=e^{inx}/\sqrt L$ in the box, counting the $(L-a)_+$ surviving pairs at distance $a$ gives

$$\langle\psi_x,\mathcal H_{A,L}\psi_x\rangle =\sum_{a\in A}(1-a/L)_+\cos(ax).$$

Its difference from $f_A(x)$ has absolute value at most the sum in (G.3). Evaluate at a minimizer of $f_A$ and use the minimum-eigenvalue variational principle again. $\square$

For a fixed set $A$, the right side tends to zero. If $L>\max A$, it equals $\sum a/L$. However, no fixed box length depending only on $N$ provides a positive certificate for all unrestricted frequencies: for $A=\{a\}$ and $a\ge L$, there are no surviving edges, so $B_L^{\mathrm{box}}=0$ although $K(A)=1$.

For comparison, if all frequencies divided by their gcd are odd, evaluation at the corresponding half-period makes every cosine equal $-1$, and $K=N$. In the lattice model the parity transformation $\psi(n)\mapsto(-1)^n\psi(n)$ changes the sign of every such hopping term. This elementary favorable case cannot be assumed for arbitrary sets.

H. A discrete grid can miss the minimum

Theorem H.1 — Exact energy does not imply an exact negative peak. For $A_N=\{1,\ldots,N\}$, let $q=2N+1$ and $x_j=2\pi j/q$, $0\le j<q$. Then

$$f_{A_N}(x_0)=N,\qquad f_{A_N}(x_j)=-\frac12\quad(1\le j<q), \tag{H.1}$$

while

$$\frac1q\sum_{j=0}^{q-1}f_{A_N}(x_j)^2=\frac N2, \qquad K(A_N)\ge\frac12+\frac q{3\pi}. \tag{H.2}$$

Proof. The terms in $D_N(x_j)$ from (F.1) run through a complete set of $q$-th roots of unity. Their sum is zero for $j\ne0$, and $q$ for $j=0$. This proves (H.1). The square average equals $(N^2+2N/4)/(2N+1)=N/2$. The final bound is (F.2). $\square$

Thus the grid depth is $1/2$ while the continuous depth grows at least linearly. This does not mean the grid has lost the information needed for exact interpolation. A trigonometric polynomial of degree $N$ is uniquely determined by these $2N+1$ values: their discrete Fourier transform recovers its coefficients. The mistake is to identify the minimum of the interpolated curve with the minimum of the samples. Interpolation is not a convex averaging rule, and values between samples can fall below every sample.

The finite cyclic model makes this discrepancy especially transparent. Let $T_0$ be the cyclic shift on $q$ coordinates. Since the residues $\pm1,\ldots,\pm N$ are exactly the nonzero residues modulo $q$,

$$\mathcal H_0=\frac12\sum_{a=1}^N(T_0^a+T_0^{-a}) =\frac{J-I}2,$$

where $J$ has every entry equal to one. Its eigenvalues are $N$, once, and $-1/2$, with multiplicity $2N$. For every integer $k\ge0$,

$$2^k\frac1q\sum_j f_{A_N}(x_j)^k =\frac{(2N)^k+2N(-1)^k}{2N+1}\in\mathbb Z. \tag{H.3}$$

The divisibility follows from $2N\equiv-1\pmod{2N+1}$. These quantities also count closed walks from a vertex in the complete graph. Therefore positivity of moment matrices, the correct second moment, and integrality of the doubled-signal moments can coexist with a grid depth of only $1/2$. Those conditions alone do not enforce the exact integer relations in (C.1).

H.1. Twisting the boundary restores the continuous spectrum

For a real $\phi$, define a unitary shift on the same basis by

$$T_\phi e_j=e_{j+1}\ (j<q-1),\qquad T_\phi e_{q-1}=e^{i\phi}e_0.$$

Then $T_\phi^q=e^{i\phi}I$. Its eigenvalues are the $q$ distinct roots $z_j=\exp(i(2\pi j+\phi)/q)$. For any admissible $A$, retain the actual integer powers and define

$$\mathcal H_\phi=\frac12\sum_{a\in A}(T_\phi^a+T_\phi^{-a}).$$

Its eigenvalues are exactly

$$E_j(\phi)=f_A\left(\frac{2\pi j+\phi}{q}\right).$$

As $\phi$ ranges over $[0,2\pi]$, these arguments cover the whole circle. Hence

$$\min_{0\le\phi\le2\pi}\lambda_{\min}(\mathcal H_\phi) =\min_x f_A(x)=-K(A). \tag{H.4}$$

For the interval example, $\phi=\pi$ and $j=1$ already give the deep witness $x=3\pi/q$. For frequencies exceeding $q$, reducing exponents modulo $q$ without keeping the accumulated factor $e^{i\phi}$ would be an error. The formula requires the actual powers.

Equation (H.4) is an exact reformulation, not a proof that some boundary twist always yields energy below $-c\sqrt N$. That last assertion is still the original unresolved lower-bound problem. The physical use of tunable twisted boundary conditions has established precedents; for example, Bardyn, Huber, and Zilberberg, New Journal of Physics 16, 123013 (2014), arXiv:1312.6894. Our formulas follow directly from the finite shift defined above.

H.2. Certified errors for a general sampling grid

For an integer $M\ge1$, sample the original function at $2\pi j/M$, and let $K_M=-\min_{0\le j<M}f_A(2\pi j/M)$. Define $S_1=\sum_{a\in A}a$ and $S_2=\sum_{a\in A}a^2$. Then

$$K_M\le K(A)\le K_M+ \min\left(\frac{\pi S_1}{M},\frac{\pi^2S_2}{2M^2}\right). \tag{H.5}$$

To prove this, take a true minimizer $x_*$ and its nearest grid point, at circular distance at most $\pi/M$. The derivative satisfies $|f_A'|\le S_1$, giving the first error bound by the mean value theorem. At a minimizing point, $f_A'(x_*)=0$; Taylor's theorem and $|f_A''|\le S_2$ give the second. The grid minimum is no greater than the value at this particular grid point, and no smaller than the true minimum. These two comparisons prove (H.5). The lower endpoint may be replaced by $\max(1,K_M)$ using Proposition A.1. Unlike Theorem E.1, these sampling bounds explicitly depend on the sizes of the frequencies.

I. Why integer amplitudes do not preserve the conjecture

A tempting extension replaces equal amplitudes by positive weights and asks for an absolute $c>0$ such that

$$-\min_x\sum_{a=1}^Nw_a\cos(ax) \ge c\left(\sum_{a=1}^Nw_a^2\right)^{1/2}. \tag{I.1}$$

The normalization reflects the RMS, whose square is $\sum w_a^2/2$. This extension is false even for positive integer weights.

Proposition I.1 — The Fejér amplitude barrier. Put $w_a=1-a/(N+1)$, $1\le a\le N$. Then

$$-\min_x\sum_{a=1}^Nw_a\cos(ax)=\frac12, \quad \sum_{a=1}^Nw_a^2=\frac{N(2N+1)}{6(N+1)}. \tag{I.2}$$

Proof. Expanding a finite squared modulus gives

$$\frac1{N+1}\left|\sum_{j=0}^Ne^{ijx}\right|^2 =1+2\sum_{a=1}^N\left(1-\frac a{N+1}\right)\cos(ax)\ge0.$$

At each nontrivial $(N+1)$-st root of unity, the geometric sum vanishes, so the bound $-1/2$ is attained. For the norm, set $j=N+1-a$ and use $\sum_{j=1}^Nj^2=N(N+1)(2N+1)/6$. $\square$

Multiplication by $N+1$ produces integer amplitudes $v_a=N+1-a$. Their negative peak equals $(N+1)/2$, and

$$\frac{(N+1)/2}{\sqrt{\sum v_a^2}} =\sqrt{\frac{3(N+1)}{2N(2N+1)}}\longrightarrow0. \tag{I.3}$$

This disproves (I.1) for every proposed positive absolute constant. A fixed amplitude quantum merely multiplies both numerator and denominator and cannot remove the obstruction.

These weights cannot be relabeled as distinct unit-amplitude frequencies: repeated sinusoids at the same frequency combine into a larger coefficient. The original assumption is equal nonzero coefficients at distinct frequencies, which is much stronger than requiring a discrete amplitude step. Likewise, a quantum occupation number does not by itself specify the signed mean field of a harmonic component; the state and its phase information matter. No theorem for (A.1) automatically extends to arbitrary music with unequal amplitudes and phases.

J. A positive Riesz-product certificate

Define

$$P_A(x)=\prod_{a\in A}(1-\cos(ax)),\qquad Z_A=\mathbb E P_A, \qquad B_R(A)=-\frac{\mathbb E[f_AP_A]}{Z_A}. \tag{J.1}$$

The product is nonnegative and positive outside a finite union of zeros, so $Z_A>0$. It follows immediately that $B_R(A)\le K(A)$. This proof does not establish that $B_R\ge0$ for every set; $\max(0,B_R)$ remains a valid certificate even if a negative value occurs. The use of positive Riesz products is classical, and no new general inequality for them is claimed here.

Put

$$Q_A(z)=\prod_{a\in A}(2-z^a-z^{-a}) =\sum_jq_jz^j.$$

Its coefficients are integers with $q_j=q_{-j}$, and $Q_A(e^{ix})=2^NP_A(x)$. Extracting constant coefficients proves

$$Z_A=\frac{q_0}{2^N},\qquad B_R(A)=-\frac{\sum_{a\in A}q_a}{q_0}. \tag{J.2}$$

An independent implementation can use the identity $Q_A(e^{ix})=|\prod_{a\in A}(1-e^{iax})|^2$ and integer autocorrelation. No sampling of trigonometric values is needed for either formula.

For $A=\{1,4,\ldots,4^{N-1}\}$, signed digit representations with digits $-1,0,1$ are unique. Indeed, subtracting two representations leaves digits between $-2$ and 2; the largest nonzero term has magnitude at least $4^m$, while the total possible magnitude of all smaller terms is $2(4^m-1)/3<4^m$. Thus the only constant contribution in $P_A$ is one and the coefficient at each $e^{\pm i4^jx}$ is $-1/2$. Consequently

$$Z_A=1,\qquad B_R(A)=N/2. \tag{J.3}$$

Short additive relations change this calculation. For $A=\{1,2,3\}$, exact multiplication gives $q_0=6$ and $(q_1,q_2,q_3)=(0,-3,-2)$. Hence $B_R=5/6<\sqrt3/2$, since $25/36<3/4$. The conjectured coefficient $1/2$ for this detector is therefore false.

A stronger finite example is

$$A=\{1,2,3,4,5,7,9,11,13\}.$$

Here $q_0=20$, and the coefficients at the listed frequencies are

$$(4,-2,-7,-10,-2,4,-1,-3,-4).$$

Their sum is $-21$, so $B_R=21/20$ and $B_R/\sqrt9=7/20$. This excludes any larger universal coefficient for this particular detector.

Finite computational result. The archived exact enumeration of all $2^{18}-1=262{,}143$ nonempty subsets of $\{1,\ldots,18\}$ finds the minimum ratio $B_R/\sqrt{|A|}$ to be exactly $7/20$. Comparisons use the exact sign and the rational number $B_R^2/|A|$, not rounded square roots. A second coefficient method was compared for all 4,095 nonempty subsets of $\{1,\ldots,12\}$. This is a finite-domain minimum. It neither proves $7/20$ valid for unrestricted sets nor rules out a smaller positive universal coefficient. It is also a result about $B_R$, not the actual extremum $K$.

K. Negative moments, quantiles, and a saturating detector

Let $Y_A=(f_A)_-=\max(-f_A,0)$. Every probability density $\rho\ge0$ with $\mathbb E\rho=1$ yields a valid bound $-\mathbb E(\rho f_A)\le K(A)$. Many ideas in the investigation amount to selecting such a density, or to measuring a norm or quantile of $Y_A$. Their validity as lower bounds must be distinguished from their ability to retain a uniform square-root scale.

K.1. The first absolute moment and the fourth moment

Since $\mathbb E f_A=0$, its positive and negative parts have equal integrals. Therefore

$$K(A)\ge\mathbb E Y_A=\frac12\|f_A\|_1. \tag{K.1}$$

Hölder’s inequality, applied to $|f|^{2/3}|f|^{4/3}$, gives

$$M_2\le\|f\|_1^{2/3}M_4^{1/3}, \quad K(A)\ge\frac{M_2^{3/2}}{2\sqrt{M_4}} =\frac{N^{3/2}}{4\sqrt2\sqrt{M_4}}. \tag{K.2}$$

A bound $M_4/M_2^2\le C$ by an absolute constant would make (K.2) a square-root certificate on that class. It is false as a uniform assumption: (F.5) gives $M_4/M_2^2\sim4N/3$ for intervals.

There is an arithmetic caution in interpreting the fourth moment. Define the ordered counts

$$E(A)=\#\{(a,b,c,d)\in A^4:a+b=c+d\},$$

$$T(A)=\#\{(a,b,c,d)\in A^4:a+b+c=d\}.$$

Separating two-positive/two-negative and three-positive/one-negative signed tuples in (C.1) yields

$$M_4=\frac38E(A)+\frac12T(A). \tag{K.3}$$

The second term cannot simply be omitted. There are six sign placements in the first case and eight in the second, followed by division by $2^4$.

On the interval family, the elementary estimate

$$|D_N(x)|\le\min(L,\pi/|x|),\qquad0<|x|\le\pi, \tag{K.4}$$

follows from $\sin(|x|/2)\ge|x|/\pi$. Splitting its integral at $\pi/L$ gives

$$\|D_N\|_1\le1+\log L,\qquad \mathbb E Y_{A_N}\le\frac12+\frac14\log L. \tag{K.5}$$

Hence the certificate (K.1), divided by $\sqrt N$, tends to zero on a family whose true minimum satisfies (F.2). The obstruction concerns the detector, not the conjecture.

K.2. Higher negative moments

For $0<p<\infty$, set $L_p^-(A)=(\mathbb E Y_A^p)^{1/p}$. For $p<1$ this is a quasi-norm. On a probability space, Jensen’s inequality gives

$$L_p^-(A)\le L_q^-(A)\le K(A)\quad(0<p<q), \qquad \lim_{p\to\infty}L_p^-(A)=K(A). \tag{K.6}$$

To prove the limit, every set $\{Y_A>K-\eta\}$ has positive measure; its contribution gives $L_p^-(A)\ge(K-\eta)\mu(\{Y_A>K-\eta\})^{1/p}$. Let $p\to\infty$, then $\eta\downarrow0$.

For fixed $p$, the interval family has the orders

$$L_p^-(A_N)= \begin{cases} \Theta_p(1),&0<p<1,\\ \Theta(\log N),&p=1,\\ \Theta_p(N^{1-1/p}),&p>1. \end{cases} \tag{K.7}$$

Here constants may depend on the fixed $p$, and the orders refer to $N\to\infty$. We give the details because a change of order at $p=1$ matters.

For $p>1$, (K.4) implies

$$\mathbb E|D_N|^p \le L^{p-1}+\frac{L^{p-1}-1}{p-1}.$$

Since $|f_{A_N}|^p\le(|D_N|^p+1)/2$, this proves the upper bound. For a lower bound, if $N\ge2$, use the disjoint arcs

$$E_N=\left[\frac{8\pi}{3L},\frac{10\pi}{3L}\right] \cup\left[-\frac{10\pi}{3L},-\frac{8\pi}{3L}\right]. \tag{K.8}$$

On the positive arc, $Lx/2\in[4\pi/3,5\pi/3]$, so $\sin(Lx/2)\le-1/2$, whereas $\sin(x/2)\le5\pi/(3L)$. The negative arc follows by evenness. Thus

$$\mu(E_N)=\frac2{3L},\qquad Y_{A_N}\ge\frac{3L}{20\pi}\quad\text{on }E_N. \tag{K.9}$$

Its contribution yields the matching lower order for every $p>1$.

For $p=1$, the upper bound is (K.5). To obtain the logarithmic lower bound, for $N\ge6$ put $m_0=\lfloor N/3\rfloor$ and take the arcs

$$J_k=\left[\frac{(12k+7)\pi}{3L}, \frac{(12k+11)\pi}{3L}\right],\qquad0\le k<m_0,$$

together with their reflections. They lie within $[-\pi,\pi]$. On each positive arc, $\sin(Lx/2)\le-1/2$ and $Y_{A_N}\ge1/(2x)$. Integration and $\log(b/a)\ge(b-a)/b$ give

$$\mathbb E Y_{A_N}\ge\frac2\pi \sum_{k=0}^{m_0-1}\frac1{12k+11} \ge\frac1{6\pi}\log\left(1+\frac{12m_0}{11}\right) \ge\frac1{6\pi}\log\left(1+\frac{2N}{11}\right).$$

For $0<p<1$, subadditivity of $u^p$ and (K.4) give

$$\mathbb E Y_{A_N}^p\le2^{-p}\left(1+\frac1{1-p}\right).$$

A uniform positive lower bound follows from a fixed mass of moderately negative values. On any interval of length $\ell$, the set where $\sin y\le0$ has length at least $\ell/2-\pi/2$: a primitive of $\mathbf1_{\{\sin y\le0\}}-1/2$ has oscillation $\pi/2$. On $x\in[\pi/2,\pi]$, substituting $y=Lx/2$ yields a suitable subset of length at least $\pi/4-\pi/L\ge\pi/20$ for $L\ge5$. Reflecting it gives

$$\mu\{f_{A_N}\le-1/2\}\ge1/20. \tag{K.10}$$

For $N=1$ the measure is $1/3$, so (K.10) remains valid. Consequently $L_p^-(A_N)\ge(1/2)(1/20)^{1/p}$, completing (K.7).

It follows that every fixed $p<2$ gives $L_p^-(A_N)/\sqrt N\to0$. For $p=2$, the interval family has the correct scale; it neither establishes nor refutes a uniform negative-second-moment lower bound for all frequency sets.

K.3. A fixed probability of seeing the trough is too demanding

For $0<p<1$, define the upper-tail depth quantile

$$q_A(p)=\sup\{t\ge0:\mu(Y_A\ge t)\ge p\}.$$

It is nonincreasing in $p$, satisfies $0\le q_A(p)\le K(A)$, and tends to $K(A)$ as $p\downarrow0$. The last claim follows from the positive measure of every strict near-maximum neighborhood of $Y_A$.

For intervals and $t>1/2$, (F.1) implies that $f_{A_N}(x)\le-t$ requires $|x|\le\pi/(2t-1)$, because $|f_{A_N}(x)|\le1/2+\pi/(2|x|)$. Therefore

$$\mu(f_{A_N}\le-t)\le\min\left(1,\frac1{2t-1}\right), \qquad q_{A_N}(p)\le\frac{1+1/p}{2}. \tag{K.11}$$

Every fixed $p>0$ gives a bounded quantile as $N$ grows. In contrast, (K.9) proves

$$q_{A_N}\left(\frac2{3L}\right)\ge\frac{3L}{20\pi}.$$

A fraction of time of order $1/N$ can carry a trough of order $N$. Choosing a fixed fraction of observation time is therefore a substantially stronger requirement than the existence of a deep point.

K.4. The zero-threshold Fermi–Dirac detector

For $\beta\ge0$, introduce the logistic weight

$$w_\beta(t)=\frac1{1+e^{\beta t}},\qquad B_\beta^{\mathrm{FD}}(A)= -\frac{\mathbb E[f_Aw_\beta(f_A)]}{\mathbb E w_\beta(f_A)}. \tag{K.12}$$

Only the mathematical shape of the Fermi–Dirac occupation function at zero chemical potential is borrowed. The waveform is not declared a fermionic energy, and $\beta$ is a parameter rather than a measured temperature.

For all admissible $A$, $0\le B_\beta^{\mathrm{FD}}\le K(A)$. The upper bound follows from a normalized positive weight. For the lower bound, use $w_\beta(t)=(1-\tanh(\beta t/2))/2$ and $\mathbb E f_A=0$; the negative numerator becomes $\frac12\mathbb E[f_A\tanh(\beta f_A/2)]\ge0$. At $\beta=0$ the value is zero. Since the zero set of $f_A$ has measure zero, bounded convergence gives

$$\lim_{\beta\to\infty}B_\beta^{\mathrm{FD}}(A) =\frac{\mathbb E Y_A}{\mu(f_A<0)}. \tag{K.13}$$

This is the average depth of the negative part, not its largest depth.

Proposition K.1 — Failure uniformly over the tuning parameter. For every $N\ge1$,

$$0\le\sup_{\beta\ge0}B_\beta^{\mathrm{FD}}(A_N) \le20+10\log(2N+1). \tag{K.14}$$

Proof. On the set in (K.10), $w_\beta(f_{A_N})\ge1/2$, so its expectation is at least $1/40$, uniformly in $\beta$. Since $0\le w_\beta\le1$, the negative numerator is at most $\mathbb E Y_{A_N}$. Apply (K.5) and multiply by 40. $\square$

Thus no selection of $\beta$, even one depending on the entire set, turns this zero-threshold family into an all-set square-root certificate. The restriction to zero threshold is crucial. A changing chemical-potential parameter defines a different family and is not excluded by this proof.

L. Gibbs weights and an entropy budget

Saturating weights need not adequately emphasize a rare deep trough. Exponential weights behave differently. Set $X=-f_A$, so $\mathbb E X=0$, $\mathbb E X^2=N/2$, and $\max X=K(A)$. Define

$$g(\beta)=\log\mathbb E e^{\beta X},\qquad \rho_\beta=e^{\beta X-g(\beta)},\qquad\beta>0. \tag{L.1}$$

The normalized density $\rho_\beta$ has expectation one. Differentiation under the integral is justified because $X$ is bounded, and gives

$$g'(\beta)=\mathbb E(\rho_\beta X),\qquad g''(\beta)=\operatorname{Var}_{\rho_\beta}(X)>0. \tag{L.2}$$

Strict positivity holds because $X$ is nonconstant and $\rho_\beta$ is everywhere positive. For finite $\beta>0$, Jensen’s inequality and nonconstancy give

$$0<\frac{g(\beta)}\beta<K(A).$$

This bound increases with $\beta$, since

$$\frac{d}{d\beta}\frac{g(\beta)}\beta =\frac{\mathbb E\rho_\beta\log\rho_\beta}{\beta^2}>0.$$

Moreover, it tends to $K(A)$. For each $\eta>0$, the set $E_\eta=\{X\ge K-\eta\}$ has positive measure, and

$$\frac{g(\beta)}\beta\ge K-\eta+ \frac{\log\mu(E_\eta)}\beta.$$

Take $\beta\to\infty$, then $\eta\downarrow0$. Again, a convergent description of the extremum is not a lower bound for its size.

L.1. Entropy-constrained optimization

For a density $\rho\ge0$ with $\mathbb E\rho=1$, let

$$D(\rho)=\mathbb E(\rho\log\rho),\qquad0\log0=0.$$

For $h\ge0$, put

$$b_A(h)=\sup\{\mathbb E(\rho X):D(\rho)\le h, \ \rho\ge0,\ \mathbb E\rho=1\}. \tag{L.3}$$

This is an optimization over a redistribution of the original time measure. Entropy measures the cost of concentrating that measure; it is not extra physical energy in the waveform.

Theorem L.1 — Exact dual formula and attainment. For $h>0$, there is a unique $\beta_h>0$ such that

$$\beta_hg'(\beta_h)-g(\beta_h)=h. \tag{L.4}$$

The optimizer in (L.3) is $\rho_{\beta_h}$, and

$$b_A(h)=g'(\beta_h) =\inf_{\beta>0}\frac{h+g(\beta)}\beta. \tag{L.5}$$

Also $b_A(0)=0$, $b_A$ is nondecreasing and concave, and $b_A(h)\to K(A)$ as $h\to\infty$.

Proof. Relative entropy is nonnegative by convexity of $u\log u$. Applied relative to the positive density $\rho_\beta$, this says

$$0\le D(\rho\|\rho_\beta) =D(\rho)-\beta\mathbb E(\rho X)+g(\beta).$$

It proves the upper bound in (L.5) for every admissible $\rho$. The entropy of $\rho_\beta$ is

$$D_\beta=\beta g'(\beta)-g(\beta),\qquad D_\beta'=\beta g''(\beta)>0,$$

and tends to zero with $\beta\downarrow0$.

We verify that $D_\beta\to\infty$. The nonconstant trigonometric polynomial $X$ attains its maximum at finitely many points: otherwise the polynomial equation obtained after multiplication by a suitable power of $e^{ix}$ would have infinitely many roots. Therefore $p_\eta=\mu(E_\eta)\to0$ as $\eta\downarrow0$. For a fixed $\eta>0$, the Gibbs mass outside $E_\eta$ is at most $p_{\eta/2}^{-1}e^{-\beta\eta/2}$, hence tends to zero. If the Gibbs mass of $E_\eta$ is $a$, Jensen’s inequality on this set and its complement gives

$$D_\beta\ge a\log(a/p_\eta) +(1-a)\log((1-a)/(1-p_\eta)) \ge a\log(1/p_\eta)-\log2.$$

First let $\beta\to\infty$ for fixed $\eta$, then let $\eta\downarrow0$. This proves divergence. Continuity and strict increase now imply a unique solution of (L.4). Substitution of $\rho_{\beta_h}$ into the relative-entropy inequality makes it an equality, establishing attainment and (L.5).

At zero entropy only the uniform density is possible, so $b_A(0)=0$. Increasing the budget enlarges the feasible set. A mixture of densities feasible at two budgets is feasible at the same mixture of budgets, by convexity of entropy; its objective is the corresponding mixture. This proves concavity. Finally, the density $\mathbf1_{E_\eta}/p_\eta$ has entropy $\log(1/p_\eta)$ and objective at least $K-\eta$, proving the limit as the budget grows. $\square$

These variational identities belong to the established Gibbs/relative-entropy framework. Perlaza and Bisson study changes of expectation under changes of probability measure and their connections to Gibbs measures, information projections, and relative-entropy identities PerlazaBisson2025. This is prior context for the reweighting interpretation. The cosine-specific estimates and attainment argument needed here are proved explicitly above and below; this presentation does not claim to originate the Gibbs variational principle.

L.2. A rare-event construction

Suppose $E$ has measure $p\in(0,1)$, with $X\ge d>0$ on $E$. For any $h\ge0$, let

$$\lambda=\min\left(1,\frac h{\log(1/p)}\right),\qquad \rho=(1-\lambda)+\lambda\frac{\mathbf1_E}{p}.$$

Convexity gives $D(\rho)\le\lambda\log(1/p)\le h$. Since $\mathbb E X=0$, the objective is at least $\lambda d$. Hence

$$b_A(h)\ge d\min\left(1,\frac h{\log(1/p)}\right). \tag{L.6}$$

Applying (K.9) to consecutive frequencies gives, for $N\ge2$,

$$b_{A_N}(h)\ge\frac{3(2N+1)}{20\pi} \min\left(1,\frac h{\log(3(2N+1)/2)}\right). \tag{L.7}$$

Thus a fixed positive entropy budget can retain an order $N/\log N$ lower bound on this family, although fixed quantiles and the saturated detector lose the square-root scale. The construction adds only a small probability weight to the narrow deep event.

L.3. A universal upper bound and its direction

For every admissible $A$ with $N\ge2$,

$$b_A(h)\le\left(h+\frac12\right)\frac N{\log N}. \tag{L.8}$$

To prove this, use the increasing function

$$j(u)=\frac{e^u-1-u}{u^2} =\int_0^1(1-t)e^{tu}\,dt,$$

with $j(0)=1/2$. Because $X\le N$, for $\beta>0$,

$$e^{\beta X}\le1+\beta X+ \frac{e^{\beta N}-1-\beta N}{N^2}X^2.$$

Taking expectations and using $\log(1+v)\le v$ yields

$$g(\beta)\le\frac{e^{\beta N}-1-\beta N}{2N}.$$

At $\beta=\log N/N$ this is less than $1/2$, and (L.5) gives (L.8). Together with (L.7), it proves $b_{A_N}(h)=\Theta_h(N/\log N)$ for every fixed $h>0$.

The upper bound (L.8) holds for every set; the matching lower order has been proved only for consecutive frequencies. It would be an unjustified reversal to use (L.8) as evidence of a universal lower bound. The sought statement $b_A(h_0)\ge c\sqrt N$ with two absolute constants $h_0,c>0$ remains unproved here.

L.4. Why variance alone cannot close the argument

For $N\ge2$, consider a different random variable $V$ that equals $-1$ with probability $N/(N+2)$ and $N/2$ with probability $2/(N+2)$. Direct arithmetic gives

$$\mathbb E V=0,\qquad\mathbb E V^2=N/2,\qquad |V|\le N, \quad -\min V=1.$$

It has the first two moments and amplitude bound of the cosine model, but no square-root negative depth. It is not an exhibited cosine sum, so it is not a counterexample to Chowla. It shows exactly why an argument retaining only mean, variance, and a pointwise amplitude bound cannot suffice; additional integer-frequency structure is indispensable.

M. Frequency scaling and finite observation time

For every positive integer $r$,

$$f_{rA}(x)=f_A(rx),\qquad K(rA)=K(A). \tag{M.1}$$

The second identity holds because $x\mapsto rx$ maps the circle onto itself. Thus the gcd can be divided out without changing $K$. The raw largest frequency cannot by itself describe the negative depth: $\{1,2\}$ and $\{r,2r\}$ have the same value $9/8$, from

$$\cos x+\cos2x =2(\cos x+1/4)^2-9/8. \tag{M.2}$$

This invariance does not mean that the time required to observe a trough, or an approximation error based on a fixed time window, is independent of the frequency scale.

For positive real frequency vectors $\omega=(\omega_1,\ldots,\omega_N)$, temporarily extend the model and define

$$K_T(\omega)=-\min_{0\le t\le T}\sum_{i=1}^N\cos(\omega_it), \qquad T\ge0.$$

This quantity may be negative for a very short window; the definition is a signed minimum functional, not its positive part. For any two vectors $\omega,\nu$,

$$|K_T(\omega)-K_T(\nu)| \le T\sum_i|\omega_i-\nu_i|. \tag{M.3}$$

Indeed, cosine is 1-Lipschitz, so the two waveforms differ uniformly on the window by at most the right side; minima differ by no more than the uniform difference.

Over unbounded time, small frequency changes can behave differently. For irrational $\alpha>0$,

$$-\inf_{t\in\mathbb R}(\cos t+\cos\alpha t)=2. \tag{M.4}$$

At odd multiples of $\pi$, the first term is $-1$, while the irrational rotation allows the second to approach $-1$. Here is the elementary density fact being used: among $m+1$ fractional parts $0,\alpha,\ldots,m\alpha$, two lie within $1/m$; subtraction, and sign reversal if necessary, gives a nonzero orbit step $\theta\in(0,1/m)$. Its multiples before passing one approximate every point of the unit interval within $\theta$. Letting $m$ grow proves density of the two-sided rotation orbit. A fixed translation of that dense orbit, as required for odd multiples, is still dense. Exact equality of both cosines to $-1$ would force $\alpha$ to be a ratio of odd integers, so an infimum rather than a minimum is needed.

Consequently, $K_\infty(1,2+\eta)=2$ for irrational $\eta$ near zero, but $K_\infty(1,2)=9/8$. If a finite window achieves depth at least $2-\delta$, where $0<\delta<7/8$, (M.3) forces

$$T\ge\frac{7/8-\delta}{|\eta|}. \tag{M.5}$$

A change in the infinite-time extremum may require an observation time diverging as the detuning shrinks. The integer-frequency problem, a real-frequency extension, and a finite-duration audio measurement are therefore distinct mathematical questions.

N. What has and has not been established

The proof architecture gives several independently checkable outcomes, but it does not remove the central missing estimate. The following distinctions should be retained in every public summary.

The archived programs provide exact coefficient arithmetic, finite comparisons, and checks of stated identities. Such checks are useful for detecting errors in constants, signs, and normalization. They do not replace a quantified proof over unrestricted integer frequencies. Conversely, a short proof for an infinite family can establish a genuine limitation of a method even when a numerical search over that family would be inconclusive.

The specialized uniform approximation theorem is presented as a checkable result of this project, with novelty unverified. An independent specialist should compare it with the existing pushforward and polynomial-concentration literature before it is described as a new research contribution. An internal verification pass is not external peer review, and publication of these materials does not imply endorsement by a journal or a mathematical institution.

The mathematical work also does not establish an audible improvement, a processing algorithm suitable for every audio signal, or a performance guarantee for Bass Plus. The connection to sound motivates the model and helps identify physically meaningful distinctions—amplitude versus squared amplitude, phase-sensitive fields versus intensities, continuous minima versus samples—but those distinctions do not expand the hypotheses of the original problem.

Citation keys used in this appendix

The publication package supplies full bibliographic entries. The core methodological references are:

Lasserre2021 Jean B. Lasserre, Connecting optimization with spectral analysis of tri-diagonal matrices, Mathematical Programming 190, 795–809 (2021); DOI: 10.1007/s10107-020-01549-3; arXiv:1907.09784. Prior art for the pushforward hierarchy and its eigenvalue interpretation.

SlotLaurent2021 Lucas Slot and Monique Laurent, Near-optimal analysis of Lasserre’s univariate measure-based bounds for multivariate polynomial optimization, Mathematical Programming 188, 443–460 (2021); DOI: 10.1007/s10107-020-01586-y; arXiv:2001.11289. Prior analysis of the univariate measure-based bounds and their convergence.

SlotLaurent2022 Lucas Slot and Monique Laurent, Improved convergence analysis of Lasserre’s measure-based upper bounds for polynomial minimization on compact sets, Mathematical Programming 193, 831–871 (2022); DOI: 10.1007/s10107-020-01468-3; arXiv:1905.08142. Prior art for Chebyshev and needle-polynomial concentration estimates.

NISTDLMF National Institute of Standards and Technology, Digital Library of Mathematical Functions, Section 18.5, “Explicit Representations,” equation 18.5.1. Source for the standard Chebyshev representation used in (E.3).

PerlazaBisson2025 Samir M. Perlaza and Gaetan Bisson, Variations on the Expectation Due to Changes in the Probability Measure, Entropy 27(8), 865 (2025); DOI: 10.3390/e27080865. Context for changes of expectation, Gibbs measures, and relative entropy in Section L.

The reference dates in these keys refer to the bibliography’s publication identification; the linked preprints can have earlier submission dates. The historical investigations and source programs underlying this appendix are retained elsewhere in the package for reproducibility, rather than silently rewritten as if every exploratory statement had been correct from its first appearance.