# Unconventional approaches 16–20: independent working checks

18 October 2026 UTC. These notes check concrete proposals, not entire areas of
mathematics. They do not prove or disprove Chowla's conjecture. No novelty is
claimed. All integrals use normalized Haar measure on the circle.

For a finite nonempty A subset Z_{>0}, write N=|A|,
f(x)=sum_(a in A) cos(ax), and K=-min f. The common target is K>=c sqrt(N).

## 16. Let a hypothetical counterexample disappear in a weak limit

**Idea.** Assume a counterexample sequence K_j/sqrt(N_j)->0 and rescale
u_j=f_j/sqrt(N_j). A nonnegative limiting wave of mean zero must vanish.
Perhaps this contradicts its fixed energy.

**Exact check.** We have

    integral u_j=0, integral u_j^2=1/2,
    u_j>=-k_j, where k_j=K_j/sqrt(N_j)->0,
    ||u_j||_1=2 integral (u_j)_- <=2k_j ->0.

Thus u_j really does vanish in L1 and in probability. But L2 energy need not
pass to this limit. Define the ordinary Fejer kernel

    F_M(x)=(1/M)|sum_(r=0)^(M-1) exp(irx)|^2

and v_M=sqrt(3)(F_M-1)/(2sqrt(M)). Then

    min v_M >=-sqrt(3)/(2sqrt(M)),
    ||v_M||_1 <=sqrt(3)/sqrt(M),
    ||v_M||_2^2=(2M^2-3M+1)/(4M^2) ->1/2.

Here integral F_M^2=(2M^2+1)/(3M), exactly. These are finite, smooth waves
whose energy survives in increasingly narrow positive peaks while the
function tends to zero in L1. The coefficients are fractional, not the
allowed equal coefficients. They refute the proposed weak-limit inference,
not Chowla. A successful concentration argument must retain the exact
coefficient quantization in the concentration profiles; that step is open.

**Status:** the automatic weak-limit contradiction is false. A quantitative
profile argument preserving the discrete spectrum remains undeveloped.

## 17. Treat the shifted wave as information and bound its entropy

**Idea.** The probability density q=1+f/K has an unusually rigid Fourier
spectrum. Perhaps a Shannon entropy bound limits the number of frequencies.

**Exact check.** Its second moment and order-two relative Renyi entropy are

    integral q^2=1+N/(2K^2),
    D_2(q)=log(1+N/(2K^2)).

A universal upper bound for D_2 on these particular densities is exactly
another formulation of Chowla. Shannon entropy D_1=integral q log q satisfies
D_1<=D_2, not the reverse inequality needed by this shortcut.

Even a uniform bound D_1<=1 does not control D_2 for positive smooth
densities. For integer M>=3 set epsilon=1/log(M) and

    q_M=(1-epsilon)+epsilon F_M.

By convexity of x log x,

    D_1(q_M)<=epsilon D_1(F_M)<=epsilon log M=1,

because 0<=F_M<=M and integral F_M=1. But exactly

    integral q_M^2
      =1+epsilon^2*((2M^2+1)/(3M)-1)
      ->infinity.

The Fourier coefficients in this countermodel taper; they are not a
Chowla counterexample. The failure shows why generic information bounds
cannot replace a theorem about the exact zero/one frequency support.

**Status:** the Shannon-to-energy inference is false. A universal Renyi-2
bound for the original densities is unproved and equivalent to the target.

## 18. Use an uncertainty principle to prohibit hidden narrow peaks

**Idea.** A sum with only N frequencies should not be able to hide its
negative excursion on a negligible set. Apply a sparse Remez inequality
whose constant does not depend on the maximum frequency.

**Checked theorem.** The Turan–Nazarov inequality for m distinct purely
imaginary exponential terms gives, for a measurable E in an interval I,

    sup_I |p| <= (C |I|/|E|)^(m-1) sup_E |p|,

with absolute C. Source: Nazarov's original article, and Theorem 2.1 of
Bownik–Speegle's author-hosted paper:
https://www.mathnet.ru/eng/aa397
https://pages.uoregon.edu/mbownik/papers/51.pdf

**Actual application.** Since integral f_+=integral f_-<=K, the set
E={f<=2K} occupies at least half the circle. On E, |f|<=2K. There are 2N
exponential terms in f and its supremum in absolute value is N. Therefore

    N <=2K(2C)^(2N-1),
    K >=N/[2(2C)^(2N-1)].

This is a valid frequency-independent estimate. Its exponential loss in N
makes it much weaker than even the elementary K>=1. Choosing a different
constant threshold changes constants, not this obstruction. No stronger
Remez-to-square-root transition has been proved here.

**Status:** valid but quantitatively insufficient. This does not rule out
other uses of uncertainty principles.

## 19. Remove the trigonometry and demand an algebraic square certificate

**Idea.** Set y=cos x. Then f(x)=p_A(y)=sum_(a in A) T_a(y), where T_a is
the Chebyshev polynomial. Nonnegativity of K+p_A on [-1,1] has a
sum-of-squares representation. Could a bounded-degree identity handle all
sets with N terms at once?

**Exact reduction.** For a polynomial of even degree 2d, nonnegativity on
[-1,1] is equivalent to

    K+p_A(y)=s(y)^2+(1-y^2)t(y)^2,
    deg s<=d, deg t<=d-1.

For odd degree 2d+1 the representation is

    K+p_A(y)=(1+y)s(y)^2+(1-y)t(y)^2,
    deg s,deg t<=d.

Theorem 2.2, Nie–Demmel, *Shape optimization of transfer functions*:
https://mathweb.ucsd.edu/~njw/PUBLICPAPERS/TrnShapOptCoef.pdf

The required ordinary polynomial degree depends on max A. A bound on this
degree depending on N alone cannot hold before frequency compression:
with N=1 and A={D}, the left-hand side 1+T_D has degree D, arbitrarily large.
For even D the exact certificate is

    1+T_D(y)=2 T_(D/2)(y)^2.

It needs only one square but that square has unbounded degree. Small rank
or a small number of squares therefore does not bound the original degree
or supply the desired relation between N and K.

**Status:** exact certification for a prescribed set remains valid. The
proposed fixed-degree universal shortcut is false. No coefficient-count
bound of the required strength follows just from the square representation.

## 20. Interpret the wave as a passive causal filter and invoke a waterbed law

**Idea.** With P_A(z)=sum_(a in A)z^a, the condition Re(K+P_A)>=0 gives a
positive-real response. Transfer-function identities may force negative
gain somewhere when too many unit taps are present.

**Exact calculation.** If Re P_A>=-K, then 2K+P_A has positive real part
and is zero-free on the closed disk. For

    T(z)=2K/(2K+P_A(z)),

log|T| is harmonic on a neighborhood of the closed disk and T(0)=1.
Consequently

    integral log|T(exp(ix))| dx=0.

This is the logarithmic balance one would try to use as a waterbed
constraint. However, zero-freeness alone already gives the identity,
without the required pointwise positive-real bound.

Take A={1,...,N} and put the candidate K=1, P=z+...+z^N. Then

    (1-z)(2+P)=2-z-z^(N+1).

The polynomial 2+P has no zero on the closed unit disk: strict triangle
inequality excludes the open disk, and equality on the circle forces z=1,
where 2+P=N+2>0. Thus the logarithmic balance holds for this constant K for
every N. Nonetheless, at x=3pi/(2N), a Riemann sum gives

    f_A(x)/N -> integral_0^1 cos(3pi t/2)dt =-2/(3pi).

The true K is therefore at least (2/(3pi)+o(1))N. The Bode-type mean-log
identity by itself cannot certify pointwise passivity or a square-root
bound. The positive-real pointwise condition retains the original problem.

**Status:** a precise logarithmic-balance shortcut is disproved. General
control-theoretic methods are not ruled out; they need an additional
constraint exploiting the exact unit impulse-response coefficients.

## Check summary

All five entries have a concrete mathematical check. Entries16,17,19,20
reject specific proposed implications; entry18 obtains only an insufficient
estimate. No square-root theorem, asymptotic Chowla counterexample, or new
general exponent is asserted.