A Log-Free $n^{1/5}$ Bound for Chowla's Cosine Problem

Abstract: For a finite set $S$ of positive integers, put $K(S):=-\min_{x\in\mathbb T}\sum_{s\in S}\cos(2πsx)$. Bedert recently proved the uniform lower bound $K(S)\geq |S|<sup>{1/5-o(1)}$. We remove the subpolynomial loss and prove that $K(S)\geq c|S|<sup>{1/5}$ for an absolute constant $c&gt;0$. The proof combines two estimates from Bedert's argument with an exact averaging identity for the asymmetric boundaries of additive intersections. This identity replaces the multiplicative-amplification step responsible for the logarithmic loss.

• With rigorous proofs, the paper establishes a new log-free polynomial lower bound, $\mathcal{K}(N)\gg N^{1/5}$, for Chowla's cosine problem.
• The new result removes the logarithmic factor $(\log N)^4$ from preceding estimates that started with a logarithmic loss in a multiplicative-amplification argument.
• By eliminating the logarithmic factor, the paper adopts a global additive-triple count to directly derive a significant boundary of size |B_t|\geq \frac{T}{3|A|} for at least one t∈A, demonstrating why this logarithmic factor disappears in the exact bounded region.

The paper proves a log-free polynomial lower bound for Chowla’s cosine problem. For a finite nonempty set $S$ of positive integers, define

$$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$

and let

$\mathcal K(N)=\inf_{|S|=N}K(S).$

The main theorem establishes

$\mathcal K(N)\gg N^{1/5},$

with an absolute, effectively computable implied constant. This removes the logarithmic loss from Bedert’s preceding estimate $\mathcal K(N)\gg N^{1/5-o(1)}$ (Bedert, 5 Sep 2025). The exponent remains below Chowla’s conjectured square-root scale $\mathcal K(N)\gg N^{1/2}$.

Position within Chowla’s cosine problem

Chowla’s problem asks how negative a cosine sum must become when its frequencies form an arbitrary $N$-element set of positive integers. The conjecture that $\mathcal K(N)\to\infty$ was subsequently strengthened to the assertion that the optimal order is $\sqrt N$. A Sidon-difference construction gives the matching upper bound $\mathcal K(N)\ll\sqrt N$, so the unresolved issue is the lower bound.

The historical progression summarized in the paper runs from logarithmic estimates obtained through the Littlewood $$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$0 problem to superlogarithmic and eventually polynomial bounds. Roth established

$$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$1

while the resolution of the Littlewood $$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$2 conjecture yielded $$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$3. Bourgain and Ruzsa subsequently crossed the logarithmic barrier, with Ruzsa proving a lower bound of the form $$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$4.

Polynomial growth was obtained only recently. Jin, Milojević, Tomon, and Zhang proved an $$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$5 estimate (Jin et al., 3 Sep 2025), and Bedert successively improved the exponent, reaching $$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$6 in version 3 of (Bedert, 5 Sep 2025). The remaining factor arose from a logarithmic loss in a multiplicative-amplification argument. The present paper isolates that loss and eliminates it without improving the exponent.

Symmetrization and the analytic inputs

The proof passes from a positive-frequency set $$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$7 to the symmetric set

$$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$8

Writing

$$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$$9

one has

$\mathcal K(N)=\inf_{|S|=N}K(S).$0

Thus a lower bound for $\mathcal K(N)=\inf_{|S|=N}K(S).$1 immediately gives a lower bound for $\mathcal K(N)=\inf_{|S|=N}K(S).$2.

The central parameter is

$\mathcal K(N)=\inf_{|S|=N}K(S).$3

so that $\mathcal K(N)=\inf_{|S|=N}K(S).$4 on $\mathcal K(N)=\inf_{|S|=N}K(S).$5. The argument retains two estimates from Bedert’s work.

First, a Roth-type additive-triple estimate states that if $\mathcal K(N)=\inf_{|S|=N}K(S).$6 and $\mathcal K(N)=\inf_{|S|=N}K(S).$7, then

$\mathcal K(N)=\inf_{|S|=N}K(S).$8

Applied with $\mathcal K(N)=\inf_{|S|=N}K(S).$9, this produces many additive configurations whenever $\mathcal K(N)\gg N^{1/5},$0 is sufficiently large relative to $\mathcal K(N)\gg N^{1/5},$1.

Second, for

$\mathcal K(N)\gg N^{1/5},$2

Bedert’s asymmetric-boundary estimate gives

$\mathcal K(N)\gg N^{1/5},$3

for every nonzero $\mathcal K(N)\gg N^{1/5},$4. Consequently, if one can produce a single $\mathcal K(N)\gg N^{1/5},$5 with $\mathcal K(N)\gg N^{1/5},$6 substantially larger than $\mathcal K(N)\gg N^{1/5},$7, then $\mathcal K(N)\gg N^{1/5},$8 must already be large. The role of the new contribution is to derive such a $\mathcal K(N)\gg N^{1/5},$9 directly from the global additive-triple count.

The total-boundary identity

For

$\mathcal K(N)\gg N^{1/5-o(1)}$0

the paper proves the inequality

$\mathcal K(N)\gg N^{1/5-o(1)}$1

This is the decisive combinatorial improvement.

The proof gives a more precise formula. Let $\mathcal K(N)\gg N^{1/5-o(1)}$2 and define

$\mathcal K(N)\gg N^{1/5-o(1)}$3

together with

$\mathcal K(N)\gg N^{1/5-o(1)}$4

Then

$\mathcal K(N)\gg N^{1/5-o(1)}$5

Since $\mathcal K(N)\gg N^{1/5-o(1)}$6, one obtains

$\mathcal K(N)\gg N^{1/5-o(1)}$7

The identity is established by expressing each boundary indicator as

$\mathcal K(N)\gg N^{1/5-o(1)}$8

After grouping $\mathcal K(N)\gg N^{1/5-o(1)}$9 according to the signs of their absolute values, each sign class contributes the square of a difference between two membership indicators: one for $\mathcal K(N)\gg N^{1/2}$0 and one for $\mathcal K(N)\gg N^{1/2}$1. The total boundary is therefore an exact quadratic count rather than the output of an iterative amplification process.

This point is structurally important. Bedert’s method produced a large boundary through multiplicative amplification and incurred a factor $\mathcal K(N)\gg N^{1/2}$2. The new identity instead averages all boundaries and converts the complete additive-triple count into a boundary of size

$\mathcal K(N)\gg N^{1/2}$3

for at least one $\mathcal K(N)\gg N^{1/2}$4. When the Roth-type estimate gives $\mathcal K(N)\gg N^{1/2}$5, this becomes

$\mathcal K(N)\gg N^{1/2}$6

Thus the logarithmic factor disappears at the exact step where it arose in Bedert’s argument.

The constant $\mathcal K(N)\gg N^{1/2}$7 in the averaging inequality is not merely an artifact of the proof. For $\mathcal K(N)\gg N^{1/2}$8, the ratio between the total boundary and $\mathcal K(N)\gg N^{1/2}$9 tends to $N$0. Hence the elementary inequality $N$1 is asymptotically sharp for this family.

Derivation of the $N$2 exponent

Let $N$3 and retain

$N$4

The proof first observes that $N$5. Indeed, $N$6 is nonnegative and its Fourier coefficient at any $N$7 equals $N$8, so positivity implies

$N$9

There are then two cases. If $\mathcal K(N)\to\infty$0, the trivial inequality $\mathcal K(N)\to\infty$1 gives

$\mathcal K(N)\to\infty$2

and therefore $\mathcal K(N)\to\infty$3.

In the complementary case $\mathcal K(N)\to\infty$4, the Roth–Bedert estimate yields

$\mathcal K(N)\to\infty$5

The total-boundary identity gives

$\mathcal K(N)\to\infty$6

so some $\mathcal K(N)\to\infty$7 satisfies

$\mathcal K(N)\to\infty$8

The asymmetric-boundary estimate then implies

$\mathcal K(N)\to\infty$9

or equivalently

$\sqrt N$0

Hence

$\sqrt N$1

for an absolute constant $c_0&gt;0.</p> <p>Returning to $\sqrt N$2, one has $\sqrt N$3 and

$\sqrt N$4

Therefore

$\sqrt N$5

Taking the infimum over all $\sqrt N$6-element sets $\sqrt N$7 proves

$\sqrt N$8

The exponent is transparent from the proof: the additive-combinatorial input supplies a boundary of order $\sqrt N$9, while the analytic boundary estimate bounds every such boundary by $\mathcal K(N)\ll\sqrt N$0. Balancing these quantities gives $\mathcal K(N)\ll\sqrt N$1, hence the exponent $\mathcal K(N)\ll\sqrt N$2.

Nature of the improvement

The paper’s contribution is not a stronger exponent but a sharper passage between two existing estimates. Bedert’s argument already contained the two ingredients needed for the $\mathcal K(N)\ll\sqrt N$3 bound: a lower bound for additive triples and an upper bound for asymmetric intersections. The logarithmic loss entered only when converting the additive information into one large boundary through multiplicative amplification.

The exact identity replaces this conversion mechanism with a global average: $\mathcal K(N)\ll\sqrt N$4 No asymptotic estimate or density increment is used in this step. The only loss is the constant factor $\mathcal K(N)\ll\sqrt N$5, which is shown to be essentially optimal.

The resulting theorem is therefore a genuine strengthening of the known uniform bound: $\mathcal K(N)\ll\sqrt N$6 The improvement is uniform in the frequency set and has an effectively computable absolute constant, although the paper does not optimize that constant.

Limitations and open questions

The proof depends quantitatively on the estimate

$\mathcal K(N)\ll\sqrt N$7

As long as this fourth-power dependence is unchanged, the present boundary argument naturally yields only the fifth-power relation $\mathcal K(N)\ll\sqrt N$8. Improving the exponent beyond $\mathcal K(N)\ll\sqrt N$9 through this framework would therefore require either a stronger boundary estimate or additional information forcing many of the sets $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$00 to be simultaneously large.

The total-boundary identity itself controls only the sum of the boundary sizes. The proof extracts one large $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$01 by averaging and does not exploit any distributional information beyond that. Whether the additive structure supplies enough uniformity among the boundaries to improve the exponent remains unresolved.

Consequently, the theorem leaves a substantial gap to the conjectured lower bound $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$02. It also does not address whether the $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$03 boundary estimate is quantitatively optimal in the relevant regime, or whether a different structural consequence of the nonnegativity condition $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$04 could produce a stronger relation between $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$05 and $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$06.

Conclusion

The paper establishes the log-free bound

$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$07

for Chowla’s cosine problem. Its main innovation is the exact identity

$K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$08

which converts a global additive-triple count directly into a large asymmetric boundary and removes the $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$09 loss in the preceding $K(S)=-\min_{x\in \mathbb R/\mathbb Z}\sum_{s\in S}\cos(2\pi sx),$10 estimate (Bedert, 5 Sep 2025). The argument clarifies precisely which part of the proof controls the current exponent and isolates the boundary estimate as the principal quantitative obstruction to further improvement.