Observe that for any Schwartz function $f \in \mathcal{S}(\mathbb{R})$ having
$$
f(0) = \widehat{f}(0) = 1
$$
and
$$
f, \widehat{f} \geq 0 \quad \textrm{outside of} \quad [-1,1],
$$
the following ridiculous argument based on the prime number theorem yields the strict upper bound
$$
\int f(t) \log{\frac{1}{|t|}} \, dt < 1 + \gamma = 1.57\ldots.
$$
[*Proof.* Just take the $X \to \infty$ limit in the identity
$$
\frac{1}{X}\sum_{k \in \mathbb{Z} \setminus \{0\}} f\Big( \frac{k}{X} \Big) \log{|k|} = \sum_{n \leq X} \Big( \frac{\widehat{f}(0)}{n} - \frac{f(0)}{X}\Big)\Lambda(n) + \sum_{n \leq X} \frac{\Lambda(n)}{n} \sum_{s \in \mathbb{Z} \setminus \{0\}} \widehat{f} \Big( \frac{sX}{n} \Big) \\ + \frac{1}{X} \sum_{n > X} \Lambda(n) \sum_{s \in \mathbb{Z} \setminus \{0\}} f\Big( \frac{sn}{X} \Big).
$$
Our conditions on $f, \widehat{f}$ imply that the second and third sums on the right-hand side have only non-negative terms, while the first sum of the right-hand side is evaluated by the logarithmic form of the prime number theorem: $S(X) := \sum_{n \leq X} \Big( \frac{1}{n} - \frac{1}{X} \Big) \Lambda(n) = \log{X} - 1 - \gamma + o(1)$. ]

This begs the question of whether such an argument could conceivably be reversed. Any $f$ as above places an 'explicit' upper bound of $\log{X} - \int f(t) \log{\frac{1}{|t|}} \, dt + O_f((\log{X})/X)$ on the prime number sum $S(X)$, whence my

**Question.** *Is $1+\gamma$ the supremum value of $\int f(t) \, \log{\frac{1}{|t|}} \, dt$ under the given conditions $f(0) = \widehat{f}(0) = 1$ and $f, \widehat{f} \geq 0$ outside of $[-1,1]$? Or may the $1+\gamma$ bound be improved?*

Also, quite independently of such a motivation, I would be curious to see any alternative proofs of an upper bound on $\int_{\mathbb{R}} f(t) \, \log{\frac{1}{|t|}} \, dt$ by an absolute constant (under the above conditions on the Fourier pair $f, \widehat{f}$), even if these are weaker than the $1+\gamma$ bound here.

Of course, the normalization of the Fourier transform here is the following one: $$ \widehat{f}(y) = \int_{\mathbb{R}} f(x) e^{- 2\pi i xy} \, dx. $$

*An application.* Here is an application added of the observed inequality. For $g : [0,1] \to \mathbb{R}^{\geq 0}$ any non-negative continuous function with $\int_0^1 g(t) \, dt = 1$ and $S := \int_0^1 g^2(t) \, dt$, we may apply the preceding to (an extension by zero and smoothing of) $f(t) :=S^{-1} \cdot (g * g)(t/S)$. Indeed, we have $f(0) = S^{-1} \cdot (g * g)(0) = 1$ and $\widehat{f}(0) = |\widehat{g}(0)|^2 = 1$, and both $f(t)$ and its Fourier transform $\widehat{f}(t) = |\widehat{g}(t)|^2$ are everywhere non-negative. The conclusion reads:

**Special case.** *The inequality
$$
\int_0^1 \int_0^1 g(x) g(y) \log{\frac{1}{|x-y|}} \, dx \, dy - \log \int_0^1 |g(t)|^2 \, dt < 1 + \gamma = 1.57\ldots
$$
takes place for every continuous non-negative function $g \in C([0,1],\mathbb{R}^{\geq 0})$ on $[0,1]$ of unit integral: $\int_0^1 g(t) \, dt = 1$*.

May one give a direct proof of the last elementary inequality, with any absolute constant bound in place of $1+\gamma$? And may one prove a strictly smaller bound?