TITLE: The supremum value of $\int f(t) \log{\frac{1}{|t|}} \, dt$ for normalized Fourier pairs non-negative outside of $[-1,1]$
QUESTION [10 upvotes]: 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?

REPLY [8 votes]: One can take the continuum limit of your proof as $X \to \infty$, again using the prime number theorem, to obtain a proof that does not involve primes at all:
$$ \int f(t) \log \frac{1}{|t|}\ dt = \gamma - \sum_{\sigma = \pm 1} \int_0^\infty f(\sigma t) (\log t + \gamma)\ dt $$
$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \int_0^\infty f(\sigma t) (\log \frac{t}{\varepsilon} + \gamma)\ dt  + \log \frac{1}{\varepsilon}$$
$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \int_0^\infty f(\sigma t) (\sum_{0 < s < t/\varepsilon} \frac{1}{s})\ dt + \log \frac{1}{\varepsilon}$$
$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \sum_{s>0} \frac{1}{s} \int_{\varepsilon s}^\infty f(\sigma t)\ dt + \log \frac{1}{\varepsilon}$$
$$ = \gamma - \lim_{\varepsilon \to 0} \sum_{\sigma = \pm 1} \sum_{s>0} \int_{\varepsilon}^\infty f(\sigma s t)\ dt + \log \frac{1}{\varepsilon}$$
$$ = \gamma - \lim_{\varepsilon \to 0}  \int_{\varepsilon}^\infty \sum_{s \in \mathbb{Z} \backslash \{0\}} f(s t)\ dt + \log \frac{1}{\varepsilon}$$
$$ = \gamma - A - \lim_{\varepsilon \to 0}  \int_{\varepsilon}^1 \sum_{s \in \mathbb{Z} \backslash \{0\}} f(s t)\ dt + \log \frac{1}{\varepsilon}$$
$$ = \gamma - A - \lim_{\varepsilon \to 0}  \int_{\varepsilon}^1 (\sum_{s \in \mathbb{Z} \backslash \{0\}} f(s t) - \frac{1}{t})\ dt $$
$$ = \gamma - A - \lim_{\varepsilon \to 0}  \int_{\varepsilon}^1 (\frac{1}{t} \sum_{s \in \mathbb{Z} \backslash \{0\}} \hat f(s/t) - 1)\ dt $$
$$ = \gamma + 1 - A - B$$
where
$$ A := \int_1^\infty \sum_{s \in \mathbb{Z} \backslash \{0\}} f(st)\ dt$$
$$ = \int_{|t| \geq 1} f(t) (\sum_{1 \leq s \leq |t|} \frac{1}{s})\ dt$$
and
$$ B := \int_0^1 \sum_{s \in \mathbb{Z} \backslash \{0\}} \hat f(s/t)\ \frac{dt}{t}$$
$$ = \int_1^\infty \sum_{s \in \mathbb{Z} \backslash \{0\}} \hat f(st)\ \frac{dt}{t}$$
$$ = \int_{|t| \geq 1} \hat f(t) \frac{\lfloor |t| \rfloor}{t}\ dt.$$
[In the language of distributions, what this identity is saying I think is that the distributional Fourier transform of $\lfloor |t| \rfloor/t - 1$ is $\gamma - \log \frac{1}{|t|} 1_{|t| \leq 1} - \sum_{1 \leq s \leq |t|} \frac{1}{s}$.]
Since $A,B$ are clearly non-negative, this gives your inequality.  This also shows that one is within $o(1)$ of equality if and only if one simultaneously has
$$ \int_{|t| \geq 1} f(t) (1 + \log |t|)\ dt = o(1)$$
and
$$ \int_{|t| \geq 1} \hat f(t)\ dt = o(1).$$
By the Hahn-Banach theorem, these estimates are incompatible with the hypotheses $f(0)=\hat f(0)=1$, $f(t), \hat f(t) \geq 0$ for $|t| \geq 1$ if and only if there exist non-negative measurable functions $a(t), b(t)$ supported on $|t| \geq 1$ with $\sup_t \frac{a(t)}{1+\log |t|}, \sup_t b(t) < \infty$ and numbers $\alpha,\beta$ not summing to zero, such that
$$ \alpha f(0) + \beta \hat f(0) = \int_{\mathbb R} f(t) a(t)\ dt + \int_{\mathbb R} \hat f(t) b(t)\ dt $$
for all Schwartz $f$, or equivalently that
$$ \alpha \delta + \beta = a + \check b$$
in the sense of tempered distributions, where $\delta$ is the Dirac delta.  But the right-hand side is continuous at the origin, so $\alpha$ must vanish; the Fourier transform of the right-hand side has a continuous antiderivative at the origin, so $\beta$ must vanish, contradiction.  This shows that one can make $A$ and $B$ simultaneously $o(1)$, so $1+\gamma$ is in fact optimal.  (But the invocation of the Hahn-Banach theorem makes it difficult to explicitly construct $f$ that come close to equality!)
One can solve this equation as follows.  By Lemma 3 of
Amrein, W.O.; Berthier, A.M., On support properties of Lsup(p)-functions and their Fourier transforms, J. Funct. Anal. 24, 258-267 (1977). ZBL0355.42015,
one can find, for any $R>0$, a non-zero function $f \in L^2({\bf R})$ such that $f$ and $\hat f$ both vanish on $[-R,R]$ (this is basically because the compact operator $1_{[-R,R]} {\mathcal F} 1_{[-R,R]}$ is a strict contraction on $L^2$, which in turn follows from the uncertainty principle that a function and its Fourier transform cannot be simultaneously compactly supported), in fact Proposition 6 gives an infinite-dimensional space of such functions.  By convolving $f$ by a suitable approximation to the identity, and then multiplying by the Fourier transform of a suitable approximation to the identity (and shrinking $R$ slightly), one can make $f$ Schwartz.
When one takes a second antiderivative of $f$, one obtains a new Schwartz function $f_1$ which is equal to a linear function $a+\beta x$ on $[-R,R]$, while the Fourier transform still vanishes on $[-R,R]$.  If $a=0$ (which can be achieved due to the infinite dimensional nature of the space of $f$), one can divide by $x$ and obtain a further Schwartz function $f_2$ that is equal to a constant $\beta$ on $[-R,R]$, while the Fourier transform is equal to a constant $\alpha$ on $[-R,R]$.  This gives the identity
$$ \alpha \delta + \beta = (\beta - f) + (\alpha - \hat f)^{\vee}$$
 I think one can work a little harder to ensure that $\alpha,\beta$ can be arbitrary real numbers while simultaneously keeping $a=0$, and in particular can have non-zero sum (otherwise by Hahn-Banach there would be a way to express some nontrivial combination of polynomials on a halfline and Fourier transforms of polynomials on a halfline as functions supported on $[-R,R]$ plus a function with Fourier transform supported on $[-R,R]$, which should be easy to rule out by the argument in strikethrough).  This gives a constraint of the desired form (taking $R=1$).  So some improvement to $1+\gamma$ is in fact possible.
Update: here are some details on the "working a little harder".  For $f$ Schwartz with both $f$ and $\hat f$ vanishing on $[-R,R]$, one can write $\beta$ as the inner product of $f$ with $1_{(-\infty,0]}$ and $a$ as the inner product of $-1_{(-\infty,0]} x$.  If $a$ vanishes, one can write $\alpha$ as the inner product of $f$ with a function $\phi$ whose Fourier transform is equal to a constant multiple of $1_{(-\infty,0]}(x) / x^2$ outside of $[-R,R]$ and is smooth in $[-R,R]$.  So supposing for contradiction that there is a non-trivial constraint between $\alpha$ and $\beta$ when $a=0$, there must exist some non-trivial linear combination $g$ of $1_{(-\infty,0]}$, $1_{(-\infty,0]} x$, and $\phi$ such that all Schwartz functions $f$ with both $f$ and $\hat f$ vanishing on $[-R,R]$ are orthogonal to $g$.  In particular, if $f \in L^2$ with $f$ and $\hat f$ vanishing on $[-2R,2R]$, and $\psi_1, \psi_2$ are suitable approximations to the identity (let's say real symmetric), then $(f \hat{\psi_1}) * \psi_2$ is orthogonal to $g$, or equivalently $f$ is orthogonal to $(g * \psi_2) \hat \psi_1$.  Taking limits as $\psi_2$ approaches the Dirac delta, we conclude that $f$ is orthogonal to $g \hat \psi_1$.  Taking duals, this means that we have a decomposition $g \hat \psi_1 = g_1 + \hat g_2$ where $g_1,g_2$ are $L^2$ function supported in $[-2R,2R]$.  This implies in particular that $g \hat \psi_1$ extends to a holomorphic function on ${\bf C} \backslash [-2R,2R]$.  Dividing by $\hat \psi_1$ (which one can choose to be non-zero at any given complex number), we conclude that $g$ extends to a holomorphic function on ${\bf C} \backslash [-2R,2R]$ (the extension is independent of $\psi_1$ by analytic continuation).
The function $x \phi''(x)$ has a test function for a Fourier transform with nonzero integral, so $\phi''(x)$ (as a distribution) is equal to a Schwartz function plus a non-zero multiple of $p.v. 1/x$, and extends to an entire function plus a non-zero multiple of $1/x$ away from the origin.  This implies that $\phi$ is extends holomorphically to ${\bf C}$ with $[-2R,2R]$ and the negative imaginary axis (removed).  On the other hand, by uniqueness of analytic continuation, any non-trivial multiple of $1_{(-\infty,0]}$ and $1_{(-\infty,0]} x$ cannot be extended to this region.  These facts are only consistent if $g$ is a scalar multiple of $\phi$ alone.  But $\phi$ has nontrivial monodromy around $[-2R,2R]$ (it behaves like the sum of an entire function and the multivalued function $z \log z$), while $g$ does not, giving the required a contradiction.
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