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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?

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    $\begingroup$ Just an observation that if $f(x) =\max(0,1-|x|)$ then the integral is $1.5$. So $1+\gamma$ is not far from the answer! $\endgroup$ – Lucia Nov 13 '17 at 2:31
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    $\begingroup$ @Lucia: Indeed, and also, it is worth adding that this $1.5$ lower bound can be improved. In the special case below the line, this example amounts to taking $g(t) \equiv 1$ on $[0,1]$. If instead we choose there $g(t) := (-30t^2 + 30t + 2)/7$ on $[0,1]$, we get the improved value of $977/588-\log{54/49} = 1.5644\ldots$. (I think this is best possible for a quadratic choice of $g$.) $\endgroup$ – Vesselin Dimitrov Nov 13 '17 at 2:50
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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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    $\begingroup$ The uncertainty principle will kick in if one has some control on $f$ in some function space norm (e.g. $L^2$ norm, $L^p$ norm, Sobolev norm, etc.). It's only because one has no norm control here that one can evade the uncertainty principle. It's odd though that this argument seems to give "one half of the prime number theorem" in some sense. When combined with Brun-Titchmarsh to estimate the error terms, it seems to give an elementary proof of PNT that is slightly different from the Erdos-Selberg one. $\endgroup$ – Terry Tao Nov 13 '17 at 6:05
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    $\begingroup$ Hang on, there is an issue with the last part of my argument - $\check b$ and the antiderivative of $\hat a$ are not in fact continuous at the origin, so I can't yet rule out a non-trivial solution to the equation $\alpha \delta + \beta = a + \check b$. I'll have to think about this. $\endgroup$ – Terry Tao Nov 13 '17 at 6:23
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    $\begingroup$ As $f, \widehat{f}$ are assumed non-negative on $\mathbb{R} \setminus [-1,1]$, aren't we implicitly dealing with their $L^1$ norms on $\mathbb{R} \setminus [-1,1]$? $\endgroup$ – Vesselin Dimitrov Nov 13 '17 at 7:25
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    $\begingroup$ You're right - the two seminorms we have here are strong enough to combine to a norm that allows the uncertainty principle to kick in, so in fact my answer should be reversed - some improvement to $1+\gamma$ is in fact possible. I've updated the answer accordingly. $\endgroup$ – Terry Tao Nov 13 '17 at 15:51
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    $\begingroup$ I see. I was expecting this, since otherwise sharpness would contain 'for free' a proof of the PNT and raise the possibility of quantiative refinements, which seems just too good. It is the uncertainty principle that prevents us from reaching PNT this way. I will now study this technique and paper in detail, but it makes sense. This is extremely helpful, thank you very much! $\endgroup$ – Vesselin Dimitrov Nov 13 '17 at 16:33

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