5
$\begingroup$

If the answer to this question is in affirmative, then this would yield a good answer to this question.

Let $f\colon \mathbb C\to\mathbb C$ be an entire function whose values on the real line are real and bounded from below by $$ f(x)\geq \sqrt{\frac{\sinh x}{x}}, \qquad 0\neq x\in\mathbb R. $$ [EDIT] Also, we have $|f(x+iy)|\leq C\exp(|x|/2)$ and, moreover, $f(x+iy)\to0$ as $y\to\infty$ for each separate $x$.

Is it true that $f(0)\geq 2/\sqrt\pi$?

Remarks. 1. If so, this estimate is sharp and achieved, e.g., on the analytic continuation of $$ f(x)=e^{x/2}\sqrt{\frac8{\pi x}}\int_0^{\sqrt {x/2}}\exp(-t^2) \,\mathrm{d}t =\frac{e^{x/2}}{\sqrt{\pi x}}\int_0^{x}\frac{e^{-s/2}}{\sqrt s} \,\mathrm{d}s, \qquad x>0. $$ (The fact that this function satisfies the inequality follows from the connection to the question mentioned above.)

2. It is worth nothing to mention that one may assume $f$ to be even.

3. Clearly, $\sqrt{\sinh z/z}$ is not an entire function, as $\sinh z$ has many simple zeroes.

4. Only the first additional constraint $|f(x+iy)|\leq C\exp(|x|/2)$ does not exclude the counterexample $\cosh(z/2)$; thanks to Michael Engelhardt who noticed that I need some extra conditions.

$\endgroup$
5
  • $\begingroup$ @FedorPetrov: Sorry, forgot a factor of 2... $\endgroup$ Commented Oct 1, 2019 at 15:04
  • 2
    $\begingroup$ Isn't $\cosh x$ a counterexample? $\endgroup$ Commented Oct 1, 2019 at 15:46
  • $\begingroup$ @MichaelEngelhardt: Thanks! Clearly, it is, and even $\cosh (x/2)$ works. I tried to impose some additional constraints $f$ should satisfy... $\endgroup$ Commented Oct 1, 2019 at 16:15
  • 1
    $\begingroup$ @AlexandreEremenko: So what's wrong? Surely, the function is not unique. What I need os to find the minimal possible value of $f(0)$ for such an $f$. $\endgroup$ Commented Oct 2, 2019 at 4:18
  • $\begingroup$ @AlexandreEremenko,: Well, in fact there is something wrong with your example. It does not fit for large $x$, as $\sinh(x/2)\leq\sqrt{\frac12\sinh x}$. $\endgroup$ Commented Oct 2, 2019 at 10:58

1 Answer 1

5
$\begingroup$

The answer is no. Let $f$ be the function described under Remark 1 of the OP. By continuity, there exists a $\delta >0$ such that $f(x) - \sqrt{(\sinh x)/x} > 0.05$ on the real axis for $|x|\leq \delta $ (note that, at $x=0$, the aforementioned difference is $2/\sqrt{\pi } -1 = 0.128$).

Consider furthermore the function $$ g(z)=\epsilon \left[\frac{2\sinh (z/2)}{z}-(1+\kappa )\frac{\pi^{2} \cosh (z/2)}{z^2+\pi^{2} } \right] $$ $g$ satisfies the required constraints $|g(x+iy)| \leq C\exp (|x|/2)$ and $g(x+iy)\rightarrow 0$ as $y\rightarrow \infty$ for each separate $x$. On the real axis, $g(x)\geq 0$ for $\kappa =0$; we can adjust to a $\kappa > 0$ such that $g(x)\geq 0$ for $|x|\geq \delta $ but $g(x)<0$ for $|x|<\delta $. We can furthermore adjust $\epsilon $ such that $|g(x)|<0.05$ for $|x|< \delta $. With such a choice of $\kappa $ and $\epsilon $, the function $f+g$ is a counterexample to the OP's question, $f(0)+g(0) < 2/\sqrt{\pi } $ while satisfying all the given constraints.

$\endgroup$
1
  • $\begingroup$ Thanks a lot! Quite instructive. $\endgroup$ Commented Oct 3, 2019 at 5:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .