Bounds for analytic circles

Question

It is known that for certain particular entire functions $f(s)$ of first order, in the circle $|s| = p$, if $\epsilon$ is a positive number as small as desired, the following bound holds:

$$|f(s)| = O(e^{\epsilon p})$$

A classical entire function of first order is Riemann's xi function, written as:

$$\xi(s) = \frac{1}{2} s(s-1)\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)$$

Is it easy to show that Riemann's hypothesis(RH), or Lindeloph's hypothesis(LH), imply this bound? If this bound is assumed, would both these hypothesis follow? It is known that RH implies LH, but whether LH implies RH is still a deep unknown. Consider the function $f(\xi(s)) = f(\xi)$ defined by

$$f(\xi) - \xi\left(\frac{s-i}{2}\right) - \xi\left(\frac{i+s}{2}\right) = 0$$

An interesting question is whether, in the circle $|s| = p$, the mentioned bound $|f(\xi)| = O(e^{\epsilon p})$ holds. The function $f(\xi)$ is interesting in that it is easy to show that all its zeros are real.

@bojonnson thanks. You're right. My aim was to see if the bound held in the critical strip, and $\xi(s)$ was mixed with $\Xi(t)$. Since @GHfromMO makes key points, it's best not to amend my question, so other's can see the error and understand the thread. Strangely enough, the question arouse in my head from the following raw observation: It is a fascinating elementary fact, first seen by Jensen, that given a Bernoulli polynomial $P_{2k}(s)$ and a Bernoulli number $B_k$, the map $\Xi(t)=\Xi(-t)$ shows that the critical strip is a cut of the complex plane $s$ where the difference $P_{2K}(s) - B_K(-1)^k$, divided symmetrically into $4k(-1)^k$ slices, creates $4k(-1)^k$ finite slices that can be mapped by a suitable infinite elliptic series running from zero to infinity on the real axis. Hence $f(\Xi)$ can be written for complex $z$ as:

$$f(\Xi(z))=\int_{0}^{\infty}g(t) \cos(zt) \ dt$$

Here $g(t)$ is defined by $$g(t)= (h''(t) - h(t))(e^t -e^{-t})$$ and $h(t)$ is

$$4h(t) = e^t\theta(e^t)$$

$\theta(x)$ is the Jacobi elliptic theta function:

$$\theta(x) = \sum_{n=-\infty}^{\infty} e^{-\pi x^2 n^2}$$

Answer 1

It looks like @boJonsson made a typo, mixing up $\xi(s)$ with $\Xi(t)$, defined, for $s=1/2+it$, as:

$$\Xi(t) = \frac{1}{2} s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$$

All the zeros in the following function $f(\Xi)$ are real:

$$f(\Xi(t)) - \Xi\left(\frac{t+i}{2}\right) - \Xi\left(\frac{t-i}{2}\right) = 0$$

If you assume the bound $O(e^{\epsilon p})$ for the Riemann xi function $\xi(s)$, where $s$ is on the critical line and $|s| < p$, it seems highly likely that Lindelöf Hypothesis(LH) would follow. Both the specific bound for $\xi(s)$ and the Lindelöf Hypothesis are implied by the Riemann Hypothesis (RH). However, it seems subtle whether the bound directly implies RH, or LH.

That the bound and LH imply each other, might follow from the fact that the bound $S_1(t)=o(t)$ implies LH and vice versa. It is an interesting question whether $|f(\Xi)| = O(e^{\epsilon p})$ holds for $f(\Xi)$ in $0<Im(t)<1/2$, $|t| <p$.

Answer 2

The functions $\xi(s)$ and $$f(s):=\xi\left(\frac{s-i}{2}\right) + \xi\left(\frac{i+s}{2}\right)$$ grow faster than exponentially on the positive axis, hence they do not satisfy the first bound. This follows from Stirling's approximation for $\Gamma(s)$ and the fact that $\zeta(s)$ tends to $1$ as $\Re s\to\infty$.