$|\zeta(1/2 + it)|^2 \geq \frac{\log(t)}{\log \log(t)}$

Question

I've tried to solve this exercise for hours but I didn't managed to figure it out.

Show that there exists a sequence $t \to \infty$ for which $$|\zeta (1/2 + it)|^2 \geq \frac{\log(t)}{\log\log(t)}$$

Hint : Look at $\lim_{T \to \infty}{\frac{1}{2T}\int_{-T}^{T}{|\zeta(\sigma+it)|^2 dt}}$

I have some sort of ideas but haven't managed to make them work.

I guess the arguments should be something like this :

If $\mu$ is a Radon measure then

$$\lim_{T \to \infty}{\frac{1}{2T}{\int_{-T}^{T}{|\hat{\mu}|^2(t)dt} }} =\sum_{t\in\mathbb{R}}{|\mu(\{t\})|^2}$$

So that we find

$$\lim_{T \to \infty}{\frac{1}{2T}{\int_{-T}^{T}{|\zeta(\sigma+it)|^2(t)dt} }} =\sum_{n=1}{(\frac{1}{n^\sigma})^2} =\zeta(2\sigma)$$

So basically when $\sigma = 1/2$ the limit goes to $\infty$.

I think that if there wasn't such a sequence then the limit should go to $\infty$ slower than it actually does, but I have no idea about how to show it. Another guess is that if there wasn't such a sequence then one would be able to find some bounds on $|\zeta(1/2(1+\epsilon)+it)|^2$ which would led by the previous formula to a bound on $|\zeta(1+\epsilon)| \sim \frac{1}{\epsilon}$ and maybe this bound is clearly not true, to of course if I knew a bound on $|\zeta'(s)|$ around the critical line then it would be very easy to do so, but I don't know anything about such a bound.

The problem is also that I'm working near the critical line where I can't use the classic representation of $\zeta$.

Answer 1

Note that $\lim_{T \to \infty} \frac{\int_T^{2T}|\zeta(1/2+it)|^2dt}{T\log T}=1$ so for any function $V(t) \to \infty$ monotonically, let $E(T)$ the set of $t\in [T,2T]$ for which $|\zeta(1/2+it)|^2 \ge \frac{\log t}{V(t)}$

Assume that for some large $T$ where $V(T) >10$ and $\frac{\int_T^{2T}|\zeta(1/2+it)|^2dt}{T\log T} >1/2$ we have $m(E(T))=0$ so the complement $F(T)$ has full measure $T$ and $|\zeta(1/2+it)|^2 \le \frac{\log t}{V(t)} \le \frac{\log 2T}{V(T)}$ for $t \in F(T)$. Then $$\frac{1}{2} \le \frac{\int_T^{2T}|\zeta(1/2+it)|^2dt}{T\log T}=\frac{\int_{F(T)}|\zeta(1/2+it)|^2dt}{T\log T} \le \frac{\log 2T}{V(T)\log T} < \frac{\log 2T}{10\log T}$$ so $5 \log T < \log 2+ \log T$ which of course is impossible.

Hence for all large enough $T$ we have $m(E(T))>0$ so there is $t_V \in [T, 2T]$ st $|\zeta(1/2+it_V)|^2 \ge \frac{\log t}{V(t)}$

Choosing $V=\log \log, t_N=t_{\log \log}$ for the intervals $[3^N, 2\times 3^N]$ which are disjoint, gives us the result.

Note that it is a nontrivial Theorem of Selberg that $|\zeta(1/2+it)| \ge e^{(\log \log t)^{1/2-\epsilon}}$ about half the time in $[T,2T]$ and of course $|\zeta(1/2+it)| \le e^{-(\log \log t)^{1/2-\epsilon}}$ the other half the time (so on the critical line $|\zeta|$ is almost always either very large or very small), while pretty much within $1$ of any $t \in [T,2T]$ there are values $|\zeta(1/2+iu)|>(\log T)^{1-\epsilon}$. Both the above results are proven essentially by a (considerably more elaborate) argument similar to the above.