I am hesitating to write this answer since I know too little about the whole subject. However it is already a couple of years ago that I stumbled on a interesting paper on the arXiv https://arxiv.org/pdf/1703.03827.pdf by Vladimir Blinovsky .This paper seems to me like a simple and appealing idea to tackle the problem. It just seems strange for me that this idea has not attracted much publicity since it has been lying around on the arXiv for a quite long time. As a matter of fact this idea may be even described in plain words without resorting to mathematics. Let me do this now.
Let us fix the notation:
\begin{equation}
\zeta(s) := \sum\limits_{n=0}^\infty \frac{1}{n^s} \quad (A)
\end{equation}
where $s\in {\mathbb C}$ and $Re[s] > 1$.
By analytic continuation (as originally carried out by Riemann https://en.wikipedia.org/wiki/File:Ueber_die_Anzahl_der_Primzahlen_unter_einer_gegebenen_Gr%C3%B6sse.pdf by deforming the integration contour in the complex plane and using Cauchy theorem) we obtain the functional equation:
\begin{equation}
\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s) \quad (B)
\end{equation}
for $s\in {\mathbb C}$ and $s\neq1$.
Let us define the squared module of the zeta function as follows:
\begin{equation}
K(\sigma,T) := \left| \zeta(s)\right|^2
\end{equation}
where $s=\sigma + \imath T$. Clearly since $\zeta(s)$ is continuous and smooth the same holds for the function $K$, i.e. it too is continuous and smooth.
Now comes the important part. What Blinovsky claims in his paper is that the function $K(\sigma,T)$ above is convex in the variable $\sigma$. In other words he claims that
\begin{equation}
\frac{\partial^2 K(\sigma,T)}{\partial \sigma^2} \ge 0 \quad (C)
\end{equation}
for $\sigma \in (0,1)$ and $T$ being big enough, i.e. being bigger than some number which is independent of $\sigma$. Note that the property (C) along with functional equation (B) immediately implies the Riemann hypothesis being true. Indeed let us assume that there is some stray zero off the critical line at some $(\sigma = \sigma_0,T)$ where $\sigma_0 \in (1/2,1)$. Then from the functional equation there must be another zero at $(1-\sigma,T)$ where $1-\sigma_0 \in (0,1/2)$. But this would actually mean that the set $\left\{\xi \in (1-\sigma_0,\sigma_0) | K(\xi,T)\right\}$ is either strictly negative which it cannot be since per definition the function $K$ is non-negative or that the set in question is identically equal to zero which again is impossible because the function is smooth per definition.
Now the question appears how does Blinovsky prove the convexity? He starts from a certain integral representation of the zeta function and then by changing variables appropriately and then differentiating with respect to $\sigma$ twice he ends up with a following neat expression . We have:
\begin{equation}
\frac{\partial^2 K(\sigma,T)}{\partial \sigma^2} = \frac{8 }{\pi T} \left(\int\limits_0^1 f(h,T) \cos\left( \frac{\pi h}{2}\right)^2 dh - \frac{1}{2} \int\limits_0^1 f(h,T) dh \right)
\end{equation}
where $f(h,T)$ is some complicated function expressed through an improper integral and an infinite sum, a function to long to be written down in here without the benefit of any insight. Now since $\int\limits_0^1 \cos(\pi h/2)^2 dh = 1/2$ proving the convexity property is equivalent to the Fortuin–Kasteleyn–Ginibre (FKG) inequality https://en.wikipedia.org/wiki/FKG_inequality . Since the squared cosine is monotonically decreasing in $h\in(0,1)$ the only thing one needs to do is to prove that $f(h,T)$ too is monotonically decreasing in the domain in question. Blinovsky proceeds to show that by looking at the derivative $\partial_h f(h,T)$ and proving that it is negative for all $h \in(0,1)$ and for $T$ being big enough.
Having said all this my question would be is this approach to this problem sound or instead is there anything internally flawed in here that won't make this thread worth pursuing at all.
I would really appreciate some feedback. Thank you in advance for it.
