Proof (claimed) for Riemann hypothesis on ArXiv Has anyone noticed the paper On the zeros of the zeta function and eigenvalue problems by M. R. Pistorius, available on ArXiv?
The author claims a proof of RH, and also a growth condition on the zeros.
It was posted two weeks ago, and I expected it would have been shot down by now. Has there been any discussion or attempt at verification of this preprint?
 A: I had a go reading through the paper and I think I found the error. The main argument in the paper can be summarized as follows:

The Riemann $\Xi$-function $\Xi(t) = \xi\left(\frac{1}{2} + it\right)$ satisfy $\Xi(t) = \Xi(0)\nu_t(\pi/2)$ where $$\nu_t(x) = \int_0^\infty \cos(t(y+\cos(x))\Phi(y){\rm d}y$$ and $\Phi$ is related to the Jacobi $\theta$-function. This is a result by Riemann and holds true. The author then notes that when $t$ is such that $\Xi(t) = 0$ then $\nu_t(x)$ satisfy the Sturm–Liouville (SL) problem 
  $$\left(\frac{\nu_t'(x)}{\sin(x)}\right)' + t^2\sin(x)\nu_t(x) = 0,~~~\nu_t'(0) = 0,~~~\nu_t(\pi/2) = 0$$
  This is also true. The proof is completed by appealing to a theorem that says that this problem only has real eigenvalues. If this holds then it follows that $\Xi(t) = 0\implies t\in\mathbb{R}$ which is the Riemann hypotesis.

The error is in the last step. It is indeed true that a regular SL problem only has real eigenvalues, however this is not a regular SL problem as $\frac{1}{\sin(x)}$ has a pole at $x=0$. In this case there is no guarantee that the eigenvalues have to be real and we can show this explicitly with a simple counter-example: the function $\nu_t(x) = \sin(t\cos(x))$ satisfy the SL problem above for any complex number $t$.

Given how much interest this question has generated, which I take to mean that many people thought (or perhaps just hoped) this was a potentially viable proof, I think it’s useful to talk a bit about why it had to be wrong. Personally I have only been in academia for $\sim 10$ years, but I have already managed to see $\sim 50$ papers$^1$ like this where a major result is proven in a few pages using elementary methods. This is just another one. One soon learns that papers like this are never correct and the reason is often this: if it could have been solved this way then it would have been solved this way many many years ago - the techniques used are just too simple. As for some useful pointers for how to judge for yourself if a paper like this has the potential for being correct I reccommend Scott Aaronsons "Ten Signs a Claimed Mathematical Breakthrough is Wrong".
$^1$ For some examples see the MSE questions [1], [2], [3]
A: I'll try to summarize my understanding of the manuscript (v2). I was unable to find any issue with the proof (Edit: see the answer by Winther), but maybe these notes will help someone in following the argument and forming their own opinion.
The Riemann hypothesis is the conjecture that the Riemann zeta function has its zeros only at the negative even integers ("trivial zeros") and the complex numbers $\frac12 + i t$ for real $t$.
The function
$$
\Xi(t) := \xi(\tfrac12 + i t),
\quad
\xi(s) := \tfrac{s(s-1)}{2} \pi^{-s/2} \Gamma(\tfrac{s}{2}) \zeta(s)
,
$$
serves as a proxy to the $\zeta$-function, because as it says here,


*

*the zeros of $\xi$ are all located on the strip $(0,1) + i \mathbb{R}$, 

*for real $t$, one has: $\Xi(t) = 0$ iff $\zeta(\frac12 + i t) = 0$,

*$\Xi(t)$ is real for real $t$.


In particular, $\Xi$ does not share the trivial zeros with $\zeta$. 
Hence,
$$ \label{e:RH} \tag{RH}
\Xi(t) = 0
\quad\Rightarrow\quad
t \in \mathbb{R}
$$
is what the author sets out to prove (p.1).
To do that, the author introduces the family of functions 
$$
[0, \tfrac\pi2] \to \mathbb{C},
\quad
x \mapsto v_\Psi(x, t),
$$
parameterized by a scalar $t \in \mathbb{C}$ and a function $\Psi : \mathbb{R}_+ \to \mathbb{R}_+$
as
$$
v_\Psi(x, t) 
:=
\int_0^\infty \cos( t (y + \cos(x)) ) \Psi(y) \, dy
.
$$
He observes (p.4) that
for a particular choice of $\Psi$,
there holds
$$ \label{e:xiv} \tag{1}
\Xi(t) = \Xi(0) \, v_\Psi(\tfrac\pi2, t)
\quad
\forall t \in \mathbb{C}
.
$$
The key argument is then this:
$$ \label{e:key} \tag{2}
\text{If $t \in \mathbb{C}$ satisfies $v_\Psi(\tfrac\pi2, t) = 0$ then $t^2$ is real.}
$$
Note that in that case, either


*

*$t \in i \mathbb{R}$, which is incompatible with \eqref{e:xiv} because $\Xi(t) \neq 0$ for such $t$ (see above), or

*$t \in \mathbb{R}$.


So: \eqref{e:xiv}-\eqref{e:key} imply \eqref{e:RH}.

The particular function $\Psi$ for which \eqref{e:xiv} holds
is readily available (apparently; I'm no expert and would welcome a reference, but I have "verified" numerically).
Since 
$$
\Xi(t) = \int_0^\infty \cos(t y) 2 \Phi(y) \,dy
,
$$
where
$$
\Phi(y) = 
2 \pi e^{\frac52 y} \sum_{n \geq 1} 
(2 \pi e^{2 y} n^2 - 3)
n^2
e^{-n^2 \pi e^{2 y}}
,
$$
the choice
$$
\Psi := 2 / \Xi(0) \, \Phi 
$$
gives \eqref{e:xiv}.
To observe is 


*

*the super-exponential decay of $\Psi(y)$ as $y \to \infty$ due to the double exponential in the sum, 

*measurability of $\Psi$, 

*$\Psi \geq 0$, and

*$\int_0^\infty \Psi(y) \,dy = 1$.



It remains to discuss the key argument \eqref{e:key}.
The author's strategy is to prove this for any function $\Psi$ that satisfies the above four observations; the particular shape of $\Psi$ is therefore irrelevant for this argument. This is the subject of Lemma 3 (p.3):
Suppose $t \in \mathbb{C}$ is such that
$$
v_\Psi(\tfrac\pi2, t) = 0.
$$
Fix this $t$. To show: $t^2$ is real.
The idea is to show that the function
$$
f :
[0, \tfrac\pi2] \to \mathbb{C},
\quad
x \mapsto v_\Psi(x, t)
$$


*

*satisfies a Sturm--Liouville eigenvalue problem with eigenvalue $t^2$, and

*apply a classical result that says that the eigenvalue is necessarily real.


The Sturm--Liouville eigenvalue problem is
$$
\text{SL}_f := (f' / r)' + ( t^2 p + q ) f = 0 \text{  on  } (0, \tfrac\pi2),
\quad
f'(0) = 0,
\quad
f(\tfrac\pi2) = 0
,
$$
where
$$
r = p = \sin
\quad\text{and}\quad
q = 0
.
$$
Note that our $f$ is (complex-valued but) $C^2$ due to the good integrability of $\Psi$.
Some elementary manipulations show that the eigenvalue problem is indeed satisfied, including the boundary conditions (the one at $\frac\pi2$ by assumption).
The author cites Theorem 8.3.1 in [F.V. Atkinson, Discrete and Continuous Boundary Problems. Academic Press, New York, London,
1964] which says that under those conditions, the eigenvalue $t^2$ is real (the Sturm--Liouville operator is self-adjoint, after all).
The theorem is quoted almost verbatim in the manuscript.
Sidenote:
Testing the eigenvalue problem with $\bar{f}$ and integrating would immediately show that $t^2$ is real, but the theorem also gives the summability $\sum_{t \in \Xi^{-1}(0)} |t|^{-1-\epsilon} < \infty$ of the countably many zeros of $\Xi$.

Edit: The answer by Winther points to where the problem could be. Indeed,
$$
f'(x) = t \sin(x) \int_0^\infty \sin(t(y + \cos(x))) \Psi(y) \, dy
$$
behaves like $f' \propto \sin$ at $x = 0$, so the following computation is justified (all terms are finite; upper boundary term is zero):
$$
0 =
\int_0^{\frac\pi2}
\text{SL}_f \, \bar{f}
=
 - \frac{f'(0^+)}{\sin(0^+)} \bar{f}(0^+)  
+
\int_0^{\frac\pi2} \{ -r^{-1}|f'|^2 + t^2 p |f|^2\}
.
$$
If the boundary term was zero (or just real), we could conclude that $t^2$ is real.
However:
$$
\frac{f'(0^+)}{\sin(0^+)} \bar{f}(0^+) = t \times (\text{a real nonzero constant})
$$
could be complex,
so $t^2 \in \mathbb{R}$ cannot be concluded.
