Riemann hypothesis: is Bender-Brody-Müller Hamiltonian a new line of attack?

I tried to put my own argument on this blog recently but I failed. Let me try again. 1)- The original definition of $\zeta(z,x) $ is

$$\zeta(z,x) = \sum_{n=1}^\infty \frac{1}{(n+x)^z} $$

Then $\zeta(z,x)-\zeta(z,x-1)=-1/x^z$. Then, using the integral representation, this relation extends by analiticity to the maximal domain in which both sides are defined and holomorphic. For our purpose, it is enough to consider $x>0$ and $\Re{z}>0$. Let $\zeta_z$ denote the map $x\to \zeta(z,x)$. Then, using the generator $A$ of the dilation operator, it is easy to check that the eigenvalues have the form $1/2+\imath E$. For $A=(pq+qp)/2=-\imath xd/dx-\imath 1/2$. This gives $A(1/x^z)=\imath (z-1/2)(1/x^z)$. Going from $A$ to $H$ is done using the operator $\Delta$, leading to the formal equation $H\zeta_{1/2+\imath E}=E\zeta_z$. The Dirichlet b.c. at the origin forces E to satisfy $\zeta(1/2+\imath E)=0$.

2)- The first problem I see is that $\zeta_z$ does not seem to belong to the Hilbert space ${\mathcal H}=L^2(0,\infty)$. Using a classical method by Hadamard

$$\int_0^\infty |\zeta(z,x)|^2 dx=\frac{1}{1-2\Re{z}}\zeta(2\Re{z}-1)$$

which converges for $\Re{z}>1$. Using the integral representation, a similar firmula can be obtained, but I failed to show that square integrability hold for $\Re{z}=1/2$. Hence I see a problem here.

3)- The other problem is the definition of $\Delta$. Let $S$ denote the translation operator, defined non rigorously by $Sf(x)=f(x-1)$. If restricted to $\mathcal H$, it is not unitary, because it is defined only for $x>1$. We can defined it by imposing $Sf(x)=0$ for $x\in [0,1]$. Then it is only a partial isometry. For $S^\ast f(x)=f(x+1)$ leading to $S^\ast S=I$, but $S S^\ast =P$ is the projection onto $L^2(1,\infty)$. Using these notations $\Delta= I-S=(S^\ast-I)S$. But we get a problem here: the function $\zeta_z$ is defined for $x>-1$, and the extension to the interval $(-1,0]$ is explicitly used in 1)-. So we cannot use $S$. But then what is the operator $e^{-\imath \hat{p}}$ used by the authors?

4)- If $\hat{p}$ is the usual operator

$$\hat{p}=-\imath \frac{d}{dx}$$

then there is a problem with its domain of definition. On $L^2(\mathbb R)$, it is selfadjoint as can be seen by using Fourier transform. But on $\mathcal H$, it is not. This is a classical exercise found in the very old book of Courant-Hilbert. Namely one can always define it in the set of $L^2$ functions with $L^2$ derivative, vanishing at $x=0$. Then it is symmetric. If so its adjoint is defined on the same space but without the vanishing at $x=0$. Not only the adjoint is not symmetric but its set of eigenvalues is the open lower half plane. This is because if $f_z(x)=e^{zx}$, then $\hat{p}^\ast f_z=-\imath z f_z$, while $f_z\in \mathcal{H}$ for $\Re{z}<0$. The same argument shows that $+\imath$ cannot be an eigenvalue. This means that the "defect indices", namely the dimension of the eigenspaces with eigenvalues $\pm\imath$, are not equal. Then, the von Neumann theorem show that the operator $\hat{p}$ has no selfajoint extension. If not selfadjoint, the definition of its exponential becomes a problem, because the functional calculus is nit defined in general.

5)- The previous argument can be rephrased in terms of the operator $S$. Its adjoint admits a lot of eigenvalues, namely the points inside the unit disk.

In conclusion, the sloppyness of the definitions used but the authors leads to a complete mess. Nothing is correct in this paper.

As long as physicists use algebra, or algorithmic arguments, they can find outstanding results. But when it comes to analysis, they may loose their judgment, and grave mistakes show up at the corner. Analysis is not easily amenable to algoritmic descriptions. And this is precisely where the power of Mathematics lies: by manipulating infinities, Mathematics goes way beyond the Church-Turing definition of computability. And what is Analysis if not manipulating infinities, through limits, convergence and the likes?

I didn't get everything written in this paper.

But keep in mind this line of attack has some good chances to be way too simple : all this works the same for $F(s) = \alpha L(s,\chi_5)+\overline{\alpha}L(s,\overline{\chi_5})= 2\sum\limits_{n=1}^\infty \Re(\alpha \;\chi_5(n)) n^{-s}$ where $\chi_5$ is the non-real character modulo $5$ and $\alpha \in \mathbb{C}$.

$F(s)$ has the same kind of integral representation and functional equation as $\zeta(s)$, so we can write for it the same kind of differential operator. But the RH obviously fails for $F(s)$ (it doesn't have an Euler product)