The Riemann hypothesis as a problem in analysis The recent post("Long-standing conjectures in analysis ... often turn out to be false") prompted me to think about a question which I have not given much though before: to what extent the Riemann hypothesis (RH) may be regarded as a problem in analysis. It may actually be not as silly as it sounds. 
The particular side of it I am curious about is the following. The 
Theorem : $\zeta(s)\neq 0$ for $\Re s>1$.
may be considered a very weak consequence of RH. However, this statement is only trivial in the context of number theory. One may ask, is it possible to give it a "purely analytic" proof, without using Euler product and other stuff related to the  primes? Apparently, it is not possible to formulate this as a precise mathematical question because all the theories used to formalize analysis contain a good deal of arithmetic, but it should be precise enough for practical purposes. (You know number theory when you see it.)
Most likely such a proof does not exist yet, but some readers may know more about the relevant things then I do. I am honestly curious because while it definitely looks tough, it may be not entirely implausible. If such a proof is found, it might give us a fresh look on the old problem.
 A: First, let me say that a proof just of $\zeta(s) \not= 0$ for ${\rm Re}(s) > 1$ that does not in any straightforward way also yield a larger nonvanishing half-plane should not be regarded as a "fresh look" at anything as deep as RH. In fact, since the nonvanishing of $\zeta(s)$ on the line ${\rm Re}(s) = 1$ is equivalent to the Prime Number Theorem, it is not realistic that an approach to nonvanishing on ${\rm Re}(s) > 1$ alone is going to get you anything more since you don't just stumble into a new proof of the Prime Number Theorem. 
With that in mind, consider the following integral representation of $\zeta(s)$ for ${\rm Re}(s) > 1$, which follows from partial summation on the Dirichlet series for $\zeta(s)$ (it is in Serre's Course in Arithmetic, for example): 
$$
\zeta(s) = s\int_1^\infty \frac{[x]}{x^{s+1}}\,dx = \frac{s}{s-1} + s\int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx.
$$
If $\zeta(s) = 0$ at an $s$ with ${\rm Re}(s) > 1$, then by the above formula
$$
\frac{1}{1-s} = \int_1^\infty \frac{\{x\}}{x^{s+1}}\,dx.
$$
Taking absolute values of both sides and using $|\{x\}|<1$ and writing $s=\sigma + it$, 
$$
\frac{1}{|1-s|} < \frac{1}{\sigma},
$$
so $\sigma < |1-s|$. Squaring both sides, $\sigma^2 < (1-\sigma)^2 + t^2$, which is the same as $\sigma < (1+t^2)/2$. Combining this with the restriction $1 < \sigma$ puts all $s$ with ${\rm Re}(s) > 1$ and $\zeta(s) = 0$ in a restricted range in the half-plane ${\rm Re}(s) > 1$, but not into the empty set. Perhaps by not using something as simple as $|\{x\}| < 1$ in the estimate on the integral someone can restrict the range of possible $s$ even further.
A: Andrew Booker is of the opinion that it is the nonvanishing of zeta (or an L-function) to the right of the critical strip which is more fundamental than the Euler product.  See Slide 10 of his recent talk https://heilbronn.ac.uk/wp-content/uploads/2018/07/Booker-talk.pdf
