I was wondering whether we can prove that $\zeta$ has at least one zero on the critical line $\mathrm{Re}(s)=1/2$.
For instance, Hardy proved in 1914 that $\zeta$ has infinitely many zeros on the critical line. See also here for the critical strip. I'm asking a much easier question, which to prove that $\zeta$ has at least one zero on the critical line.
The point of my question is that I don't need a constructive answer. I know that we have many examples of zeros with $\mathrm{Re}(s)=1/2$, but we can only compute approximations of these zeros (see here), because the imaginary parts of these zeros are conjecturally transcendental numbers, and probably algebraically independent with usual constants as $\pi,e$, etc (so that there is probably no closed formula...). So my question is different from that one, because I'm talking about any "known" or any "specific" zero, or any kind of numerical computation/approximation.
My question is similar but more specific than this one, which is just asking about the existence of at least one non-trivial zero. I would like an argument from analytic number theory (without approximations nor numerical computations) which shows the existence of a zero of $\zeta$ with $\mathrm{Re}(s)=1/2$.
I hope that my question is clear enough. Thank you.
It is because $\zeta(s)$ has a functional equation that we can prove some of its zeros have real part exactly $1/2$.
The functional equation implies $$\Lambda(t) = \pi^{-(1/2+it)/2} \Gamma((1/2+it)/2) \zeta(1/2+it)$$ is $\mathbb{R} \to \mathbb{R}$, thus it has a zero at every sign change. Proving $\Lambda(t)$ changes of sign around $t \approx 14.15$ is not hard. See this plot, together with some bounds for the approximation it is a proof that $\Lambda(t)$ changes of sign.
Usually we look instead at the Hardy Z-function $Z(t) = \frac{\Lambda(t)}{|\pi^{-(1/2+it)/2} \Gamma((1/2+it)/2)|} = \zeta(1/2+it) e^{i \vartheta(t)}$ because it doesn't decay as fast as $\Lambda(t)$.
In practice $\Lambda(t) = \lambda(1/2+it)$ where $$\lambda(s) = \frac{1}{s-1} - \frac{1}{s}+ \int_1^\infty (x^{s/2-1}+x^{(1-s)/2-1})\theta(x)dx$$ where $\theta(x) = \sum_{n=1}^\infty e^{-\pi n^2 x}$. Approximating $\theta(x)$ by its first few terms gives an approximation with an arbitrary precision to $\Lambda(t)$.