Consider an integral equation: $$\int_{-\infty}^{+\infty}\frac{e^{-\sigma y}f(y)}{e^{e^{x-y}}+1}dy=0$$, where $\sigma\in(\frac{1}{2},1)$
Salem proved that this equation has no bounded solution other than trivial $f(y)=0$ iff Riemann hypothesis is true.
Suppose we want to construct a counterexample, that is , a non zero function $f(y)$ satisfying the integral equation. My question is should $f(y)$ be independent of both the parameters $x$ and $\sigma$ present in the kernel of the integral equation? In other words, we need to construct a function $f(y)$ which works for all $x$ and $\sigma$?
The actual equivalence is the following: one fixes $\sigma >1/2$; then the existence of $\phi$ (nontrivial) bounded, measurable function of one real variable (say $y$ though of course the name of the integration variable is irrelevant as long as it's not $x$ or $\sigma$ eg $f(y)=y$ or whatever), solving the above for all $x$ real is equivalent (by Wiener famous Tauberian theorem and a little functional analysis and measure theory, noting that by duality, the existence of $\phi$ is essentially equivalent to the fact that the linear space generated by the translates, $K_{\sigma}(x-y), x$ arbitrary, regarded as functions of $y$, are not dense in $L^{1}(\mathbb R(y))$) to the fact that the Fourier transform of the kernel $$K_{\sigma}(u)=\frac{u^{\sigma}}{e^{e^u}+1}$$ vanishes at $\sigma+it$ for some $t$ real; but this transform is essentially (up to constants) $\Gamma(\sigma+it)\eta(\sigma+it)$ where $\eta$ is the usual Dirichlet eta obtained from the alternating zeta (so $\eta(s)=(1-2^{1-s})\zeta(s)$).
In particular the existence of such a $\phi$ is equivalent to the fact that $\eta$ hence $\zeta$ has a zero with abscissa $\sigma$ so to the negation of RH.
The simple details above and Wiener's Theorem are presented for example in Broughan's reference book on Equivalents of the RH volume $2$ chapter $8$