The function I have in mind is $f(s)=\phi(t)\cdot|\zeta(s)|$, where $s=\sigma + it$, $|\cdot|$ is the modulus, and $\frac{1}{2}<\sigma<1$ is fixed. Here $\phi(t)$ is a positive, increasing function of $t>0$, or possibly constant ($\phi(t)=1$). Let us denote as $L_\sigma$ the following quantity:
$$L_\sigma=\liminf_{t\rightarrow\infty} f(\sigma + it).$$
I can see the following potential scenarii:
- If $\phi(t)$ does not grow fast enough, say $\phi(t)=1$, then maybe $L_\sigma=0$. This would add some additional challenge to proving RH (the Riemann Hypothesis), yet this scenario by itself does not disprove RH.
- If $\phi(t)$ grows fast enough, say $\phi(t)=\exp(t)$, then maybe $L_\sigma>0$. Yet this scenario by itself does not prove RH. It would just simply be a much weaker result than RH.
I was wondering what are the most recent results concerning these types of $\liminf$.
Fix $1/2<\sigma < 1$ and $1\leq \phi(t)$ as a continuous function with your desired properties. Define $g_{\epsilon}$ as the polynomial approximation of $\epsilon*[\phi(\frac{1}{1-t})]^{-2}$ on the interval $[\sigma-i,\sigma+i]$. Since $\phi$ is increasing in $t$, we see that $g_\epsilon(1)=0$ for each $\epsilon \rightarrow 0$. Take $U$ to be the compact subset of $1/2<\sigma < 1$ containing our interval with no larger imaginary parts. We see that $f_\epsilon$ is analytic on the interior of $U$. Using Universality of the zeta function, we can arbitrarily approximate each $g_\epsilon$ by $\zeta$ on $U+it_{univ}$ for some $t_{univ}>0$ depending on said approximation. In fact, there are infinitely such $t_{univ}$ that can be selected to be arbitrarily large. We can then see that $$f(t)= \phi(t)*|\zeta(\sigma+it)| \approx \phi(t)*g_\epsilon(\sigma +i(t-t_{univ,\epsilon})) \approx \epsilon\phi(t)[\phi(\frac{1}{1-(t-t_{univ,\epsilon})})]^{-2}$$ Taking $t\rightarrow t_{univ,\epsilon}$, we see that $f$ can be arbitrarily small.
Sorry for a lot of missing details, but the universality of the zeta function in the critical strip should imply your limit infinum is zero for any choice of $\phi$.