It looks like @boJonson made a typo, mixing up $\xi(s)$ with $\Xi(t)$, defined, for $s=1/2+it$, as:

$$\Xi(t) = \frac{1}{2} s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$$

All the zeros in the following function $f(\Xi)$ are real:

$$f(\Xi(t)) - \Xi\left(\frac{t+i}{2}\right) - \Xi\left(\frac{t-i}{2}\right) = 0$$

If you assume the bound $O(e^{\epsilon p})$ for the Riemann xi function $\xi(s)$, where $s$ is on the critical line and $|s| < p$, it seems highly likely that Lindelöf Hypothesis(LH) would follow. Both the specific bound for $\xi(s)$ and the Lindelöf Hypothesis are implied by the Riemann Hypothesis (RH). However, it seems subtle whether the bound directly implies RH, or LH.

That the bound and LH imply each other, might follow from the fact that the bound $S_1(t)=o(t)$ implies LH and vice versa. It is an interesting question whether $|f(\Xi)| = O(e^{\epsilon p})$ holds for $f(\Xi)$ in $0<Im(t)<1/2$, $|t| <p$.

It looks like @boJonsson made a typo, mixing up $\xi(s)$ with $\Xi(t)$, defined, for $s=1/2+it$, as:

$$\Xi(t) = \frac{1}{2} s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$$

All the zeros in the following function $f(\Xi)$ are real:

$$f(\Xi(t)) - \Xi\left(\frac{t+i}{2}\right) - \Xi\left(\frac{t-i}{2}\right) = 0$$

If you assume the bound $O(e^{\epsilon p})$ for the Riemann xi function $\xi(s)$, where $s$ is on the critical line and $|s| < p$, it seems highly likely that Lindelöf Hypothesis(LH) would follow. Both the specific bound for $\xi(s)$ and the Lindelöf Hypothesis are implied by the Riemann Hypothesis (RH). However, it seems subtle whether the bound directly implies RH, or LH.

That the bound and LH imply each other, might follow from the fact that the bound $S_1(t)=o(t)$ implies LH and vice versa. It is an interesting question whether $|f(\Xi)| = O(e^{\epsilon p})$ holds for $f(\Xi)$ in $0<Im(t)<1/2$, $|t| <p$.

It looks like @boJonson made a typo, mixing up $\xi(t)$ with $\Xi(t)$, defined, for $s=1/2+it$, as:

$$\Xi(t) = \frac{1}{2} s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$$

All the zeros in the following function $f(\Xi)$ are real:

$$f(\Xi(t)) - \Xi\left(\frac{t+i}{2}\right) - \Xi\left(\frac{t-i}{2}\right) = 0$$

If you assume the bound $O(e^{\epsilon p})$ for the Riemann xi function $\xi(s)$, where $s$ is on the critical line and $|s| < p$, it seems highly likely that Lindelöf Hypothesis(LH) would follow. Both the specific bound for $\xi(s)$ and the Lindelöf Hypothesis are implied by the Riemann Hypothesis (RH). However, it seems subtle whether the bound directly implies RH, or LH.

That the bound and LH imply each other, might follow from the fact that the bound $S_1(t)=o(t)$ implies LH and vice versa. It is an interesting question whether $|f(\Xi)| = O(e^{\epsilon p})$ holds for $f(\Xi)$ in $0<Im(t)<1/2$, $|t| <p$.

It looks like @boJonson made a typo, mixing up $\xi(s)$ with $\Xi(t)$, defined, for $s=1/2+it$, as:

$$\Xi(t) = \frac{1}{2} s(s-1) \pi^{-\frac{s}{2}} \Gamma\left(\frac{s}{2}\right) \zeta(s)$$

All the zeros in the following function $f(\Xi)$ are real:

$$f(\Xi(t)) - \Xi\left(\frac{t+i}{2}\right) - \Xi\left(\frac{t-i}{2}\right) = 0$$

If you assume the bound $O(e^{\epsilon p})$ for the Riemann xi function $\xi(s)$, where $s$ is on the critical line and $|s| < p$, it seems highly likely that Lindelöf Hypothesis(LH) would follow. Both the specific bound for $\xi(s)$ and the Lindelöf Hypothesis are implied by the Riemann Hypothesis (RH). However, it seems subtle whether the bound directly implies RH, or LH.

That the bound and LH imply each other, might follow from the fact that the bound $S_1(t)=o(t)$ implies LH and vice versa. It is an interesting question whether $|f(\Xi)| = O(e^{\epsilon p})$ holds for $f(\Xi)$ in $0<Im(t)<1/2$, $|t| <p$.