What are the consequences of an ineffective proof of the Riemann Hypothesis? Suppose a proof came out (and was verified by credible peer review) of the following statement:

There is a $T_0$ such that for all $t>T_0$, all zeros $\zeta(\beta+it)=0$ have $\beta=1/2.$

where $T_0$ is totally ineffective. What interesting consequences would this partial result have?

Of course you could ask this sort of question for all kinds of weakenings/strengthenings/relatives of RH:

*

*Zero-density estimates (which already has its own questions here and here)

*Density Hypothesis

*Lindelöf Hypothesis

*Generalized Riemann hypotheses for various L-functions

*Grand Lindelöf Hypothesis

But so far all the uses I have seen of $\zeta$ zeros has been in the strip $0<T<T_0$ and I wondered if that was convenience (where we've checked) or more than that.
 A: As a strengthening of what @KConrad commented, it would imply that the density of nontrivial zeros on the critical line is 100% in each horizontal strip of height 1, which is not useless: this is equivalent to the Lindelöf Hypothesis, which states that $\zeta \left( \frac{1}{2} + i t \right) = \mathcal{O}_{\varepsilon} \left( 1 + \lvert t \rvert^{\varepsilon} \right)$.
One example of a consequence of the Lindelöf Hypothesis (which is exactly much easier to prove directly from your non-effective Riemann Hypothesis using the explicit formula) is that the prime gaps satisfy $p_{n + 1} - p_n \leq \sqrt{p_n} \log \left( p_n \right)^2$, improving the best current unconditional result by Baker-Harman-Pintz of $p_{n}^{0.525}$.
The exponent of the logarithm might be a bit less, but I decided to err on the side of caution. By the way, this is still very far from the conjectured upper bound $p_{n}^{\varepsilon}$ (and in fact it is relatively widely believed that the gap is at most $C \log \left( p_n \right)^2$ for some absolute constant $C$).
However, the Lindelöf Hypothesis appears in estimating arithmetic sums, as many counting problems can be transformed into a zeta integral. Let me illustrate by a simple example. We will prove (conditionally) the following:
$$\sum_{n = 1}^{N} d (n) = n \log n + (2 \gamma - 1) n + \mathcal{O}_{\varepsilon} \left( n^{1/2 + \varepsilon} \right)$$
where $d(n)$ is the number of divisors of $n$ and $\gamma$ is the Euler-Mascheroni constant. This is usually proved via the Dirichlet hyperbola method (and with good reason), and we get a slightly weaker error term than usual, but this is just to illustrate the technique. Recall the classical inverse Mellin transform
$$\intop_{c - i \infty}^{c + i \infty} x^s \frac{\mathrm{d} s}{s} = 1_{x > 1} + \frac{1}{2} 1_{x = 1}$$
for any $c > 0, \ x \in \mathbb{R}$. Taking $c > 1$, and interchanging summation and integration we get (up to an error of $\frac{d (n)}{2}$, which is negligible)
$$\sum_{n = 1}^{N} d(n) = \intop_{c - i \infty}^{c + i \infty} \sum_{n = 1}^{\infty} d(n) \left( \frac{N}{n} \right)^{s} \frac{\mathrm{d} s}{s} = \intop_{c - i \infty}^{c + i \infty} \zeta \left( s \right)^2 \frac{N^s \mathrm{d} s}{s}$$
Now, shift the contour to $\mathrm{Re(s) = \frac{1}{2}}$. The residue picked up at the pole $s = 1$ is exactly $N \log N + (2 \gamma - 1) N$, so all we have left to proveis to show that the expression
$$N^{\frac{1}{2}} \intop_{-\infty}^{\infty} \zeta \left( \frac{1}{2} + i t \right)^2 N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$
is $\mathcal{O}_{\varepsilon} \left( N^{1/2 + \varepsilon} \right)$, or equivalently that the integral is $\mathcal{O}_{\varepsilon} \left( N^{\varepsilon} \right)$.
Here is the point where I tell you that I actually lied beforehand: it turns out that using the full inverse Mellin transform is, although very elegant, not necessarily the best choice to get a good analytic bound. What is usually done is approximate it by integrating not from $c - i \infty$ to $c + i \infty$, but from $c - i N$ to $c + i N$, where $c$ is say something like $1 + \frac{1}{\log N}$. I don't remember the details off the top of my head (they appear for example in Montgomery's book, and in a few expositions of proofs of the Prime Number Theorem), so just trust me here when I say that it is sufficient to bound the integral
$$\intop_{- N}^{N} \zeta \left( \frac{1}{2} + i t \right)^{2} N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$
But now (and here we finally use the Lindelöf Hypothesis!) we can bound pointwise this integral, and get that it is $\mathcal{O}_{\varepsilon} \left( N^{\varepsilon} \right)$ as required.

This example, although somewhat stupid, shows the power of the Lindelöf Hypothesis. Indeed, see Tao's answer The relationship between the Dirichlet Hyperbola Method, the prime counting function, and Mertens function, where he points out the fundamental difference between arithmetic functions with zeta in their denominator (whose behaviour is controlled very much by the zeroes of zeta) and arithmetic functions with zeta in the numerator. Despite that, we still managed to use information about the zeroes of zeta to get a nontrivial estimate.
A: One would have an ineffective but strengthened version of the Prime Number Theorem.  A consequence of this  would be there need to be some $\epsilon>0$ such that there's no zero in the strip with real part between $1$ and $1-\epsilon$. So one would get that $$\pi(x) = \mathrm{Li}(x) + O(x^{1-\epsilon} (\log x)^m)$$ for some $m$, or
$$\pi(x) = \mathrm{Li}(x) + O(x^{1-\frac{\epsilon}{2}})$$ if one wishes to avoid the log factors.
