# 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 and I wondered if that was convenience (where we've checked) or more than that.

• It would imply the density of nontrivial zeros on the critical line is $100\%$, but I think that consequence is useless for anything else. There are only finitely many zeros in each bounded region, so by a compactness argument the real parts of all the nontrivial zeros would have a uniform upper bound $1-\varepsilon$ for some positive $\varepsilon$. For that to imply weakened forms of most applications of GRH you need the same $\varepsilon$ for infinitely many $L$-functions, but your ineffectivity condition is inadequate for that. Aug 13, 2021 at 15:00
• One interesting consequence would be huge media coverage, and a shower of prizes. Aug 13, 2021 at 16:11
• My professor once said that he rather beliefs that there may be nontrivial zeroes not on the critical line but not "that much", i.e., with density 0. What kind of difference would that make for the discussion? As again with respect to @KConrad statement this would again mean 100% density of non-trivial zeroes on the critical line... Aug 14, 2021 at 12:49
• Not what you're looking for, but it's conceivable (though unlikely) that an ineffective proof of RH may have no practical consequences at all, because the minimum such $T_0$ may be bigger than the number of particles in the observable universe. That is, it may be possible that RH holds in the observable universe despite your proof. Aug 15, 2021 at 4:26
• @user21820 It sounds like maybe you have in mind an ineffective disproof of RH; i.e., a proof that there exists a zero off the line, but with no effective upper bound on the size of the smallest such zero. Aug 15, 2021 at 13:22

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.

• Can you clarify how the statement 'the density of nontrivial zeros on the critical line is 100% in each vertical strip of width 1' is different from 'the density of nontrivial zeros on the critical line is 100%'? Aug 13, 2021 at 16:47
• And how does the former imply Lindelof? It seems the latter does not imply anything; mathoverflow.net/questions/161442/… Aug 13, 2021 at 23:09
• @Stopple: Backlund (1919) and Littlewood (1924) proved that the Lindelöf Hypothesis is equivalent to $N(\sigma,T+1)-N(\sigma,T)=o(\log T)$ for any $\sigma>1/2$ under $T\to\infty$. This is clearly the case when only finitely many nontrivial zeros of $\zeta(s)$ are off the critical line. (BTW "Random" clearly meant horizontal strips of width $1$ instead of vertical ones.) Aug 14, 2021 at 0:29
• When I wrote in a comment to the OP that "I think that consequence is useless for anything else" I had in mind exactly the link given by Stopple (which I was not able to locate at the time I was writing the earlier comment). I was not saying the assumption of finitely many zeros had no known consequence, but rather the $100\%$ hypothesis in its general form (not assumed to be due specifically to finitely many counterexamples to RH for the zeta-function) has no known consequence. Aug 14, 2021 at 4:30
• @GHfromMO Thanks, I edited the question to reflect this. Aug 14, 2021 at 13:59

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.

• You can omit $(\log x)^c$, since it is smaller than a constant times $x^{\epsilon/2}$. Aug 13, 2021 at 16:09
• @GHfromMO Yes, that requires a different epsilon in the big-Oh than from the strip. I suppose though that since we're in an ineffective context, that doesn't make a difference. Although now I'm curious: can we actually construct an explicit constant c such that if there are no zeros with real part between $1$ and $1-\epsilon$, then $\pi(x) = \mathrm{Li}(x) + O(x^{1-\epsilon})$. Aug 13, 2021 at 17:01
• I meant $O(x^{1-\epsilon/2})$, not $O(x^{1-\epsilon})$, which is false as you observe. The $\epsilon$ should be the same quantity at all occurrences in your post. About your question: if $\zeta(s)$ has a zero of multiciplity $m$ with real part $1-\epsilon$, then the error term is $\Omega_{\pm}(x^{1-\epsilon}(\log x)^{m-1})$, I think. So multiplicity influences the correct exponent of $\log x$ that you deleted now. Aug 13, 2021 at 17:37
• @GHfromMO It was deleted by Charles, not me, I think based on your comment. I think you are correct, so we can't probably make a c there explicit without having some control over the multiplicity. I'll edit again to include both versions. Aug 13, 2021 at 17:56