The Hardy Z-function and failure of the Riemann hypothesis - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:10:54Z http://mathoverflow.net/feeds/question/50245 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50245/the-hardy-z-function-and-failure-of-the-riemann-hypothesis The Hardy Z-function and failure of the Riemann hypothesis David Hansen 2010-12-23T16:23:50Z 2011-04-14T14:17:19Z <p>David Feldman <a href="http://mathoverflow.net/questions/50186/" rel="nofollow">asked</a> whether it would be reasonable for the Riemann hypothesis to be false, but for the Riemann zeta function to only have finitely many zeros off the critical line. I very rashly predicted that this question would be essentially as hard as the Riemann hypothesis itself. However, on further reflection, I stumbled upon a natural and reasonable conjecture which has a serious bearing on whether this dichotomy holds, which I have never seen in print.</p> <p>So, let $f:\mathbf{R}\to \mathbf{R}$ be continuous. There are various notions of quasi-periodic and almost periodic function in the literature. The following (quite weak) one is more than enough for my purposes:</p> <blockquote> <p>Definition. A function $f: \mathbf{R} \to \mathbf{R}$ is <em>locally quasiperiodic</em> if, for every bounded interval $I \subset \mathbf{R}$ and every $\delta>0$, there exists an unbounded sequence $t_n \in \mathbf{R}$ such that $\mathrm{sup}_{t\in I} |f(t+t_n)-f(t)|&lt;\delta$.</p> </blockquote> <p>For example, finite trigonometric polynomials $\sum a_j \sin{(b_jt+c_j)}$ are locally quasiperiodic. </p> <p>Back to the zeta function: the Hardy $Z$-function is defined as $Z(t)=\pi^{-it/2}\frac{\Gamma(1/4+it/2)}{|\Gamma(1/4+it/2)|}\zeta(1/2+it)$. The functional equation for the zeta function immediately implies that $Z(t)$ is real-valued, and by construction we have $|Z(t)|=|\zeta(1/2+it)|$. One of the nice things about the $Z$-function is that it turns out to be computable in fairly efficient ways (the Riemann-Siegel formula), and it reduces the problem of finding zeros of zeta on the critical line to finding sign changes of the $Z$-function. In fact, the $Z$-function knows about the Riemann hypothesis: If the $Z$-function has a negative local maximum or a positive local minimum, then the Riemann hypothesis is false; see e.g. Section 8.3 of Edwards's book. I don't believe the converse to this is known, so let's call such an extrema a <em>strong failure of the Riemann hypothesis</em>.</p> <p>Now, I don't believe that the $Z$-function itself is locally quasiperiodic, because the density of its zeros should grow as $t$ grows, and it should wiggle "faster and faster" accordingly; more precisely, the number of zeros in an interval $[t,t+h]$ for $h$ fixed should be $\sim \frac {h}{2\pi}\log{t}$ as $t\to\infty$. However, rescaling in a naive manner, let's consider instead $Z(\frac{t}{\log{t}})$. This should have $\sim \frac{h}{2\pi}$ zeros in an interval $[t,t+h]$ for $h$ fixed and $t \to \infty$, and I see no reason not to believe that</p> <blockquote> <p>Conjecture A. The function $Z(\frac{t}{\log{t}})$ is locally quasiperiodic.</p> </blockquote> <p>My main reason for enunciating this is that the truth of Conjecture A implies that if there is one strong failure of the Riemann hypothesis, then there are infinitely many strong failures. This is actually pretty evident; take $I$ a small interval containing the relevant bad local extrema and take $\delta$ small enough so the intervals $I+t_n$ contain bad local extremas of the same type.</p> <p>It's not obvious to me whether Conjecture A is at all accessible by current technology. For example, I don't a single example of an <em>unbounded</em> function which is <em>provably</em> locally quasiperiodic. I would <em>love</em> to see such an example (I've tried and failed to construct one). Also, it seems natural to ask whether there is some simple characterization of locally quasiperiodic functions in terms of properties of their (distributional) Fourier transforms. Is such a characterization reasonable to expect?</p> http://mathoverflow.net/questions/50245/the-hardy-z-function-and-failure-of-the-riemann-hypothesis/50250#50250 Answer by Terry Tao for The Hardy Z-function and failure of the Riemann hypothesis Terry Tao 2010-12-23T17:43:01Z 2010-12-23T18:12:48Z <p>This doesn't quite sound like the right conjecture here, because Z is known to go to infinity on the average, by Selberg's central limit theorem (see my <a href="http://terrytao.wordpress.com/2009/07/12/selbergs-limit-theorem-for-the-riemann-zeta-function-on-the-critical-line/" rel="nofollow">blog post</a> on this topic). But this is easy to fix by working with a projective notion of local quasiperiodicity in which one divides $f(t)$ or $f(t+t_n)$ by an $n$-dependent scaling factor. In that case, one is basically asking for the zero process of the zeta function to be recurrent, and this would be predicted by the GUE hypothesis. However, I doubt that this question will be resolved before the GUE hypothesis itself is settled.</p> <p>EDIT: Note though that there are other hypotheses than the GUE hypothesis that also lead to a recurrent zero process, such as the <a href="http://www.aimath.org/WWN/lrmt/articles/html/117a/" rel="nofollow">Alternative hypothesis</a>, which is linked to the existence of infinitely many Siegel zeroes. I suppose it is <em>a priori</em> conceivable that some sort of dichotomy might be set up in which recurrence is obtained by completely different means in each case of the dichotomy (as is the case with proofs of multiple recurrence in ergodic theory) but I am personally skeptical that one could really handle all the cases without making enough progress on understanding zeta to solve much more difficult and prominent conjectures about that function. (In particular, with this approach one would have to first eliminate the possibility of having only finitely many zeroes off the critical line, leading us back to the original conjecture that motivated the one here.)</p> http://mathoverflow.net/questions/50245/the-hardy-z-function-and-failure-of-the-riemann-hypothesis/56882#56882 Answer by David Bernier for The Hardy Z-function and failure of the Riemann hypothesis David Bernier 2011-02-28T06:49:02Z 2011-02-28T06:54:54Z <p>I'd have another suggestion to replace your $Z(t/\log(t))$ : there's the Riemann-Siegel Theta function described in Harold Edwards' book, $\theta(t)$. The Gram points satisfy: $\theta(g_n) = n\pi$, $n$ = 1, 2, 3, ... So the idea is to look at $W(\alpha) = Z(\theta^{-1}(\alpha))$ . That way, if $g_n$ is the $n$th Gram point, $\theta^{-1}(n\pi) = g_n$ and<br> $W(n\pi) = Z(\theta^{-1}(n\pi)) = Z(g_n) = (-1)^n \zeta(1/2 + i g_n)$ .</p> <p>Cf. `Gram's Law' at MathWorld: .</p> <p>Perhaps there's a way to rescale $W(.)$ vertically from $Z(.)$ to get identical square-integrals over corresponding intervals say $[g_n, g_{n+1}]$ for Z and $[n\pi, (n+1)\pi]$ for $W(.)$ ...</p> http://mathoverflow.net/questions/50245/the-hardy-z-function-and-failure-of-the-riemann-hypothesis/61200#61200 Answer by Marc Palm for The Hardy Z-function and failure of the Riemann hypothesis Marc Palm 2011-04-10T10:09:05Z 2011-04-10T11:39:54Z <p>Dear David, I want to state first some prelimnary remarks. Do you know about the universality of the Riemann Zeta function? The behaviour in the region $1/2 &lt; \Re s &lt; 1$ of the Riemann zeta function is chaotic. In fact, given $\epsilon >0$, for any compact region $D$ in $1/2 &lt; \Re s &lt; 1$ ans any non vanishing bounded holomorphic function $f$ on $D$, there exists a sequence $T_n = \Omega(n)$ such that $$\sup_{z \in D} | f(z) -\zeta(z + iT_n)| &lt; \epsilon,$$ or even stronger the measure of all such $T$ has lower positive density. Here are the precise statements <a href="http://en.wikipedia.org/wiki/Zeta_function_universality" rel="nofollow">http://en.wikipedia.org/wiki/Zeta_function_universality</a>.</p> <p>Perhaps a related fact: The Riemann hypothesis holds if and only if $f$ can be replaced here by the Riemann Zeta function! Proof for $&lt;=$: Assume that $\zeta$ can be replaced for $f$ anywhere and $RH$ fails once say for a point in some $D$, then $\zeta$ would approximate itself on this $D$ arbitrary good in linear time. This produced every time a zero by Rouche's principle, which are far to many zeros by contradicting density results for zeros. Poof For $=>$: If $\zeta$ fulfils RH, we can replace $f$ by $\zeta$. </p> <p>So since the rescaling factor for $Z$ is pretty regular, hence you can deduce this quasiperiodic property in the region $1/2 &lt; Re s &lt;1$ directly from the property known for the Riemann Zeta function. (Approximate a continous function by an entire via the theorem of Mergelyan). On the critical line Joern Steuding &amp; Co. have presented some results last year also for $Re s = 1/2$, which probably imply conjecture $A$. </p> http://mathoverflow.net/questions/50245/the-hardy-z-function-and-failure-of-the-riemann-hypothesis/61702#61702 Answer by Mike Battaglia for The Hardy Z-function and failure of the Riemann hypothesis Mike Battaglia 2011-04-14T14:17:19Z 2011-04-14T14:17:19Z <p>Is it worth anything that since the prime-indicator function will only spike at integer values, that its Fourier transform is periodic? Think about how the prime-indicator and prime-counting functions are related, and how the Fourier and Mellin transforms are related.</p>