Provable zero-free region for any entire function that analytically is similar to zeta(s) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:03:12Z http://mathoverflow.net/feeds/question/111321 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111321/provable-zero-free-region-for-any-entire-function-that-analytically-is-similar-to Provable zero-free region for any entire function that analytically is similar to zeta(s) Ralph Furmaniak 2012-11-03T00:02:57Z 2012-11-10T22:13:00Z <p>Is there an entire function $f:\mathbb C\rightarrow\mathbb C$ such that for some $\delta>0$:</p> <ol> <li>$f(z)$ is bounded when $\Re z>1+\delta$</li> <li>$f(z)$ is unbounded when $\Re z=1$</li> <li>$f(z)$ grows polynomially in vertical strips, ie for all $\sigma$ there is $C_\sigma$ so that $|f(\sigma+i t)|\ll|t|^{C_\sigma}$</li> <li>$f(z)$ does not vanish when $\Re z>\frac12$ (provably!).</li> </ol> <p>Conjecturally there is a very rich family: $L$-functions, but (4) is unproven.</p> <p>If you drop (2), $1+e^{-z}$ works, or $\zeta(1/2+i t)$, or Wang zeta functions</p> <p>If you drop (3), the Selberg zeta functions works, or $\exp(L(s))$</p> <p>Edit: Note that $\zeta(s)$ is not entire so you can instead look at $\zeta(s)(s-1)/(s+2)$ or change the question to allow finitely many singularities, where the bounds are taken away from the singularities.</p> http://mathoverflow.net/questions/111321/provable-zero-free-region-for-any-entire-function-that-analytically-is-similar-to/112031#112031 Answer by fedja for Provable zero-free region for any entire function that analytically is similar to zeta(s) fedja 2012-11-10T22:13:00Z 2012-11-10T22:13:00Z <p>OK, shameless cheating, as promised.</p> <p>Part 1. Let's start with something.</p> <p>We need a function bounded in $\Re z>1$ and growing not too fast on each vertical line whose zeroes are somewhere on the left. The first thing that comes to mind is $1$. No zeroes anywhere in sight, beautiful control on vertical lines. All that is lacking is the unboundedness on $\Re z=1$.</p> <p>Part 2. Push it up!</p> <p>We now want to add some bumps on the uneventful road $\Re z=1$. It is natural to add one bump a time. We have two options for bumping: addition and multiplication. Since we want to control the zeroes without trouble, we'll use multiplication. So, we'll be looking for an infinite product.</p> <p>Part 3. A tiny little bump.</p> <p>Take some entire function $g$ bounded by $1$ in the right half-plane, tending to $0$ at infinity in any right half-plane, and attaining its maximum of absolute value in $\Re z\ge 1$ at $1$, where it is real and positive. Denote $g(1)=a>0$. The exact choice doesn't matter. I'll take $g(z)=\frac{1-e^{-z}}{z}$. Put $F(z)=1+g(z)$. Now the ride along the line $\Re z=1$ is not that smooth anymore: you have to ascend to a small hill at $1$. However, at infinity everything levels to $1$ uniformly in any right half-plane. Also, if there are any zeroes, they all have non-positive real parts. </p> <p>Part 4. Amplify the bump (being naive and fair)</p> <p>Just raise $F$ to a high power $N$. You'll get as huge bump as you want. The problem is that it also becomes huge well to the right of $1$.</p> <p>Part 5. Discriminate against numbers with the large real part.</p> <p>Replace $g(z)$ by $g(z)e^{-n^2z}$. Of course the value at $1$ will suffer enormously, but everything with real part greater than $1$ will suffer much more (which is the whole point of any true discrimination).</p> <p>Part 6: Amplify with discrimination. </p> <p>Raise $F(z)=1+g(z)e^{-nz}$ to the power $N$. We'll get $(1+ae^{-n^2})^N$ at $1$ but only at most $(1+ae^{-n^2-2n})^N$ for $\Re z>1+\frac 2{n}$. Choose $N\approx a^{-1}e^{n^2+n}$. We'll get about $e^{n}$ at $1$ and at most $1+2e^{-n}$ to the right of $1+\frac 2n$. Now it is quite a bump, and it is next to invisible just a tiny bit to the right of $1$.</p> <p>Part 7: Ship it up the line to satisfy the local regulations.</p> <p>Replace $F(z)^N$ with $F(z-iy_n)^N$ with large $y_n$ to satisfy the polynomial growth restriction in $\Re z>-n$: let's even make $|F(z)^N-1|&lt;2^{-n}(1+|z|)^{2^{-n}}$ in $\Re z>-n$. Remember that though our bump function is huge, it is still bounded in any right half plane and levels to $1$ at infinity there. We also have $|F(z)|^N\le 1+2e^{-n}$ when $\Re z>1+\frac 2n$ regardless of the shipment.</p> <p>Part 8. Put the production and shipment of bumps on the conveyor belt with $n=1,2,3,...$, and enjoy the product.</p> <p>Of course, this is as shameless, abominable, and mostly illegal as any manufacturing under loose government regulations. Every loophole that could be exploited in the formulation of the problem has been exploited. So, please, do not accept or upvote. Instead, think of how to tighten the regulations to force someone to do honest work. :)</p>