# Provable zero-free region for any entire function that analytically is similar to zeta(s)

Is there an entire function $f:\mathbb C\rightarrow\mathbb C$ such that for some $\delta>0$:

1. $f(z)$ is bounded when $\Re z>1+\delta$
2. $f(z)$ is unbounded when $\Re z=1$
3. $f(z)$ grows polynomially in vertical strips, ie for all $\sigma$ there is $C_\sigma$ so that $|f(\sigma+i t)|\ll|t|^{C_\sigma}$
4. $f(z)$ does not vanish when $\Re z>\frac12$ (provably!).

Conjecturally there is a very rich family: $L$-functions, but (4) is unproven.

If you drop (2), $1+e^{-z}$ works, or $\zeta(1/2+i t)$, or Wang zeta functions

If you drop (3), the Selberg zeta functions works, or $\exp(L(s))$

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.

• Also $\zeta(s)-(s-1)^{-1}$ have zeros on $\sigma>1/2$, example: one near 0.50000132378059385590 + 99997.458357552926821 I. We must recongize that until now we can not give any example of an entire (or even meromorphic with a finite number of poles) function satisfying conditions (1), (2), (3) and (4). – juan Nov 6 '12 at 18:14
• Oops, you're right, Juan. The right way to get rid of the pole Of zeta is to multiply by $(s-1)/(s+2)$, which will also work in general, pairing up poles and zeros – Ralph Furman Nov 6 '12 at 18:25
• @Ralph: How would $L$-functions behave bizarrely under GRH, at around height e^{e^{e^3}}}? – GH from MO Nov 6 '12 at 18:32
• $e^{e^{e^3}}\approx 2\times 10^{229\,520\,860}$ – juan Nov 6 '12 at 18:44
• @Ralph Furmaniak. But I expect that any disk with center at 5/4+i T with radius 1 and T big enough contains points with zeta(s)=1. You have not explained the criptic form of your constant $e^{e^{e^3}}$. I would like to know the reason of this election. This said I am not so sure that there is a function satisfying your conditions. All my examples ends with sup beta = sigma limit where the function is not bounded if beta + i gamma are the zeros. Of course when one is in mood of being skeptical. – juan Nov 7 '12 at 7:29

OK, shameless cheating, as promised.

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 $\Re z=1$.

Part 2. Push it up!

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.

Part 3. A tiny little bump.

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.

Part 4. Amplify the bump (being naive and fair)

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 $1$.

Part 5. Discriminate against numbers with the large real part.

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).

Part 6: Amplify with discrimination.

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 $1$.

Part 7: Ship it up the line to satisfy the local regulations.

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 $|F(z)^N-1|<2^{-n}(1+|z|)^{2^{-n}}$ 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.

Part 8. Put the production and shipment of bumps on the conveyor belt with $n=1,2,3,...$, and enjoy the product.

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. :)

• Sorry, but reading the last paragraph put a smile on my face, and compelled me to first actually read the answer, and then upvote you despite your request. – Frank Thorne Nov 11 '12 at 2:22
• That is indeed an answer and fits all of my requirements. The one thing I should have perhaps added is that the function should be order 1 in the sense of growing like $exp(z^{1+\epsilon})$, so that it has a nice Hadamard Product and can be analyzed in terms of its zeros. Now I need to understand how we can make your function look, in particular the growth rate as a whole. The zeros of $F(z)$ are near the 0 axis but moving away at a rate of $\log T$. Regardless of the $y_n$ there will be infinitely many in horizontal strips, Lthough the exact choice of $y_n$ will effect the density – Ralph Furman Nov 11 '12 at 3:35
• Growth rate as a whole is fine if you measure it just with $e^|z|^{1+\epsilon}$: each factor is merely exponential in $|z|$ (without $\varepsilon$), so shipping it even further up does the job. The location of zeroes, on the other hand, is very far from the one you want to see. You may want to elaborate on that a bit. There are some other features that you may want to think of before formalizing the next challenge, so take your time :). – fedja Nov 11 '12 at 3:57
• Manufacturing under loose government regulation is not a priori entirely bad, even if it is shameless, abominable, and mostly illegal :-) – Peter Samuelson Nov 11 '12 at 5:24
• @Ralph @fedja I have upvoted the solution. I do not see anything illegal. Or do you think anything not relevant to RH is illegal? Of course the example have too many zeros, and with very large multiplicity. I think the solution must be accepted and deserves the bounty. And still I would like to see a solution with the number of zeros rho = beta + i gamma with -T < t < T bounded by C T log T. Perhaps in the line. – juan Nov 11 '12 at 12:11