A question about zeros of Tate type integral

Fix a positive integer $n$. Fix a continuous character $\chi$ of $\mathbb{R}^*$ with the form $\chi(x)=sign(x)|x|^t$ for some complex number $t$. If $\phi$ is a Schwartz function on $\mathbb{R}$, let $s$ be a complex variable. Consider the following integral $$F(s)=\int_{\mathbb{R}^*}\chi(x)|x|^{ns}\phi(x)\frac{dx}{x}$$

This is a Tate-type integral at real place. $F(s)$ converges in some right half complex plane and has a meromorphic continuation to the whole plane. The poles of $F(s)$ are responsible for the poles of the $L$ factor of some character.

But here I'm concerning the zeros of $F(s)$ at some left half plane. My question is: may I find some Schwartz function $\phi$ so that $F(s)$ is not identically zero, but has infinitely many zeros of the form $s_0, s_0-\frac{2}{n}, s_0-\frac{4}{n}, s_0-\frac{6}{n}....$?

Thank you.

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2 Answers

The Beta function $B(s,s_0)={\Gamma(s)\Gamma(s_0)\over\Gamma(s+s_0)}$ has the correct outcome (infinitely many zeros and poles of the proper type). But it is the Mellin transform of a discontinuous function (with compact support): $$B(s,s_0)=\int_0^\infty (1-x)^{s_0-1}\mathbf 1_{(0,1)}(x)x^s\ {dx\over x}$$ (here $\mathbf 1_{(0,1)}$ is the characteristic function of the open unit interval, and we assume $s$ and $s_0$ are greater than zero), instead of the Mellin transform of a Schwartz function.

Can we do better than that? Maybe. We would expect there to be a correspondence between the smoothness of $\phi$ and the decay of $F(s)$ in vertical strips (like in Paley-Wiener theorems or Sobolev embedding), so maybe something that decays faster than $B(s,s_0)$ (which decays like $|s|^{-{\rm Re}(s_0)}$) would correspond to a smoother function. You could also try using a mollifier to smooth out the kernel for the Beta function, though this would require either translating everything into the Fourier transform setting or using the Mellin-compatible version of convolution.

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Under your assumption, then $F(s)$ is an analytic function in a neighborhood of $s_0$, and then by the uniqueness theorem it must be zero in that neighborhood, hence in the whole plane by applying the same theorem to meromorphic functions.

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The zeros aren't converging to s_0 (although I thought that too when I first read it). Consecutive zeros differ by 2/n with n fixed. I admit it would have been clearer if the OP had put in a fourth zero rather than typing "..." after three of them. – KConrad Sep 5 '11 at 15:23
Oh, I see. The way it is written is quite confusion. – anonymous Sep 5 '11 at 16:26
Sorry for the confusion. I've added a fourth zero there. – user1832 Sep 5 '11 at 19:47