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No. This follows from the Phragmen-Lindelof Theorem. EDIT. Consider the function of a complex variable $z$, $$F(z)=\int_R\phi(t)e^{itz}dt.$$ This function is bounded on the real axis by $\|\phi\|_1$. It is also of exponential type, $\log|F(z)|\leq O(z),\; z\to\infty$. Your condition says that $F(z)$ is bounded on the line $\{ z=x-ix:x\in R\}$. Then ...

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$$\Phi_2(\nu)=\int_0^{2\pi}\Phi_2(\nu\cos\phi,\nu\sin\phi)\nu d\phi$$ in words, you start from a function defined on the two-dimensional plane, and you average over the angular coordinate to arrive at a function that depends only on the radial coordinate.

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The Fourier transform of a function with compact support is automatically holomorphic. This support can be deduced from growth conditions on the former, at least in the case when it is a euclidean ball and the function is square integrable. This is the celebrated Payley-Wiener theorem, upon which there is a comprehensive Wikipedia article for starters.

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Alain Connes proved a similar statement in his Acta paper in 1999, but his method shows above presentation is true only with zeros on critical line, (actually he showed more, it is sum of powers of x^it(logx)^j, where i is square root of -1, j is discrete index, "i.t" is the imaginary part of zero on the critical line ) still to show there is no other ...

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The solution pattern behind is like this: Using a transform is like changing your point of view. In some cases the problem might get such easy under the new point of view, that you are able to solve the problem there and then you take the obtained solution and transform back to your original point of view. Here we might try to solve a differential ...

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Since $M^{1,1}(\mathbb R)$ is a Banach algebra, $P(f(t))$ is in $M^{1,1}(\mathbb R)$ whenever $\|f\|_{M^{1,1}}$ is smaller than the radius of convergence of the series $P$. Here you have to use a sub-multiplicative norm $\|\;\|_{M^{1,1}}$. On a Banach algebra you can always change to an equivalent sub-multiplicative norm.

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