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My question is if it is possible to find a compactly supported smooth function $\varphi:\mathbf{R}\to \mathbf{R}$ s.t. the following integration $\int_{\mathbf{R}}\varphi(t)e^{itx}e^{tx}dt$ stays bounded for $x\in \mathbf{R}$.

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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 Phragmen-Lindelof says that $F$ must be bounded, thus constant. But this is impossible if $\phi$ is smooth.

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  • $\begingroup$ In en.wikipedia.org/wiki/Phragmén–Lindelöf_principle, the Phragmén-Lindelöf Principle is about analytic functions. Maybe add an explanation showing how it applies here. $\endgroup$ – Gerald Edgar Dec 18 '13 at 19:43
  • $\begingroup$ Sorry I don't quite get it, could you say a bit more? How do we get the contradiction from the Phragmen–Lindelof principle? It just says the function is bounded, right? $\endgroup$ – Shaoming Dec 18 '13 at 19:46
  • $\begingroup$ The Fourier transform of $\phi$ is an entire function with exponential growth which is bounded on the real axis. If it is also bounded on another line through the origin, then the Phragmen-Lindelof Principle forces it to be constant. This allows only the trivial case $\phi=0$. $\endgroup$ – Michael Renardy Dec 18 '13 at 22:39
  • $\begingroup$ I don't see how $F$ being bounded implies $F$ being constant. For example $e^{-z^2}$ is bounded on the cone with forms an angle less than 45 degrees with the horizontal axis, but is not constant. $\endgroup$ – Shaoming Dec 19 '13 at 14:27

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