Let $f(x)=e^{-\frac{x^2}{2}}$ ($x\in\mathbb{R}$), and $g\in C^{\infty}(\mathbb{R})$ with $|g(x)|=O(e^{-k|x|^{\gamma}})$ as $|x|\to\infty$, for $k>0$, $\gamma>0$. Let $h=f*g$, the convolution of $f$ and $g$. I guess that the following statement is true: $|h(x)|=O(e^{-M|x|^{\xi}})$ as $|x|\to\infty$, for some $M>0$, $\xi>0$, but I still have not proven. Can someone help me? Thank so much.
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$\begingroup$ If I'm not mistaken, proving the estimate $\frac12(x-y)^2+k|y|^\gamma\geq M(|x|^\xi+|y|^\alpha)$ for some $M,\xi,\alpha>0$ (and all $x,y\in\mathbb R$) would verify your guess. Does this look feasible? $\endgroup$– Joonas IlmavirtaCommented Aug 28, 2014 at 12:33
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$\begingroup$ Thank for your hint. I have tried to prove the inequality which you give, but until now I have not succeeded. How to prove your inequality ? $\endgroup$– CaoCommented Aug 28, 2014 at 15:18
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$\begingroup$ Frankly, I don't know. (I'll try to see if I can think of something, but I can't promise anything.) But the short calculation I made suggests that my inequality would be a very natural lemma to use. If my inequality fails, I don't really know how to approach the original problem. $\endgroup$– Joonas IlmavirtaCommented Aug 28, 2014 at 18:16
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By the definition of the convolution, $$ |h(x)| \lesssim \int_{-\infty}^{\infty} \exp\left(-k|y|^{\gamma}-\frac{(x-y)^2}{2}\right)\, dy . $$ We can now split this into two parts: $|x-y|>|x|^{1/2}$ and $|x-y|\le |x|^{1/2}$. Then the first integral is $O(e^{-|x|/2})$, and the second one is $O(e^{-c|x|^{\gamma}})$ (because $|y|\gtrsim |x|$ here for large $x$).
This can of course be done more carefully if you want optimized bounds.
answered Aug 28, 2014 at 23:04
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$\begingroup$ I understood on your hint. Thank you very much. $\endgroup$– CaoCommented Aug 29, 2014 at 0:02