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Is it true that $$p(x)\,e^{-x^2} \leq A e^{-Bx^2},$$ for any $x\in \mathbb{R}$ and some positive constants $A$ and $B$ where $p$ is any polynomial of a given degree? I guess $A$ and $B$ depend on the degree of $p$ of course.

Comment: I even think one can assume $A$ is the biggest coefficient of $p$ and here $B$ gets smaller as the degree of $p$ gets bigger.

Thanks a lot!

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    Do you need that the inequality holds for any $x$ large enough or for any $x\geq 0$? – 2017-02-06
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    What's different between $e^{-x^2}$ and $\exp \left\{-x^2\right\},$ – 2017-02-06
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    I have edited it now :) – 2017-02-06
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    Don't think $A$ can always be $1$ (consider $x=0$ and let $p(x)$ have constant term $>1$). Other than that I think it's true. – 2017-02-06
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    Yes I realised that right now hehe, I know, sorry. – 2017-02-06
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    $p=x$ for what $A$ and $B$, $xe^{(B-1)x^2}\leq A$ – 2017-02-06
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    For $B=1/2$ and $A=1$ for instance. – 2017-02-06

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For any $\varepsilon>0$, $f(x)=p(x)\,e^{-\varepsilon x^2}$ is a continuous function whose limits at $\pm\infty$ are zero.
It follows that such a function has an absolute maximum over the real line and $ f(x)\leq M$ can be re-written in the wanted form (with $A=M$ and $B=1-\varepsilon$).

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    I don't understand what you mean by "f can be re-written", you mean that the estimate is possible for some $A$ and $B$ right? – 2017-02-06
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    @Martingalo: assume that $\varepsilon=\frac{1}{2}$ and $M=100$. Then $$p(x) e^{-x^2/2} \leq 100 $$ is equivalent to $$ p(x) e^{-x^2} \leq 100 e^{-x^2/2}$$ i.e. to $A=100=M$ and $B=\frac{1}{2}=1-\varepsilon$. – 2017-02-06
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    Do you see how this constant $M$ depends on the degree of $p$? Let us call it $n$? Is it a power function of $n$? – 2017-02-06
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    @Martingalo: you may compute by yourself what is the absolute maximum of $x^{2m} e^{-\varepsilon x^2}$ over the real line and draw the conclusion. – 2017-02-06
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    Aha! well in any case I guess $B_n$ converges to 0. Yes, thanks a lot! – 2017-02-06
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    @Martingalo: that simply follows by considering a truncated Taylor series for the exponential function, $p(x)=\sum_{n=0}^{N}\frac{x^{2n}}{n!}$. – 2017-02-06