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Let $\left\{f_n\right\}_{n\in\mathbb{N}}\subset \mathcal{S}(\mathbb{R}), f\in L^2(\mathbb{R})$ with $\lim f_n=f$ in $L^2$ and $p(x)$ an polynomial. Since $\mathcal{S}$ is closed under the product of polynomials, for each $n$, $p(x)f_n(x)\in\mathcal{S}$.

Question 1. $p(x)f(x)\in L^2$?

I thought I could get out of inequality $|p(x)f(x)|_{2}\leq |p(x)(f(x)-f_n(x)|_{2}+|p(x)f_n(x)|_{2}$ but I see that it doesn't work.

eraldcoil
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1 Answers1

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Take any $f \in L^{2}$. There there exists a sequence $(f_n)$ of $C^{\infty}$ functions with compact support such that $f_n \to f$ in $L^{2}$. Each $f_n \in \mathcal S(\mathbb R)$. So your question amounts to asking if $p(x)f(x) \in L^{2}$ whenever $f(x)\in L^{2}$ and $p$ is a polynomial. This is clearly false: Take $f(x)=\frac 1 {x^{2}}$ for $x>1$, $0$ elsewhere and $p(x)=x$.

geetha290krm
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