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Suppose I have some $f\in L^1$, such that $x^k f(x)$ is also integrable. Now I have some $\{f_n\}\subset\mathcal{C}^\infty$ satisfying $f_n\rightarrow f$ in $\lVert\cdot\rVert_1$. Is it true that $x^k f(x)$ can be approximated by $x^k f_n(x)$ in $L^1$? Of course one needs that $x^k f_n\in L^1$ for all $n$. More specifically $f_n=j_{1/n}*f$, where $j$ is some mollifier. I have a proof for my special case, but it is ugly and I don't like it. Maybe Vitali's Theorem? But I have the suspicion that it doesn't work. I don't trust my intuition in that one. What do you think?

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What do you don't like in the proof? –  Davide Giraudo Apr 22 '12 at 22:19
    
It didn't provide full generality and used bad looking estimations. It just wasn't elegant. –  user13655 Apr 25 '12 at 22:25
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up vote 1 down vote accepted

Take $\{\alpha_n\}\subset \mathbb R_{>0}$ a sequence such that $n\alpha_n\to 0$ but $n^2\alpha_n$ doesn't converge to $0$ (for example, $\alpha_n=n^{-3/2}$).

Put $f_n(x):=\alpha_n\exp\left(-\frac{x^2}{n^2}\right)$. Then $f_n\in L^1$, $x\mapsto x^kf_n(x)\in L^1$ for all $n$ and with the substitution $nt=x$ $$\lVert f\rVert_{L_1}=\int_{-\infty}^{+\infty}\alpha_n\exp\left(-\frac{x^2}{n^2}\right)dx=n\alpha_n\int_{-\infty}^{+\infty}\exp(-t^2)dt$$ which converges to $0$. But if we fix $k\geq 1$ an integer then, with the same substitution $$\int_{-\infty}^{+\infty}x^kf_n(x)dx=n\alpha_n\int_{-\infty}^{+\infty}n^k\exp(-t^2)dt$$ which doesn't converge to $0$.

So the result expressed in the title holds for particular cases of $f_n$, but not in general.

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