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It is known that if $g_n: \mathbb{R} \rightarrow \mathbb{R}$, $n=1,2,...$, is in $C_c^{\infty}(\mathbb{R})$, $ \int_\mathbb{R} g_n(x)dx=1$, $supp(g_n) \subset (-r_n,r_n)$,where $0, then for arbitrary locally integrable $f: \mathbb{R} \rightarrow \mathbb{R}$ the convolution $f*g_n$ is smooth and ,if $f$ continuous, $f*g_n(x) \rightrightarrows f(x)$ on compact subsets of $\mathbb{R}$.

Let now $$g_n(x)=\frac{1}{\pi} \frac{r_n}{r_n^2+x^2},$$ $x\in \mathbb{R}$, $n \in \mathbb{N}$, where $0

Is it also true that if $f$ integrable and continuous, then $f*g_n(x) \rightrightarrows f(x)$ on compact subsets of $\mathbb{R}$?

Thanks in advance!

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    For smoothness, show by induction on $d$ that for a fixed $n$, the derivative of $g_n$ of order $d$ can be expressed as $\frac{r_n}{\pi}\frac{p_d(x)}{(r_n^2+x^2)^{d+1}}$ where $p_d$ is a polynomial of degree at most $d$. Then you will be able to show that we can apply dominated convergence theorem.2012-03-17

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