# Proof that a convolution with $g(x)=\frac{1}{\pi} \frac{r}{r^2+x^2}$ is smooth

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 $0<r_n \rightarrow 0$, 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 $0<r_n \rightarrow 0$. How to show that for integrable $f: \mathbb{R} \rightarrow \mathbb{R}$ the convolution $f*g_n$ is smooth? (It is a part of exercise 10 in Chapter 9 of W. Rudin's book Real and complex analysis.)

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

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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. – Davide Giraudo Mar 17 '12 at 10:47

Another way of proving of smoothness is to use theorem 9.8 [Rudin, Real and complex analysis]

Let $h_\lambda(x)=\sqrt{\frac{2}{\pi}} \frac{\lambda}{\lambda^2+x^2}$, for $\lambda>0, x\in \mathbb{R}$, and let $H_\lambda(t)=e^{-\lambda |t|}$.

Then $$H_\lambda (t)=\frac{1}{\sqrt{2\pi}} \int_\mathbb{R} h_\lambda(x) e^{-itx}dx$$ $$h_\lambda(x)=\frac{1}{\sqrt{2\pi}} \int_\mathbb{R} H_\lambda (t) e^{itx} dt$$ ( the function $H_\lambda$ is the Fourier transform of $h_{\lambda}$, and the function
$h_{\lambda}$ is the inverse Fourier transform of $H_\lambda$).

Theorem. Let a function$f: \mathbb{R} \rightarrow \mathbb{R}$ be integrable. Then $$(f*h_\lambda)(x)=\frac{1}{2\pi}\int_\mathbb{R} H_\lambda (t) {\cal F} (f)(t) e^{ixt} dt,$$ where ${\cal F} (f)(t)=\frac{1}{\sqrt{2 \pi}} \int_R f(x) e^{-itx} dx$ - is the Fourier transform of $f$.

(This Theorem follows immediately by Fubini theorem and fact that $h_\lambda$ is the inverse Fourier transform of function $H_\lambda$).

The function under integral in the formula for convolution $f*h_\lambda$ is continuous with respect to $x,t$ and has derivatives of all order with respect to $x$, which (since ${\cal F}(f)$ is bounded) are bounded by integrable functions $t \mapsto C_k t^k e^{-\lambda |t|}$, where $C_k$, for $k=0,1,2...$, are constants. Hence $$(f*h_\lambda)^{(k)}(x)=\frac{1}{2 \pi} \int_R H(\lambda t) {\cal F} (f)(t) (it)^k e^{ixt} dt$$ and consequently the convolution is smooth.

In the same way, taking $h_\lambda(x)=\frac{1}{\lambda} e^{-\frac{x^2}{2\lambda^2}}$, and $H_\lambda(t)=e^{-\frac{\lambda^2 t^2}{2}}$, for $\lambda >0$, $x \in \mathbb{R}$, may show that $f* h_\lambda$ is smooth.

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