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Let $f\in L^1(\mathbb{R})$ be of compact support and $\psi(x)=C \exp(-(1-x^2)^{-1})$ where $C$ is chosen so that $\int_{\mathbb{R}} \psi =1$. Show that the convolution $f*\psi(x)=\int_{\mathbb{R}} f(x-y)\psi(y) dy$ is infinitely differentiable.

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With a change of variable you obtain $$ \int_{\mathbb{R}}f(x-y)\psi(y)dy = \int_{\mathbb{R}}f(y)\psi(x-y)dy. $$ Now, using the fact that $\psi\in\mathscr{C}^{\infty}$ ($\psi$ and all its derivatives are bounded because $\mathrm{supp}(\psi)\subset\mathbb{R}$) and $f\in L^1$ with compact support, the Dominated Convergence Theorem gives the differentiability you need. In particular $$ D_x\left[ \int_{\mathbb{R}}f(y)\psi(x-y)dy\right] = \int_{\mathbb{R}}f(y)D_x\psi(x-y)dy. $$

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First prove the following statement.

If $g$ is any compactly supported smooth function on $\mathbb{R}$, then $f*g$ is differentiable, and $(f*g)' = f*(g')$.

To do this, you appeal directly to the limit definition of the derivative.$$ (f*g)'(x) = \lim_{h\to 0}\frac{(f*g)(x + h) - (f*g)(x)}{h} = \lim_{h\to 0}\int f(y)\left\{\frac{g(x+h-y) - g(x-y)}{h}\right\}\,dy.$$ The quantity in braces converges uniformly (in $y$) to $g'(x-y)$ as $h\to 0$, so you can pass the limit inside the integral to get that $(f*g)'(x)$ exists and is equal to $$\int f(y)g'(x - y)\,dy = f*(g')(x).$$

This proves the statement. Repeated application of it proves that the derivatives $(f*\psi)^{(n)}$ exist and are equal to $f*(\psi^{(n)})$.

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Why can you pass the limit inside? Don't you need DCT? –  newbie May 26 '13 at 22:09
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Use that $$ \widehat{d/dx f(x)} = 2 \pi i \xi \hat{f}(\xi)$$ where $\hat{}$ denotes the Fourier transform and that $\widehat{f \ast g} = \hat{f} \cdot \hat{g}$.

Then $$ \widehat{(d/dx (f \ast g)(x))} = 2 \pi i \xi \hat{f} \cdot \hat{g}$$

Applying the inverse Fourier transform to both sides with either $(2 \pi i \hat{f}) \cdot \hat{g}$ or $ \hat{f} \cdot (2 \pi i \hat{g})$ gives you $$ d/dx (f \ast g) = (d/dx f) \ast g = f \ast (d/dx g)$$

From which you get that $f \ast \psi$ is smooth since $\psi$ is.

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