convolution with $C^{\infty}$ produces $C^{\infty}$

Question

Problem: So I have the following function in $\mathbb{R^p}$$$f_{\sigma}\left(\boldsymbol{x}\right)=\dfrac{1}{\left(2\pi\right)^{d/2}}\int_{\mathbb{R}^{p}}f\left(\boldsymbol{z}\right)\exp\left(-\dfrac{1}{2\sigma^{2}}\left|\boldsymbol{z}-\boldsymbol{x}\right|^{2}\right)d\boldsymbol{z}$$ and I would like to show that it infinitely differentiable. where $f$ is a bounded function with compact support

Attempted: So first notice that $$Df_{\sigma}\left(\boldsymbol{x}\right)=\dfrac{1}{\left(2\pi\right)^{d/2}}\lim_{x\in B,m\left(B\right)\rightarrow0}\dfrac{1}{m\left(B\right)}\int_{\mathbb{R}^{p}}f\left(\boldsymbol{z}\right)\exp\left(-\dfrac{1}{2\sigma^{2}}\left|\boldsymbol{z}-\boldsymbol{x}\right|^{2}\right)d\boldsymbol{z}$$which by rearrange terms on the RHS gives $$\dfrac{1}{\left(2\pi\right)^{d/2}}\lim_{x\in B,m\left(B\right)\rightarrow0}\int_{\mathbb{R}^{p}}f\left(\boldsymbol{z}\right)\dfrac{1}{m\left(B\right)}\exp\left(-\dfrac{1}{2\sigma^{2}}\left|\boldsymbol{z}-\boldsymbol{x}\right|^{2}\right)d\boldsymbol{z}$$. I think I am supposed to to Dominated Convergence Theorem here. But I don't see what is dominating this function, i.e, the satisfying condition for me to invoke DCT..

Answer 1

I suggest you try to prove the following first (I'll retrict myself to functions on $\mathbb R$)

If $f\in L^\infty(\mathbb R)$ and has compact support and $g\in C^1(\mathbb R)$, then the function $h$ defined so that $h(x)=\int_{\mathbb R}f(z)g(x-z)dz$ is in $C^1(\mathbb R)$ and in fact $h'(x)=\int_{\mathbb R}f(z)g'(x-z)dz$.

You can do this by showing that for all $x_0\in\mathbb R$ the limit of the quotient $\frac{h(x)-h(x_0)}{x-x_0}$ as $x\to x_0$ exists and is equal to the expected thing. You will use the DCT to do this.

After that, you can prove by induction that

if $f\in L^\infty(\mathbb R)$ and has compact support and $g\in C^\infty(\mathbb R)$, then the function $h$ defined so that $h(x)=\int_{\mathbb R}f(z)g(x-z)dz$ is in $C^\infty(\mathbb R)$.

and give an explicit formula for the derivatives, if you want.