I want to show that $$\frac{d}{dx}(f*g)=(\frac{d}{dx}f)*g$$ where $f(x)=\frac{1}{\sqrt{x}}e^{-\frac{1}{x}}$, $g(y)$ is continuous and bounded. the convolutions are improper integrals.
I'm now here $$\frac{d}{dx}(f*g)=\frac{d}{dx}\int_{-\infty}^{+\infty}f(x-y)g(y)dy$$ $$=\lim_{h\to 0}\int_{-\infty}^{+\infty}\frac{f(x-y+h)-f(x-y)}{h}g(y)dy$$ $$=\lim_{h\to 0}\int_{-\infty}^{+\infty}f'(x-y+c)g(y)dy$$
by Mean Value Theorem, where $0<c<h$.
to switch the limit sign and integral sign, I need to know that the inside of improper integral is uniformly convergent. So I tried this: $$|f'(x-y+c)g(y)-f'(x-y)g(y)|$$ $$\le M|f'(x-y+c)-f'(x-y)| $$ $$=Mc|f''(x-y+C)|$$ where $0<C<c$ again by MVT.
and I found that $f''$ is not bounded. I'm not sure that even I can apply MVT and even if it is uniformly convergent, whether I can switch them. Any help will be much appreciated.
