I have a question that I have seen a lot on this site but always with Lebesgue Theory, which I have not learned so I get kinda confused by all the notation and am looking for a easier more straightforward solution. The question is:
Suppose $g$ is any integrable function which need not be continuous or differentiable. Suppose $f$ is a smooth function with derivatives of all orders on $\mathbb R$. Show that $(f*g)(x)\in C^{\infty}(\mathbb R)$, or in other words, continuous and differentiable in all orders. Does $f$ have compact support?
Now, I firstly try to show continuity. I wrote as $y\rightarrow z$, we need $|f*g(z)-f*g(y)|\rightarrow 0$. The way I tried to solve this is by showing: $$|f*g(z)-f*g(y)| = \left|\int_{-\infty}^\infty f(z-x)\cdot g(x)-f(y-x)\cdot g(x) dx\right| = \left|\int_{-\infty}^\infty (f(z-x)-f(y-x))\cdot g(x) dx\right|$$ $g$ is integrable, thus must have a supremum, and so we get: $$|f*g(z)-f*g(y)| \leq ||g(x)||_{\infty}\cdot\int_{-\infty}^\infty |f(z-x)-f(y-x)|dx$$ Now the integrand above will tend to $0$ as $y\rightarrow z$ because of the continuity of $f$ no? Also, was wondering if a bounded function required compact support? Picturing it in my head, it does seem to make sense, but I was wondering if it was mandatory. If anyone could help me out it would be greatly appreciated!