I know that this question has been "answered" before, but I am struggling to understand the solutions previously given.
Let $f$ and $g$ be in $L^1(R, L, m)$
a) Show that $f*g$ is well defined for a.e. $x \in R$. That is show $f(x-y)g(y)$ is integrable for all x.
b) Show that $ ||f*g||_1 \le||f||_1||g||_1$
I have the following so far:
For part a) $|\int f(x-y)g(y)dy|\le \int |f(x-y)||g(y)|dy \le ||f||_\infty ||g||_1$ by Holder's Inequality?
$||f*g||_1 = \int\int f(x-y)g(y)dydx = \int\int f(x-y)g(y)dxdy$ by Fubini Tonelli
= $\int g(y)\int f(x-y)dxdy = \int g(y)dy \int f(x)dx = ||f||_1||g||_1 $ by translation invariance.
First in part b, why is it an inequality instead of an equality? Is this a typo?
I have seen arguments that the proof of b implies a too, or simply saying "apply Tonelli's theorem" to prove part a, but have yet to find anything concrete for an answer.
Actually, you do prove $b$ before proving $a$. Let $F(x,y)=f(x-y)g(y)$. $F$ is clearly measurable, and $|F|$ is clearly integrable over $\mathbb{R}^2$, with integral $\|f\|_1\|g\|_1$ (Fubini-Tonelli theorem ensures that all manipulations are legit on the double integral of $|F|$ so you can integrate wrt $x$ first, make the change of variables $x=x’+y$).
Then you use Fubini’s theorem to prove that $x \longmapsto \int{F(x,y)dy}$ is well-defined and absolutely convergent ae and is in $L^1(\mathbb{R})$ with norm at most $\|f\|_1\|g\|_1$.