Proof of convolution inequality

Question

I have to prove that if $f$, $g$ $\in L^1(\mathbb{R^n})$ then $\operatorname{dom}\left(f*g\right)$ is a set of full measure and: $\left\|f*g\right\|_{L^{1}} \le \left\|f\right\|_{L^1} \left\|g\right\|_{L^1}$

Any help would be appreciated.

Answer 1

Start with $\|f*g\|_{L^1}$ and write it up as a double integral, then interchange the order of integration and see what you can get out of it. I recommend assuming they are both nonnegative first, to give you a feeling for the problem. And don't try to go for too much rigor on the first pass; just try to find out where you're going. Once you have the inequality you seek, go back and try to justify the steps rigourously.

Answer 2

$\left\|f*g\right\|_{L^{1}}=\int_{R^n}|\int_{R^n}f(x-y)g(y)dy|dx \le \int_{R^n}\int_{R^n}|f(x-y)g(y)|dydx = $ $\int_{R^n}|g(y)|\int_{R^n}|f(x-y)|dxdy=\int_{R^n}|g(y)|\int_{R^n}|f(t)|dtdy = \left\|g\right\|_{L^{1}} \left\|f\right\|_{L^{1}}$

Answer 3

It might be edifying to consider that the convolution product turns the Banach space $L^{1}(\mathbb{R}^{n})$ into a commutative and associative algebra, for which your desired inequality holds. That is, $[L^{1}(\mathbb{R}^{n}),*]$ is a commutative Banach algebra. More commonly, we consider the convolution kernel $K$. That is, we suppose $K(x) \in L^{1}(\mathbb{R}^{n})$. Then the linear mapping $ f \rightarrow K*f$ is a bounded map on $L^{1}(\mathbb{R}^{n})$ with operator norm $ \leq \left\Vert K \right\Vert_{L^{1}}$, i.e. $\left\Vert K*f \right\Vert_{L^{1}} \leq \left\Vert K \right\Vert_{L^{1}}\left\Vert f \right\Vert_{L^{1}}$.