Why is a convolution of an $L^1$ and an $L^p$ functions well-defined?

I was reading the wikipedia article about convolution and I found this one:

If $f \in L^1(\mathbb{R}^n)$ and $g\in L^p(\mathbb{R}^n)$, then $||f\ast g||_p \leq ||f||_1||g||_p$.

But to $f\ast g$ be well-defined, $y\mapsto f(x-y)g(y)$ must be of $L^1$. Right? But why? How do I prove this?

Solution 1:

One form of Minkowski's integral inequality states that if $1\leq p\leq\infty$ and if $h(x,y)$ is a measurable function such that $h(\cdot,y)\in L^p$ for almost every $y$ and $y\mapsto \|h(\cdot,y)\|_p$ is in $L^1$, then $h(x,\cdot)\in L^1$ for almost every $x$, and the function $x\mapsto \int_{\mathbb{R}^n}h(x,y)\;dy$ is in $L^p$, with $$ \biggl\|\int_{\mathbb{R}^n}h(\cdot,y)\;dy\bigg\|_p\leq \int_{\mathbb{R}^n}\|h(\cdot,y)\|_p\;dy.$$

In our case, set $h(x,y)=f(y)g(x-y)$. Ths $h$ satisfies the hypotheses above since $g\in L^p$ and $f\in L^1$, so we know that $y\mapsto f(y)g(x-y)$ is in $L^1$ for almost every $x$, and $f\ast g\in L^p$, with $$ \|f\ast g\|_p=\biggl\|\int_{\mathbb{R}^n}h(\cdot,y)\;dy\biggr\|_p\leq \int_{\mathbb{R}^n}\|h(\cdot,y)\|_p\;dy=\int_{\mathbb{R}^n}|f(y)|\cdot\|g\|_p\;dy=\|f\|_1\|g\|_p.$$