- \(\star\): Show that \[f,g \in L^1 \implies f\ast g \in L^1 \qtext{and} \norm{f\ast g}_1 \leq \norm{f}_1 \norm{g}_1.\]
Solution
Let \(f,g \in L^1(\RR)\).
1. \(f \ast g\) is defined almost everywhere and belongs to \(L^1(\RR)\).
1.1. \(\int\!\int |f(x-y)g(y)|\,dy\,dx = \norm{f}_1 \norm{g}_1 < \infty\).
The function \((x,y)\mapsto f(x-y)g(y)\) is measurable on \(\RR^2\). By Tonelli’s theorem, \(\int\!\int |f(x-y)g(y)|\,dy\,dx = \int |g(y)| \int |f(x-y)|\,dx\,dy = \int |g(y)| \norm{f}_1\,dy = \norm{f}_1 \norm{g}_1\), since \(\int |f(x-y)|\,dx = \norm{f}_1\) for each fixed \(y\) by translation invariance of Lebesgue measure.
1.2. Q.E.D.
By step 1, \((x,y)\mapsto f(x-y)g(y)\) is in \(L^1(\RR^2)\). By Fubini’s theorem, \(\int f(x-y)g(y)\,dy\) converges absolutely for a.e. \(x\), and \(x \mapsto \int f(x-y)g(y)\,dy\) is measurable and integrable.
2. \(\norm{f\ast g}_1 \leq \norm{f}_1 \norm{g}_1\).
\(\norm{f\ast g}_1 = \int |f\ast g(x)|\,dx \leq \int\!\int |f(x-y)g(y)|\,dy\,dx = \norm{f}_1 \norm{g}_1\), where the last equality is step 1.
3. Q.E.D.
Step 1 gives \(f\ast g \in L^1\) and step 2 gives the norm bound.