Let \(f,g\in L^2(\mathbb R)\). Prove that \[ (f*g)(x)=\int_{\mathbb R}f(y)g(x-y)\,dy \] defines a bounded continuous function on \(\mathbb R\).
Solution
1. The convolution is bounded, with\[ \|f*g\|_\infty\le\|f\|_2\|g\|_2. \]
For each \(x\), the Cauchy--Schwarz inequality in $L^2$ and translation invariance of the \(L^2\) norm give \[ |(f*g)(x)|\le \|f\|_2\,\|g(x-\cdot)\|_2=\|f\|_2\|g\|_2. \] Taking the supremum over \(x\) proves the claim.
2. The convolution is uniformly continuous.
For \(h\in\mathbb R\), \[ (f*g)(x+h)-(f*g)(x) =\int_{\mathbb R}f(y)\bigl(g(x+h-y)-g(x-y)\bigr)\,dy. \] Applying Cauchy–Schwarz again gives, uniformly in \(x\), \[ |(f*g)(x+h)-(f*g)(x)| \le\|f\|_2\,\|g(\cdot+h)-g\|_2. \] By continuity of translation in $L^p$, the final norm tends to \(0\) as \(h\to0\). The bound is independent of \(x\), so \(f*g\) is uniformly continuous, hence continuous.
3. Q.E.D.
Step 1 proves boundedness and step 2 proves continuity.