Let \(1\le p,q\le\infty\) satisfy \[ \frac1p+\frac1q=1. \] If \(f\in L^p(\RR^d)\) and \(g\in L^q(\RR^d)\), then the convolution \[ (f*g)(x)\coloneqq\int_{\RR^d}f(x-y)g(y)\,dy \] converges absolutely for every \(x\in\RR^d\), satisfies \[ \sup_{x\in\RR^d}\abs{(f*g)(x)}\le \norm{f}_p\norm{g}_q, \] and is uniformly continuous on \(\RR^d\).
For every \(x\in\RR^d\), the function \(y\mapsto f(x-y)\) is measurable with \(\norm{f(x-\cdot)}_p=\norm{f}_p\) by invariance of Lebesgue measure under \(y\mapsto x-y\). Hölder’s inequality gives \[ \int_{\RR^d}\abs{f(x-y)g(y)}\,dy \le \norm{f(x-\cdot)}_p\norm{g}_q=\norm{f}_p\norm{g}_q, \] so the integral converges absolutely and \(\abs{(f*g)(x)}\le\norm{f}_p\norm{g}_q\).
Suppose first that \(p<\infty\). For \(h\in\RR^d\) and every \(x\), Hölder’s inequality and the same invariance give \[ \abs{(f*g)(x+h)-(f*g)(x)} \le \norm{f(\,\cdot+h)-f}_p\,\norm{g}_q . \] Translation is continuous in \(L^p(\RR^d)\) for \(1\le p<\infty\), so the right-hand side, which does not depend on \(x\), tends to \(0\) as \(h\to0\). Hence \(f*g\) is uniformly continuous.
If \(p=\infty\), then \(q=1\). The substitution \(y\mapsto x-y\) gives \[ (f*g)(x)=\int_{\RR^d}f(y)g(x-y)\,dy, \] so \[ \abs{(f*g)(x+h)-(f*g)(x)} \le \norm{f}_\infty\,\norm{g(\,\cdot+h)-g}_1 \] for every \(x\), and \(\norm{g(\,\cdot+h)-g}_1\to0\) as \(h\to0\) by continuity of translation in $L^1$. Hence \(f*g\) is uniformly continuous in this case as well.