Let \(\phi\in C_c^\infty(\mathbb R^n)\) satisfy \(\int\phi=1\), and define \[ \phi_\varepsilon(x)=\varepsilon^{-n}\phi(x/\varepsilon). \] Prove that if \(1\le p<\infty\) and \(f\in L^p(\mathbb R^n)\), then \[ f*\phi_\varepsilon\to f \qquad\text{in }L^p. \] Show that the analogous statement fails for \(p=\infty\).
Solution
For \(1\le p<\infty\), changing variables \(y=\varepsilon z\) gives \[ (f*\phi_\varepsilon)(x)-f(x) =\int_{\mathbb R^n}\phi(z)\bigl(f(x-\varepsilon z)-f(x)\bigr)\,dz. \] By Minkowski’s integral inequality, \[ \|f*\phi_\varepsilon-f\|_p \le\int |\phi(z)|\,\|f(\cdot-\varepsilon z)-f\|_p\,dz. \] For each fixed \(z\), translation continuity in \(L^p\) gives \[ \|f(\cdot-\varepsilon z)-f\|_p\to0. \] The integrand is bounded by \(2\|f\|_p|\phi(z)|\), which is integrable in \(z\). Dominated convergence therefore yields \[ \|f*\phi_\varepsilon-f\|_p\longrightarrow0. \]
For \(p=\infty\), take \(f=\mathbf1_{[0,\infty)}\) in one dimension. Every \(f*\phi_\varepsilon\) is continuous. But no continuous function \(g\) can satisfy \[ \|g-f\|_\infty<\frac12. \] Indeed, such an essential-supremum bound and continuity would force \(g(x)<1/2\) for every \(x<0\) and \(g(x)>1/2\) for every \(x>0\), contradicting continuity at \(0\). Hence \[ \|f*\phi_\varepsilon-f\|_\infty\ge\frac12 \] for every \(\varepsilon\), so convergence in \(L^\infty\) fails.