- Show that if \(\phi\) is an approximate identity, then \[\norm{f\ast \phi_t - f}_1 \converges{t\to 0}\to 0.\]
Solution
Let \(f \in L^1(\RR^n)\), let \(\phi \in L^1(\RR^n)\) with \(\int \phi = 1\), and put \(\phi_t(x) = t^{-n}\phi(x/t)\). For \(h \in \RR^n\) let \(\omega(h) \coloneqq \int \abs{f(x - h) - f(x)}\,dx\).
1. \(f \ast \phi_t(x) - f(x) = \int \phi(y)\big(f(x - ty) - f(x)\big)\,dy\) for a.e. \(x\).
Substituting \(u = ty\) in \(f \ast \phi_t(x) = \int f(x - u)\phi_t(u)\,du\) gives \(\int f(x - ty)\phi(y)\,dy\), and \(f(x) = \int f(x)\phi(y)\,dy\) because \(\int \phi = 1\).
2. \(\|f \ast \phi_t - f\|_1 \le \int |\phi(y)|\,\omega(ty)\,dy\).
Take absolute values in step 1, integrate in \(x\), and exchange the order of integration by Tonelli’s theorem.
3. \(\omega(h) \le 2\|f\|_1\) for every \(h\), and \(\omega(h) \to 0\) as \(h \to 0\).
The bound is the triangle inequality with translation invariance of the integral. The limit is continuity of translation in \(L^1\).
4. \(\int |\phi(y)|\,\omega(ty)\,dy \to 0\) as \(t \to 0\).
For each \(y\), \(\omega(ty) \to 0\) as \(t \to 0\) by step 3, and \(|\phi(y)|\,\omega(ty) \le 2\|f\|_1|\phi(y)|\), which is integrable. The dominated convergence theorem applies along every sequence \(t_k \to 0\).
5. Q.E.D.
Steps 2 and 4.