Let \(\phi\in L^1(\RR^n)\) such that \(\int \phi = 1\) and define \(\phi_t(x) = t^{-n}\phi(\inverseof{t} x)\). Show that if \(f\) is bounded and uniformly continuous then \(f\ast \phi_t \converges{t\to 0}\to f\) uniformly.
Solution
If \(\|f\|_\infty = 0\) then \(f \ast \phi_t = 0 = f\); assume \(\|f\|_\infty > 0\).
1. \(\int \phi_t = 1\) and \(\int|\phi_t| = \|\phi\|_1\) for every \(t > 0\).
Substitute \(y = x/t\).
2. \((f \ast \phi_t)(x) - f(x) = \int \phi_t(y)\,(f(x-y) - f(x))\,dy\).
\((f \ast \phi_t)(x) = \int f(x-y)\phi_t(y)\,dy\) and \(f(x) = f(x)\int\phi_t\) by step 1.
3. Given \(\eps > 0\), there are \(R > 0\) with \(\int_{|z| > R}|\phi(z)|\,dz < \frac{\eps}{4\|f\|_\infty}\) and \(\eta > 0\) with \(|f(u) - f(v)| < \frac{\eps}{2\|\phi\|_1}\) whenever \(|u - v| < \eta\).
The first holds because \(\phi \in L^1\), by dominated convergence applied to \(|\phi|\chi_{\theset{|z| > R}}\) as \(R \to \infty\). The second is uniform continuity of \(f\).
4. For \(0 < t < \eta/R\) and every \(x\), \(|(f \ast \phi_t)(x) - f(x)| < \eps\).
Split the integral of step 2 at \(|y| = tR\). For \(|y| \le tR < \eta\), step 3 bounds \(|f(x-y) - f(x)|\) by \(\frac{\eps}{2\|\phi\|_1}\), and step 1 bounds the integral of \(|\phi_t|\), so this part is at most \(\eps/2\). For \(|y| > tR\), bound \(|f(x-y) - f(x)|\) by \(2\|f\|_\infty\) and substitute \(z = y/t\): \(\int_{|y| > tR}|\phi_t(y)|\,dy = \int_{|z| > R}|\phi(z)|\,dz\), so this part is less than \(\eps/2\).
5. Q.E.D.
Step 4 gives \(\|f\ast\phi_t - f\|_\infty \le \eps\) for \(0 < t < \eta/R\).