Show that if additionally \(\abs{\phi(x)} \leq c(1 + \abs{x})^{-n-\eps}\) for some \(c,\eps>0\), then this converges is almost everywhere.
Show that is \(f\) is bounded and uniformly continuous and \(\phi_t\) is an approximation to the identity, then \(f\ast \phi_t\) uniformly converges to \(f\).
Solution
Let \(\phi \in L^1(\RR^n)\) with \(\int \phi = 1\), and put \(\phi_t(x) \coloneqq t^{-n}\phi(x/t)\) for \(t > 0\), so that \(\int \phi_t = 1\) and \(\norm{\phi_t}_1 = \norm{\phi}_1\). Part (a) concerns \(f \in L^1(\RR^n)\) and the convergence \(f \ast \phi_t \to f\) as \(t \to 0\). Let \(Mf\) denote the Hardy–Littlewood maximal function of \(f\).
1. There is a constant \(A\) with \(\sup_{t>0}\abs{f \ast \phi_t(x)} \leq A\, Mf(x)\) for every \(x\) and every \(f \in L^1\).
\(\Psi(x) \coloneqq c(1+\abs x)^{-n-\eps}\) is radial, radially decreasing, and integrable, and \(\abs\phi \leq \Psi\). For a radial, radially decreasing, integrable \(\Psi\), \(\sup_{t>0}\abs{f\ast\Psi_t(x)} \leq \norm{\Psi}_1\, Mf(x)\); this is the standard maximal inequality for such kernels, proved by writing \(\Psi\) as an increasing limit of positive combinations of normalized indicators of balls centered at \(0\). Since \(\abs{f\ast\phi_t} \leq \abs f \ast \Psi_t\), take \(A = \norm\Psi_1\).
2. For \(g \in C_c(\RR^n)\), \(g \ast \phi_t \to g\) uniformly as \(t \to 0\).
\(g\) is bounded and uniformly continuous, so this is part (b) below.
3. For \(f \in L^1\) let \(\Omega f(x) \coloneqq \limsup_{t \to 0}\abs{f\ast\phi_t(x) - f(x)}\). Then \(\Omega f \leq A\, M(f-g) + \abs{f - g}\) for every \(g \in C_c(\RR^n)\).
\(f\ast\phi_t - f = (f-g)\ast\phi_t - (f-g) + (g\ast\phi_t - g)\). The last term tends to \(0\) by step 2, and the first two are bounded by \(A\,M(f-g)\) and \(\abs{f-g}\) by step 1.
4. \(\Omega f = 0\) a.e.
Fix \(\alpha > 0\) and \(\eta > 0\), and choose \(g \in C_c(\RR^n)\) with \(\norm{f-g}_1 < \eta\). By step 3, the weak type \((1,1)\) bound \(m\theset{M h > \beta} \leq \frac{3^n}{\beta}\norm h_1\), and Chebyshev’s inequality, \[ m\theset{\Omega f > 2\alpha} \leq m\theset{A\,M(f-g) > \alpha} + m\theset{\abs{f-g} > \alpha} \leq \frac{3^n A + 1}{\alpha}\,\eta. \] Since \(\eta\) is arbitrary, \(m\theset{\Omega f > 2\alpha} = 0\) for every \(\alpha > 0\).
5. Q.E.D.
Step 4 says \(f\ast\phi_t(x) \to f(x)\) for a.e. \(x\).
- Let \(f\) be bounded and uniformly continuous.
1. \(f\ast\phi_t(x) - f(x) = \int \phi_t(y)\big(f(x-y) - f(x)\big)\,dy\).
This uses \(\int \phi_t = 1\).
2. Given \(\eps > 0\), there is \(\delta > 0\) with \(\int_{\abs y<\delta} \abs{\phi_t(y)}\,\abs{f(x-y)-f(x)}\,dy \leq \eps/2\) for every \(x\) and \(t\).
By uniform continuity choose \(\delta\) with \(\abs{f(x-y) - f(x)} < \eps/(2\norm\phi_1)\) for \(\abs y < \delta\), and use \(\int \abs{\phi_t} = \norm\phi_1\).
3. For this \(\delta\), \(\int_{\abs y\geq\delta} \abs{\phi_t(y)}\,\abs{f(x-y) - f(x)}\,dy \leq 2\norm f_\infty \int_{\abs z \geq \delta/t} \abs{\phi(z)}\,dz\), which tends to \(0\) as \(t \to 0\) uniformly in \(x\).
Bound \(\abs{f(x-y) - f(x)}\) by \(2\norm f_\infty\) and substitute \(y = tz\). The tail integral of \(\abs\phi \in L^1\) tends to \(0\) by dominated convergence.
4. Q.E.D.
By steps <1>1–<1>3, \(\abs{f\ast\phi_t(x) - f(x)} < \eps\) for every \(x\) once \(t\) is small.