- Prove continuity in \(L^p\): If \(f\) is uniformly continuous then for all \(p\), \[\norm{\tau_h f - f}_p \converges{h\to 0}\to 0.\]
Solution
Let \(f\colon \RR^n \to \CC\) be uniformly continuous, let \(\tau_h f(x) \coloneqq f(x - h)\), and let \(\omega(r) \coloneqq \sup\theset{\abs{f(x) - f(y)} : \abs{x - y} \leq r}\). Uniform continuity says \(\omega(r) \to 0\) as \(r \to 0\). For \(1 \leq p < \infty\) assume \(f \in L^p(\RR^n)\), so that \(\norm{\tau_h f - f}_p\) is finite.
1. \(\norm{\tau_h f - f}_\infty \leq \omega(\abs h) \to 0\).
\(\abs{f(x-h) - f(x)} \leq \omega(\abs h)\) for every \(x\).
2. For \(1 \leq p < \infty\), \(\norm{\tau_h f - f}_p \to 0\).
Fix \(\eps > 0\) and choose \(R\) with \(\int_{\abs x > R} \abs f^p < \eps\). Let \(\abs h \leq 1\) and \(B = \theset{\abs x \leq R + 1}\). Off \(B\), \(\abs{x - h} > R\), so by \(\abs{a - b}^p \leq 2^{p-1}(\abs a^p + \abs b^p)\), \[ \int_{\RR^n \setminus B} \abs{f(x-h) - f(x)}^p\,dx \leq 2^{p-1}\left(\int_{\abs y > R}\abs{f(y)}^p\,dy + \int_{\abs x > R}\abs{f(x)}^p\,dx\right) < 2^p\eps. \] On \(B\), step 1 gives \(\int_B \abs{f(x-h) - f(x)}^p\,dx \leq \omega(\abs h)^p\, m(B)\), which is less than \(\eps\) for \(\abs h\) small. So \(\norm{\tau_h f - f}_p^p < (2^p + 1)\eps\) for \(\abs h\) small.
3. Q.E.D.
Step 1 is the case \(p = \infty\) and step 2 is the case \(1 \leq p < \infty\).