- Let \(1\le p<\infty\) and \(f \in L ^ { p } ( \mathbb { R } ^ { n } )\) . Show that
\[ \operatorname* { l i m } _ { h \to 0 } \| f ( x - h ) - f ( x ) \| _ { L ^ { p } } = 0 . \]
Explain why the corresponding statement is false for \(p=\infty\).
Solution
1. Prove the result first for compactly supported continuous functions.
Let \(g\in C_c(\mathbb R^n)\). Since \(g\) is uniformly continuous, \[ \sup_{x\in\mathbb R^n}|g(x-h)-g(x)|\longrightarrow0 \] as \(h\to0\).
For all sufficiently small \(h\), the support of \[ g(\,\cdot-h)-g \] is contained in one fixed compact set \(K\). Hence \[ \|g(\,\cdot-h)-g\|_p \le m(K)^{1/p} \|g(\,\cdot-h)-g\|_\infty \longrightarrow0. \]
2. Pass to arbitrary \(L^p\) functions by density.
Fix \(f\in L^p(\mathbb R^n)\) and \(\varepsilon>0\). Since \(C_c(\mathbb R^n)\) is dense in \(L^p(\mathbb R^n)\) for \(1\le p<\infty\), choose \(g\in C_c(\mathbb R^n)\) such that \[ \|f-g\|_p<\frac\varepsilon3. \] Translation preserves the \(L^p\) norm, so \[ \|(f-g)(\,\cdot-h)\|_p=\|f-g\|_p. \] Therefore \[ \begin{aligned} \|f(\,\cdot-h)-f\|_p &\le \|(f-g)(\,\cdot-h)\|_p +\|g(\,\cdot-h)-g\|_p +\|g-f\|_p\\ &<\frac{2\varepsilon}{3} +\|g(\,\cdot-h)-g\|_p. \end{aligned} \] By step 1, the final term is below \(\varepsilon/3\) for all sufficiently small \(h\). Hence \[ \boxed{ \|f(\,\cdot-h)-f\|_p\longrightarrow0 \qquad(h\to0).} \]
3. Show that the theorem fails in \(L^\infty\).
Take \[ f=\mathbf1_{[0,\infty)} \] on \(\mathbb R\). For every \(h\ne0\), \[ \|f(\,\cdot-h)-f\|_\infty=1, \] because the two characteristic functions differ on an interval of positive measure. Thus translations need not be norm-continuous in \(L^\infty\).
4. Q.E.D.
Steps <1>1–<1>2 prove translation continuity for \(1\le p<\infty\), and step 3 gives the requested \(L^\infty\) counterexample.