Show that $\lim _{r \to 0} \|T_rf−f\|_{L_p} =0.$

Question

I am having a hard time with the following real analysis qual problem. Any help would be awesome.

Suppose that $f \in L^p(\mathbb{R})$, where $1\leq p< + \infty$. Let $T_r(f)(t)=f(t−r)$. Show that $\lim_{r \to 0} \|T_rf−f\|_{L_p} =0$, that is $$ \lim_{r\to 0} \left( \int_{\mathbb R} |f(t-r) - f(t)|^p \mathrm d t\right)^{1/p} =0.$$

Answer 1

This can be handled using the definition of the Lebesgue integral plus some facts about Lebesgue measure like regularity.

First, we prove the result when $f$ is the characteristic function of an open set $O$ of finite measure. When $O$ is an interval, this is clear. It is also the case when it's a finite disjoint union of open intervals. In the general case, for a fixed $\varepsilon$, we take $O_\varepsilon$ which is a finite disjoint union of open intervals of finite measure such that $O_\varepsilon\subset O$ and $\lambda(O\setminus O_\varepsilon)\lt \varepsilon$. Then $$ \lVert \chi_{O+r}-\chi_O\lVert_p\leqslant \underbrace{\lVert\chi_{O+r}-\chi_{O_\varepsilon+r}\lVert_p}_{=\lVert\chi_{O}-\chi_{O_\varepsilon}\lVert_p}+\lVert \chi_{O_\varepsilon+r}-\chi_{O_\varepsilon}\rVert_p+\varepsilon^{1/p}, $$ hence $$ \limsup_{r\to 0}\lVert \chi_{O+r}-\chi_O\lVert_p\leqslant 2\varepsilon^{1/p}. $$ Once this is done for an open set, we obtain by outer regularity that the result holds when $f$ is the characteristic function of any Borel subset of finite measure.

Then we approximate by simple functions, noticing that $\lVert T_r\rVert_{L^p\to L^p}=1$ for each $r$.

Answer 2

Hint: Prove it first with $C^{\infty}_{c}$ functions.

Answer 3

Suppose that $f$ is a continuous function with compact support (notation: $f \in C_c(\mathbb{R})$). Then $f$ is in particular uniformly continuous, i.e. for any $\varepsilon>0$ there exists $r_0>0$ such that

$$|f(t-r)-f(t)| < \varepsilon$$

for any $r<r_0$ and $t \in \mathbb{R}$. Obviously, this implies

$$\|T_r f - f\|_{L^p}^p = \int |f(t-r)-f(t)|^p \, dt \leq \varepsilon^p \cdot \lambda(\text{supp} \, f).$$

Here $\lambda$ denotes the Lebesgue measure and $\text{supp} \, f$ the support of $f$. Since $\text{supp} \, f$ is compact, hence $\lambda(\text{supp} \, f)<\infty$, we conclude that $\|T_r f-f\|_{L^p} \to 0$ as $r \to 0$.

Now let $f \in L^p(\mathbb{R})$. Since the continuous functions with compact support are dense in $L^p(\mathbb{R})$, there exists a sequence $(f_k)_k \subseteq C_c(\mathbb{R})$ such that $f_k \to f$ in $L^p$. Now note that

$$\|T_r f_k-T_r f\|_{L^p} = \|f_k-f\|_{L^p}, \tag{1}$$

this follows directly from the translational invariance of the Lebesgue measure. Hence,

$$\begin{align} \|T_r f-f\|_{L^p} &\leq \|T_r f - T_r f_k\|_{L^p} + \|T_r f_k - f_k\|_{L^p} + \|f_k-f\|_{L^p} \\ &\stackrel{(1)}{=} 2 \|f_k-f\|_{L^p} + \|T_r f_k - f_k\|_{L^p}. \end{align}$$

Now the claim follows if we let $k \to \infty$ and $r \to 0$.

Remark In fact, one can even show that the mapping $$\mathbb{R} \ni r \mapsto f(\cdot-r)=T_rf \in L^p$$ is uniformly continuous. Here, we proved continuity in $r=0$.