# Showing the following integral is 0. [duplicate]

This question already has an answer here:

Let $f$ be in $L^1(\mathbb{R}).$ Show that

$$\lim_{a\to 0}\int_{\mathbb{R}}|f(x)-f(x-a)|dx=0.$$

As $f\in L^1(\mathbb{R})$, we have $\int_{\mathbb{R}}|f(x)|dx<\infty.$ Define $f_n(x):=f(x-a_n)$. For each fixed $n$, $f_n(x)\in L_1(\mathbb{R}).$

I don't know how to solve this kinda of problems. Should I apply Fatou's lemma or the dominated convergence theorem? But I'm not evey sure if $f$ is bounded. I appreciate any help for this problem.

I don't undstrand the proof provided in the mentioned link. I'm very new in measure theory and I think there must be some easier solution for this question.

## marked as duplicate by C. Falcon, Leucippus, Siong Thye Goh, Yujie Zha, user296602 Aug 9 '17 at 4:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 2 Answers

Step 1. Define $T_a : L^1(\mathbb{R}) \to L^1(\mathbb{R})$ by

$$T_a f (x) = f(x-a).$$

Then this function has the following properties:

1. $T_a$ is linear.
2. $\|T_a f\|_{L^1} \leq \|f\|_{L^1}$ for all $f \in L^1(\mathbb{R})$ and for all $a$.
3. For each $g \in C_c(\mathbb{R})$, i.e. if $g$ is continuous and compactly supported, we have $T_a g \to g$ in $L^1$ as $a \to 0$.

Probably only the 3rd property requires explanation. Assume that $g$ vanishes outside $[-R, R]$. Then $g$ is uniformly continuous on $[-R, R]$ and hence on $\mathbb{R}$. Then for each $\epsilon > 0$, there exists $\delta \in (0, 1)$ such that $|g(x) - g(y)| < \frac{\epsilon}{2R+2}$ whenever $|x - y| < \delta$. Then for $|a| < \delta$, we have

$$\| T_a g - g \|_{L^1} = \int_{-R-1}^{R+1} |g(x) - g(x-a)| \, dx \leq \int_{-R-1}^{R+1} \frac{\epsilon}{2R+2} \, dx = \epsilon.$$

This proves the 3rd property. (Or the bounded convergence theorem gives a one-line proof for this property.)

Step 2. In this step, forget everything about the specific definition of $T_a : L^1(\mathbb{R}) \to L^1(\mathbb{R})$ and recall only the 3 properties listed above. Recapitulating, they are

1. $T_a$ is linear.
2. $\|T_a f\|_{L^1} \leq \|f\|_{L^1}$ for all $f \in L^1(\mathbb{R})$ and for all $a$.
3. For each $g \in C_c(\mathbb{R})$, we have $T_a g \to g$ in $L^1$ as $a \to 0$.

Then for any $f \in L^1(\mathbb{R})$ and for any $g \in C_c(\mathbb{R})$, we have

\begin{align*} \| T_a f - f \|_{L^1} &\leq \| T_a f - T_a g \|_{L^1} + \| T_a g - g \|_{L^1} + \| g - f \|_{L^1} \\ &\leq \| T_a g - g \|_{L^1} + 2\| g - f \|_{L^1}. \end{align*}

You may recognize this as our good old $3\epsilon$-argument. Then taking $\limsup$ as $a \to 0$ gives

$$\limsup_{a\to 0} \| T_a f - f \|_{L^1} \leq 2\| g - f \|_{L^1}.$$

But since $C_c(\mathbb{R})$ is dense in $L^1(\mathbb{R})$ in $L^1$-norm, we can send $g \to f$ in $L^1$ and thus the RHS can be made arbitrarily small. Therefore $\limsup_{a\to 0} \| T_a f - f \|_{L^1} = 0$ and hence

$$\lim_{a\to 0} \| T_a f - f \|_{L^1} = 0.$$

• Thanks, why are you saying $\|T_af\|_{L^1}\leq \|f\|_{L^1}$ – Parisina Aug 9 '17 at 0:54
• @Parisina Because why not? Putting jokes aside, notice that $\|T_af\|_{L^1}\leq \|f\|_{L^1}$ is a uniform estimate on $a$. In this proof, this was essential for the approximation argument; showing that approximating $f$ by $g$ in $L^1$ also produces an approximation of $T_a f$ by $T_a g$ in $L^1$ uniformly in $a$. To see the importance of this, assume contrarily that we only have non-uniform bound, say $\|T_a f\|_{L^1} \leq \frac{1}{|a|}\|f\|_{L^1}$, and see how this breaks down the proof of Step 2. – Sangchul Lee Aug 9 '17 at 0:57

Hint: Note that $\lim_{a\rightarrow0}\frac{f(x)-f(x-a)}{a}=f^\prime(x)$. Can you convert your limit to something involving an integral of $f^\prime(x)$ ?

• How can I know if $f$ is differentiable? I'm ever not sure if $f$ is continuous. – Parisina Aug 9 '17 at 0:33