Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

Let $f \in L^1(\mathbb{R})$, for $a\in \mathbb{R}$ let $f_a(x)=f(x-a)$, prove that: $$\lim_{a\rightarrow 0}||f_a -f ||_1=0$$

I know that there exists $g\in C(\mathbb{R})$ s.t $||f-g||_1 \leq \epsilon$, this is also true for $f_a$ and $g_a$.

Now I have the next estimation:

\begin{align*} ||f_a-f||_1 &= \int_{\mathbb{R}} |f(x)-f(x-a)| dx \\ &= \int |f(x)-g(x)+g(x)-g(x-a)+g(x-a)-f(x-a)| \\ &\leq ||f-g||_1 + ||f_a-g_a||_1 + \int |g(x)-g(x-a)| \end{align*} my question is: I can argue that $\lim_{a\rightarrow 0} \int |g(x)-g(x-a)| = \lim_{a \rightarrow 0} \lim_{T\rightarrow \infty} |g(x_0)-g(x_0-a)| 2T$, now I think that I can change the order of the limits here, but I am not sure why?

P.s $x_0$ is some point in $[-T,T]$, and the above is valid from the intermediate integral theorem for continuous functions, right?


share|cite|improve this question

marked as duplicate by Live Forever, MATHEMATIKER, BLAZE, MathOverview, Alex M. Nov 22 '15 at 14:16

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.

We can prove more generally that $\displaystyle \lim_{t\to 0} \| f_t - f\|_p = 0$ for $f\in L^p(\mathbb{R}^n), 1\leq p < \infty.$ The idea is to prove it first for the characteristic function of a rectangle where it is almost obvious, and note that $f$ can be uniformly approximated by linear combination of characteristics functions of rectangles. – Ragib Zaman Jun 12 '12 at 14:54
up vote 1 down vote accepted

What you need to do is to be a little more picky in your choices. You can choose $g$ not only continuous, but also in $L^1$ and compactly supported.

You have already reduced the problem to proving it for $g$. If you know that $g$ is compactly supported, then there exists a $T$ such that $$ \int_{\mathbb{R}}|g(x)-g(x-a)|=\int_{-T}^T|g(x)-g(x-a)| $$ and then you don't need to take a limit on $T$.

(By the way, I don't think you can exchange the limits at the end like you want)

share|cite|improve this answer

You can't change the order of the limits since with the initial one it's $+\infty$, except particular case, and with the second one it's $0$.

But in fact, you can approximate $f$ in $L^1$ by continuous function with compact support $g$ (i.e., the closure of the set of $x$ such that $g(x)\neq 0$ is compact). To see that, use $\psi_n$ a continuous function with compact support, with is $1$ on $[-n,n]$ and $0$ outside $[-n-1,n+1]$, for example using distance of a point from these sets) and $\psi_n g$ to get what you want.

Once you have $g$ continuous with support contained in $[-R,R]$, use the fact that $g$ is uniformly continuous to get, for a fixed $\varepsilon>0$, $\delta$ such that if $|a|\leq \delta$ and $x\in\Bbb R$ then $|g(x+a)-g(x)|\leq \varepsilon$. Hence for $|a|\leq \min(\delta,1)$, $$\left|\int_{\Bbb R}(g(x+a)-g(x))dx\right|\leq \int_{[-R-1,R+1]}|g(x+a)-g(x)|dx\leq 2(R+1)\varepsilon.$$

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.