continuity in $L_p$ norm

Question

I was reading a textbook on Sobolev space and encountered the following claim:

Suppose $f\in L_p([0,1])$ for some $1\leq p<\infty$, meaning that $\int_0^1{|f(x)|^p dx}<\infty$. Then $$ \lim_{h\to 0} \int_{|h|}^{1-|h|}\big|f(x+h)-f(x)\big|^p dx = 0. $$

Can anybody gives me a hint on how to prove this? It is obvious that the claim does not hold for $p=\infty$, though.

It suffices to show that $$ \lim_{h\to 0} \int_{0}^{1}\big|f(x+h)-f(x)\big|^p dx = 0. $$ — , Aug 30 '17 at 02:23
@Jack although that does require some sort of fudge to deal with the arguments of the functions not being in $f$'s domain, i.e. $[0,1]$. — Chappers, Aug 30 '17 at 12:31

score 4 · Accepted Answer · answered Aug 30 '17 at 01:24

4

The limit is clearly true if $f$ is continuous since then $f$ is uniformly continuous ($[0,1]$ is compact). The continuous functions are dense in $L^p$ for $1 \le p < \infty$ - this is a fact whose proof you can look up in an analysis book.

The corresponding statement for $p=\infty$ would be that $||f(\cdot+h)-f(\cdot)||_{\infty,[h,1-h]} \to 0$ as $h \to 0^+$. But consider $f(x) = 1$ on $[0,.5]$ and $f(x) = 0$ on $(.5,1]$. Then $||f(\cdot+h)-f(\cdot)||_{\infty,[h,1-h]} = 1$ for all small $h$.

answered Aug 30 '17 at 01:24

mathworker21

34,399

1

Thank you! very simple and elegant answer. – Yining Wang Aug 30 '17 at 01:33
When did we assume $f$ was continuous? – Carl Aug 31 '17 at 00:20
1

@Carl We never did – mathworker21 Aug 31 '17 at 00:45
sorry - just took a glance haha – Carl Aug 31 '17 at 11:44

Chappers · Answer 2 · 2017-08-30T02:12:19.260

2

Outline: it is easy to check that the set $\mathscr{F}$ of $f$ for which the result holds is closed under pointwise addition and multiplication using the triangle inequality. Another use of the triangle inequality shows that $\mathscr{F}$ is closed.

One can easily show using the triangle inequality and Dominated Convergence that indicator functions of intervals are in $\mathscr{F}$. Closedness then implies that indicator functions of measurable sets are in $\mathscr{F}$. Linearity implies that linear combinations of such functions are in $\mathscr{F}$, and these are dense in $L_p([0,1])$, so the result follows.

edited Aug 30 '17 at 02:12

answered Aug 30 '17 at 01:23

Chappers

67,606

Thanks for your answer. However, I'm not sure I get it because even showing point-wise convergence $\lim_{h\to 0}f(x+h)=f(x)$ could be hard, because $f\in L^p([0,1])$ might not be continuous? It does help if one uses the fact that continuous functions are dense in $L^p([0,1])$ for $1\leq p<\infty$. – Yining Wang Aug 30 '17 at 01:34
I'm confused why this answer was upvoted. – mathworker21 Aug 30 '17 at 01:50
@YiningWang I've improved the answer. – Chappers Aug 30 '17 at 02:12
@Chappers where does your proof fall apart for $L^\infty$? – mathworker21 Aug 30 '17 at 03:16
@mathworker21 In that case $\mathscr{F}$ doesn't contain indicator functions of intervals. – Chappers Aug 30 '17 at 12:22

score 0 · Answer 3 · answered Feb 26 '20 at 11:40

In the comments, @Jack said that it suffices to show that $\lim_{h\to 0} \int_{0}^{1}\big|f(x+h)-f(x)\big|^p dx = 0.$

In case someone doesn't know why, I want to explain a bit.

Suppose we have that $\lim_{h\to 0} \int_{0}^{1}\big|f(x+h)-f(x)\big|^p dx = 0.$ (Actually this is true)

Let $\varepsilon \gt 0.$ Then $\exists \,\delta\gt0$ s.t. $\lvert \int_{0}^{1}\big|f(x+h)-f(x)\big|^p dx\rvert\lt \varepsilon$ for all $h\in(-\delta,\delta).$

Then we have that, $$\lvert \int_{\lvert h\rvert}^{1-\lvert h\rvert}\big|f(x+h)-f(x)\big|^p dx\rvert\le \lvert \int_{0}^{1}\big|f(x+h)-f(x)\big|^p dx\rvert\lt \varepsilon\text{ for all }h\in(-\delta,\delta)$$ because the integrand is nonnegative.

That's $$\lvert \int_{\lvert h\rvert}^{1-\lvert h\rvert}\big|f(x+h)-f(x)\big|^p dx\rvert\lt \varepsilon\text{ for all }h\in(-\delta,\delta)\,,$$

i.e. $$\lim_{h\to 0} \int_{|h|}^{1-|h|}\big|f(x+h)-f(x)\big|^p dx = 0.$$

While as @Chappers pointed out in the comments, there will be some problem regarding the domain of $f(x+h)$. I don't know if the domain problem can be resolved by extending the function $f(x)$ outside the domain $[0,1]$. — Sam Wong, Feb 26 '20 at 11:53

continuity in $L_p$ norm

3 Answers3