0
$\begingroup$

Let $f\in L_p$, so

$$\lVert f \rVert_p <+\infty \Longrightarrow \int\vert f\rvert^p\ d\mu=K<\infty$$

For a demonstration of an exercise I assumed that,

$$\int \lvert f \rvert^p\ \chi_E\ d\mu\le\int K\chi_E\ d\mu=K\mu(E)$$

But a colleague questioned me about that statement, and I ended up not so sure about it anymore, because it’s an old proof and I don’t remember exactly what I thought when I wrote this. I was unable to demonstrate this statement or find a counterexample.

So, my question is whether the above implication is true or not.

$\endgroup$
  • 2
    $\begingroup$ Try it with $p=1$ and an $f$ that takes two different positive values. $\endgroup$ – Robert Israel Oct 28 at 18:27
2
$\begingroup$

The inequality can be written as $$\frac{1}{\mu(E)}\int_E|f|\le\|f\|$$ (Put $|f|^p$ instead, if you wish.)

That the average of a function on a subset need not be less than the total function (or even the total average) is obvious: Take $f(x)=\chi_{[0,1/2)}+2\chi_{[1/2,1)}$ and $E=[1/2,1)$, $$2>\frac{2+1}{2}$$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

No it's not. Consider for example $$f(x)=\sum_{n=1}^{\infty} n \chi_{[\frac{1}{n^3+1}, \frac{1}{n^3})}$$ for $x\in (0,1)$ and $f\equiv 0$ elsewhere.

Then $f$ is in $L^1$, and lets say it has norm $\infty>K>0$. Choose $n$ large enough so that $n>K$. Then for $E=[\frac{1}{n^3+1}, \frac{1}{n^3})$ you have that $$\int_E f > \int_E K$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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