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

Question

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.

Answer 1

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 $E=[1/2,1)$, $$2>\frac{2+1}{2}$$

Answer 2

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 $$\int_E f > \int_E K$$