If $f^p\in L^1([0,1])$ it's bounded a.e.

Question

We know that being Lebesgue integrable does not imply boundedness of the function (e.g. $g(x)=\frac{1}{\sqrt x}$). However function in $L^p$ spaces are functions with some decay conditions.

Suppose we have a function $f$ such that for all $p\ge 1$, $f^p\in L^1([0,1])$, can we infer $f$ is bounded from above and below? if not can one bound it only from one direction?

Answer 1

No, you can't infer that.

Take $f(x) = \ln(x)$

$$\forall p \geq 1, \int_0^1 \ln(x)^p dx < \infty $$

edit : but if $\exists M, \forall p \geq 1, \|f\|_p < M$, then $f\in L^\infty$.

Indeed, suppose $M=1$, take $\epsilon >0$, and consider $\{f > 1+\epsilon\}$

You have $\mu ( \{f > 1+\epsilon\} ) (1+\epsilon)^p\leq 1$

But $\lim_{p\to \infty} \frac{1}{(1+\epsilon)^p} = 0$, so $\mu ( \{f > 1+\epsilon\} )=0$ and $f$ is bounded.