If all $L^p$ norms are bounded then the $L^\infty$ is bounded

Question

Suppose that $||f||_p \le K$ for all $1 \le p <\infty$ for some $K>0$. How to show that the essential supremum exists and bounded by $K$ that i s$||f||_\infty \le K$?

I know how to prove that if $f \in L^\infty$ then \begin{align} lim_{p \to \infty} ||f||_p=||f||_\infty \end{align} but this already assume that $f \in L^\infty$ in this question we have to show that $f$ has an essential supremum. To be more precise I don't think I can use a technique when I define \begin{align} A_\epsilon =\{ x | \ |f(x)|>||f||_{\infty}-\epsilon \} \end{align}

I feel like here we have to use some converges theorem. Thanks for any help

Answer 1

Assume $\| f \|_\infty=\infty$. Let $M>0$. Define $A_M=\{ |f| \geq M \}$. Then $\mu(A_M)>0$. Take $p$ so large that $\mu(A_M)^{1/p} \geq 1/2$, then $\| f \|_p \geq (\mu(A_M) M^p)^{1/p} \geq M/2$. Since $M$ was arbitrary, $f$ is not uniformly bounded in $L^p$, and your result follows by contraposition.

This is essentially the argument suggested by John Ma in the comments, but decoupling $M$ and $p$.

Answer 2

I am going to assume that $f\geq 0$ (otherwise replace $f$ with $|f|$). Consider $$g = \min(f,K + 1)\in L^{\infty}.$$ Then $\|g\|_p\leq \|f\|_p\leq K$ hence $$\|g\|_\infty = \lim_{p \rightarrow \infty} \|g\|_p \leq K$$ so $$ \min(f,K + 1) \leq K \ \ \ a.e. \implies f\leq K a.e.$$