functional analysis - If all $L^p$ norms are bounded then the $L^infty$ is bounded

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

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 $p$.