Showing $f\in L^\infty$ in a finite measure space I am studying for a qualifying exam, and I seem to have difficulty working problems involving $L^p$-spaces. An explanation for the following problem would be very helpful!
Let $(X, \Sigma, \mu)$ be a finite measure space and let $f$ be a real-valued measurable function on $X$. Prove that $f\in L^\infty(\mu)$ if and only if $f\in L^p(\mu)$ for every $1<p<\infty$ and $\sup_p ||f||_p <\infty$.
I think that if $f\in L^\infty(\mu)$, then we have 
$$||f||_p \leq \left( \int_X ||f||_\infty^p d\mu \right)^{1/p} = ||f||_\infty \mu(X)^{1/p} <\infty$$
and so $f\in L^p$ for all $1<p< \infty$. However, I and stuck on showing $\sup_p ||f||_p <\infty$ and the other direction.
 A: If $\mu(X) \le 1$, $\mu(X)^{1/p} \le 1$.  If $\mu(X) > 1$, $\mu(X)^{1/p} \le \mu(X)$.
For the other direction, suppose $f \notin L^\infty$.  Then for every $N$ there is a set $A$ with $\mu(A) > 0$ such that $|f| > N$ on $A$.  Then 
$\int |f|^p \; d\mu \ge N^p \mu(A)$ and $\|f\|_p \ge N \mu(A)^{1/p}$.
Take
$p$ large enough that $\mu(A)^{1/p} > 1/2$...  
A: You are right on the fact that if $f\in L^\infty$ then from $$\|f\|_p\leq\|f\|_\infty\mu(X)^{1/p}\quad\forall p>1$$ and from this you get that $f$ is in $L^p$ for all $p>1$ and all the p-norms have the same bound! Basically $\sup_p\|f\|_p<\|f\|_\infty\mu(X)$ because $\mu(X)^{1/p}<\mu(X)$ forall $p>1$ if $\mu(X)>1$ or $\sup_p\|f\|_p<\|f\|_\infty$ if $\mu(X)<1$.
The other way around follows the same logic and using Holder inequality to get $$\|f\|_p=(\int_x |f|^p\cdot 1\, \mathrm{d}\mu)^{1/p}\leq \|f\|_q (\mu(X))^{1/p}$$
Basically to say that if we substitute $\mu$ with $\nu=\mu/{\mu(X)}$ to get a probability measure $\nu$ ($\nu(X)=1$), the norm is an increasing and continuous function in p converging to $\|f\|_\infty$ if $f$ is bounded a.e.. So if we ask that the collection of all the norms is bounded on some $f$ then $f$ itself is in $L^\infty$ and represents the bound for the supremum. (i.e.$\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$)
