Another property of the function $\phi : p \mapsto \int |f|^p d\mu$. Let $(X,\mu)$ be a finite measure space and $f:X \to \mathbb{R}$ a real valued measurable function. 
Define $E=\{p : \int |f|^p <\infty\}$ and $\phi : E \to [0,\infty]$ by $\phi: p \mapsto \int |f|^p d\mu$.
I'm trying to prove that for all $0<p<q\le\infty \space$:
$$\space ||f||_p=||f||_q \iff ||f||_\infty=C \text{ for some constant}$$
Clearly $\phi(t)=C^t$ for $t \in[p,q]$. Which makes the case where $q=\infty$ easy since $E=(0,\infty]$ and $||f||_\infty=\lim_{p\to \infty} (\int |f|^p d\mu)^{1/p}=C$. 
But for $q<\infty$ i'm really stuck, any help would be appreciated. 
(These $L_p$ inequalities are a nasty businesses...) 
 A: We may assume without loss that $\mu$ is a probability measure $P$. (Check that if you know the result for probability measures, considering $\mu/\mu(X)$ will give the result.)
For probability measures we have Jensen's inequality: if $\phi$ is convex and $E|X|<\infty$ then $\phi(E[X]) \leq E [\phi(X)]$. Taking $\phi(x) = x^{q/p}$ with $q>p$ this gives $E[X^p]^{1/p} \leq E[X^q]^{1/q}$. When can equality hold? We claim only if $X$ is a.s. constant.
Suppose $\phi$ is strictly convex so $\phi(x) > \phi(y) + (x-y)\phi_+'(y)$ for all $x\neq y$.
Set $y=E[X]$, so
$$\phi(x) > \phi(E[X]) + (x-E[X])\phi_+'(E[X])$$ for all $x \neq E[X]$.
Suppose $X$ is not a.s. constant.
Then the strict inequality holds on a set of $\mu_X$-strictly positive measure (and of course the not strict inequality holds everywhere), where $\mu_X = P\circ X^{-1}$ is the distribution of $X$. Integrating with respect to $\mu_X$ then gives
$E\phi(X) > \phi(E[X]) + 0$ which completes the proof.
Since $x\longmapsto x^{q/p}$ is strictly convex if $q>p$, we actually have $E[X^p]^{1/p} < E[X^q]^{1/q}$ for  $p < q$.
