I'm attempting the problem 3.4 in Rudin's Real and Complex Analysis. The problem is the following:
Suppose $f$ is a complex measurable function on $X$, $\mu$ is a positive measure on $X$, and $\phi(p) = \int_X |f|^{p} d\mu$ for $0 < p < \infty$. Let $E = \{p: \phi(p) < \infty\}$. Show that if $r < p < s$ and $r, s \in E$, then $p \in E$.
I'm stuck at trying to transform $\phi(x) = \int_X |f|^{p} d\mu$ into something else so I can use the finiteness of $\phi(s)$ and $\phi(r)$ to bound $\phi(p)$.
I tried to apply some inequalities, but the only inequality we have seen is Hölder's which sadly doesn't apply here.
Is there something else I'm missing? Any hint would be helpful!
Actually, Hölder's inequality is the correct inequality to apply. Since $r < p < s$ there is a number $0 < t < 1$ satisfying $p = rt + s(1-t)$ (you can write out exactly what $t$ is by solving this equation and showing that $0 < t < 1$.)
Note $1/t > 1$ and its Hölder conjugate is $1/(1-t)$.
You can write $|f|^p = |f|^{rt} |f|^{s(1-t)}$ and apply Hölder's inequality with exponents $1/t$ and $1/(1-t)$: $$ \int_X |f|^p \, d\mu = \int_X |f|^{rt} |f|^{s(1-t)} \, d\mu \le \left( \int_X |f|^r \, d\mu \right)^t \left( \int_X |f|^s \, d\mu \right)^{1-t}.$$
In the notation of the problem you get $\phi(p) \le \phi(r)^t \phi(s)^{1-t}$.