If $rI am having trouble doing problem 3.4 from rudin's Real and Complex analysis. Given a measurable space $X$ and a measurable function $f$, it defines a function $\phi(p)=\int_X |f|^p$,
and defines $E=\{p:\phi(p)<\infty\}$, and assumes that $||f||_\infty>0$.
It asks to prove that $E$ is connected and that $\log(\phi)$ is convex and $\phi$ is continuous, which I have already done, but then it asks to prove that if $r<p<s$, then $$||f||_p\le \max\{||f||_r,||f||_s\},$$ or in other words, that $$\phi(p)^{1/p}\le \max\{\phi(r)^{1/r},\phi(s)^{1/s}\},$$ and that is where I am stuck.
I have tried to use the convexity of $\log(\phi)$ to prove it but It hasn't lead me anywhere.
 A: Assume $0< p < \infty$ and let $p=\theta r + (1-\theta)s$ for some $\theta\in(0,1)$. Then,
$$
\log\left(\phi(\theta r + (1-\theta)s)\right)^{\frac{1}{p}} = \frac{1}{p}\log\left(\phi(\theta r + (1-\theta)s)\right) \leqslant \frac{1}{p}\Big(\theta\log(\phi(r)) + (1-\theta)\log(\phi(s))\Big)\,.
$$
Therefore, 
$$
\phi(p)^{\frac{1}{p}} \leqslant \phi(r)^{\frac{\theta}{p}}\phi(s)^{\frac{1-\theta}{p}}\,.\tag{1}
$$
If $\phi(r)^{\frac{1}{r}} \leqslant \phi(s)^{\frac{1}{s}}$, then $(1)$ implies
$$
\phi(p)^{\frac{1}{p}} \leqslant \phi(r)^{\frac{r\theta}{rp}}\phi(s)^{\frac{s(1-\theta)}{sp}} \leqslant \phi(s)^{\frac{r\theta}{sp}}\phi(s)^{\frac{s(1-\theta)}{sp}} = \phi(s)^{\frac{1}{s}}
$$
Similarly, if $\phi(r)^{\frac{1}{r}} \geqslant \phi(s)^{\frac{1}{s}}$, we have
$$
\phi(p)^{\frac{1}{p}} \leqslant \phi(r)^{\frac{r\theta}{rp}}\phi(s)^{\frac{s(1-\theta)}{sp}} \leqslant \phi(r)^{\frac{r\theta}{rp}}\phi(r)^{\frac{s(1-\theta)}{rp}} = \phi(r)^{\frac{1}{r}}
$$
We have shown

$$
\phi(p)^{\frac{1}{p}} \leqslant \max\left\{\phi(r)^{\frac{1}{r}},\ \phi(s)^{\frac{1}{s}}\right\}\,.
$$

