TITLE: The function $\phi(p)=\|f\|_{L^p}^p$ is convex
QUESTION [5 upvotes]: Fix an arbitrary function $f\in L^p([0,1])$ and define
  $$\phi(p)=\|f\|_{L^p}^p$$ for $p\in [1,\infty)$. Prove $\phi$ is
  convex.

Comments: This is a standard property of $L^p$ spaces, but no good reference is available online. This question aims to fix that.

REPLY [2 votes]: Remarks: The argument below proves the following.

Let $1<r<\infty$ and $f\in L^r(\Omega, \Sigma, \mu)$ for some probability space $(\Omega, \Sigma, \mu)$. Then, the function $\phi:[1,r]\to$$\mathbb R$ defined by $\phi(p)=\|f\|^{p}_{L^p}$ is convex.


Choose $\theta\in[0,1]$ and $p,q\in[1,r]$. Then, $\theta p+(1-\theta)q\in[1,r]$. Note that, by the monotonicity of the $L^p$-norm, since $f\in L^{r}$ then $f\in L^p, L^{\theta p+(1-\theta)q}\text{ and }L^q$. Moreover, by the arithmetic-mean/geometric-mean inequality,
$$
\phi(\theta p+(1-\theta)q)=\|f\|_{L^{\theta p+(1-\theta)q}}^{\theta p+(1-\theta)q}=\int_{\Omega}|f|^{\theta p+(1-\theta)q}d\mu=\int_{\Omega}|f|^{\theta p}|f|^{(1-\theta)q}d\mu\\ \leqslant\int_{\Omega}\theta |f|^{p}+(1-\theta)|f|^{q}d\mu=\theta \int_{\Omega}|f|^{p}d\mu+(1-\theta)\int_{\Omega}|f|^{q}d\mu=\theta\|f\|_{L^{p}}^{p}+(1-\theta)\|f\|_{L^{q}}^{q}=\theta \phi(p)+(1-\theta)\phi(q)\,.
$$