The function $\phi(p)=\|f\|_{L^p}^p$ is convex 
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. 
 A: You have
$$ \phi(p) = \int_0^1 \lvert f(x) \rvert^p \, dx $$
$p \mapsto x^p = e^{p\log{x}}$ is smooth for $x \geqslant 0$, so we can differentiate twice with respect to $p$,
$$ \phi''(p) = \int_0^1 \lvert f(x) \rvert^p (\log{\lvert f(x) \rvert})^2 \, dx, $$
which is obviously nonnegative. (This is consistent even for $f$ possessing zeros if we make the standard definition $0\log{0}=0$.)

Had to think about this one for a while, but the reason that the integral exists is that $|f|^p (\log{|f|})^2 \in L^{p+\varepsilon}[0,1] $ for any $\varepsilon>0$, so you can show the second derivative is exists and is positive on an open interval $(p_1,p_2)$ provided $ f \in L^{p_2} $, which is sufficient for convexity.
A: 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,
$$
\phi(\theta p+(1-\theta)q)=\|f\|_{L^{\theta p+(1-\theta)q}}^{\theta p+(1-\theta)q}=\int_{\Omega}|f|^{\theta p+(1-\theta)q}d\mu=\int_{\Omega}|f|^{\theta p}|f|^{(1-\theta)q}d\mu\\ \leqslant\int_{\Omega}\theta |f|^{p}+(1-\theta)|f|^{q}d\mu=\theta \int_{\Omega}|f|^{p}d\mu+(1-\theta)\int_{\Omega}|f|^{q}d\mu=\theta\|f\|_{L^{p}}^{p}+(1-\theta)\|f\|_{L^{q}}^{q}=\theta \phi(p)+(1-\theta)\phi(q)\,.
$$
