TITLE: Proof of Hölder inequality by differentiation
QUESTION [10 upvotes]: I need a reference where we can read a proof of the inequality $\|f\|_r\leq \|f\|_p^{1-\theta}\|f\|_q^\theta$ where $\frac{1}{r}=\frac{1-\theta}{p}+\frac{\theta}{q}$ for $L^p$-spaces of a measure space with the folowing method: 
differentiate $p\to \log \| f\|_{\frac1{p}}$ twice and observe that the result is positive.
Remark: the inequality is equivalent to the Hölder inequality.

REPLY [5 votes]: Perform the differentiation:
$$
\begin{align}
\frac{\mathrm{d}^2}{\mathrm{d}p^2}\log\left(\int|f(x)|^p\;\mathrm{d}x\right)
&=\frac{\mathrm{d}}{\mathrm{d}p}\frac{1}{\int|f(x)|^p\;\mathrm{d}x}\int\log(|f(x)|)\;|f(x)|^p\;\mathrm{d}x\\
&=-\frac{1}{\left(\int|f(x)|^p\;\mathrm{d}x\right)^2}\left(\int\log(|f(x)|)\;|f(x)|^p\;\mathrm{d}x\right)^2\\
&\phantom{=}+\frac{1}{\int|f(x)|^p\;\mathrm{d}x}\int\log(|f(x)|)^2\;|f(x)|^p\;\mathrm{d}x\tag{1}
\end{align}
$$
and Jensen's inequality says that because $x\mapsto x^2$ is convex,
$$
\frac{1}{\int|f(x)|^p\;\mathrm{d}x}\int\log(|f(x)|)^2\;|f(x)|^p\;\mathrm{d}x\ge\left(\frac{1}{\int|f(x)|^p\;\mathrm{d}x}\int\log(|f(x)|)\;|f(x)|^p\;\mathrm{d}x\right)^2\tag{2}
$$
Equations $(1)$ and $(2)$ show that
$$
\frac{\mathrm{d}^2}{\mathrm{d}p^2}\log\left(\int|f(x)|^p\;\mathrm{d}x\right)\ge0
$$
thus, $p\mapsto\log\left(\int|f(x)|^p\;\mathrm{d}x\right)$ is a convex function.  Therefore, since $\frac{1}{r}=\frac{1-\theta}{p}+\frac{\theta}{q}$, we get
$$
\frac{1}{r}\log\left(\int|f(x)|^r\;\mathrm{d}x\right)\le\frac{1-\theta}{p}\log\left(\int|f(x)|^p\;\mathrm{d}x\right)+\frac{\theta}{q}\log\left(\int|f(x)|^q\;\mathrm{d}x\right)
$$
which becomes
$$
\|f\|_r\leq \|f\|_p^{1-\theta}\|f\|_q^\theta
$$