p-norm of a function Let $f\in L^1(\mu)\cap L^\infty(\mu)$. I have proved for any $1<p<\infty$, $f\in L^p(\mu)$, $w(p)=||f||_p$ is continuous w.r.t. $p$, and  $\lim_{p\to \infty}||f||_p=||f||_\infty$.
Is $w(p)$ differentiable? Will $w(p)$ be a concave or convex function w.r.t. $p$ when $p$ sufficiently large?
 A: There is some convexity, don't know if this helps you, but: 
Let $p_0, p_1 \in [1,\infty]$, $\theta \in [0,1]$, $\frac 1p = \frac{1-\theta}{p_0} + \frac{\theta}{p_1}$. Then for $f \in L^{p_0}\cap L^{p_1}$, by Hölder
\begin{align*}\def\norm#1#2{\left\|#1\right\|_{#2}}\def\abs#1{\left|#1\right|}
  \norm fp &= \norm{\abs{f}^{1-\theta}\abs{f}^\theta}p \\
           &\le \norm{\abs f^{1-\theta}}{p_0/(1-\theta)}\norm{\abs{f}^\theta}{p_1/\theta}\\
           &\le \norm f{p_0}^{1-\theta}\norm{f}{p_1}^\theta
\end{align*}
Taking logarithms, we have 
$$ \log \norm fp \le (1-\theta)\log \norm f{p_0}+ \theta\log\norm f{p_1} $$
Hence
$$ [0,1] \to \mathbb R, \quad r \mapsto \log \norm f{1/r} $$
is convex.
A: Adding to martini's answer: From 
$$
\|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta
$$
one finds using Young's inequality in the form
$$
ab \le \theta a^{\frac1\theta} + (1-\theta)b^{\frac1{1-\theta}}
$$
the estimate
$$
\|f\|_p \le \|f\|_{p_0}^{1-\theta}\|f\|_{p_1}^\theta \le (1-\theta)\|f\|_{p_0} + \theta\|f\|_{p_1}.
$$
Using the definition of $p$ it is
$$
\|f\|_{\frac1{\frac{1-\theta}{p_0} + \frac\theta{p_1}}} \le (1-\theta)\|f\|_{p_0} + \theta\|f\|_{p_1}.
$$
Hence the mapping
$$
1/p \mapsto \|f\|_{1/(1/p)}
$$
is convex.
