# Derivative of $t \mapsto \Vert f+tg \Vert_p^p$

Suppose $(X,\mathcal A, \mu)$ is a measure space and let $f,g\in L^p(X)$ be real-valued functions, $p\in(1,+\infty)$. Let us define $$F:\mathbb{R} \ni t \mapsto \int_X \vert f(x)+tg(x) \vert^p d\mu = \Vert f+tg \Vert_p^p$$ Prove that $F$ is differentiable and compute $F'(0)$.

Have you got any ideas? I've tried different things, quite unsuccessfully. I think that the derivability of $F$ is an application of dominated convergence theorem (but I can't see how exactly).

What about $F'(0)$? $$\lim_{t \to 0}\frac{F(t)-F(0)}{t}=\lim_{t \to 0} \frac{\int_X \vert f+tg\vert^p-\vert f \vert^pd\mu}{t}$$ but I do not know how to go on. Thanks in advance for your kind help.

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It is enough to show that the function $F$ is differentiable on every open subset of $\mathbb{R}$. So let $r>0$, and $$\phi: X\times(-r,r) \to [0,\infty],\ \phi(x,t)=s(f(x)+tg(x))=:\phi^x(t),$$ where $$s: \mathbb{R} \to [0,\infty),\ s(t)=|t|^p.$$ Since $s$ is differentiable, and $$s'(t)=\begin{cases} p|t|^{p-2}t &\text{ for } t \ne 0\\ 0 &\text{ for } t=0 \end{cases},$$ it follows that for every $x$ in $$\Omega:=\{x \in X:\ |f(x)|<\infty\}\cap\{x \in X:\ |g(x)|<\infty\}$$ the function $\phi^x$ is differentiable and $$(\phi^x)'(t)=\partial_t\phi(x,t)=g(x)s'(f(x)+tg(x)) \quad \forall\ t \in (-r,r).$$ Therefore $$|\partial_t\phi(x,t)| \le G_r(x):=\max(1,r^{p-1})|g(x)|(|f(x)|+|g(x)|)^{p-1} \quad \forall\ (x,t) \in \Omega\times(-r,r)$$ Thanks to Hölder's inequality we have $$\int_XG_r\,d\mu=\max(1,r^{p-1})\int_X|g|(|f|+|g|)^{p-1}\le \max(1,r^{p-1})\|g\|_{L^p(X)}\|(|f|+|g|)\|^{p-1}_{L^p(X)},$$ i.e. $G_r \in L^1(X)$

Given $t_0 \in (-r,r)$ and a sequence $\{t_n\} \subset (-r,r)$ with $t_n \to t_0$ we set $$\tilde{\phi}_n(x,t_0)=\frac{\phi(x,t_0)-\phi(x,t_n)}{t_0-t_n} \quad \forall x \in \Omega, n \in \mathbb{N}.$$ Then $$\lim_n\tilde{\phi}(x,t_0)=\partial_t\phi(x,t_0) \quad \forall\ x\in \Omega.$$ Thanks to the MVT there is some $\alpha=\alpha(t_0,t_n) \in [0,1]$ such that $$|\tilde{\phi}_n(x,t_0)|=|\partial_t\phi(x,\alpha t_0+(1-\alpha)t_n)|\le G_r(x) \quad \forall\ x \in \Omega, n \in \mathbb{N}.$$ Applying the dominated convergence theorem to the sequence $\{\tilde{\phi}_n\}$ we get for every $t_0 \in (-r,r)$: $$\int_X\partial_t\phi(x,t_0)d\mu(x)=\int_X\lim_n\tilde{\phi}_n(x,t_0)d\mu(x)=\lim_n\int_X\tilde{\phi}_n(x,t_0)=\lim_n\frac{F(t_0)-F(t_n)}{t_0-t_n}= F'(t_0).$$ In particular we have $$F'(0)=\int_X g(x)s'(f(x))=\int_X fg|f|^{p-2}.$$

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Hint: let $F(t,x):=|f(x)+tg(x)|^p$. Check that we can take the derivative under the integral.
We have $\partial_tF(t,x)=|g(x)|\cdot |f(x)+tg(x)|^{p-1}\operatorname{sgn}(f(x)+tg(x))$, which is, locally in $t$, bounded in $t$ by an itnegrable function of $x$.
Hi Davide, thank you for your answer. I am quite suprised: I am wondering why you suggest considering $c(s)=\vert x \vert^s$. The functions involved in the problem are not exponential... Anyway, I'm still thinking about your hint. Thanks again. –  Romeo Dec 6 '12 at 21:32