My question is:
What is the statement of this "Lagrange Theorem" used in the last step below?
Let $\Omega \subset \Bbb{R}^N$ be an open set, $N \geq 3$. For $p \in (1, \infty)$ define a functional $J: W_0^{1, p} (\Omega) \longrightarrow \Bbb{R}$ by $$ J(u) = \int_\Omega |\nabla u|^p \ dx. $$
We want to show that $J$ is differentiable and that $$ J'(u)v = p \int_\Omega |\nabla u|^{p - 2} \nabla u \cdot \nabla v \ dx, \quad v \in W_0^{1, p} (\Omega) $$
The book I am reading (Badiale and Serra, Semilinear Elliptic Equations for Beginners) proceeds as follows:
Let $\varphi: \Bbb{R}^N \longrightarrow \Bbb{R}$ defined by $\varphi(x) = |x|^p$. It is a $C^1$ function with $\nabla \varphi (x) = p|x|^{p - 2}x$, so for all $x, y \in \Bbb{R}^N$ we have that $$ \lim_{t \to 0} \frac{\varphi(x + ty) - \varphi(x)}{t} = p |x|^{p - 2}x \cdot y $$ (it is just the directional derivative). As a consequence, $$ \lim_{t \to 0} \frac{|\nabla u(x) + t \nabla v(x)|^p - |\nabla u (x)|^p}{t} = p |\nabla u(x)|^{p - 2}\nabla u(x) \cdot \nabla v (x) \quad \text{ a.e. in } \Omega. $$ Now, and here lies my doubt, the book argues that by Lagrange's Theorem there exists $\theta \in \Bbb{R}$ such that $|\theta| \leq |t|$ and $$ \left| \frac{|\nabla u+ t \nabla v|^p - |\nabla u|^p}{t} \right| \leq p \left||\nabla u + \theta \nabla v|^{p - 2} (\nabla u + \theta \nabla v) \cdot \nabla v\right|. $$
Any references or hints will be the most appreciated.
Thanks in advance and kind regards.
Let $$ f(t)=|\nabla u+ t \nabla v|^p$$ and then $$ f(t)=((\nabla u+ t \nabla v)\cdot(\nabla u+ t \nabla v))^{\frac p2}=(|\nabla u|^2+2t\nabla u\cdot\nabla v+t^2|\nabla v|^2)^{\frac p2}. $$ Noting that \begin{eqnarray*} f'(t)&=&\frac p2 (|\nabla u|^2+2t\nabla u\cdot\nabla v+t^2|\nabla v|^2)^{\frac p2-1}(2\nabla u\cdot\nabla v+2t|\nabla v|^2) \\ &=&p||\nabla u+ t \nabla v|^{p-2}(\nabla u+t\nabla v)\cdot \nabla v. \end{eqnarray*} By the Mean Value Theorem, there is $\theta$ such that $$ f(t)-f(0)=tf'(\theta),|\theta|<|t| $$ or $$ \frac{|\nabla u+ t \nabla v|^p - |\nabla u|^p}{t}=p|\nabla u+ \theta \nabla v|^{p-2}(\nabla u+\theta\nabla v)\cdot \nabla v. $$