Question about derivative and duality product in a Banach space

Question

I have a $C^1$ function $J: X\to \mathbb{R}$ where X is a Banach space Why is this identity true ? $$ J(u)-J(v)=\langle J'(t u +(1-t)v), u-v\rangle ,t\in (0,1), u,v\in X. $$

Answer 1

Not sure which derivative is defined there, but if it were the Frechet one, then the following is true, look at Real and Functional Analysis, Serge Lang.

\begin{align*} J(u)-J(v)&=J(v+(u-v))-J(v)\\ &=\int_{0}^{1}J'(v+t(u-v))(u-v)dt\\ &=\int_{0}^{1}J'(tu+(1-t)v)dt\cdot(u-v)\\ &=\left<\int_{0}^{1}J'(tu+(1-t)v)dt,u-v\right>. \end{align*}

Answer 2

Let us define $\varphi : [0,1] \to \mathbb R$ via $$\varphi(t) = J( u + t \, (v - u) ).$$

If $J$ is Gateaux-differentiable, then $\varphi$ is differentiable and you can apply the ordinary mean value theorem.