# Understanding step in proof that $|J(v)-J(u)-\langle \nabla J(u),v-u\rangle|\leq \frac{\mu}{2}||v-u||^2$ involviing integral of inner-product.

In this proof that:

For $$J:\mathbb{E}\to\mathbb{R}$$ $$\mu$$-lipschitz differentiable. Have $$\forall u,v \in \mathbb{E}$$

$$|J(v)-J(u)-\langle \nabla J(u),v-u\rangle|\leq \frac{\mu}{2}||v-u||^2$$

The very first step is given as

$$J(v)=J(u)+\int_{0}^{1} \langle \nabla J(u+t(v-u)),v-u\rangle dt$$

But I cannot for the life of me get my head around this. Can anyone give me a clue?

## 1 Answer

For every functional $$\ell \in \mathbb E^*$$ you can define the real-valued function $$\varphi : [0,1] \to \mathbb R$$ via $$\varphi(t) := \langle \ell, J(u + t \, (v - u))\rangle .$$

This function is differentiable and you can apply the fundamental theorem of calculus.