# Fundamental Theorem of Calculus in Descent Lemma

I'm trying to understand a proof of the descent lemma, which says that if f is a continuously differentiable function over $\mathbb{R}^n$ with L-Lipschitz continuous gradient. Then, for any $x, y \in \mathbb{R}^n$:

$f(y) \leq f(x) + \nabla f(x)^T(y-x)+ \frac{L}{2}||x-y||^2$

By using the Fundamental Theorem of Calculus:

$\int_a^b f(x) dx = F(b) - F(a) \qquad F' = f$

And by defining:

$g(t) = f(x + t(y-x))$

Which gives:

$g(0) = f(x) \qquad g(1)=f(y)$

The proof starts by using the Fundamental Theorem of Calculus to write:

$f(y) - f(x) = \int_0^1 \langle \nabla f(x+t(y-x)), y-x\rangle dt$

However I don't get why there's a inner product in the integration, to my understanding shouldn't it be:

$f(y) - f(x) = \int_0^1 \nabla f(x+t(y-x)) (y-x) dt$

• $x,y \in \Bbb R^n$. How do you interpret $\nabla f \cdot (y-x)$ if not as an inner product? – bames Feb 23 '18 at 10:40

Your notation is unclear. Anyway, by the chain rule, $$\frac{d}{dt} f(x+t(y-x)) = \nabla f(x+t(y-x)) \cdot (y-x),$$ where the dot denotes the inner product.