# Is the gradient of a convex continuosly differentiable function also a monotonically non-decreasing vextor function?

The definition of monotonically non-decreasing is:

Suppose h: R^n -> R^n, and if

then h is said to be monotonically non-decreasing.

I know that if a function is convex and continuously differentiable then we have:

Then can I add these two inequality together to prove the assumption?

• h is R^n -> R^n because f is R^n -> R and the gradient of f is thus R^n -> R^n. So that the gradient of f can be inner product with n-dimension vector. – Parting Dec 8 '16 at 22:17

• Yes, in those steps I'm just using some basic properties of inner products, such as $\langle u, -v \rangle = \langle -u, v \rangle$ and $\langle u, v \rangle + \langle w, v \rangle = \langle u + w, v \rangle$. – littleO Dec 8 '16 at 22:09