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

Question

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?

Answer 1

Yes, by adding your two inequalities, we discover that \begin{align} &\quad f(x) + f(y) \geq f(x) + f(y) + \langle \nabla f(y) - \nabla f(x),x - y \rangle\\ \iff & \quad \langle \nabla f(x) - \nabla f(y), x - y \rangle \geq 0. \end{align}

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Edit: Here are a few intermediate steps shown in more detail. \begin{align} \langle \nabla f(y),x - y \rangle + \langle \nabla f(x), y - x) \rangle &= \langle \nabla f(y),x - y \rangle + \langle -\nabla f(x), x - y) \rangle \\ &= \langle \nabla f(y) - \nabla f(x), x - y \rangle. \end{align}

Also, \begin{align} & 0 \geq \langle \nabla f(y) - \nabla f(x), x - y \rangle \\ \implies & 0 \leq - \langle \nabla f(y) - \nabla f(x), x - y \rangle \\ \implies & 0 \leq \langle \nabla f(x) - \nabla f(y), x - y \rangle. \end{align}