Convex function: $\nabla f(x)\cdot v=0\implies\nabla f(x+\epsilon v)\cdot v>0$?

Question

$\nabla f(x)\cdot v=0$

$f$ is convex. $\epsilon$ is a small positive number

Is it true that $\nabla f[x+\epsilon v]\cdot v>0$?

Seems to be true graphically but is there a simple proof?

This is a question about gradient.

Answer 1

The comment by daw gives you an idea how to do this. Using that, we obtain:

Because $f$ is convex, $\nabla f$ is monotone, meaning that $\langle\nabla f(x)-\nabla f(x+\varepsilon v),-\varepsilon v\rangle\geq0$, which gives $0=\varepsilon\langle\nabla f(x),v\rangle \leq \varepsilon\langle\nabla f(x+\varepsilon v),v\rangle$.

The inequality cannot be made strict, as e.g. $f\equiv0$ shows