If $0\notin\partial f(x_k)$ then any $-\xi\in-\partial f(x_k)$ is a descent direction.

Question

Let $f:R^n\to R$ a convex function, let's supposse $0\notin\partial f(x_k)$. I think any $-\xi\in -\partial f(x_k)$ is a descent direction, i.e., exists $T>0$ such that for all $t\in(0,T]$ $$f(x_k-t\xi)\leq f(x_k).$$ I have tried to prove that, but I don't get it. I have the following results: If $x^*$ is the minimun of $f$ then $$0>\xi^T(x^*-x_k).$$ Also, for $t>0$ $$f(x_k-t\xi)\geq f(x_k)-t||\xi_k||^2. $$ Now, if what I think is false, please, mention me a counterexample.

Answer 1

Let $f(x) = \max(100x_1+x_2,-100x_1+x_2)$. Then $\partial f(0) = \operatorname{co} \{ (100,1), (-100,1) \}$, but $d=(-100,-1)$ is not a descent direction at $x=0$ since $0+10000t-t > 0$ for all $t>0$ and so $f(0+td)>0$ for all $t>0$.

Note that the nearest point to $\partial f(0)$ is $(0,1)$ and $d=-(0,1)$ is a descent direction.