# Is a strictly convex function "Lipschitz from below"

Let $$f:\mathbb{R^n} \rightarrow \mathbb{R}$$. It is known that if $$f$$ is convex then $$f$$ is locally Lipschitz, i.e. for every $$\epsilon>0$$ and $$x^* \in \mathbb{R}^n$$, there exists $$M>0$$ such that:

$$\Vert f(x)-f(x^*) \Vert \leq M\Vert x-x^*\Vert \qquad \forall x \in N_{\epsilon}(x^*)$$

However, if $$f$$ is strictly convex, is it also the case that for every $$\epsilon>0$$ and $$x^* \in \mathbb{R}^n$$, there exists $$M>0$$ such that:

$$M\Vert x-x^*\Vert \leq \Vert f(x)-f(x^*) \Vert \qquad \forall x \in N_{\epsilon}(x^*)$$

Or sort of "Lipschitz from below" (I don't know if there's terminology for this concept).

If you allow $$M=0$$, then trivially yes.
If you require $$M>0$$, then no. For example take $$n=1$$, $$f(x)=x^2$$, and then $$\epsilon=3$$, $$x^*=-1$$, $$x=1$$.
• @ghiufhe: Only for $n=1$. In higher dimensions every neighborhood of anywhere will contain points with the same values of $f$. Oct 22, 2018 at 16:32