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 $\epsilon=3$, $x^*=-1$, $x=1$.