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).


1 Answer 1


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

  • $\begingroup$ Thanks! Would it be true if I enforced that the neighborhood does not contain a global minimum? $\endgroup$
    – ghiufhe
    Oct 22, 2018 at 16:29
  • $\begingroup$ @ghiufhe: Only for $n=1$. In higher dimensions every neighborhood of anywhere will contain points with the same values of $f$. $\endgroup$ Oct 22, 2018 at 16:32

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