Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a $C^2$ convex function which is strictly positive. If $x_n$ is a sequence of points such that $f(x_n)\rightarrow 0$, show that (or give a counterexample) the gradient $\nabla f(x_n)$ also tends to zero.

  • $\begingroup$ By an accident I posted my answer twice. And don't know how to delete the second one:-) $\endgroup$ – Alexandre Eremenko Sep 19 '12 at 23:25

A counterexample is $$f=\sqrt{y^2+e^{-x}}.$$ You can verify by computing the second derivatives that this is convex. As a sequence $x_n$ you can take $(n,1/n)$. Then $f(x_n)\to 0$ but the derivative with respect to $y$ tends to 1. Thus the gradient does not tend to 0.

  • $\begingroup$ Thank you very much. Now, if we suppose the gradient map of $f$ limited (which is the case I have in mind), do we get a positive answer? This problem arose in the following setting: Suppose $\Sigma$ is a complete hypersurface in $\mathbb{R}^{n+1}$ such that the position vector is everywhere transverse to it and $\Sigma$ is locally strongly convex (whith the position vector pointing to the convex side). Then I was trying to show that the property of $\Sigma$ being asymptotic (or not) to the boundary of the convex cone $\mathcal{C}$ which it generates is equivalent to the property of its $\endgroup$ – Henrique Sep 20 '12 at 20:43
  • $\begingroup$ (continuing....) conormal image $\Sigma^*\subset(\mathbb{R}^{n+1})^*$ be closed in $(\mathbb{R}^{n+1})^*$. $\endgroup$ – Henrique Sep 20 '12 at 20:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.