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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.

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By an accident I posted my answer twice. And don't know how to delete the second one:-) – Alexandre Eremenko Sep 19 '12 at 23:25
up vote 1 down vote accepted

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.

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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 – Henrique Sep 20 '12 at 20:43
(continuing....) conormal image $\Sigma^*\subset(\mathbb{R}^{n+1})^*$ be closed in $(\mathbb{R}^{n+1})^*$. – Henrique Sep 20 '12 at 20:45

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