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In page 460 of Stephen Boyd's "Convex Optimization", he described a property of strongly convex functions:

"The inequality (9.8) (i.e. $f(y) \geq f(x) + \nabla f(x)^T (y - x) + \frac{m}{2} \|y - x\|_2^{2}$) implies that the sublevel sets contained in $S$ (i.e. $S = \{x|f(x) \leq f(x^{(0)})\}$) are bounded, so in particular, $S$ is bounded. Therefore the maximum eigenvalue of $\nabla^2 f(x)$, which is a continuous function of $x$ on $S$, is bounded on $S$"

I don't understand why the boundedness of $S$ implies the boundedness of $\nabla^2 f(x)$.

Can anyone explain it for me ? Thank you for reading my question.

Brian Borchers
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HOANG GIANG
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1 Answers1

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Let

$g(x)=\lambda_{max}(\nabla^{2}f(x))=\| \nabla^{2}f(x) \|_{2}$.

$g(x)$ is a continuous function.

$S$ is a closed and bounded subset of $R^{n}$ and thus compact.

By the extreme value theorem, $g(x)$ achieves its maximum value on $S$. Call it $M$.

Thus on $S$, $\| \nabla^{2}f(x) \|_{2}$ is bounded by $M$.

Brian Borchers
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    To shows that S is bounded, see the earlier stack exchange question: https://math.stackexchange.com/questions/993357/boundedness-of-sublevel-sets-of-convex-function-boyd-vandenberghe?rq=1 – Brian Borchers Sep 08 '18 at 22:32