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Is strongly convex function always bounded below?

If yes, can we prove it? Or could we construct an example if the answer is no?

I have tried to construct counterexamples such as the function $f(x) = x + e^x$, the function is unbounded below as $x\to-\infty$, $f(x) \to-\infty$, but the function is not strongly convex since $\nabla^2 f(x) = e^x $ doesn't satisfy that $\exists m >0, \forall x, \nabla^2 f(x)> m$.

Zhu ZengBao
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  • Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read its title. – José Carlos Santos Dec 16 '24 at 07:32
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    your definition does not match the standard one. The counterexample works for the correct definition. – daw Dec 16 '24 at 07:40
  • What's wrong with the given counterexample? $x + e^x$ has positive second derivative everywhere in $\mathbb{R}$ so it is strictly convex, according to both the usual definition and the given definition. – lily Dec 16 '24 at 08:12
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    sorry @daw and lily, I used the wrong terminology, what I want to ask is "Is strongly convex function always bounded below", "strongly convex function" is more strict than "strictly convex function". I edit the question description now. – Zhu ZengBao Dec 16 '24 at 09:47
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    Wikipedia says it does have minimum: https://en.wikipedia.org/wiki/Convex_function#Strongly_convex_functions – student91 Dec 16 '24 at 11:05
  • Thanks @student91, your information is quite helpful! – Zhu ZengBao Dec 16 '24 at 12:57
  • Thanks @daw, I agree the question in the link you provided has included my question as a subcase. Then this post can be closed! – Zhu ZengBao Dec 16 '24 at 13:08
  • You should note that your definition in the question is not the same as "strongly convex" (even ignoring the question of twice-differentiability) because the quantifiers are swapped. – lily Dec 17 '24 at 07:48
  • @lily, I didn't use the standard definition of strongly convex function, instead, a property was used: If the function $f(x)$ is twice continuously differentiable, then $f(x)$ strongly convex if and only if $\nabla^2f(x)>m>0$ for all $x$. – Zhu ZengBao Dec 18 '24 at 03:14
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    Yes but the definition in your question is wrong. $\exists m \forall x$ is not the same as $\forall x \exists m$ – lily Dec 18 '24 at 17:38
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    Thanks @lily! I revised it to $\exists m>0, \forall x, \nabla ^2f(x)>0$. – Zhu ZengBao Dec 19 '24 at 03:33

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