Is the following statement true?
Let $f:X\to \mathbb R$, $X$ normed vector space, be a continuous function that is strictly concave function over a convex set $C$, then $f$ is strictly concave over $\bar C$ (the closure of $C$).
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$\begingroup$ How about something like $x^2+xy^2$ and $C = (0,1) \times (0,1)$? $\endgroup$– Matias HeikkiläAug 22, 2020 at 20:14
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$\begingroup$ @MatiasHeikkilä That function is not concave nor convex. $\endgroup$– jjagmathAug 22, 2020 at 22:59
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2 Answers
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Hint: try the counterexample $f(x,y)=\sqrt[3]{xy}$ on $x>0$, $y>0$.
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$\begingroup$ You are right. Your counterexample works. Thanks. $\endgroup$– Condor5Aug 23, 2020 at 18:03
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I don't think that statement is correct. Consider the function $$f(x)= \begin{cases} &-x^2 & ,x>0 \\ &1 &,x=0 \end{cases}$$ and let the set $C =\{ x\,|\, x>0\} $. Then $f$ is strictly concave over $C$, but it is not concave over $Cl (C)$.
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$\begingroup$ I see my statement was ambiguous. I have edited it. $\endgroup$– Condor5Aug 23, 2020 at 17:38