0
$\begingroup$

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 $C$).

$\endgroup$
2
  • $\begingroup$ How about something like $x^2+xy^2$ and $C = (0,1) \times (0,1)$? $\endgroup$ Commented Aug 22, 2020 at 20:14
  • $\begingroup$ @MatiasHeikkilä That function is not concave nor convex. $\endgroup$
    – jjagmath
    Commented Aug 22, 2020 at 22:59

2 Answers 2

1
$\begingroup$

Hint: try the counterexample $f(x,y)=\sqrt[3]{xy}$ on $x>0$, $y>0$.

$\endgroup$
1
  • $\begingroup$ You are right. Your counterexample works. Thanks. $\endgroup$
    – Condor5
    Commented Aug 23, 2020 at 18:03
0
$\begingroup$

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 $Cl (C)$.

$\endgroup$
2
  • $\begingroup$ But this function is not continuous. $\endgroup$
    – Condor5
    Commented Aug 23, 2020 at 17:32
  • $\begingroup$ I see my statement was ambiguous. I have edited it. $\endgroup$
    – Condor5
    Commented Aug 23, 2020 at 17:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .