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The problem I am thinking about is like follows.

Suppose that $h$ is a strictly convex function on an open convex set $S$. Then, we extend $h$ continuously to the closure of $S$ that is denoted by $\bar{S}$. Let $\bar{h}$ be the continuous extension of $h$ on $\bar{S}$. Is $\bar{h}$ strictly convex on $\bar{S}$?

I think this might not be true but I have not found a counterexample. I guess there might be a function that is strictly convex on an open square, but is only convex on one of its sides.

Thanks for help.

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    $\begingroup$ This function is not even convex. You can consider $g(x)=h(x,x)=(x^2-1)^2$. $\endgroup$
    – mining
    Oct 3, 2014 at 6:32

2 Answers 2

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I believe that $f(x,y) = \dfrac{x^2}{\sqrt{1+y}}$, considered on $[0,1]^2$, is a counterexample of the type you describe.

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You may also take $f(x,y) =e^{x^2}y^2$ for $x\neq 0$.

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