Let $X\subset\mathbb{R}^n$ be a nonempty convex set such that $\text{int}(X)$ is nonempty and $f:X\rightarrow \mathbb{R}$ be a continuous function. Suppose that $f$ is strictly convex on the interior of $X$. Is $f$ strictly convex everywhere on $X$?

The result holds for $f$ convex but I suspect that it fails for strict convexity. For instance, suppose that $X$ is a square and $n=2$. It seems that we could come up with $f$ strictly concave on $\text{int}(X)$ with $f$ being a line at the boundary of $X$. I'm not sure what part of the proof fails though.

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    $\begingroup$ I can't think of a counterexample yet (I think there will be one), but the proof won't work because of the taking of limits of both sides of the inequality. Such an operation will not preserve (in general) strict inequality. $\endgroup$ Jul 10, 2018 at 14:51

1 Answer 1


No. For example, $f(x, y) = x^2y^2+y^2$ is strictly convex on the vertical half-strip $\{(x,y): |x|<1/2, y>0\}$ as its Hessian matrix is $$ \begin{pmatrix}2y^2 & 4xy \\ 4xy & 2x^2 + 2\end{pmatrix} $$ with the determinant $4x^2y^2 + 4y^2 - 16x^2y^2 = 4y^2(1-3x^2) > 0$. But on the boundary segment $\{(x, 0) : |x|<1/2\}$ the function $f$ is identically zero.


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