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.
Best Answer
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.