Minimax theorem without compactness

Question

Can we say that $$ \inf_{x \in X}\sup_{y \in Y} f(x,y) = \sup_{y \in Y}\inf_{x \in X} f(x,y)$$ when $X$ and $Y$ are convex subsets of topological vector spaces (potentially infinite dimensional), $f$ is bilinear and continuous, the suprema are always achieved, and the infima are always achieved within a compact set $X_0$? Sion's minimax theorem gives that $$ \inf_{x \in X}\sup_{y \in Y} f(x,y) = \inf_{x \in X_0}\sup_{y \in Y} f(x,y) = \sup_{y \in Y} \inf_{x \in X_0}f(x,y),$$ but I'm not sure if this helps. If my desired statement is false, a counterexample would be much appreciated.

Answer 1

Note: in the question as originally asked, $X, Y$ were the full vector spaces, not just convex subsets. This answer addresses that version.

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If for some $x$, there is a $y_x$ with $f(x,y_x) \ne 0$, then $f(x,ay_x) = af(x,y_x)$ by bilinearity. By choice of $a$ you can make $f(x,ay_x)$ as high or low as you like. Therefore $\sup_{y \in Y} f(x,y) = \infty$. If no such $y_x$ exists, then obviously $\sup_{y \in Y} f(x,y) = 0$.

Similarly for all $y$, either $\inf_{x\in X} f(x,y) = -\infty$ or $\inf_{x\in X} f(x,y) = 0$

Since $f(0,y) = 0$ for all $y$ and $f(x,0) = 0$ for all $x$, we have $$\inf_{x\in X}\sup_{y\in Y} f(x,y) = 0\\\sup_{y\in Y}\inf_{x\in X} f(x,y) = 0$$ So yes you can say they are the same, but it is actually a trivial statement when $X, Y$ are full vector spaces instead of bounded subsets, and $f$ is fully bilinear instead of convex/concave.