# Can this function be minimized?

Let $$X$$ be a locally convex TVS, and let $$A$$ and $$B$$ be convex and compact subsets of $$X$$ with $$A \subset B$$.

Let $$f: A \times B \to [0,\infty]$$ have the following properties:

(1) For all $$b \in B$$, $$f(\cdot, b)$$ is lower semicontinuous.

(2) For all $$b \in B$$, $$f(\cdot, b)$$ is convex.

(3) For all $$a \in A$$, $$f(a, \cdot)$$ is continuous.

(4) For all $$a \in A$$, the function $$f(a, \cdot)$$ is minimized uniquely at $$a$$, i.e. $$f(a,a) < f(a,b)$$ for all $$b \neq a$$.

Does it follow that for all $$b \in B$$, the function $$g_b: A \to [0,\infty]$$ defined by $$g_b(a) = f(a,b) - f(a,a)$$ attains a minimum?

Note that by (4), $$g_b$$ is bounded below by $$0$$, so $$\inf g_b(A)$$ is finite for all $$b \in B$$. If $$b \in A$$, then the result is trivial, for then $$g_b$$ is minimized (uniquely) at $$b$$, again by (4).

In general, it would be sufficient to show that $$g_b$$ is lower semicontinuous, but I have not been able to see why that should be the case.

This question is cross-posted at MSE.