Let $\mathbb {\overline B}^n\subset \mathbb R^n$ be a closed unit ball in and let $\mathbb B^k\subset \mathbb R^k$ be a open unit ball.


Suppose $F$ is a smooth function on $\mathbb {\overline B}^n\times \mathbb B^k$ that has the following properties.


1) $F(x,y)$ tends to $-\infty$ when $|y|\to 1$ for $(x,y)\in \mathbb {\overline B}^n\times \mathbb B^k$.


2) For any $y\_0\in \mathbb B^k$ the *strict* minimum of $F$ on the ball $\mathbb {\overline B}^n\times y\_0$ is attained in the interior of the ball. (Here by *strict* I mean, that the minimum of $F$ on the boundary of $\mathbb {\overline B}^n\times y\_0$ is larger than the minimum of $F$ over the whole ball.) 


**Question.** Is it true that $F$ has a critical point $(x\_0,y\_0)$ in $\mathbb {\overline B}^n\times \mathbb B^k$ such that $F$ restricted on $\mathbb {\overline B}^n\times y\_0$ has a minimum at $(x\_0,y\_0)$? If yes, I would like to know if there are some natural generalizations of this statement. If no, could one impose some additional conditions that guarantee a positive answer?



Finding a critical point on a product of two balls under some boundary conditions
I think the following is a counter example to your first question. Let $k = n = 1$, so we consider $(x,y) \in [-1,1] \times (-1,1)$. Construct $F$ such that for $y\_0 \le 0$ the minimum of $F(\,\cdot\,,y\_0)$ is attained at $x = -1/2$ with value $F(-1/2,y\_0) = \frac{-1}{y\_0+1}$, and for $y\_0 \ge 0$ construct a minimum at $x = 1/2$ with $F(1/2,y\_0) = \frac{1}{y\_0-1}$, and extend $F$ smoothly around this and growing along $x$.


I think this contradicts your requested result, since the set of minima for fixed $y\_0$ is precisely those given above, while those are not critical points since $\frac{\partial F}{\partial y} \neq 0$ there.


One way of making sure that you have a critical point, is if you have a differentiable path of minima $x(y\_0)$ for $F(\,\cdot\,,y\_0)$. Then you can restrict $F$ to that path and apply a version of the mean value theorem to conclude that $\frac{\partial F}{\partial y}(x(y\_0),y\_0) = 0$ for some $y\_0$.


