Does $f$ have a critical point if $f(x, y) \to +\infty$ on all horizontal lines and $f(x, y) \to -\infty$ on all vertical lines?

Given a function $$f:\Bbb R^2 \to \Bbb R, f \in C^\infty$$ with the property that $$\lim_{x\to+\infty}f(x,y_0) = \lim_{x\to-\infty}f(x,y_0) = +\infty \qquad \forall y_0\in \Bbb R, \\[2ex] \lim_{y\to+\infty}f(x_0,y) = \lim_{y\to-\infty}f(x_0,y) = -\infty \qquad \forall x_0\in \Bbb R.$$
Determine whether $$f(x,y)$$ necessarily has at least one critic point.

My attempt:

I suppose that such a function could be something just like: $$(x^{2n}-y^{2m})$$; anyway what I mean is that both $$f(x,y_0)$$ and $$f(x_0,y)$$ have to assume eventually the shape of a sort of "parabola".

Because of $$f\in C^{\infty}$$ then both $$f(x,y_0)$$ and $$f(x_0,y)$$ are continuous, thus:
1.$$f(x,y_0)=g(x),$$ has a global minimum and it means: $$\forall y_0 \in \Bbb R$$ there is at least one $$x^*$$ such that $$f_x(x^*,y_0)=0;$$
2.$$f(x_0,y)=h(y),$$ has a global maximum and it means: $$\forall x_0 \in \Bbb R$$ there is at least one $$y^*$$ such that $$f_y(x_0,y^*)=0.$$

If my attempt is correct until now, the last thing I need to do is observe that there is at least a couple $$(x^*,y^*)$$ such that $$f_x(x^*,y^*)=0$$ and $$f_y(x^*,y^*)=0$$. This last step is the one I stuck in.

Is there someone who can handle this? (or can propose another path to follow)

No, a function like this need not have a critical point. However, the only example I could find is a bit complicated and unfortunately not explicit. With a bit of work, one could make this explicit, but it would not be pretty. There should be a simpler example, though...

Let $$L$$ be the ray from the origin at a $$45^\circ$$ angle, i.e., defined by $$x=y \ge 0$$, and let $$U$$ be the open 1-neighborhood of $$L$$ in the plane, i.e., all points with distance less than $$1$$ to $$L$$. Then there exists a $$C^\infty$$-diffeomorphism $$\phi: U \to U\setminus L$$, which is the identity in a neighborhood of $$\partial U$$. (Note that $$U$$ and $$U\setminus L$$ are open sets, and $$\phi^{-1}$$ obviously does not extend to $$L$$.) Then we can extend $$\phi$$ to a $$C^\infty$$-diffeomorphism $$\phi: \mathbb{R}^2 \to \mathbb{R}^2 \setminus L$$ by defining it to be the identity outside $$U$$.

Now let $$g(x,y) = x^2-y^2$$, which already has the correct limits, but obviously has a critical point at $$(0,0)$$. Define $$f(x,y) = g(\phi(x,y))$$. Any horizontal or vertical line either does not intersect $$U$$ at all, or it intersects it in an interval of length $$\le 2\sqrt{2}$$, and since $$f=g$$ outside of $$U$$, we have that the limits of $$f$$ and $$g$$ along horizontal and vertical lines are the same, so $$f$$ has the desired limits. As a composition of smooth functions, it is smooth, and by the chain rule $$f$$ does not have any critical points, because $$\phi$$ does not have any, and the only critical point of $$g$$ is at $$(0,0) \in L$$, which is not in the image of $$\phi$$.

• I'm really sorry for my long absence.. Anyway, I think your answer is quite reasonable, but I really don't have such a competence to state it is correct. The goal of my question was mainly to create a point of discussion about.. So, I will be glad to accept your answer if no one have no answer anymore. Jun 29, 2020 at 15:29
• @DOmonoXYLEDyL: No worries, I was also hoping that someone would come up with a simpler construction. I might try to give it some more thought at some point, too. There should really be an explicit formula... Jun 29, 2020 at 16:59

Think about it as a zero sum game, where the payoff to the $$x$$ player is $$f(x,y)$$, and the payoff to the $$y$$ player is $$-f(x,y)$$. So $$X$$ is trying to maximize $$f$$ and $$Y$$ is trying to minimize $$f$$.

Consider, for example, the sets $$R_\delta = \{ (x,y) \in \mathbb{R}^2 : ||(x,y)|| \le \delta \}$$, or the set of balls based at some other point. Then each player has an upper hemi-continuous best reply to the other player on $$R_\delta$$, by continuity of $$f$$ and Berge's Theorem of the Maximum.

If your conjecture is true, there exists a finite $$\delta^*$$ for which there is a pure-strategy equilibrium to the game with $$(x^*,y^*)$$ in the interior of $$R_{\delta^*}$$. If it is false, either there is no pure strategy equilibrium to the game, or there is a pure-strategy eqm but it is always on the boundary of $$R_\delta$$ for any $$\delta$$. By picking a particular fixed-point theorem to get existence of an eqm point, you could then try to reverse engineer sufficient conditions for the conjecture to be true or false.

I am not sure it is either always true or false, since the assumptions are of the "coerciveness" kind: they describe what happens as $$x$$ or $$y$$ gets very large, but give no local information. So I can imagine lots of critical points in some neigborhood of, say, zero, and then the function goes off to $$\pm \infty$$ outside the neighborhood. Conversely, I can imagine that at any critical point in $$x$$, there is first-order variation in $$y$$, and vice versa (i.e., one player or the other always wants to deviate'' from the proposed $$x$$ and $$y$$ even if the other is happy with their strategy, so there is no pure-strategy Nash equilibrium).

Maybe I managed to solve the problem, so I post my answer in the eventuality someone will appreciate this: $$f(x,y)=x^2-e^{|y|}$$ could be a nice counterexample to the hypothesis that the function must have a critic point.

[EDIT]:
My counterexample doesn't satisfy the $$C^\infty$$ condition; so it's not a real answer to my question..

• This is a good counterexample. Not too sure if it meets the $C^{\infty}$ condition though, as it's not differentiable at $y=0$. May 21, 2020 at 12:30
• @K.defaoite Yes, you're right.. I totally forgot the $C^\infty$ condition. Adding that information may be an harder challenge. May 21, 2020 at 14:11
• In fact, no $C^\infty$ function of the form $f(x,y)=g(x)+h(y)$ can be a counterexample, since $g'(x_0) = 0$ for some $x_0$ and $h'(y_0) = 0$ for some $y_0,$ giving $(x_0,y_0)$ as a critical point.
– zhw.
May 25, 2020 at 18:05