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)


3 Answers 3


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 $\phi$.

  • 1
    $\begingroup$ 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. $\endgroup$ Jun 29, 2020 at 15:29
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    $\begingroup$ @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... $\endgroup$ 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.

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

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    $\begingroup$ This is a good counterexample. Not too sure if it meets the $C^{\infty}$ condition though, as it's not differentiable at $y=0$. $\endgroup$
    – K.defaoite
    May 21, 2020 at 12:30
  • $\begingroup$ @K.defaoite Yes, you're right.. I totally forgot the $C^\infty$ condition. Adding that information may be an harder challenge. $\endgroup$ May 21, 2020 at 14:11
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    $\begingroup$ 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. $\endgroup$
    – zhw.
    May 25, 2020 at 18:05

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