Let $f \in C^{\infty}(\mathbb{R^{2}})$ be a function such that: $$\lim_{y \to \pm \infty} f(x,y)=+\infty, \ \forall x \in \mathbb{R},$$ and that $$\lim_{x \to \pm \infty} f(x,y)=-\infty, \ \forall y \in \mathbb{R}.$$ With this hypothesis, does a stationary point necessarily exist? I'm looking for either a proof of the existence or a counterexample. It sounds false to me, but I didn't find a counterexample.


3 Answers 3


I believe this problem can be solved using a version of the Mountain Pass Theorem. Let $$ m = \min_{y \in \mathbb R} f(0,y)\,, $$ which exists because $\lim_{y \to \pm \infty} f(0,y) = +\infty$. Because of the other limit condition there exist $x_- < 0 < x_+$, such that $$ f(x_-,0) < m \;\text{and}\; f(x_+,0) < m\,. $$ Now consider the set $$ \Gamma = \{ \gamma \in C([0,1], \mathbb R^2 \,:\, \gamma(0) = (x_-,0),\, \gamma(1) = (x_+,0) \}\,,$$ of continuous paths between $(x_-,0)$ and $(x_+,0)$. Then a version of the Mountain Pass Theorem should allow us to conclude that $$ c = \inf_{\gamma \in \Gamma} \max_{t \in [0,1]} \gamma(t)\,,$$ is a critical value of $f$.

There are several formulations of the theorem and I do not have the right formulation at hand, but this question on MathOverflow as well as Wikipedia should contain links to the relevant literature.

  • $\begingroup$ Wow, I didn't know that theorem! Thank you very much, I'll tell you if I have some doubts :) $\endgroup$ Mar 26, 2017 at 16:49
  • $\begingroup$ It has a nice intuitive interpretation: if you have to cross a mountain range and you choose a path that requires you to ascend to the smallest maximum altitude, then the highest point of your path will be a saddle point. $\endgroup$ Mar 26, 2017 at 18:05

$\color{red}{\text{Misinterpreted OP's question, so this is merely an example!}}$ Consider the function $f: \mathbb{R}^2 \to \mathbb{R}: (x,y) \mapsto y^2 - x^2$.

Since this map is a polynomial in the variables $x,y$, it is $C^{\infty}(\mathbb{R}^2)$.

Moreover, we have that $$\lim_\limits{y \to \pm \infty}f(x,y) = \lim_\limits{y \to \pm \infty}(y^2 - x^2) = + \infty - x^2 = + \infty$$ and in the same way, it can be shown that $\lim_\limits{x \to \pm \infty}f(x,y) = - \infty$.

Since we already have that $f \in C^{\infty}(\mathbb{R}^2)$, we know the function is differentiable at each point and we find that the gradient of $f$ is equal to $(-2x, 2y)$, showing that $(0,0)$ is a stationary point.

The following pictures shows the graph of $f$. The red axis is the $x$-axis, whereas the green axis is the $y$-axis. enter image description here

  • 2
    $\begingroup$ I think OP is looking for either (i) An example of a smooth $f$ that has the stated asymptotic behavior yet no stationary point, or else (ii) A proof that every function having the stated asymptotic behavior has at least one stationary point. ;) $\endgroup$ Mar 22, 2017 at 1:04
  • $\begingroup$ Exactly! I'm looking for a counterexample or a proof. Thanks anyway :) $\endgroup$ Mar 22, 2017 at 6:14
  • $\begingroup$ @RiccardoCeccon my bad! I must have misread your question, i am really sorry. $\endgroup$
    – Student
    Mar 22, 2017 at 8:42
  • $\begingroup$ @RiccardoCeccon But I think it should be true: given that your function has to be smooth and because of the limits, I think you will always have something similar as to my example: a point where the graph bends upwards, whereas it bends downwards in some other direction. This gives us some kind of 'saddle point' and the derivative is zero in this point. (This is only what I think, so certainly no answer) $\endgroup$
    – Student
    Mar 22, 2017 at 9:13
  • $\begingroup$ I thought about it, but in my mind it is possible to make the locus of the zeros of the partial derivatives $\partial_y f$ and $\partial_x f$ to be the graph of two functions (implicit functions theorem), one from $y$ to $x$, and the other vice versa. These two graphs can have no intersection (think about $y=x+1$ and $x=y$). This is the way to construct a counterexample I was thinking about, but I didn't find a working analytic form. $\endgroup$ Mar 22, 2017 at 10:11

I think this is true. This is a sketch of what I would try here; it's too long for a comment, so I'm am putting this as an answer.

(1) For every $y_0$, there exists a $x_0$ such that $\frac{\partial f}{\partial x}=0$ at $(x_0,y_0)$, and guarantee a point $(x_1,y_1)$ same in reverse with respect to $y$.

(2) Smoothness will guarantee for any $x_2$ close to $x_0$ there will be another $y_2$ in the neighborhood of $y_0$ such $\frac{\partial f}{\partial x}=0$ at $(x_,y_2)$. From here you can construct 2 curves in the xy-plane, one of fixed points with respect to $x$, and one of fixed points with respect to $y$.

(3) Then, show that these 2 curves intersect.

I am sure that (1) and (2) work. If (3) does not work, it might give you the way to construct a counterexample.

  • $\begingroup$ This is what I was thinking about, as you can read in the comments above (under student's answer). Clearly, the two graphs, in general, need not intersect, like $y=x+1$ and $x=y.$ But I don't find how the hypothesis on the function could help :(. However, thank you! If you have other ideas, tell me!! $\endgroup$ Mar 24, 2017 at 7:43

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