Must a 'nice' function $f(x,y)$ have a stationary point inside a closed contour loop?

Question

Must a 'nice' function $f(x,y)$ have a stationary point inside a closed contour loop?

By this I mean, if we have a surface in 3 dimensions described by some function of the $x$ - $y$ plane, and we plot the contours of $f$ in this plane, if there is a closed loop of some sort must there be a stationary point (in which $ \partial f/ \partial x = \partial f / \partial y = 0$ ) ?

By 'nice' function I mean $f$ is continuous and defined at the area of interest.

Intuitively, I'd say yes as at any 2 points on the contour, $f$ must have the same value. So I would think there must be some way to extend the mean value theorem to 2 dimensions, however I can't see how exactly I would do this.

For example, pick any 2 points on the contour with the same $y$ coordinate. Applying mean value theorem, there must be a point in this $x$ interval with $ \partial f/ \partial x =0$.

Likewise for any 'slice' in the $y$ direction. However this does not guarantee that they both occur at the same point. I must need to use the fact $f$ is continuous somehow.

A rough sketch to clarify is shown below.

How could I go about proving/disproving this?

Furthermore, can this idea be extended to higher dimensions?

Answer 1

You won't necessarily get a point: take a function that's the square of the minimum distance to $x$ from a line segment $L$. Then the minimum is achieved when $x \in L$, and there are no other stationary points. One can do the same for a ball to get a 2D level set. Indeed, we can create a smooth function with the same property, by using properties of $e^{-1/x}$.

What you can do is use the extreme value theorem: $f$ is bounded inside the level contour and attains its bounds (one of which may be the contour itself, but both only if $f$ is constant).

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The mean value theorem basically does not generalise to higher dimensions: the best you can get is the mean-value inequality.