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I'm searching for a function $f: \Bbb R^2 \rightarrow \Bbb R$ which has following properties:

  • Both $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist and are continuous in $\Bbb R^2$.
  • There are infinitely many critical points* , in all of which, $f$ has local maximum.
  • $f$ is unbounded both from above and from below.

Could you give me some hints, or even better, show me explicitly the desired function?

*$(x,y)$ such that $\frac{\partial f}{\partial x}(x,y)=\frac{\partial f}{\partial y}(x,y)=0$.

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  • $\begingroup$ In one dimension, $f(x)=x^2\cos(x)$ does the job. Try $F(x, y)=f(x)f(y)$. $\endgroup$ Commented Nov 2, 2017 at 19:50
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    $\begingroup$ Your function has both minima and maxima. I'm interested in functions having exclusively maxima. $\endgroup$
    – Dagon
    Commented Nov 2, 2017 at 20:16
  • $\begingroup$ Right, I had misread, that was too easy. I'm sorry. I think that there's no one-dimensional example, as you surely already know. This example might be used as a building block for your function, but I am not sure yet about this. $\endgroup$ Commented Nov 2, 2017 at 20:28

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NOTE. In this post I consider functions having only local minima, instead of maxima. Also, this is not a rigorous solution but only a hint.

There is a similar problem which asks to find a two-variable function that is unbounded above and below and that has only one critical point that is a local minimum. I saw somewhere the solution by Tim Gowers, which is the following recipe.

Take the graph of $f(x, y)=x$ and put a finger in it, at $(x, y)=(0,0)$. This creates a local minimum that is not global. This also creates a saddle point near the point $(x, y)=(-1, 0)$. Sliding the finger in the negative $x$ direction, the saddle point moves accordingly. Passing to the limit, we can make the saddle point disappear, as if it had "moved to the $x, y$ infinity". Done.

(Unfortunately I cannot find the precise reference. It is in a comment to some blog, so it is not indexed by search engines, I think.)

You can redo this recipe countably many times, each time putting your finger far from what you have already done. This will produce an unbounded function with countably many local minima that are not global, and no more critical points.

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  • $\begingroup$ If I'm understanding your suggestion correctly, moving the saddle point to infinity violates continuity or existence of partial derivatives in that point. $\endgroup$
    – Dagon
    Commented Nov 3, 2017 at 12:46
  • $\begingroup$ @Dagon: No, you do it smoothly. You slide your finger towards infinity. The saddle point will smoothly move away from the minimum. $\endgroup$ Commented Nov 3, 2017 at 12:55

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