0
$\begingroup$

Let $\mathbb B^n$ be an open unit ball in $\mathbb R^n$ and let $F$ be a smooth function on it. Let $\frac{1}{2}\mathbb B^n\subset \mathbb B^n$ be an open ball or radius $\frac{1}{2}$. Let $\mathbb B^k$ be an open unit ball in $\mathbb R^k$, and $\frac{1}{2}\mathbb B^n\subset \mathbb B^n$ be an open ball or radius $\frac{1}{2}$.

Question. Is it true that $F$ has a critical point in $\mathbb B^n$ if there exists a smooth surjective map $\varphi:\mathbb B^n\to \mathbb B^k$, that has the following property:

1) For any $y\in \frac{1}{2}\mathbb B^k $ the strict minimum $$\min_{x\in \varphi^{-1}(y)} F(x)$$ is attained in $\mathbb B^n$ at some point ${\bf x}\in \frac{1}{2}\mathbb B^n$. In particular, $\inf F$ on $(\varphi^{-1}(y)\cap (\mathbb B^n\setminus \frac{1}{2}\mathbb B^n))$ is smaller than $\min_{x\in \varphi^{-1}(y)} F(x)=F({\bf x})$ (if such an intersection is non-empty).

2) The point ${\bf y}\in \mathbb B_k$ where the maximum of $$\max_{y\in \mathbb B^k}\min_{x\in \varphi^{-1}(y)} F(x)$$ is attained lies in $\frac{1}{2}\mathbb B^k.$

Comment. This is a largely rewritten formulation. I hope that the answer is positive if the differential of $\varphi$ is surjective on the whole ball $\mathbb B^n$

I would be grateful for a reference (or a counter-example...).

$\endgroup$
  • 1
    $\begingroup$ Doesn't property 1) suffice to show that the global minimum of $F$ is attained in the interior of $B^n$? Just set $x$ to be the image of the point where it is attained... $\endgroup$ – Ilya Bogdanov Sep 2 '16 at 20:15
  • $\begingroup$ Ilya, thanks, I corrected the question, one has to require condition 1) only for the interior of $\mathbb B^k$. In such a case condition 1) alone is not sufficient for existence of a critical point. $\endgroup$ – aglearner Sep 2 '16 at 20:53
  • 1
    $\begingroup$ Are you sure even the comment is OK? I have the following picture in mind: instead of the first ball, take a cylinder whose base is the unit ball in $x,y$ plane. Now, the mapping $\varphi$ is the trivial projection on the bases, but in between it first shrinks the disk a bit and then rotates. so that when you go from the top to the bottom, you make a few full turns. Thus the pre-images are spirals that wind around the vertical axis, which is the pre-image of the origin in $\mathbb R^2$. The function $F$ is just the $x$ coordinate. What do you think of this example? $\endgroup$ – fedja Sep 3 '16 at 0:32
  • $\begingroup$ Fedja, thank you for the example. But I don't think it works - it seems to me that on some of these vertical spirals $x$ takes minimum value at the top or at the bottom of the cylinder? $\endgroup$ – aglearner Sep 3 '16 at 9:09
  • 2
    $\begingroup$ No. That's what the "shrinking" is for (it is like squaring the radius and keeping the direction). The spirals are a bit wider in the middle part then at the endpoints, so the endpoints are beaten after one full turn. You may object that the minimum on the vertical line is not strict, but that can be easily remedied by pulling the points to the right in the middle slightly, which will move the level curves a bit to the left in the middle. $\endgroup$ – fedja Sep 3 '16 at 11:16
1
$\begingroup$

It seems to me, unfortunately, that I found my own very simple counterexample to the claim (as Fedja was suggesting)

Counter-example. Instead of $\mathbb B^n$ we will consider the square $[-2,2]\times [-2,2]\subset \mathbb R^2$. And $\mathbb B^k=[-2,2]\subset \mathbb R^1$. The map $\varphi$ is just the projection (on the $x$ axis).

The function $F$ has the following shape: $$F(x,y)=(y-1)^2(y+1)^2+xf(y).$$

Hence, $F(x,y)$ is affine on each horizontal interval. Let us define by $f(y)$ a monotonous $C^{\infty}$ function, that has the following properties:

1) $f(y)=-1$ for $y\in [-2,-1]$. 2) $f(y)=1$ for $y\ge 0$. 3) $f(y)$ is strictly increasing on $[-1,0]$, attains its single zero at $y=-\frac{1}{2}$ and $f'(-\frac{1}{2})=f''(-\frac{1}{2})=0$.

It is not hard to check that all the required conditions hold, and $F$ doesn't have any critical point in $[-2,2]\times [-2,2]$. What happens here, is that $F$ has two local minima on each interval $x=const$ at points $y=\pm 1$. One of these minima is global for $x\le 0$ and the other for $x\ge 1$...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.