For a given positive real number $x$, construct a smooth function $f:\mathbb{R}^n\to \mathbb{R}^m$ such that the Lebesgue measure of the critical points is $x$.

I was trying to do by using the bump function idea. First thing, enough to construct a function $f:\mathbb{R}^n\to \mathbb{R}$. If we construct a bump function which is taking value $1$ on a ball whose measure is $x$ then the critical values are the ball and the points where $f$ is zero. enter image description here But how to get the critical points whose measure is $x$?


Take the same kind of function which is not $0$ at infinity. If $r$ is such that $B_0(r)$ has measure $x$, the function looks like this:enter image description here

For an explicit formula in the case $f:\Bbb R^n\to \Bbb R$, take $h:\Bbb R\to \Bbb R$ such that $h(x)=1$ if $x<0$ and $h(x)=e^{-1/x^2}$ if $x\geq 0$. The function $h$ is smooth with set of critical points $\Bbb R_-$. Then take $$f(x)=h(\Vert x\Vert^2-r^2).$$ Because $$\nabla f(x)=2h^\prime(\Vert x\Vert^2-r^2)\cdot x,$$ the set of critical points of $f$ is $B_0(r)$.

| cite | improve this answer | |
  • $\begingroup$ What will be the function at least in $\mathbb{R}\to \mathbb{R}$ case? $\endgroup$ – I am pi May 12 '19 at 10:04
  • $\begingroup$ @Iampi I edited the post. $\endgroup$ – Adam Chalumeau May 12 '19 at 10:48

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.