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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 $x$?

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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 $B_0(r)$.

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  • $\begingroup$ 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

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