# Constructing a smooth function such that the measure of critical points of $f$ is a given positive real number $x$.

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. 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:
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 $$B_0(r)$$.
• What will be the function at least in $\mathbb{R}\to \mathbb{R}$ case? – I am pi May 12 '19 at 10:04