4
$\begingroup$

Let $f: \mathbb{R}^n \to \mathbb{R}$ be measurable and differentiable almost everywhere.

Is it true that $df=0 $ a.e on $\{f=0\}$?

If $f \in W^{1,p}$ then it's known to be true- but the proof* uses the fact that weak derivatives behave well w.r.t fundamental theorem of calculus, so I am guessing that withouth this assumption this should not hold in general.

*(It is Theorem 4.4 (iv) in "Measure theory and fine properties of functions", revised ed)

1 Answers 1

3

Yes, it's true. In particular $df=0$ at every Lebesgue point (point of density $1$) of $\{f=0\}$, assuming the differential exists. When $f\in W^{1,p}$ the only relevant implied information is the a.e. approximate differentiability, under which assumption (instead of a.e. classical differentiability) the following should remain true with minor modifications.

Let us consider a point $x_0\in \{f=0\}$ such that $df(x_0)\neq 0$, and let us prove that the density of $\{f=0\}$ at $x_0$ is zero.

Set $df(x_0)[y]=\xi\cdot y$ with $\xi\neq 0$. By definition of differential $$|f(x)|=|f(x)-f(x_0)|=|\xi\cdot(x-x_0)+o(|x-x_0|)|\geq |\xi\cdot (x-x_0)|-|o(|x-x_0|)|.$$ For every $\epsilon>0$ we can find a radius $r_\epsilon$ such that $o(|x-x_0|)\leq \frac\epsilon2|\xi||x-x_0|$ for $|x-x_0|0 \quad\forall x\neq x_0.$$ Now in the limit $\epsilon\to 0$ the cone $C_\epsilon$ fills a portion of $B(x_0,r_\epsilon)$ going to $1$. In particular $f\neq 0$ in a portion of the balls $B(x_0,r_\epsilon)$ that goes to $1$. This implies that the density of $\{f=0\}$ at $x_0$ is zero, and concludes the proof since a.e. point of $\{f=0\}$ is of density $1$ by the Lebesgue differentiation theorem.

If I'm not wrong it can be proved that the set of points of $\{f=0\}$ where the differential is nonzero is $(n-1)$-rectifiable, and therefore of Hausdorff dimension $n-1$ (which is stronger than Lebesgue negligible).

  • 0
    Thanks. Your argument is very nice! I have two questions though: (1) Perhaps you meant to write $C_\epsilon:=\{x:|\xi\cdot (x-x_0)|\geq |\xi| \epsilon |x-x_0|\}\cap B(x_0,r_\epsilon)$ (forgot a $|\xi|$)? (2) Do you have an easy argument for showing that in the limit $\epsilon \to 0$ the cone fills a portion of $B(x_0,r_{\epsilon})$ going to $1$? This is so non-trivial for me since the radii $r_{\epsilon}$ also change with $\epsilon$. – 2017-01-19
  • 0
    @AsafShachar (1) Yes you're right. I'll edit this. (2) Since we're talking about portions, and $C_\epsilon$ is a cone, then it is equivalent if we rescale the balls $B(x_0,r_\epsilon)$ to make them of radius $1$, and look at $C_\epsilon\cap B(x_0,1)$ . But now the complement of the cone inside the ball, $C^c_\epsilon\cap B(x_0,1)$, converges as $\epsilon\to 0$ to the strictly lower dimensional subspace $\{x:\xi\cdot (x-x_0)=0\}$, which has measure zero. – 2017-01-19
  • 0
    I slightly changed the definition of $C_\epsilon$: now it is not intersected with the ball $B(x_0,r_\epsilon)$. – 2017-01-19