$f\in C(B_1)\cap W^{1,2}(B_1\setminus \{f=0\})$ implies $f\in W^{1,2}(B_1)$?

Question

In a paper I am writing I need to show that a certain real-valued function $f\in C^0(B_1)$ belongs to the Sobolev space $W^{1,2}(B_1)$ ($B_1$ is the unit ball). So far I have been able to show that the weak derivative $Df$ exists on the open set $B_1\setminus \{f=0\}$ and $Df \mathbb{1}_{f\neq 0}\in L^2(B_1)$, so now I only have to deal with the set $\{f=0\}$. I suspect that it shouldn't matter and that the weak differential $Df$ exists on all $B_1$ and is equal to $Df \mathbb{1}_{f\neq 0}$ a.e., but I am having troubles proving it formally. Any help/hint is appreciated.

Answer 1

This should follow from the ACL (absolute continuity on lines) characterisation of Sobolev spaces, see for instance, Theorem 4.1.10 here.

Indeed, since $f \in W^{1, 2} (B_1 \setminus \{f = 0\})$, it restricts to an absolutely continuous function on the intersection of $B_1 \setminus \{f = 0\}$ with a.e. line segment parallel to each coordinate axis.

We will first show

Claim 1: $f \in W^{1,1}(B_1)$.

We will appeal to the ACL characterization, and prove the claim by showing $f$ is absolutely continuous on the intersection $K$ of $B_1$ with a.e. line segment parallel to each coordinate axis. The proof is tedious but the idea is straightforward.

Proof:

Fix a line $L$ such that $f$ is absolutely continuous on the intersection $U := L \cap (B_1 \setminus \{f = 0\})$.

Let $\varepsilon > 0$ be arbitrary, and let $\delta > 0$ be such that $\sum_k |f(y_k) - f(x_k)| < \varepsilon$ whenever $(x_k, y_k)$ are a finite collection of intervals with $x_k, y_k \in U$ such that $\sum y_k - x_k < \delta$. We claim that the same $\delta$ works for $f$ restricted to $K$.

Indeed, let $(x_i, y_i)$ be an arbitrary finite collection of intervals with $\sum y_i - x_i < \delta$. If for all $i$, we have that either both $x_i, y_i$ are in $U$, or both are in $K \setminus U$, then there is nothing to prove, since the latter terms contribute zero to the sum, while the former have total lengths less than $\delta$.

On the other hand, for each $(x_i, y_i)$ such that one endpoint is in each of $U$ and $K \setminus U$, we proceed as follows. Assume first $x_i \in U$ and $y_i \in K \setminus U$. Set $z_i := \sup \{z < y_i \, | \, z \in U\}$, and for each $i$, let $y_{i, j}$ be a sequence in $U$ approaching $z_i$ from below. By continuity of $f$, we have

$$f(y_i) - f(x_i) = f(z_i) - f(x_i) = \lim_{j \to \infty} f(y_{i, j}) - f(x_i).$$

Likewise, if $x_i \in K \setminus U$ and $y_i \in U$, we replace $x_i$ with $x_{i, j}$ approaching $\inf \{z > x_i \, | \, z \in U\}$ from above. We label the new collection of intervals so obtained by $(a_{i, j}, b_{i, j})$.

We note that for each $i, j$, the intervals $(a_{i, j}, b_{i, j})$ are shorter in length than $(x_i, y_i)$, in particular their total length is less than $\delta$. Further, $a_{i, j}, b_{i, j} \in U$ by construction. So finally we may bound

$$\sum_i |f(y_i) - f(x_i) | = \lim_{j \to \infty} \sum_{i} |f(b_{i ,j}) - f(a_{i, j})| \leq \varepsilon,$$

and we conclude the absolute continuity of $f$ on $K$ as desired. $\square$

Finally, we check that the weak derivative is $Df|_{\{f \neq 0\}} \mathbf 1_{\{f \neq 0\}}$, which will imply also that $f \in W^{1,2} (B_1)$. To see this we use the second part of the ACL characterisation which says that the strong derivatives along a.e. line agree with the weak partials a.e. Hence we reduce to checking the result in one dimension. But it is known that the derivative of an absolutely continuous function is zero a.e. on each level set; applying this to $\{f = 0\}$, we conclude the theorem.