Extending a $C^1$ function on $\mathbb R^n$ to a set of finite $\mathcal H^{n-2}$ measure

Question

Note: Here $\mathcal H^k$ denotes the $k$-dimensional Hausdorff measure.

Let $n \geq 2$ be an integer, and $E \subset \mathbb R^n$ be a set of finite $\mathcal H^{n-2}$ measure.

Suppose $f: \mathbb R^n \setminus E \to \mathbb R$ is a $C^1$ function such that for each $x \in E$, $\lim_{y \to x} f(y)$ and $\lim_{y \to x} \nabla f(y)$ exist.

Question: Does $f$ admit a $C^1$ extension to $\mathbb R^n$?

Answer 1

The claim holds under the much weaker assumption that the exceptional set satisfies $\mathcal H^{n-1}(E)=0$.

Since the limits of $f$ and $\nabla f$ exist everywhere we get two continuous functions $F\in C(\mathbb R^n)$ and $G\in C(\mathbb R^n)^n$ such that $F$ is differentiable in $\mathbb R^n\setminus E$ with $\nabla F=G$ there. We will show that $F$ is differentiable in each $x\in E$ with Fréchet derivative $G(x)$.

To verify this, we take $\varepsilon>0$ and $\delta>0$ with $\|G(x)-Gy)\|\le \varepsilon/2$ for all $\|y-x\|\le\delta$. The very nice answer to If $\mathcal{H}^{n-1}(E)=0$ then $\mathbb{R}^n\setminus E$ is connected by @saúl-rm shows that, for $\|y-x\|\le\delta/2$, there is a smooth curve $\gamma:[0,1]\to \mathbb R^n$ of length $\le 2 \|x-y\|$ with $\gamma(0)=x$, $\gamma(1)=y$, and $\gamma(t)\notin E$ for all $t\in (0,1)$. Since the fundamenal theorem of calculus only requires differentiability in the interior of the interval we obtain $$F(y)-F(x)-\langle G(x),y-x\rangle=\int_0^1\langle \nabla F(\gamma(t))-G(x),\dot\gamma(t)\rangle dt.$$ As the length of $\gamma$ is less than $2\|x-y\|\le\delta$, all $\gamma(t)$ are $\delta$-close to $x$ so that $\|\nabla F(\gamma(t))-G(x)\|=\|G(\gamma(t))-G(x)\|\le \varepsilon/2$. Using Cauchy Schwarz we get $$\|F(y)-F(x)-\langle G(x),y-x\rangle\|\le \frac{\varepsilon}{2} \int_0^1\|\dot\gamma(t)\|dt \le \varepsilon \|y-x\|$$ which is the desired differentiability of $F$ in $x$.

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The case $n=1$ only requires $E$ to be finte, i.e., that $\mathcal H^{n-1}(E)$ is finite. For $n\ge 2$ the assumption $\mathcal H^{n-1}(E)$ finite is certainly not enough for the given proof.

Answer 2

You are extending from an open set.

Related, but different, extending from a closed set:
There is a paper of H. Whitney on defining the concept $C^k$ for a function defined on a closed subset of $\mathbb R^n$, and extending such a function to a $C^k$ function on all of $\mathbb R^n$.

Whitney, H., Analytic extensions of differentiable functions defined in closed sets., Transactions A. M. S. 36, 63-89 (1934). ZBL60.0217.01.

For $E \subseteq \mathbb R^1$, the required definition of $C^1$ for $f : E \to \mathbb R$ is more complex than merely the existence of $$ f'(a) := \lim_{x \in E, x \to a} \frac{f(x) -f(a)}{x-a}\qquad\text{for } a \in E . $$ It involves (locally uniform) estimates for error terms $$ \frac{f(x)-f(y)}{x-y} - f_1(a) $$ as $x,y \to a$, $x,y \in E$, $x\ne y$ where $f_1$ is the candidate for the derivative.

Answer 3

This is not an answer, but rather a suggestion of an idea on how a possible counterexample could be constructed (if it exists).

Let $n=1$ (rather than $n\ge2$ as in the OP). Let $E$ be the Cantor subset of the interval $[0,1]$, so that $\mathcal H^1(E)=0$ (rather $\mathcal H^{n-2}(E)<\infty$ as in the OP). Let $f$ be the restriction to $\mathbb R\setminus E$ of the Cantor function $c$ (extended to $\mathbb R$ by letting $c=1$ on $(1,\infty)$ and $c=0$ on $(-\infty,0)$). Then $c$ is continuous and $E$ is nowhere dense, so that $c$ is the only possible extension $g$ of $f$ such that $g(x)=\lim_{E\not\ni y\to x}f(y)$ for all $x\in E$. Also, $f'=0$ everywhere on $\mathbb R\setminus E$, so that $f\in C^1(\mathbb R\setminus E)$. But $c$ is not $C^1$. So, there is no $C^1$ extension $g$ of $f$ such that $g(x)=\lim_{E\not\ni y\to x}f(y)$ for all $x\in E$.