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Let $n\geq2$, $0<R<\infty$ and $\Omega=B(0,R)\subset\mathbb{R}^n$. I am looking for a function $u\in C^1(\Omega\setminus\{0\})$ such that $u\in L^2(\Omega)$, $\nabla u\in L^2(\Omega)$ (where $\nabla u$ is the classical derivative which exists except at $0$) but $u\notin H^1(\Omega)$. For $n=1$, the function $u(x)=sgn(x)$ is an example. How to construct such a function in higher dimension?

Cameron Buie
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badam
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  • Maybe try $f(r,\theta) = \sqrt{r} \cos(\theta)$ on $\mathbb R^2$ with polar coordinates? I'm not whether this is $H^1$ or not at first glance, but it should be $L^2$ with $L^2$ derivative when excluding the origin. – Anthony Carapetis Oct 15 '13 at 11:19

1 Answers1

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There is no such function.

Let $\Omega$ be a domain in $\mathbb R^n$, $n\ge 2$. If $u$ is locally Lipschitz in $C^1(\Omega\setminus\{0\})$ and $u,\nabla u\in L^2(\Omega)$, then $u\in H^1(\Omega)$.

Indeed, being locally Lipschitz implies that $\nabla u$ exists a.e. (Rademacher's Theorem), and more importantly, it implies that $u$ is absolutely continuous on every closed line segment contained in $\Omega\setminus\{0\}$. Since for every direction, almost every line segment in $\Omega$ does not pass through $\{0\}$, the function $u$ has the ACL property. If a function is ACL with integrable partial derivatives, then its pointwise gradient is its weak gradient. Hence $u\in H^1(\Omega)$.

The reason you have an example in one dimension is that a single point there is a codimension $1$ subspace, which disconnects the space. The analog in higher dimensions would be a function that is smooth on the complement of a codimension $1$ hyperplane. Such as $u(x,y)=\mathrm{sgn}\,x$.

user100000
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