# What is the closure of $C_c^{\infty}(\mathbb{R}^3\setminus\left\lbrace 0\right\rbrace)$ with respect to the norm of $H^{1}(\mathbb{R}^3)$?

It is known that $$\overline{C_c^{\infty}(\mathbb{R}^3)}^{\Vert\cdot\Vert_{H^{1}(\mathbb{R}^3)}} = H^1(\mathbb{R}^3).$$

I am thinking about what happens when I consider $$C_c^{\infty}(\mathbb{R}^3\setminus\left\lbrace 0\right\rbrace)$$. It is true that also $$\overline{C_c^{\infty}(\mathbb{R}^3\setminus\left\lbrace 0\right\rbrace)}^{\Vert\cdot\Vert_{H^{1}(\mathbb{R}^3)}} = H^1(\mathbb{R}^3)?$$

Could anyone help me to understant if it is true or not? Also some refernces will be appreciated.

Thank you in advance!

• Look up the Sobolev Embedding Theorem: Is $H^1(\Bbb R^3)\subset C(\Bbb R^3)$? If no, I bet that closure is $H^1$; if yes, I bet it's $\{ f\in H^1:f(0)=0\}$. – David C. Ullrich May 27 '20 at 15:20

As David Ullrich said, look up the Sobolev Embedding Theorem. In this case $$H^1(\mathbb{R}^3)$$ is not embedded in $$C(\mathbb{R}^3)$$, and then the completion is $$H^1(\mathbb{R}^3)$$. To see this, let $$T$$ be a continuous linear functional in $$H^1(\mathbb{R}^3)$$ such that $$T(\phi) = 0$$ for every $$\phi\in C^\infty_0(\mathbb{R}^3\setminus 0)$$. Since $$\vert T(\phi)\vert\le C\Vert\phi\Vert_{H^1}\le C_K(\Vert \phi\Vert_\infty+\Vert \nabla\phi\Vert_\infty)$$, then $$T$$ is a distribution of order one. Since $$T$$ vanishes outside the origin, $$T = a\delta_0 + \sum_i b_i\partial_i\delta_0$$, for some constants $$a$$ and $$b_i$$; see Theorem 2.3.4 in Hörmander, The Analysis of Linear Partial Differential Operators, v. I, 2nd edition. However, the distribution $$T$$ cannot be continuous in $$H^1(\mathbb{R}^3)$$ ---recall the comment of David Ullrich. We prove that $$b_j = 0$$ by considering the function $$\phi := \psi x_j\vert x_j\vert^{-\frac{1}{3}}$$, where $$\psi$$ is a smooth function with compact support that equals one in a neighborhood of the origin. We mollify $$\phi$$ as usually, so that $$\phi_\varepsilon := \phi^\alpha*\zeta_\varepsilon\in C^\infty_0(\mathbb{R}^3)$$, where $$\zeta_\varepsilon\in C^\infty_0(\mathbb{R}^3)$$. We see that $$T(\varphi^\alpha) = \frac{2}{3}b_j(\psi \vert x_j\vert^{-\frac{1}{3}})*\zeta_\varepsilon \to \infty$$, but $$\Vert \varphi_\varepsilon\Vert_{H^1}$$ remains uniformly bounded. Likewise, we can use the function $$\phi = \psi\vert x\vert^{-\frac{1}{4}}$$ to see that $$a = 0$$. Hence, $$T = 0$$ and the closure is $$H^1(\mathbb{R}^3)$$.

EDIT -----

We can construct the approximations directly using smooth functions vanishing in neighborhoods shrinking to the origin.

• So the answer is that $\overline{C_c^{\infty}(\mathbb{R}^3\setminus\leftlbrace 0\rightrbrace)}^{\Vert\cdot\Vert_{H^1(\matthbb{R}^3} = H^1(\mathbb{R}^3)$, isn’t it? – C. Bishop May 27 '20 at 18:00
• @C.Bishop, Yes. – user90189 May 27 '20 at 18:19