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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!

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  • $\begingroup$ 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 $\{ f\in H^1:f(0)=0\}$. $\endgroup$ – David C. Ullrich May 27 '20 at 15:20
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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.

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  • $\begingroup$ 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? $\endgroup$ – C. Bishop May 27 '20 at 18:00
  • $\begingroup$ @C.Bishop, Yes. $\endgroup$ – user90189 May 27 '20 at 18:19

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