Closure of $C^\infty_0(\mathbb{R}^3\!\setminus\!\{0\})$ in the $H^2$-norm?

Question

It is a standard fact (e.g., Lieb-Loss, Analysis, Theorem 7.6), that the closure of $C^\infty_0(\mathbb{R}^3)$, namely the space of (complex-valued) compactly supported smooth functions on $\mathbb{R}^3$, in the $H^s$-Sobolev norm, $s\geqslant 0$, is precisely $H^s(\mathbb{R}^3)$.

Although I believe it is true, I would like to know a proof that the $H^2$-closure of $C^\infty_0(\mathbb{R}^3\!\setminus\!\{0\})$, namely the space of smooth functions whose support is a compact in $\mathbb{R}^3\!\setminus\!\{0\}$, is the space of $H^2$-functions $f$'s such that $f(0)=0$.

My understanding is that if $f\in H^2(\mathbb{R}^3)$ and $f(0)=0$, then a good sequence of approximants $f_n\in C^\infty_0(\mathbb{R}^3\!\setminus\!\{0\})$ should be given by $f_n=j_n*(f\chi_n)$ where $\chi_n(x)=\chi(nx)$ is a smooth cut-off at the origin, say, $\chi(x)=0$ for $|x|<1$ and $\chi(x)=1$ for $|x|>2$, while $j_n$ is a standard mollifier on a scale smaller than the cut-off, say, on a scale $|x|<(2n)^{-1}$. Still, I cannot control the vanishing $\|f_n-f\|_{H^2}\to 0$.

Any suggestion?

On a related note, I am vaguely aware of a class of statements of the form $W^{2,p}(\mathbb{R}^d\!\setminus\!C)=W^{2,p}_0(\mathbb{R}^d\!\setminus\!C)$ when the set $C$ is closed and with null capacity, could anyone give me a reference on that?

Answer 1

Perhaps it's easiest to work with the Fourier transforms. We are given a $g=\widehat{f}$ with $\int |g(k)|^2(1+k^4)\, dk<\infty$ and $\int g = 0$. It's now easy to approximate this by functions $g_n\in C_0^{\infty}$ with $\int g_n=0$ such that $\int |g-g_n|^2 (1+k^4)\to 0$.

This is not exactly what we want because the Fourier transforms $f_n=\check{g_n}$ are not guaranteed to be compactly supported. However, they are Schwartz functions, and these can easily be approximated in $H^2$ norm by $C_0^{\infty}$ by introducing a smooth cut-off. This will of course preserve the property that $f_n(0)=0$.

This shows that any $f\in H^2(\mathbb R^3)$ with $f(0)=0$ is in the closure of $C_0^{\infty}(\mathbb R^3\setminus 0)$. Conversely, any function in this closure vanishes at $x=0$ because $H^2$ convergence implies pointwise convergence.

Answer 2

This can be done with the way you suggested, but let's do the approximations one at a time.

Suppose that $f\in H^2$ is given. From the Sobolev inequalities, since the second derivatives of $f$ belong to $L^2$, this implies that $f$ and its first derivatives belong to $L^6$, hence Morrey's inequality shows that $f$ is $1/2$-Holder continuous.

Suppose now that, in addition, $f(0)=0$. Let $\phi_n$ be a sequence of cutoffs, supported outside $B_{1/n}$, being equal to $1$ outside $B_{2/n}$, with $|\phi_n|\leq 1$, $|\partial_i\phi_n|\leq Cn$, and $|\partial_{ij}\phi_n|\leq Cn^2$ for all $i,j=1,2,3$. Then $f\phi_n\in H^2$ for all $n$, and $f\phi_n$ is $0$ in $B_{1/n}$. Moreover, for all $i,j$, $$\int_{\mathbb R^3}|\partial_{ij}f-\partial_{ij}(f\phi_n)|^2=\int_{\mathbb R^3}|(1-\phi_n)\partial_{ij}f+\partial_if\partial_j\phi_n+\partial_jf\partial_i\phi_n+f\partial_{ij}g|^2\\ \leq C\int_{\mathbb R^3}|(1-\phi_n)\partial_{ij}f|^2+C\int_{\mathbb R^3}|\partial_if\partial_j\phi_n|^2+C\int_{\mathbb R^3}|\partial_jf\partial_i\phi_n|^2+C\int_{\mathbb R^3}|f\partial_{ij}g|^2.$$ The first integral goes to $0$, from the dominated convergence theorem. The second integral is supported in $B_{2/n}$, and, since $\partial_if\in L^6$, $$\int_{B_{2/n}}|\partial_if\partial_j\phi_n|^2\leq Cn^2\int_{B_{2/n}}|\partial_if|^2\leq Cn^2\|\partial_if\|_{L^6(B_{2/n})}|B_{2/n}|^{2/3},$$ which goes to $0$. Similarly, the third integral goes to $0$. Finally, for the fourth, Holder continuity of $f$ shows that $$\int_{B_{2/n}}|f(x)\partial_{ij}\phi_n(x)|^2\leq Cn^4\|u\|_{C^{0,1/2}(B_{2/n})}\int_{B_{2/n}}|x|\,dx\leq C\|u\|_{C^{0,1/2}(B_{2/n})}\leq C\|u\|_{W^{1,6}(B_{2/n})},$$ which converges to $0$, where we also used Morrey's inequality.

The argument above, also applied to $f$ and its first derivatives, shows that given $\varepsilon>0$, there exists an $H^2$ function $g$ which vanishes on a ball centered at $0$, such that $\|f-g\|_{H^2}<\varepsilon$.

We now approximate this $g$: suppose that $g$ vanishes in $B_{\delta}$. Consider a mollification $j_n*g$ which approximates $g$ in the $H^2$ norm; that is, given $\varepsilon>0$, there exists $n\in\mathbb N$ such that $\|j_n*g-g\|_{H^2}<\varepsilon$. By choosing $n$ large enough, we can assume that the support of $j_n$ is a subset of $B_{\delta/2}$; then, if $x\in B_{\delta/2}$ and $y\in{\rm supp}(j_n)$, we have that $|x-y|<\delta$, therefore $$j_n*g(x)=\int_{{\rm supp}(j_n)}g(x-y)j_n(y)\,dy=0,$$ since $g$ vanishes in $B_{\delta}$.

Note now that, if $M$ is sufficiently large, that we can multiply with a smooth cutoff $\phi$ that is equal to $1$ in $B_M$ and equal to $0$ outside $B_{2M}$, and obtain $\|\phi\cdot(j_n*g)-j_{n_0}*g\|_{H^2}<\varepsilon$. Then $\phi\cdot(j_n*g)$ is smooth, it is compactly supported in $\mathbb R^3\setminus\{0\}$, and also $$\|f-\phi\cdot(j_n*g)\|_{H^2}<3\varepsilon.$$ This finishes the proof.

Edit: The constant that appears in the step involving Morrey's inequality is an absolute constant. To show this, note that we do not need the full $C^{0,1/2}$ norm; instead, for all $x\in B_{2/n}$, $$|f(x)|=|f(x)-f(0)|\leq C|x|^{1/2}\|\nabla f\|_{L^6(B_{4/n})};$$ this last estimate can be found for example in Evans, right after the proof of Morrey's inequality.