# The Schwartz function and the sobolev space $W^{2,p}$

How do you prove the Schwartz functions in $\mathbb{R}^n$ are dense in the space $W^{2,p}(\mathbb{R}^n)?$

Terrence tao has a version of the proof of

The space $C_c^{\infty}(\mathbb{R}^d)$ of test function is a dense subspace of $W^{k,p}(\mathbb{R}^d)$, then the fact $\mathcal{S}(\mathbb{R}^d)$ is dense in $L^p(\mathbb{R}^d)$ is a corollary from that. I do not understand his proof. (See lemma2)

• So the proof is already there. Which parts do you not understand? – user99914 Dec 14 '15 at 11:05

The Schwartz space contains in particular $C^\infty_0(\mathbb R^n)$ and $C^\infty_0(\mathbb R^n)$ is by definition dense in $W^{k,p}_0(\mathbb R^n)$. But we have $W^{k,p}_0(\mathbb R^n) = W^{k,p}(\mathbb R^n)$. Thus the Schwartz space is dense in $W^{k,p}(\mathbb R^n)$.

The fact that $W^{k,p}(\mathbb R^n)= W^{k,p}_0(\mathbb R^n)$ can be found in Adam's Sobolev spaces (Corollary 3.23). The following is part of the proof of Theorem 3.22 in the book.

Let $f : C^\infty_0(\mathbb R^n)$ be a smooth function so that $0\le f\le 1$ and

$$f(x) = \begin{cases} 1 & \text{ when }|x|\le 1 \\ 0 & \text{ when }|x|\ge 2, \end{cases}$$

For each $\epsilon >0$, let $f_\epsilon(x) = f(\epsilon x)$. Then all derivatives of $f_\epsilon$ by bounded independent of $\epsilon <1$.

For all $u\in W^{k,p}(\mathbb R^n)$, consider $u_\epsilon = uf_\epsilon$. Then using the product rule, we have

$$\|u-u_\epsilon\|_{W^{k,p}(\mathbb R^n)} \le C \|u\|_{W^{k,p}(\Omega_\epsilon)},$$

where $\Omega_\epsilon = \{x\in \mathbb R^n : |x| \ge 1/\epsilon\}$. As $\epsilon\to 0$, the right hand side converges to $0$. Thus $u$ can be approximated by elements $u_\epsilon$ with compact support. Using mollifiers, this $u_\epsilon$ can be approximated by elements in $C^\infty_0(\mathbb R^n)$. Thus $C^\infty_0(\mathbb R^n)$ is dense in $W^{k,p}(\mathbb R^n)$.

In general it is not true that $W^{k,p}_0(\Omega) = W^{k,p}(\Omega)$.

• @ John Ma: why is $W_0^{k,p}=W^{k,p}?$ – math101 Dec 14 '15 at 5:10
• @math101 Please see the edit. – user99914 Dec 14 '15 at 6:27
• thanks I will need to spend some time on what you wrote, feel free to edit as I go along – math101 Dec 14 '15 at 6:36
• Even for extremely nice domains (e.g. balls), we have $W_0^{k,p}(\Omega) \neq W^{k,p}(\Omega)$. – PhoemueX Dec 14 '15 at 7:24
• @PhoemueX : Thanks. Edited. – user99914 Dec 14 '15 at 7:53