Denote by $W^{k,p}(\mathbb{R})$ the Sobolev space of functions in $L^p(\mathbb{R})$ whose all weak derivatives up to order $k$ belong to $L^p(\mathbb{R})$ with the norm
$$
\| f \|_{W^{k,p}(\mathbb{R})}:=\sum_{j=0}^k \| D^{(j)}f \|_{L^p(\mathbb{R})}.
$$
It is well known that the space $C_c^\infty(\mathbb{R})$ of smooth, compactly supported functions on $\mathbb{R}$ is a dense subspace of $W^{k,p}(\mathbb{R})$. It can be seen by considering, for $f\in W^{k,p}(\mathbb{R})$, the family
$$
f_{\delta,\epsilon}(x)=\eta(\delta x)f\ast \varphi_\epsilon(x),\qquad \delta>0, \epsilon>0,
$$
where $\eta\in C_c^\infty(\mathbb{R})$ is such that $\eta(x)\equiv 1$ on $(-1,1)$ and $\varphi_\epsilon(x)=\frac{1}{\epsilon}\varphi(x/\epsilon)$ for some $\varphi\in C_c^\infty(\mathbb{R})$ satisfying $\int \varphi=1$.
My question: Are compactly supported functions $C_c^\infty(0,1)$ dense in the Sobolev space $W^{k,p}(0,1)$ on the interval $(0,1)$? My concern is that if we try to use the family $f_{\delta,\epsilon}$ to approximate $f$, then the outside cutoff $\eta(\delta x)$ needs to be replaced in this setting with a smooth version of $\mathbb{1}_{(\delta, 1-\delta)}$,and derivatives of such function are large (more precisely, $n^{th}$ derivative of such a function behaves like $\delta^{-n}$). That suggests that such approximation may not be possible.
Best Answer
No, the closure of $C_c^\infty(0,1)$ is not $W^{k,p}(0,1)$ but $W^{k,p}_0(0,1)$. Informally, this is due to boundary conditions: functions from $C_c^\infty(0,1)$ are zero at the boundary, this is preserved when passing to the closure.
From Meyers–Serrin theorem we know that $C_c^\infty(\mathbb R)$ is dense in $W^{k,p}(0,1)$ (in the sense that restrictions of functions from $C_c^\infty(\mathbb R)$ or $C^\infty(\mathbb R)$ to $(0,1)$ are dense).