For $ k,p \in \mathbb{N} $, denote $ k $ order classical Sobolev space on $ (0,2\pi) $ by $ H^k(0,2\pi) $ and define \begin{equation*} \mathcal{H}^{k+p}_0:= \{ \varphi \in H^{k+p}(0,2\pi): \varphi(0) = \cdots = \varphi^{(p-1)}(0) = 0 \}. \end{equation*}
What is the closure of $ \mathcal{H}^{k+p}_0 $ in $ H^k(0,2\pi) $, if the former is dense in the latter?
When $ k = 0 $, since \begin{equation*} \mathcal{H}^p_0 = \{ \varphi \in H^p(0,2\pi): \varphi(0) = \cdots = \varphi^{(p-1)}(0) = 0\} \end{equation*} and thus, $ C^\infty_0 (0,2\pi) \subseteq \mathcal{H}^p_0 $, we have \begin{equation*} L^2(0,2\pi) \subseteq \overline{C^\infty_0 (0,2\pi)} \subseteq \overline{\mathcal{H}^p_0} \subseteq L^2(0,2\pi) \end{equation*} In the case $ k = 0 $, we have the former $ \mathcal{H}^p_0 $ dense in the latter $ L^2(0,2\pi) $.
However, for $ 0 \neq k \geq p $, the assertion is wrong. Similar to above deduction, we can see that $ \overline{\mathcal{H}^{k+p}_0 } = \mathcal{H}^k_0 $ in $ H^k(0,2\pi) $.
Notice that the case $ k < p $ is still open!