Proving that $\{\varphi\in C^{\infty}(\Omega)\,: \varphi(b) = 0\}$ is dense in $\{u\in H^1(\Omega)\,:\, u(b) = 0\}$

Question

Notation

Given $\Omega :=(a,b)\subset\mathbb{R}$ a bounded open interval, let us define the Sobolev spaces $H^1(\Omega):=\{v\in L^2(\Omega)\,: \, v'\in L^2(\Omega)\}$, $H_0^1(\Omega):=\{v\in H^1(\Omega): v(a) = v(b)= 0\}$. Further let $C^{\infty}(\Omega)$ and $C^{\infty}_0(\Omega)$ denote respectively the smooth functions on $\Omega$ in the former case and those who have compact support in the latter case.

Background

Additionaly, we define a subspace of $H^1(\Omega)$ with values that vanishes in the right boundary $H_R^1(\Omega):=\{u\in H^1(\Omega)\,:\, u(b) = 0\}$ and $S = \{\varphi\in C^{\infty}(\Omega)\,: \varphi(b) = 0\}$. The overall objective is to prove that $S$ is dense in $H_R^1(\Omega)$.

There are some previous results that i know

Lemma 1. $C_0^{\infty}(\Omega) $ is dense in $H_0^1(\Omega)$ with respecto to the norm $\|v\|_{1,\Omega} = \int_\Omega v^2$ for $v\in H^1(\Omega)$.

Theorem 8.2 (Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis) $$H^1(\Omega)\hookrightarrow C(\overline\Omega)$$ In other words, for all $v\in H^1(\Omega)$ there exists $u\in C(\overline\Omega)$ such that $u = v$ almost everywhere and $$\int_x^yv'(t) \text{d}t = u(y)-u(x),\,\,\forall x,y\in \overline \Omega $$

What i've tried so far

Remark: The objective is to prove that $S$ is dense in $H_R^1(\Omega)$. Let $v\in H_R^1(\Omega)$, is desired to build a sequence $\{\psi_n\}_{n\in\mathbb{N}}$ of elements in $S$ such that

\begin{align*} \lim\limits_{n\to+\infty} \|v-\psi_n\|_{1,\Omega} = 0 \end{align*}

I've tried two different things:

1. I first aimed to build $\{\psi_n\}_{n\in\mathbb{N}}$ from a sequence in $C_0^\infty(\Omega)$ that converges to some extension $u$ of $v$ that lies in $H_0^1(\Omega)$. So far i've tried \begin{align*} u(t) = \begin{cases}v(t) & t\in \overline \Omega, t\neq a\\ 0 & t = a\end{cases} \end{align*} With this definition might be clear that $u$ vanishes in both extreme values of $\overline\Omega$. However, this function is not necessarily continuous for $t = a$. Is there a way to ensure that this extension is in $H_0^1(\Omega)$?

2. It is clear that $H_0^1(\Omega)\subseteq H_R^1(\Omega)\subseteq H^1(\Omega)$ and $C_0^{\infty}\subseteq S$. If there's a subspace $R$ of $H_R^1(\Omega)$ that allows us to write $H_R^1(\Omega) = H_0^1(\Omega) + R$, $v$ can be decomposed as $v = v_0 + v_R$, where we can use Lemma 1, to show that there is a sequence of elements in $C_0^{\infty}$ converging to $v_0$, and then construct the desired sequence of elements in $S$. Unfortunately, I haven't been able to detect which space is the right one to perform this decomposition.

Any ideas, suggestions or corrections are well-received.

Thanks in advance.

Answer 1

Here is a sketch of the proof.

Let $u\in H^1_R$. Take $h>0$ and define the translation $$ u_h(x) = u(x+h) \quad x \in (a,b), $$ where we set $u(x)=0$ for $x\ge b$. Then $u_h\in H^1_R$ and $u_h\to u$ in $H^1$ for $h\searrow 0$.

Now we do a mollification with the standard kernel, $\phi_\epsilon = \rho_\epsilon \ast u_h$. For $\epsilon\in (0,h)$, we have $\phi_\epsilon \in S$, and $\phi_\epsilon \to u_h$ for $\epsilon \searrow 0$.