Suppose $m \neq 1$ and the space:
$$U = \{ u \in H^1(0,1) \mid u(0) = mu(1) \}$$
In trying to show that this is a closed subspace of $H^1(0,1)$, which is the sobolev space $W^{1,2}(0,1)$, I've been able to show that $U$ is closed and that if an arbitrary sequence $\{u_n\}$ in $U$ which converges to $u$, then $u \in H^1(0,1)$. However, I am struggling to show that $u(0)=mu(1)$. My best attempt has been as follows:
Since $\int|u_n' - u'|^2dx \rightarrow 0$ as $n \rightarrow \infty$, we have that
$$ \int u_n' - \int u' \le \int|u_n' - u'| \le \int|u_n' - u'|^2dx $$
then $$ (u_n(1)-mu_n(1))-(u(1)-u(0)) \rightarrow 0$$ as $n \rightarrow \infty$.
I've tried a variety of things after this point, and I can't seem to convince myself that $u(0)=mu(1)$. Any help is greatly appreciated!
One approach is to show that the endpoint evaluation functionals are continuous on $H^1$. You can use the Riesz representation for the functions, for example, where you find $g_0,g_1 \in H^1$ such that $$ f(0)= (f,g_0)+(f',g_0'),\;\; g\in H^1\\ f(1)= (f,g_1)+(f',g_1'),\;\; g \in H^1. $$ You know how to find $g_j$ because both satisfy $$ (f,g_j)+(f',g_j')=0,\;\;\; f\in\mathcal{C}_c^{\infty}(0,1). $$ So $-g_j''+g_j = 0$, which means these functions are linear combinations of $e^{-x}$ and $e^{-x}$. $g_0$ vanishes at $1$ and $g_1$ vanishes at $0$, which leads to $$ g_0(x) = C_0\sinh(x-1),\;\; g_1(x)=C_1\sinh(x), $$ where $C_j$ are constants chosen to normalized the functionals. For example, \begin{align} (f,g_0)+(f',g_0')&=(f,g_0)+fg_0'|_{0}^{1}-(f,g_0') \\ &=fg'|_0^1=-fg'|_0\\ &=f(0)C_0\sinh(1). \end{align} So $C_0=1/\sinh(1)$ does the job.
The resulting functionals $\Phi_0$ and $\Phi_1$ are continuous linear functionals, and $U$ is the null space of $\Phi_0-m\Phi_1$, which makes it closed.