The question regrading to density argument in analysis.

If you are talking about Evans & Gariepy, "Measure Theory and Fine Properties of Functions", then either you have a different edition than I have, or your page-number is of, because my version of this book only has 256 pages (before the bibliography).

Anyway, the result which I think you are looking for is the following:

Let $(T_n)_n$ be a sequence of linear maps $T_n : X \to Y$ which are uniformly bounded in the sense that $\Vert T_n \Vert \leq C$ for some fixed $C>0$ and all $n$.

Furthermore, assume that $M \subset X$ is dense with $T_n m \to T m$ for all $m \in M$, where $T : X \to Y$ is linear and bounded.

Then $T_n x \to Tx$ holds for all $x \in X$.

For the proof, let $x \in X$ and $\varepsilon > 0$ be arbitrary. By density, there is some $m \in M$ with $\Vert x - m\Vert < \varepsilon/[3(1+C + \Vert T \Vert)]$.

For this $m$, there is some $n_0 \in \Bbb{N}$ with $\Vert T_n m - Tm \Vert < \varepsilon/3$ for all $n \geq n_0$.

Hence,

\begin{eqnarray*} \Vert T_n x - Tx \Vert & \leq& \Vert T_n x - T_n m \Vert + \Vert T_n m - T m \Vert + \Vert T m - Tx \Vert \\ & \leq & C \cdot \Vert x - m\Vert + \varepsilon/3 + \Vert T \Vert \cdot \Vert x - m\Vert \leq \varepsilon \end{eqnarray*}

for all $n \geq n_0$. $\square$

To apply this in the first setting, you have to know that the sequence $(\mu_n)_n$ is bounded. In the second problem, you mention that the sequence $(u_n)_n$ is bounded in $H^1$, so that there is no problem.

In the first setting, $X = C(\Bbb{R}^n, \Bbb{R}^n)$, $Y = \Bbb{R}$ (or $\Bbb{C}$) and $M = C_c^1 (\Bbb{R}^n, \Bbb{R}^n)$, which is dense in $X$. Finally, identify each measure $\mu_n$ with the functional it induces on $X$.

In the second setting, you can take $X = H^1$, $Y = \Bbb{R}$ and $M = C_c^\infty ( \Bbb{R}^n)$. Again, identify each $u_n \in H^1$ with the linear functional it induces on $H^1$.

By the way, in the second setting, you can also apply a different (not necessarily simpler) argument: Assume that $u_n \not\to 0$ weakly in $H^1$. Then there is some $h \in H^1$ with $\langle u_n, h\rangle_{H^1} \not\to 0$, so that there is some $\varepsilon > 0$ and a subsequence $(u_{n_k})_k$ with $|\langle u_{n_k}, h\rangle_{H^1}| \geq \varepsilon$ for all $k$.

But the $H^1$ bounded sequence $(u_{n_k})_k$ has a weakly convergent subsequence $u_{n_{k_\ell}} \to v$ weakly in $H^1$. Because of $|\langle v, h\rangle_{H^1}| \leftarrow |\langle u_{n_{k_\ell}}, h\rangle_{H^1}| \geq \varepsilon$, we see $v \neq 0$.

But the embedding $H^1 \hookrightarrow L^2$ yields $u_{n_{k_\ell}} \to v$ weakly in $L^2$. But you claim that you already know $u_n \to 0$ weakly in $L^2$. By uniqueness of weak $L^2$ limits, $v = 0$, a contradiction.