$u_n \rightarrow u$ in $W^{1,2}$ implies $u_n \rightarrow u$ and $u'_n \rightarrow u$ in $L^2$

Question

I report the following excerpt from a textbook:

"By the usual density argument we can find for every $u \in X = \left\{ u \in W^{1,2}:u(-1)=u(1) \text{ and } \int_{-1}^1 u = 0 \right\}$ a sequence $u_n \in X \cap C^2([-1,1])$ so that $$ u_n \rightarrow u \in W^{1,2}(-1,1)\,.$$ Therefore, for every $\varepsilon > 0$, we can find $n$ sufficiently large so that $$\int_{-1}^1 u'\,^2 dx \geq \int_{-1}^1 u_n^2 \, dx - \varepsilon$$ and $$\int_{-1}^1 u_n^2 \, dx\geq \int_{-1}^1 u^2 \, dx - \varepsilon.$$ [...]"

My question is, whence does that conclusion come from? All i know from the convergence in $W^{1,2}$ is that $$\lVert{u_n}\rVert_{L^2}^2 + \lVert{u'_n}\rVert_{L^2}^2 \rightarrow \lVert{u}\rVert_{L^2}^2 + \lVert{u'}\rVert_{L^2}^2.$$

Answer 1

Actually $u_n\to u$ in $W^{1,2}$ is $$ \lVert{u_n-u}\rVert_{L^2}^2 + \lVert{u'_n-u'}\rVert_{L^2}^2 \to 0\ , $$ which is quite different from what you thought.

Answer 2

Convergence in $W^{1,2}$ is much stronger than that. It means $$\|u_n - u\|_{L^2}^2 + \|u_n' - u'\|_{L^2}^2 \to 0.$$