In the image below, the space $X$ is $W^{1,2}_0(\Omega)$, and $\Omega$ is a bounded open set of $\mathbb{R}^N$ so that the Sobolev imbeddings hold.
It is from the article of G. Auchmuty – Dual Variational Principles for Eigenvalue Problems.
I don't quite understand the part in red. In particular, why do we need the imbedding to be compact?
I would argue that you don't really need that, since the $L^2$-norm, being a norm, is a convex function, and it is clearly strongly continuous on $W^{1,2}_0(\Omega)$, hence it should also be weakly l.s.c., without any need for compactness.
Edit: as requested, the functional $f_1$ and $f$ are defined as follows
where $\mu$ is any positive real number.
Best Answer
This is a standard consequence of Rellich's compactness theorem; weak $W^{1,2}$ convergence implies strong $L^2$ convergence. Note the convexity argument is insufficient here, as you want weak continuity and not weak lower semicontinuity.
To see this, suppose $u_n \rightharpoonup u$ in $W^{1,2}_0(\Omega),$ so in particular we have $\{u_n\}$ is uniformly bounded in $W^{1,2},$ that is $$ \sup_{n} \lVert u \rVert_{W^{1,2}(\Omega)} < \infty. $$ Hence by the compact embedding, there is a strongly convergence subsequence $u_{n_k} \to v$ in $L^2(\Omega).$ In particular $u_n \rightharpoonup u$ and $u_n \rightharpoonup v$ weakly in $L^2(\Omega),$ so by uniqueness $u=v.$ This shows that every subsequence of $\{u_n\}$ admits a strongly $L^2$-convergence subsequence to $u$ in $L^2(\Omega),$ so strong $L^2$-convergence follows.
Strictly speaking I have only argued sequential continuity, but I assume this will suffice for your purposes.