Let $\Omega\subseteq\mathbb R^n$ be a bounded domain and $$H:=W_0^{1,2}(\Omega):=\left\{u\in L^2(\Omega):\nabla u\in L^2(\Omega)\right\}$$ be the Sobolev space, where $\nabla u$ denotes the weak derivative. Moreover, let $$\langle u,v\rangle_{1,2}:=\langle u,v\rangle_2+\langle\nabla u,\nabla v\rangle_2\;\;\;\text{for }u,v\in H\;,$$ where $\langle\;\cdot\;,\;\cdot\;\rangle_2$ is the scalar product in $L^2(\Omega)$. Let's consider the Rayleigh quotient $$R(u):=\frac{|u|_2^2}{\left\|u\right\|_2^2}\;\;\;\text{for }u\in H\setminus\left\{0\right\}\;,$$ where $|u|_2^2:=\langle\nabla u,\nabla u\rangle_2$ and $\left\|\cdot\right\|_2$ is the norm in $L^2(\Omega)$.
It's easy to verify, that there exists a minimizing sequence $(u_k)_{k\in\mathbb N}\subseteq H$ with $\left\|u_k\right\|_2=1$, $$\lim_{k\to\infty}R(u_k)=\lambda_1:=\inf_{H\setminus\left\{0\right\}}R$$ and $R(u_k)\le \lambda_1+1$. Let $\left\|\;\cdot\;\right\|_{1,2}$ be the norm induced by $\langle\;\cdot\;,\;\cdot\;\rangle_{1,2}$. By the Poincaré inequality, there is some $C>0$ such that $$\left\|u\right\|_2\le C|u|_2^2\;\;\;\text{for all }u\in H\;.\tag{1}$$ From $(1)$ it's easy to see, that $(u_k)_{k\in\mathbb N}$ is bounded with respect to $\left\|\;\cdot\;\right\|_{1,2}$. It's a well-known fact, that a bounded sequence in a Hilbert space is weakly convergent, i.e. there is some $u_0\in H$ such that $$\langle u_k,v\rangle_{1,2}\stackrel{k\to\infty}{\to}\langle u_0,v\rangle_{1,2}\;\;\;\text{for all }v\in H\;.$$ But, what I need is strong convergence in $L^2(\Omega)$, i.e. $$\lim_{k\to\infty}\left\|u_k-u_0\right\|_2=0\tag{2}\;.$$ Again, by a well-known fact, $(2)$ would follow from the weak convergence, if $$\limsup_{k\to\infty}\left\|u_k\right\|_{1,2}\le\left\|u_0\right\|_{1,2}\;.\tag{3}$$ However, I can't figure out how we can show $(3)$.
Weak convergence in $H=H^1_0$ implies strong convergence in $L^2$ because $H^1_0$ is compactly embedded in $L^2$, see: https://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem