First eigenvalue of Laplacian and Poincaré inequality Any idea on how to solve: $\int_{\Omega} |\nabla u|^2 d^n x=\lambda_1\int_{\Omega}u^2 d^n x$, with $u\in H^1_0(\Omega)$ and $\Omega\subset\mathbb{R}^n$, and $\lambda_1$ the first eigenvalue of the negative Laplacian with Dirichlet boundary conditions on $\Omega$ - I am suspecting $u$ is the eigenfunction corresponding to $\lambda_1$, but I am looking for a rigorous proof.
 A: First solution:
Let $S=\{u\in H_0^1(\Omega):\ J(u)=1\}$ where $J:H_0^1(\Omega)\to\mathbb{R}$ is defined by $$J(u)=\int_\Omega |u|^2=\|u\|_2^2$$ Define $I:S\to\mathbb{R}$ by $$I(u)=\int_\Omega |\nabla u|^2=\|u\|_{1,2}^2$$
i - $\lambda_1=\min_{u\in S} I(u)$
ii - If $u\in H_0^1(\Omega)$ satisfies $I(u)=\lambda _1$ then, $u$ is a eigenfunction associated with $\lambda_1$.
To prove ii, note that the tangent plane of $S$ in $u$ is given by $S^\perp_u=\{v\in S:\ \int_\Omega uv=0\}$. Let $\epsilon>0$ be a small number and condider the function $F:(-\epsilon,\epsilon)\times S^\perp_u\to\mathbb{R}$ defined by  $$F(t,v)=\int_\Omega\left|\nabla \left(\frac{u+tv}{\|u+tv\|_2}\right)\right|^2$$
Note that $\frac{u+tv}{\|u+tv\|_2}$ is a parametrization of $S$ in a neighbourhood of $u$. For each fixed $v\in S^{\perp}_u$ we have that $t=0$ is a minimum for the function $F(t,v)$. If you calculate the derivative, you will find that $$\frac{dF(0,v)}{dt}=\int_\Omega \nabla u\nabla v-\lambda_1\int_\Omega uv=0$$
Now you can conclude.
Second solution:
This answer assumes that the eigenvalues of the operator $-\Delta$ were characterized by means of Fredholm alternative and also that $\lambda_1$ is characterized by i. In particular, we have a spectral decomposition of $H_0^1(\Omega)$ and the multiplicity of the eigenvalues are finite.
Assume that $u\in H_0^1(\Omega)$ satisfies $\|u\|_{1,2}^2=\lambda_1$ and $\|u\|_2=1$. Write  $$u=\sum_{i=1}^\infty (u,\varphi_i)\varphi_i$$
where $\varphi_i$ are the eigenfunctions associated with $\lambda_i$ (remember that it is a orthogonal decomposition) and $(\cdot,\cdot)$ denotes inner product in $H_0^1(\Omega)$. Note that $$\lambda_1=\|u\|_{1,2}^2=(u,u)=\sum_{i=1}^\infty \lambda_i(u,\varphi_i)^2\tag{1}$$
On the other hand, remember that $\|u\|_2=\sum_{i=1}^\infty (u,\varphi_i)^2$, therefore $$\lambda_1= \sum_{i=1}^\infty \lambda_1(u,\varphi_i)^2\tag{2}$$
We combine $(2)$ and $(3)$ to conclude that 
$$\sum_{i=1}^\infty (\lambda_i-\lambda_1)(u,\varphi_i)^2=0$$
which implies that $\lambda_i=0$ for $\lambda_i>\lambda_1$. Because $\lambda_1$ has finite multiplicity, we conclude that $u=\sum_{i=1}^m (u,\varphi_i)\varphi_i$, where each $\varphi_i$ with $i=1,...,m$ is an eigenfunction associated with $\lambda_1$, hence $$-\Delta u=-\Delta \left(\sum_{i=1}^m (u,\varphi_i)\varphi_i\right)=-\sum_{i=1}^m (u,\varphi_i)\Delta \varphi_i=\lambda_1 u$$
