I have the following eigenvalue problem: \begin{equation} \begin{cases} -u''=\lambda u\\ u(0)=u(\pi)=0 \end{cases} \end{equation} and I have to prove the following min-max priciple for eigenvalues: \begin{equation} \lambda_k=\min_{S_k}\max_{u\in S_k} R(u) \ \ \ S_k\subset H^1_0(0,\pi) \ \dim(S_k)=k \end{equation} where $\lambda_k$ denotes the $k$-th eigenvector and $R(u)=\dfrac{(u',u')}{(u,u)}$ is the Rayleigh quotient.
Can someone give me an hint on how to derive this proof?
MY ATTEMPT: my idea is to use that the spectral theorem applied to the second order derivative operator $L$ asserts that there exist an orthonormal basis of $H^1_0$ of eigenfunctions $\{u_n\}$ with correspondinfg eigenvalue $\{\lambda_n\}$. Then we can observe that for any $u\in H^1_0$: \begin{equation} \|u\|^2_{L^2}=\langle u,u\rangle=\langle\sum_{n=1}^\infty\langle u,u_n\rangle u_n,u\rangle=\sum_{n=1}^\infty|\langle u,u_n\rangle|^2 \end{equation} Now we have that assuming $u$ to be twice differentiable: \begin{equation} \|u'\|^2_{L^2}=\int_0^\pi u' u' =-\int_0^\pi u'' u =\langle-\bigg(\sum_{n=1}^\infty\langle u,u_n\rangle u_n\bigg)'',u\rangle=\sum_{n=1}^\infty\lambda_n|\langle u,u_n\rangle|^2 \end{equation} Assume now that $u\in H_{k-1}=H^1_0 \cap \text{span}\{u_1,...,u_{k-1}\}^\bot$ so that $\langle u,u_j\rangle=0$ for $j=1,...,k-1$, then: \begin{equation} \|u'\|^2_{L^2}=\sum_{n=k}^\infty\lambda_n|\langle u,u_n\rangle|^2\geq\lambda_k\sum_{n=k}^\infty|\langle u,u_n\rangle|^2=\lambda_k\sum_{n=1}^\infty|\langle u,u_n\rangle|^2=\lambda_k\|u\|^2_{L^2} \end{equation} finally this implies that for any $u\in H_{k-1}$ and $u\neq 0$ we have: \begin{equation} R(u)\geq\lambda_k \Rightarrow \inf_{u\in H_{k-1}}R(u)\geq \lambda_k \end{equation} Now in particular for $u=u_k$ we obtain $R(u)=\lambda_k$ so we actually have the equality: $\min_{u\in H_{k-1}}R(u)=\lambda_k$. Similarly it is possible to prove that: \begin{equation} \max_{u\in\text{span}\{u_1,...,u_k\}} R(u)=\lambda_k \end{equation} Now we denote with $\Phi_k(H^1_0)=\{k-\text{dimensional subspaces of } H^1_0\}$, now since $\text{span}\{u_1,...,u_k\}\in\Phi_k(H^1_0)$ it must be that: \begin{equation} \lambda_k=\max_{u\in\text{span}\{u_1,...,u_k\}} R(u)\geq\min_{X\in\Phi_k(H^1_0)}\max_{u\in X}R(u) \end{equation} conversely we have that $\forall X\in\Phi_k(H^1_0)$ there exist $v\neq 0$ that belongs to $H_{k-1}\cap X$ and then: \begin{equation} R(v)\geq\min_{u\in H_{k-1}}R(u)=\lambda_k\Rightarrow\max_{v\in X}R(v)\geq\lambda_k\Rightarrow\min_{X\in\Phi_k(H^1_0)}\max_{v\in X}R(v)\geq\lambda_k \end{equation} that concludes the proof.
Does this proof look fine?