Intuitive understanding of Maximin Principle From the the book page $324$, does someone could explain to me the Theorem $2$. Maximin principle? I have a bit of difficulties to well understand how works this theorem. A simple example would be appreciate in the explication, if possible.

Theorem $2$. Maximin principle : Fix a positive integer $n \geq 2$. Fix $n-1$ arbitrary trial functions $y_1(x), \dots, y_{n-1}(x)$.
  Let $$\lambda_{n*}=\min \frac{\| \nabla w\|^2}{\|  w\|^2}$$ among all
  trial functions $w$ that are orthogonal to $y_1, \dots, y_{n-1}.$ Then
  $$\lambda_n= \max \lambda_n*$$ over all choices of the $n-1$ trial
  functions $y_1, \dots, y_{n-1}.$

 A: Let's assume we're on a bounded domain $\Omega\subset\mathbb{R}^N$ and we've fixed boundary conditions. Recall that the Laplacian $\Delta$ induces an orthogonal decomposition of $L^2(M)$ into eigenspaces $E_k$, where $E_k$ is associated to the $k^{th}$ eigenvalue $\lambda_k$, and $\Delta$ acts by scaling on $E_k$. Write $0\leq \lambda_1\leq\lambda_2\leq\cdots$ where we do not account for multiplicity. (For example, the Neumann spectrum of the unit square $[0,1]\times[0,1]$ would be $0\leq\pi^2\leq\pi^2\leq 4\pi^2\leq\cdots$.) Further, let $u_i$ be an eigenfunction corresponding to $\lambda_i$, so that $\langle u_i,u_j\rangle = \delta_{ij}$ where $\langle\cdot,\cdot\rangle$ is the $L^2$ inner product.

Think of this statement in terms of $(n-1)$-dimensional linear subspaces of $L^2(\Omega)$. Let $V$ be such a subspace. The intuition is dimension counting: the perpendicular complement of $V$ must "bleed into" eigenspaces corresponding to low-frequency eigenvalues, and the best you can do is rule out the lowest $n-1$ frequencies.

Let's formalize it this way. First, call the ratio $\mathcal{R}(v) = \|\nabla v\|^2/\|v\|^2$ the Rayleigh quotient of $v$, if it exists, for $v\in L^2$. Let $V$ be the span of $y_1,\ldots,y_{n-1}$. We're going to search for functions $v\in V^\perp$ which are also elements of $\oplus_{k\leq n} E_k$. Such a function must satisfy
$$ v = \sum_{j=1}^n a_ju_j $$
and
$$ \langle v,y_i\rangle = 0 $$
for all $i=1,\ldots,n-1$. If let $C$ be the matrix with entries $c_{ij} = \langle y_i,u_j\rangle$, we are searching for a solution $A = (a_j)$ to the equation $CA = 0$. As this system has more unknowns than equations, by rank-nullity the matrix $C$ must have a nontrivial kernel.
Denote by $\mu$ the infimum of $\mathcal{R}$ on $V^\perp$. Then 
$$ \mu\|v\|^2 \leq \|\nabla v\|^2 = \sum_{j=1}^n \lambda_ja_j^2 \leq \lambda_n\|v\|^2 $$
(Verify this for yourself with an integration by parts.)
Therefore we have $\mu\leq\lambda_n$. The subspace $V$ spanning $\{u_1,\ldots,u_{n-1}\}$ achieves equality. This establishes the desired result.
