Finite-dimensional conforming Ritz values satisfy min–max

Choose an $a$-orthonormal basis ${\psi }_1,\dots ,{\psi }_N$ of $V_h$ by [F3]. For $u_h\in V_h$, define $K_h{u}_h:={\sum }_{i=1}^{N}b(u_h,{\psi }_i){\psi }_i\mathrm{.}$ If $v_h={\sum }_{i=1}^N{c}_i{\psi }_i$, then $a(K_h{u}_h,v_h)={\sum }_{i}b(u_h,{\psi }_i)\overline{c_i}=b(u_h,v_h)$, since $b$ is conjugate-linear in its second argument and the basis is orthonormal. This finite coordinate formula also makes $K_h$ linear in $u_h$.

For $u_h,v_h\in V_h$, $a(K_h{u}_h,v_h)=b(u_h,v_h)=\overline{b(v_h,u_h)}=\overline{a(K_h{v}_h,u_h)}=a(u_h,K_h{v}_h),$ so $K_h$ is self-adjoint in the energy inner product. Also $a(K_h{u}_h,u_h)=b(u_h,u_h)>0$ when $u_h\mathrm{\ne }0$, so $K_h$ is positive and injective. To verify compactness from the compact-closure definition, use the $a$-orthonormal basis ${\psi }_1,\dots ,{\psi }_N$ from step 1.1. The matrix entries $m_{ij}:=a(K_h{\psi }_j,{\psi }_i)$ give, for $x={\sum }_j{c}_j{\psi }_j$, $\mathrm{\parallel }{K}_{h}x{\mathrm{\parallel }}_a^2={\sum }_i\mathrm{\mid }{\sum }_j{m}_{ij}{c}_j{\mathrm{\mid }}^2\le ({\sum }_{i,j}\mathrm{\mid }{m}_{ij}{\mathrm{\mid }}^2){\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$, so $K_h$ is bounded. The coordinate map identifies $V_h$ isometrically with ${\mathbb{R}}^N$ in the real case and ${\mathbb{R}}^{2N}$ in the complex case. The closure of the image of its closed unit ball is therefore closed and bounded in finite-dimensional Euclidean coordinates, hence compact by Heine--Borel [F7]; the unit-ball criterion in [F7] makes $K_h$ compact.

The inner product norm on $V_h$ is complete by [F8], so the compact self-adjoint spectral theorem [F4] applies to $K_h$. It gives an $a$-orthonormal eigenbasis ${\varphi }_{h,1},\dots ,{\varphi }_{h,N}$ with positive eigenvalues ordered ${\mu }_{h,1}\ge \cdots \ge {\mu }_{h,N}>0$. The equation $K_h{\varphi }_{h,j}={\mu }_{h,j}{\varphi }_{h,j}$ is equivalent to the Ritz equation with ${\lambda }_{h,j}={\mu }_{h,j}^{-1}$, so these are the positive Ritz eigenvalues in increasing order. For $v_h={\sum }_{j=1}^N{c}_j{\varphi }_{h,j}$, $a(v_h,v_h)={\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2,b(v_h,v_h)=a(K_h{v}_h,v_h)={\sum }_j{\mu }_{h,j}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\mathrm{.}$

On $W_h:=\mathrm{span}\{{\varphi }_{h,1},\dots ,{\varphi }_{h,m}\}$, ${\sum }_j{\mu }_{h,j}\mathrm{\mid }{c}_j{\mathrm{\mid }}^2\ge {\mu }_{h,m}{\sum }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$, with equality for $v_h={\varphi }_{h,m}$. Thus ${\max }_{v_h\in W_h\setminus \{0\}}a(v_h,v_h)\mathrm{/}b(v_h,v_h)={\mu }_{h,m}^{-1}$, so the minimum in the statement is at most ${\lambda }_{h,m}$.

Let $W_h$ be any $m$-dimensional subspace of $V_h$. The linear map $v_h\mapsto (a(v_h,{\varphi }_{h,1}),\dots ,a(v_h,{\varphi }_{h,m-1}))$ has a nonzero kernel by [F5]; when $m=1$ its target is the zero-dimensional space and the kernel is all of $W_h$. A nonzero vector in that kernel has only coefficients $c_j$ with $j\ge m$ in step 3.1. Since ${\mu }_{h,j}\le {\mu }_{h,m}$ for those indices, its quotient is at least ${\mu }_{h,m}^{-1}$. Hence the maximum on every $W_h$ is at least ${\lambda }_{h,m}$. That maximum is attained by [F6], and the span in step 4.1 attains the outer minimum.