Existence, uniqueness, linearity, and boundedness of the Ritz projection

If $V_h=\{0\}$, define $R_{h}u=0$; the Ritz equation, uniqueness, and norm bound are immediate. Otherwise fix a finite ordered basis $(e_1,\dots ,e_m)$ of $V_h$. Form the square matrix $K_{ij}=a(e_j,e_i)$ and vector $b_i=a(u,e_i)$. If $Kc=0$ and $x={\sum }_j{c}_j{e}_j$, then $a(x,e_i)=0$ for each $i$, hence $a(x,x)={\sum }_i{c}_{i}a(x,e_i)=0$. Coercivity [F2] gives $x=0$, so basis independence gives $c=0$. Thus $K$ has zero kernel and, being square, is invertible. Its unique solution $Kc=b$ defines the unique $R_{h}u$ satisfying [F1].

Testing [F1] with $v_h=R_{h}u$ gives $\alpha \mathrm{\parallel }{R}_{h}u{\mathrm{\parallel }}_V^2\le a(R_{h}u,R_{h}u)=a(u,R_{h}u)\le M\mathrm{\parallel }u{\mathrm{\parallel }}_V\mathrm{\parallel }{R}_{h}u{\mathrm{\parallel }}_V\mathrm{.}$ If $R_{h}u=0$, the bound is immediate; otherwise divide by $\alpha \mathrm{\parallel }{R}_{h}u{\mathrm{\parallel }}_V$.

For $u,w\in V$ and real scalars $c,d$, bilinearity shows that $c{R}_{h}u+d{R}_{h}w$ satisfies [F1] for $cu+dw$. Uniqueness from step 1.1 gives $R_h(cu+dw)=c{R}_{h}u+d{R}_{h}w$. The norm bound from step 2.1 proves boundedness, so $R_h$ is the required linear map.