The Lax--Milgram theorem

Assume first $H\mathrm{\ne }\{0\}$. Then the given bound and coercivity constant satisfy $\alpha \le M$ and $M>0$: choosing $u\mathrm{\ne }0$ and normalising, $\alpha \le \mathrm{Re}a(u,u)\le \mathrm{\mid }a(u,u)\mathrm{\mid }\le M\mathrm{\parallel }u{\mathrm{\parallel }}^2$, so $M\ge \alpha >0$; hence $\rho :=\alpha \mathrm{/}{M}^2$ satisfies $0<\rho <2\alpha \mathrm{/}{M}^2$.

Uniqueness: if $u$ satisfies $a(u,v)=0$ for every $v$, then testing $v=u$ gives $\alpha \mathrm{\parallel }u{\mathrm{\parallel }}^2\le \mathrm{Re}a(u,u)=0$, so $u=0$. If $u_1,u_2$ are two solutions of $a(u,\cdot )=F$, then first-slot linearity gives $a(u_1-u_2,v)=0$ for all $v$, so $u_1=u_2$.

The conjugate functional: $G(v):=\overline{F(v)}$ is a bounded linear functional with $\mathrm{\parallel }G\mathrm{\parallel }=\mathrm{\parallel }F\mathrm{\parallel }$, so by Riesz representation there is a unique $w\in H$ with $G(v)=(v,w)$ for all $v$, that is $F(v)=(w,v)$ for all $v$, and $\mathrm{\parallel }w\mathrm{\parallel }=\mathrm{\parallel }F\mathrm{\parallel }$.

Existence: fix $\rho$ as in step 1.1 and define $T(u):=u-\rho (Au-w)$. By [F3] the map $T$ is a strict contraction of the complete space $H$ with constant $q_{\rho }<1$, so by the Banach fixed-point theorem it has a fixed point $u\in H$; then $u=u-\rho (Au-w)$ gives $Au=w$, and hence $a(u,v)=(Au,v)=(w,v)=F(v)$ for every $v\in H$.

Estimate: for the solution $u$ of step 3.1, $\alpha \mathrm{\parallel }u{\mathrm{\parallel }}^2\le \mathrm{Re}a(u,u)=\mathrm{Re}F(u)\le \mathrm{\mid }F(u)\mathrm{\mid }\le \mathrm{\parallel }F\mathrm{\parallel }\mathrm{\parallel }u\mathrm{\parallel }$; if $u\mathrm{\ne }0$ divide by $\mathrm{\parallel }u\mathrm{\parallel }$ to get $\alpha \mathrm{\parallel }u\mathrm{\parallel }\le \mathrm{\parallel }F\mathrm{\parallel }$, and if $u=0$ the inequality holds trivially.

Degenerate space: if $H=\{0\}$, then the only element is $0$, the only functional is $0$ and it is the value at the unique solution $u=0$ with $\alpha \mathrm{\parallel }u\mathrm{\parallel }=0=\mathrm{\parallel }F\mathrm{\parallel }$; uniqueness is immediate. The real bilinear case is the same argument with conjugation read as the identity, and $a$ need not be symmetric.