= Lax-Milgram theorem
{c}
The theorem supplies solvability for a coercive <weak formulation>.
A <bounded bilinear form> $a$ on a real <Hilbert space> $H$ that is a <coercive bilinear form> represents every bounded <linear functional> uniquely: for each $\ell\in H'$, there is a unique $u\in H$ with $a(u,v)=\ell(v)$ for every $v\in H$.
For a proof, the <Riesz representation theorem> gives a <bounded linear operator> $T$ and a <vector> $g$ with $a(u,v)=\langle Tu,v\rangle$ and $\ell(v)=\langle g,v\rangle$. Coercivity and the <Cauchy-Schwarz inequality> give $\|Tu\|\geq\alpha\|u\|$, so $T$ is injective and its image is closed: convergence of $Tu_n$ makes $u_n$ a <Cauchy sequence>, whose limit supplies the image limit. If $v$ belongs to the <orthogonal complement> of the image, then $a(u,v)=0$ for every $u$; setting $u=v$ and using coercivity forces $v=0$. Thus the image is also dense and equals $H$. There is consequently a unique $u$ with $Tu=g$, and $\|u\|\leq\|\ell\|/\alpha$. The complex version uses a bounded sesquilinear form with $\operatorname{Re}a(u,u)\geq\alpha\|u\|^2$.
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