= Solution
{c}
The <Lax-Milgram theorem> assumes that $H$ is a real <Hilbert space> and $B:H\times H\to\mathbb R$ is a <bounded bilinear form> and a <coercive bilinear form>: there are $M,\alpha>0$ such that
$$
|B[u,v]|\leq M\|u\|_H\|v\|_H,\qquad
B[u,u]\geq\alpha\|u\|_H^2.
$$
Then for every <continuous linear functional> $\ell\in H'$ there is a unique $u\in H$ satisfying
$$
\boxed{B[u,v]=\ell(v)\quad(v\in H),\qquad
\|u\|_H\leq\alpha^{-1}\|\ell\|_{H'}.}
$$
Symmetry of $B$ is not required.
For the proof, the <Riesz representation theorem> provides a <bounded linear operator> $T:H\to H$ and $g\in H$ with $B[u,v]=(Tu,v)_H$ and $\ell(v)=(g,v)_H$. Coercivity and the <Cauchy-Schwarz inequality> imply
$$
\alpha\|u\|_H^2\leq(Tu,u)_H\leq\|Tu\|_H\|u\|_H,
\qquad\|Tu\|_H\geq\alpha\|u\|_H.
$$
Thus $T$ is injective and its range is closed: if $Tu_j$ converges, the displayed inequality makes $u_j$ a <Cauchy sequence>; <completeness> gives $u_j\to u$ and $Tu_j\to Tu$. If $w$ lies in the <orthogonal complement> of the range, then $B[v,w]=0$ for all $v$. Taking $v=w$ gives $\alpha\|w\|^2\leq B[w,w]=0$, so $w=0$. A closed subspace with zero <orthogonal complement> is all of $H$, hence $T$ is surjective. The unique solution is $T^{-1}g$, and the same inequality gives the asserted norm bound.
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