= Solution
Let $V$ be a real <Hilbert space>, $a:V\times V\to\mathbb R$ a <bounded bilinear form>, and $\ell\in V'$ a <bounded linear functional>. Assume constants $C<\infty$, $\alpha>0$ satisfy
$$
|a(v,w)|\le C\|v\|_V\|w\|_V,\qquad a(v,v)\ge\alpha\|v\|_V^2\qquad(v,w\in V).
$$
The second hypothesis says that $a$ is a <coercive bilinear form>. The <Lax-Milgram theorem> then supplies a unique <weak solution> $u\in V$ with $a(u,v)=\ell(v)$ for every $v\in V$, and $\|u\|_V\le\|\ell\|_{V'}/\alpha$.
For a finite-dimensional trial <linear subspace> $V_h\subset V$, the same boundedness and coercivity constants apply to the restricted form. Since a <finite-dimensional subspace is closed>, $V_h$ is itself a <Hilbert space>. Thus there is a unique <Galerkin method> solution $u_h\in V_h$ satisfying $a(u_h,v_h)=\ell(v_h)$ for every $v_h\in V_h$. The <Galerkin orthogonality> $a(u-u_h,v_h)=0$ gives the <Céa lemma>:
$$
\boxed{\|u-u_h\|_V\le\frac C\alpha\inf_{v_h\in V_h}\|u-v_h\|_V}.
$$
In particular, approximation by the trial spaces implies <numerical convergence>. For a complex <Hilbert space>, replace bilinearity by a <sesquilinear form>, take $\ell$ antilinear in the test variable, and require $\operatorname{Re}a(v,v)\ge\alpha\|v\|_V^2$.
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