Lax–Milgram theorem

Remarks

The form $B$ need not be symmetric. By the Riesz representation theorem, there are a bounded linear operator $A$ on $H$ with $B(u, v) = \ip{Au}{v}$ and an element $w$ with $f(v) = \ip{w}{v}$. Coercivity gives $\alpha \norm{u} \le \norm{Au}$, so $A$ is injective with closed range. If $v$ is orthogonal to the range, then $B(v, v) = \ip{Av}{v} = 0$, so $v = 0$; hence $A$ is onto and $u = A^{-1} w$. The bound follows from $\alpha \norm{u}^2 \le B(u, u) = f(u) \le \norm{f}_{H'} \norm{u}$. When $B$ is symmetric, the theorem is the Riesz representation theorem for the inner product $B$.

Example: let $U \subset \R^n$ be a bounded open set and $g \in L^2(U)$. On $H = H^1_0(U)$, the Sobolev space, take $B(u, v) = \int_U \nabla u \cdot \nabla v \, dx$ and $f(v) = \int_U g v \, dx$. The Poincaré inequality shows that $B$ is coercive, so there is a unique weak solution $u \in H^1_0(U)$ of $-\Delta u = g$ with $u = 0$ on $\partial U$.

Coercivity cannot be dropped. On $H = \ell^2$, let $S$ be the right shift and $B(u, v) = \ip{Su}{v}$, which is bounded. For $f(v) = \ip{e_1}{v}$, a solution would satisfy $S u = e_1$, which is impossible since the first coordinate of $Su$ is $0$.

Going deeper

For a student

This is the standard existence theorem for weak solutions of linear elliptic equations in divergence form. One checks boundedness from bounds on the coefficients and coercivity from uniform ellipticity together with the Poincaré inequality, possibly after adding a multiple of $u$ to the equation. The proof is also the basis of finite element error analysis. Restricting the problem to a finite-dimensional subspace gives the Galerkin approximation, and the constants $M$ and $\alpha$ control its error. The Babuška version replaces coercivity by an inf-sup condition and allows different trial and test spaces.