The Representation Theorem and the Lax-Millgram theorem

Theorem (Riesz Representation Theorem). Let $H$ be a Hilbert space and $\phi \in H^*$ . Then there exists unique $u \in H$ such that $\phi(v) = (u,v)$ for all $v \in H$ .

Proof. Let $\phi \in H^*$ and $\ker\phi$ its kernel. Note that $\ker \phi$ is closed since $\phi$ is continuous. If $\phi \equiv 0$ then $u=0$ . If not, choose $u_0 \not \in \ker \phi$ , set

u_1 = \frac{u_0 - P_{\ker \phi} u_0}{|u_0 - P_{\ker \phi} u_0|},

and

u = \phi(u_1) u_1.

Observe that $u \in (\ker \phi)^\perp$ . Thus, for all $v \in H$ ,

\phi(v) = \phi(v - P_u v + P_u v) = \phi(P_u v) = \phi\left(\frac{(u,v)}{(u,u)} u\right) = (u,v).

□

Theorem (Banach Fixed Point). Let $X$ be a nonempty, complete metric space, and $S: X \to X$ a contraction, i.e. there exists $0<\alpha<1$ such that $d(Sx,Sy) \le \alpha d(x,y)$ for all $x,y \in X$ . Then, $S$ has a unique fixed point in $X$ .

Proof. Choose $x \in X$ and consider the sequence $x_n = S^n x$ . For any $1\le n \le m$ we have

d(x_m,x_n) \le d(x_m, x_{m-1}) + \dots + d(x_{n+1},x_n) < d(Sx,x) (\alpha^m + \dots + \alpha^n) < d(Sx,x) \alpha^n \frac{1}{1-\alpha}.

Thus $x_n$ is a Cauchy sequence, and by completeness, $\lim x_n$ exists. Set $u = \lim x_n$ . By continuity of $S$ ,

S u = S\lim x_n = \lim S x_n = \lim x_{n+1} = u,

so $u$ is a fixed point. Finally, if we have another fixed point $v \in X$ , then

d(u,v) = d(Su, Sv) \le \alpha d(u,v)

so $d(u,v) = 0$ .

□

We will now prove the celebrated Lax-Millgram Theorem:

Theorem (Lax-Millgram) Let $a: H \times H \to \mathbb{R}$ be a continuous, coercive, billinear form on $H$ . Given $\phi \in H^*$ , there exists unique $u \in H$ such that

a(u,v) = \phi(v)

for all $v \in H$ . Furthermore, if $a$ is symmetric, then $u$ is the unique element such that

\frac{1}{2} a(u,u) - \phi(u) = \min_{v \in H} \frac{1}{2} a(v,v) - \phi(v).

Definition. A billinear form $a: H \times H \to \mathbb{R}$ is coercive if there exists $\alpha > 0$ such that

a(v,v) \ge \alpha |v|^2 \: \text{for all } v \in H.

We say $a$ is continuous if there exists $C > 0$ such that

a(u,v) \le C |u||v| \: \text{for all } u,v \in H.

Proof. By the Riesz-Representation, there exists $f \in H$ such that $\phi(v)=(f,v)$ for all $v \in H$ . Similarly, for any fixed $u \in H$ the map $v \mapsto a(u,v)$ is a continuous linear functional on $H$ , so there exists a unique $\beta_u \in H$ such that $a(u,v) = (\beta_u, v)$ for all $v \in H$ . Observe that $u \mapsto \beta_u$ is a linear map since $a$ is linear in the first component. Denote this linear map by $A u$ , so

\begin{equation} (Au,u) = a(u,u). \end{equation}

We wish to find a unique $u \in H$ such that

\begin{equation} (f- Au, v) = (f,v) - (Au,v) = \phi(v) - a(u,v) = 0. \end{equation}

For any $\lambda > 0$ we must have

(\lambda f - \lambda Au + u - u, v) = 0 \: \text{for all } v \in H.

Thus,

u = \lambda f - \lambda Au + u.

Hence, it suffices to show the operator $S v = \lambda f - \lambda Av + v$ has a fixed point for some $\lambda > 0$ . We compute

\begin{align*} &|Sv - Sw|^2 = |v-w - \lambda A(v-w)|^2 \\ &=|v-w|^2 - 2\lambda (v-w, A(v-w)) + \lambda^2 |A(v-w)|^2 \\ &\le |v-w|^2 (C^2 \lambda^2 -2\alpha \lambda +1). \end{align*}

If $0 < \lambda< \frac{2\alpha}{C^2}$ , then $S:H \to H$ is a contraction map, so it has a unique fixed point by the Banach fixed point theorem. This establishes part 1 of the theorem.

Suppose also that $a$ is symmetric. Then, $a$ defines an inner product on $H$ . Furthermore, the norms induced by $a$ and the standard inner product are equivalent by continuity and coerciveness. Thus, $H$ is also a Hilbert space under the inner product given by $a$ . By the Riesz Representation Theorem, there exists $g \in H$ such that $\phi(v) = a(g,v)$ for all $v \in H$ . By part 1, $a(g-u,v) = 0$ . But then, $u \in H$ is the unique element such that

a(g-u,g-u) = \min_{v \in H} a(g-v, g-v) = \min_{v \in H} a(g,g) - 2 a(g,v) + a(v,v).

Hence,

\min_{v \in H} \frac{1}{2} a(v,v) - \phi(v) = \frac{1}{2} a(u,u) - \phi(v).

□