9. Riesz representation theorem and symmetric variational problems

Let \(u \in V\). Then, we can define the related continuous linear functional \(l_u(\cdot) \in V^\ast\) by

\[ l_u (v) := (u,v)_V \qquad \forall \, v \in V. \]

The opposite is also true:

Riesz Representation Theorem:

Any continuous linear functional \(l(\cdot)\) on a Hilbert space \(V\) can be represented uniquely as

\[ l(v) = (u_l,v)_V \]

for some \(u_l \in V\). Furthermore, we have

\[ \| l \|_{V^\ast} = \| u_l \|_V. \]

Proof: First, we show uniqueness. Assume that \(u_1 \neq u_2\) both fulfill \(l(v) = (u_1,v)_V = (u_2,v)_V\). This leads to the contradiction

\[\begin{align*} 0 &= l(u_1-u_2) - l(u_1-u_2) \\ &= (u_1,u_1-u_2) - (u_2,u_1-u_2) = \| u_1 - u_2 \|^2. \end{align*}\]

Next, we construct the \(u_l\). For this, define \(S := \operatorname{ker} l\). This is a closed subspace.

Case 1: \(S^\bot = \{ 0 \}\). Then, \(S = V\), i.e., \(l = 0\). So take \(u_l = 0\).

Case 2: \(S^\bot \neq \{ 0 \}\). Pick some \(0 \neq z \in S^\bot\). There holds \(l(z) \neq 0\) (otherwise, \(z \in S \cap S^\bot = \{ 0 \}\)). Now define

\[ u_l := \frac{l(z)}{\|z\|^2} z \qquad \in S^\bot \]

Then

\[\begin{align*} (u_l,v) &= (\underbrace{u_l}_{S^\bot}, \underbrace{v - l(v)/l(z) z}_S) + (u_l,l(v)/l(z)z) \\ &= l(z) / \| z\|^2 (z,l(v)/l(z) z) \\ &= l(v) \end{align*}\]

Finally, we prove \(\|l\|_{V^\ast} = \| u_l \|_V\):

\[ \|l\|_{V^\ast} = \sup_{0 \neq v \in V} \frac{l(v)}{\|v\|} = \sup_v \frac{(u_l,v)_V}{\|v\|_V} \leq \| u_l \|_V \]

and

\[ \| u_l \| = \frac{l(z)}{\|z\|^2} \|z\| = \frac{l(z)}{\|z\|} \leq \| l \|_{V^\ast}. \]

\(\Box\)

9.1. Symmetric variational problems

Take the function space \(C^1(\Omega)\), and define the bilinear form

\[ A(u,v) := \int_\Omega \nabla u \nabla v \, dx + \int_\Gamma u v \, ds \]

and the linear form

\[ f(v) := \int_\Omega f v \, dx \]

The bilinear form is non-negative, and \(A(u,u) = 0\) implies \(u = 0\). Thus \(A(\cdot,\cdot)\) is an inner product, and provides the norm \(\|v\|_A := A(v,v)^{1/2}\). The normed vector space \((C^1, \|.\|_A)\) is not complete. Define

\[ V := \overline{C^1(\Omega)}^{\|.\|_A}, \]

which is a Hilbert space per definition. If we can show that there exists a constant \(c\) such that

\[ f(v) = \int_\Omega f v \, dx \leq c \, \| v \|_A \qquad \forall \, v \in V \]

then \(f(.)\) is a continuous linear functional on \(V\). We will prove this later. In this case, the Riesz representation theorem tells that there exists an unique \(u \in V\) such that

\[ A(u,v) = f(v). \]

This shows that the weak form has a unique solution in \(V\).

Next, take the finite dimensional (\(\Rightarrow\) closed) finite element subspace \(V_h \subset V\). The finite element solution \(u_h \in V_h\) is defined by

\[ A(u_h,v_h) = f(v_h) \qquad \forall \, v_h \in V_h. \]

This means

\[ A(u-u_h, v_h) = A(u,v_h) - A(u_h,v_h) = f(v_h) - f(v_h) = 0. \]

\(u_h\) is the projection of \(u\) onto \(V_h\), this means that the error is minimal within all elements from the finite element space:

\[ \| u - u_h \|_A \leq \| u - v_h \|_A \qquad \forall \, v_h \in V_h \]

The error \(u - u_h\) is orthogonal to \(V_h\).