Riesz Representation Theorem
Table of Contents
1. Riesz Representation Theorem #
Let $H$ be a Hilbert space over $\mathbb{R}$ or $\mathbb{C}$, and $T$ be a bounded linear functional on $H$ (a bounded operator from $H$ to the field $\mathbb{R}$ or $\mathbb{C}$, where $H$ is defined over that field). The following is known as the Riesz Representation Theorem:
If $T$ is a bounded linear functional on the Hilbert space $H$, then there exists $g \in H$ such that for every $f \in H$, we have: $$ T(f) = \langle f, g \rangle. $$
Moreover, $|T| = |g|$ (here $|T|$ denotes the operator norm of $T$, while $|g|$ is the Hilbert space norm of $g$).
Now, let’s prove this theorem.
Assume that $H$ is separable for now. The proof for any Hilbert space is not much more difficult, but the separable case nicely uses ideas we have developed related to Fourier analysis. Additionally, we will work over $\mathbb{R}$.
Since $H$ is separable, we can choose an orthonormal basis $\phi_j$, $j \geq 1$, for $H$. Let $T$ be a bounded linear functional and set $a_j = T(\phi_j)$. For $f \in H$, set $c_j = \langle f, \phi_j \rangle$, and define $$ f_n = \sum_{j=1}^{n} c_j \phi_j. $$
Since the $\phi_j$ form a basis, we know that $|f - f_n| \to 0$ as $n \to \infty$.
Since $T$ is linear, we have: $$ T(f_n) = \sum_{j=1}^{n} a_j c_j. \tag{1} $$
Since $T$ is bounded, assume with norm $|T| < \infty$, we have: $$ |T(f) - T(f_n)| \leq |T| |f - f_n|. \tag{2} $$
Because $|f - f_n| \to 0$ as $n \to \infty$, we conclude from equations (1) and (2) that: $$ T(f) = \lim_{n\to\infty} T(f_n) = \sum_{j=1}^{\infty} a_j c_j. \tag{3} $$
In fact, the sequence $a_j$ must be square-summable. To see this, first note that since $|T(f)| \leq |T| |f|$, we have: $$ \left|\sum_{j=1}^{\infty} c_j a_j\right| \leq |T| \left(\sum_{j=1}^{\infty} c_j^2\right)^{1/2}. \tag{4} $$
Equation (4) must hold for every square-summable sequence $c_j$ (since any such $c_j$ corresponds to some element in $H$). Fix a positive integer $N$ and define the sequence $c_j = a_j$ for $j \leq N$, $c_j = 0$ for $j > N$. Clearly, such a sequence is square-summable, and equation (4) gives us: $$ \left(\sum_{j=1}^{N} a_j^2\right)^{1/2} \leq |T|. \tag{5} $$
Thus, $a_j$ is square-summable, as the sequence of partial sums is bounded above.
Since $a_j$ is square-summable, the function $g = \sum_{j} a_j \phi_j$ is well-defined as an element of $H$, and $T(f) = \sum_{j} a_j c_j = \langle f, g \rangle$. Finally, equation (5) shows that $|g| \leq |T|$. But from the Cauchy-Schwarz inequality, we also have $|T(f)| = |\langle f, g \rangle| \leq |f| |g|$ or $\frac{|T(f)|}{|f|} \leq |g|$, implying $|T| \leq |g|$, hence $|T| = |g|$. The proof is complete.
2. Application to PDE #
This example illustrates how functional analysis methods are used in PDEs (although the example is for an ODE). Consider the ODE: $$ -f’’(x) + b(x)f(x) = q(x) \tag{6} $$
on the interval $0 < x < 1$, with $b(x) \geq \delta > 0$ for some $\delta$; assume the functions $b$ and $q$ are continuous on $[0, 1]$. We want to find a solution to equation (6) with $f’(0) = f’(1) = 0$ (other boundary conditions could also be applied). If we multiply (6) by a $C^1$ function $\phi$ and integrate the first term, $-f’’\phi$, by parts from $x = 0$ to $x = 1$, we obtain: $$ \int_0^1 (f’(x)\phi’(x) + b(x)f(x)\phi(x)),dx = \int_0^1 q(x)\phi(x),dx. \tag{7} $$
Equation (7) must hold for every $\phi \in C^1([0, 1])$, if $f$ is a $C^2(0, 1)$ solution of equation (6) that is continuous on $[0, 1]$. Conversely, if for a $C^2$ function $f$, we find that (7) holds for every $\phi$, then $f$ must be a solution of equation (6), because if we “undo” the integration by parts in (7), we get: $$ \phi(1)f’(1) - \phi(0)f’(0) + \phi(x)(-f’’(x) + b(x)f(x)) = \phi(x)q(x) $$ for every $\phi$.
A familiar PDE argument then shows that $f’(0) = f’(1) = 0$ and equation (6) must hold.
We will show that there is a unique solution to equation (7). Such a “solution” does not necessarily need to be twice differentiable as required by equation (6), but it will satisfy equation (7). Equation (7) is often called the “weak” form of the problem.
Define an inner product: $$ \langle g, h \rangle = \int_0^1 (g’(x)h’(x) + b(x)g(x)h(x)),dx $$
on the space $C^1([0, 1])$, and let $H$ denote the completion of this space. This is essentially the procedure used on the third problem of the first exam; the presence of $b(x)$ makes no difference. (Note that we must use $b \geq \delta > 0$ to ensure that $\langle \cdot, \cdot \rangle$ is indeed an inner product, so that $|g| = \sqrt{\langle g, g \rangle} = 0$ if and only if $g \equiv 0$.) The space $H$ is a Hilbert space and can be understood (if needed) as a subspace of $C([0, 1])$.
Define a functional $T : H \to \mathbb{R}$ by: $$ T(\phi) = \int_0^1 q(x)\phi(x),dx $$
You can easily check that $T$ is bounded on $H$ (using Cauchy-Schwarz). From the Riesz Representation Theorem, it follows that there must exist a function $f \in H$ such that: $$ T(\phi) = \langle f, \phi \rangle $$
for every $\phi \in H$. This is exactly equation (7), the weak form of the ODE!
The function $f$ satisfying equation (7) lies in $H$. Under the conditions on $b$ (specifically, $b \geq \delta > 0$ and $|b|_\infty < \infty$ since $b \in C([0, 1])$), the function $f$ lies in the same space defined in the third problem of the first exam. Specifically, $f$ is a continuous function. Proving that $f$ is actually twice differentiable requires more work, along with additional assumptions about the function $q$.
References #
[1] (Original) The Riesz Representation Theorem, MA 466, Kurt Bryan