User:Caliburn/s/spectral/Herglotz Representation Theorem

Theorem

Let $\mathbb D = \set {z \in \C : \cmod z < 1}$.

Let $C : \mathbb D \to \C$ be a Carathéodory function.

Then there exists a finite Borel measure $\nu$ such that:

$\ds \map C z = i \map \Im {\map C 0} + \int_{\closedint {-\pi} \pi} \frac {e^{i \phi} + z} {e^{i \phi} - z} \rd \map \nu \phi$

Further:

$\map \nu {\closedint {-\pi} \pi} = \map \Re {\map C 0}$

Proof

Let $C : \mathbb D \to \C$ be a Carathéodory function.

Let $r \in \openint 0 1$.

Let $z \in \C$ with $\cmod z < r$.

We first show that:

$\ds \map C z = \frac 1 {4 \pi i} \int_{\cmod \zeta = r} \paren {\frac {\zeta + z} {\zeta - z} + \frac {r^2 \zeta^{-1} + \overline z} {r^2 \zeta^{-1} - \overline z} } \frac {\map C \zeta} \zeta \rd \zeta$

First consider $z = 0$.

Setting $z = 0$ in the right hand side we have:

$\ds \frac 1 {4 \pi i} \int_{\cmod \zeta = r} \paren {\frac \zeta \zeta + \frac {r^2 \zeta^{-1} } {r^2 \zeta^{-1} } } \frac {\map C \zeta} \zeta \rd \zeta = \frac 1 {2 \pi i} \int_{\cmod z = r} \frac {\map C \zeta} \zeta$

From the Cauchy Integral Formula, we have:

$\ds \frac 1 {2 \pi i} \int_{\cmod z = r} \frac {\map C \zeta} \zeta = \map C 0$

Hence the formula holds for $z = 0$.

Now take $z \ne 0$.

Noting that the integrand is meromorphic, we use the Residue Theorem.

The poles of:

$\ds \frac {\zeta + z} {\zeta \paren {\zeta - z} } \map C \zeta$

are located at $\zeta = 0$ and $\zeta = z$, both contained in $\set {\zeta : \cmod \zeta < r}$.

Since $z \ne 0$, these are distinct and the poles are both simple.

By Residue at Simple Pole, we have:

$\ds \Res {\frac {\zeta + z} {\zeta \paren {\zeta - z} } \map C \zeta} 0 = \lim_{\zeta \mathop \to 0} \frac {\zeta + z} {\zeta - z} \map C \zeta = \map C 0$

and:

$\ds \Res {\frac {\zeta + z} {\zeta \paren {\zeta - z} } \map C \zeta} z = \lim_{\zeta \mathop \to z} \frac {\zeta + z} \zeta \map C \zeta = 2 \map C z$

We move to the second term.

We have:

$\ds \frac {r^2 \zeta^{-1} - \overline z} {r^2 \zeta^{-1} + \overline z} = \frac {\frac {r^2} {\overline z} - \zeta} {\frac {r^2} {\overline z} + \zeta}$

We have:

$\ds \cmod {\frac {r^2} {\overline z} } = \frac {r^2} {\cmod z} > \frac {r^2} r = r$

since $\cmod z < r$.

Hence the pole $\dfrac {r^2} {\overline z}$ is not contained in $\set {\zeta : \cmod \zeta \le r}$.

Hence the only pole of:

$\ds \frac {r^2 \zeta^{-1} - \overline z} {\zeta \paren {r^2 \zeta^{-1} + \overline z} } \map C \zeta$

contained in $\set {\zeta : \cmod \zeta \le r}$ is $0$ which is contained in the interior.

We compute the residue:

$\ds \Res {\frac {r^2 \zeta^{-1} - \overline z} {\zeta \paren {r^2 \zeta^{-1} + \overline z} } \map C \zeta} 0 = \lim_{\zeta \mathop \to 0} \frac {r^2 \zeta^{-1} - \overline z} {r^2 \zeta^{-1} + \overline z} \map C \zeta = -\frac {\overline z} {\overline z} \map C 0 = -\map C 0$

By the Residue Theorem, we therefore obtain:

\(\ds \frac 1 {4 \pi i} \int_{\cmod \zeta = r} \paren {\frac {\zeta + z} {\zeta - z} + \frac {r^2 \zeta^{-1} + \overline z} {r^2 \zeta^{-1} - \overline z} } \frac {\map C \zeta} \zeta \rd \zeta\)

\(=\)

\(\ds \frac 1 {4 \pi i} \int_{\cmod \zeta = r} \paren {\frac {\zeta + z} {\zeta - z} } \frac {\map C \zeta} \zeta + \frac 1 {4 \pi i} \int_{\cmod \zeta = r} \paren {\frac {r^2 \zeta^{-1} + \overline z} {r^2 \zeta^{-1} - \overline z} } \frac {\map C \zeta} \zeta \rd \zeta\)

\(\ds \)

\(=\)

\(\ds \frac {2 \pi i} {4 \pi i} \paren {\map C 0 + 2 \map C z} + \frac {2 \pi i} {4 \pi i} \paren {-\map C 0}\)

\(\ds \)

\(=\)

\(\ds \map C z\)

as desired.

By Reciprocal of Complex Number in terms of Conjugate and Modulus, we have:

$\ds \frac {\cmod z^2} z = \overline z$

and hence for $\zeta \in \C$ with $\cmod \zeta = r$ we have:

$\ds \frac {r^2} \zeta = \overline \zeta$

Hence by Product of Complex Conjugates, we have:

$\ds \overline {\frac {\zeta + z} {\zeta - z} } = \frac {\overline {\zeta + z} } {\overline {\zeta - z} } = \frac {r^2 \zeta^{-1} + \overline z} {r^2 \zeta^{-1} - \overline z}$

Hence from Sum of Complex Number with Conjugate, we have:

$\ds \frac 1 {4 \pi i} \int_{\cmod \zeta = r} \paren {\frac {\zeta + z} {\zeta - z} + \frac {r^2 \zeta^{-1} + \overline z} {r^2 \zeta^{-1} - \overline z} } \frac {\map C \zeta} \zeta \rd \zeta = \frac 1 {2 \pi i} \int_{\cmod \zeta = r} \map \Re {\frac {\zeta + z} {\zeta - z} } \frac {\map C \zeta} \zeta \rd \zeta$

Define $\gamma : \closedint {-\pi} \pi \to \set {\zeta : \cmod \zeta = r}$ by:

$\map \gamma \phi = r e^{i \phi}$

for each $\phi \in \closedint {-\pi} \pi$.

By the definition of a contour integral, we obtain:

$\ds \map C z = \frac 1 {2 \pi i} \int_{\cmod \zeta = r} \map \Re {\frac {\zeta + z} {\zeta - z} } \frac {\map C \zeta} \zeta \rd \zeta = \frac 1 {2 \pi i} \int_{-\pi}^\pi \map \Re {\frac {r e^{i \phi} + z} {r e^{i \phi} - z} } \frac {\map C {r e^{i \phi} } } {r e^{i \phi} } \paren {i r e^{i \phi} } \rd \phi = \frac 1 {2 \pi} \int_{-\pi}^\pi \map \Re {\frac {r e^{i \phi} + z} {r e^{i \phi} - z} } \map C {r e^{i \phi} } \rd \phi$

We therefore have, taking the real part:

$\ds \map \Re {\map C z} = \frac 1 {2 \pi} \int_{-\pi}^\pi \map \Re {\frac {r e^{i \phi} + z} {r e^{i \phi} - z} } \map \Re {\map C {r e^{i \phi} } } \rd \phi$

Now, for each Borel set $A \subseteq \R$, define:

$\ds \map {\nu_r} A = \int_A \map {1_{\closedint {-\pi} \pi} } \phi \frac {\map \Re {\map C {r e^{i \phi} } } } {2 \pi} \rd \phi$

Since $C$ is a Carathéodory function and since $r < 1$, we have $\cmod {r e^{i \phi} } < 1$ and hence $\map \Re {\map C {r e^{i \phi} } } \ge 0$.

Hence $\nu_r$ is a measure from Measure with Density is Measure.

From the Change of Measures Formula for Integrals, we have:

$\ds \map \Re {\map C z} = \frac 1 {2 \pi} \int_{-\pi}^\pi \map \Re {\frac {r e^{i \phi} + z} {r e^{i \phi} - z} } \map \Re {\map C {r e^{i \phi} } } \rd \phi = \int_{\closedint {-\pi} \pi} \map \Re {\frac {r e^{i \phi} + z} {r e^{i \phi} - z} } \rd \map {\nu_r} \phi$

We now look to show that the measures $\nu_r$ have:

$\ds \sup_{r \in \openint 0 1} \map {\nu_r} \R < \infty$

The concern is that $\map C {r e^{i \phi} }$ can in principle blow up as $r \to 1^-$.

This blowup is in effect controlled by the finiteness of the integral:

$\ds 0 \le \int_{\closedint {-\pi} \pi} \map \Re {\frac {r e^{i \phi} + z} {r e^{i \phi} - z} } \rd \map {\nu_r} \phi = \map \Re {\map C z} < \infty$

for $\cmod z < r$, as well as its independence of $r$.

Firstly, we compute:

\(\ds \map \Re {\frac {\zeta + z} {\zeta - z} }\)

\(=\)

\(\ds \map \Re {\frac {\paren {\zeta + z} \paren {\overline \zeta - \overline z} } {\paren {\zeta - z} \overline {\zeta - z} } }\)

\(\ds \)

\(=\)

\(\ds \map \Re {\frac {\zeta \overline z + z \overline \zeta - \zeta \overline z - z \overline z} {\cmod {\zeta - z}^2} }\)

Product of Complex Number with Conjugate

\(\ds \)

\(=\)

\(\ds \map \Re {\frac {\cmod \zeta^2 + 2 i \map \Im {z \overline \zeta} - \cmod z^2} {\cmod {\zeta - z}^2} }\)

Product of Complex Number with Conjugate, Difference of Complex Number with Conjugate

\(\ds \)

\(=\)

\(\ds \frac {\cmod \zeta^2 - \cmod z^2} {\cmod {\zeta - z}^2}\)

In particular, with $\zeta = r e^{i \theta}$:

$\ds \map \Re {\frac {r e^{i \phi} + z} {r e^{i \phi} - z} } = \frac {r^2 - \cmod z^2} {\cmod {\zeta - z}^2}$

Set $z = \dfrac 1 2$.

Then we have $\cmod z < r$ and:

$\ds \map \Re {\map C {\frac 1 2} } = \int_{\closedint {-\pi} \pi} \map \Re {\frac {r e^{i \phi} + \frac 1 2} {r e^{i \phi} - \frac 1 2} } \rd \map {\nu_r} \phi$

while by the Triangle Inequality, we have:

$\ds \map \Re {\frac {r e^{i \phi} + \frac 1 2} {r e^{i \phi} - \frac 1 2} } = \frac {r^2 - \frac 1 4} {\cmod {\zeta - z}^2} > \frac {\paren {\frac 3 4}^2 - \paren {\frac 1 2}^2} {\paren {\frac 1 2}^2 + 1} = \frac 1 4$

Hence by Measure is Monotone, we have:

$\ds \map \Re {\map C {\frac 1 2} } = \int_{\closedint {-\pi} \pi} \map \Re {\frac {r e^{i \phi} + \frac 1 2} {r e^{i \phi} - \frac 1 2} } \rd \map {\nu_r} \phi \ge \frac 1 4 \map {\nu_r} {\closedint {-\pi} \pi}$

Hence $\ds \map {\nu_r} \R = \map {\nu_r} {\closedint {-\pi} \pi} \le 4 \map \Re {\map C {\frac 1 2} } < \infty$ for all $\ds r > \frac 3 4$.

On the other hand, from Continuous Function on Compact Space is Bounded, $\map \Re C$ is bounded on $\set {\zeta : \cmod \zeta \le \frac 3 4}$, say by $M$.

We therefore have:

$\ds \map {\nu_r} \R = \int_{-\pi}^\pi \frac {\map \Re {\map C {r e^{i \phi} } } } {2 \pi} \rd \phi \le \frac {2 \pi} {2 \pi} M = M < \infty$

for $\ds r \le \frac 3 4$, by Measure is Monotone.

Hence for all $r \in \openint 0 1$, we have:

$\ds \map {\nu_r} {\closedint {-\pi} \pi} \le \sup \set {4 \map \Re {\map C {\frac 1 2} }, M} < \infty$