g'(\lambda) &= \int_{\R_+} \frac{1}{p^2+1} \left[\frac{2\lambda p^2}{1+\lambda^2p^2} - \frac 2{\lambda+1}\right] dp
\\&= \int_{\R_+} \frac{2\lambda}{\lambda^2-1}\left[\frac{1}{p^2+1} - \frac{1}{1+\lambda^2p^2}\right] - \frac 2{\lambda+1}\frac{1}{p^2+1} \,dp
\\&= \left[\frac{2\lambda}{\lambda^2-1} - \frac{2}{\lambda+1}\right]\int_{\R_+} \frac{1}{p^2+1} \,dp - \frac{2}{\lambda^2-1} \int_{\R_+} \frac{\lambda}{1+\lambda^2p^2} \,dp
\\&=\frac{2}{\lambda^2-1} \int_{\R_+} \frac{1}{p^2+1} \,dp - \frac{2}{\lambda^2-1} \int_{\R_+} \frac{1}{1+p^2} \,dp = 0.
\end{align*}

\begin{align*}
g'(1) &= \int_{\R_+} \frac{1}{p^2+1} \left[\frac{2p^2}{1+p^2} - \frac 2{1+1}\right] dp = \int_{\R_+} \frac{p^2-1}{(p^2+1)^2} \,dp
\\&= \int_\R \frac{e^{2x}-1}{(e^{2x}+1)^2} \cdot e^x \,dx = \int_\R \frac{e^x-e^{-x}}{(e^x+e^{-x})^2} \,dx = 0,
\end{align*}

where the last equality follows from the fact that the integrand is skew-symmetric. This shows that \(g' \equiv 0\) and therefore \(g\) is constant. Since

\begin{align*}
g(0) = \int_{\R_+} \frac{1}{p^2+1} \left[\log\left(1+0\right) - 2\log\left(0+1\right)\right] dp = 0
\end{align*}

this implies that \(g \equiv 0\). This yields that, for \(\lambda \in \R_+\), we have

\[0 = g(\lambda) = \int_{\R_+} \frac{1}{p^2+1} \left[\log\left(1+\lambda^2p^2\right) - 2\log\left(\lambda+1\right)\right] dp,\]

\[\int_{\R_+} \frac{1}{p^2+1} \log\left(1+\lambda^2p^2\right) dp = 2\log\left(\lambda+1\right) \int_{\R_+} \frac{1}{p^2+1} dp = \pi \log\left(\lambda+1\right)\]

and therefore

\begin{align*}
\int_\R \frac{\lambda}{p^2+\lambda^2} \log\left(1+p^2\right) dp &= 2\int_{\R_+} \frac{\lambda}{p^2+\lambda^2} \log\left(1+p^2\right) dp
\\&= 2\int_{\R_+} \frac{\lambda^2}{(\lambda p)^2+\lambda^2} \log\left(1+(\lambda p)^2\right) dp
\\&= 2 \int_{\R_+} \frac{1}{p^2+1} \log\left(1+\lambda^2p^2\right) dp = 2\pi \log\left(\lambda+1\right).
\end{align*}

This, by lem:OuterSymmetric, yields

\begin{align*}
\Out(K)(i\lambda) &= \exp\left(\frac{1}{\pi} \int_\R \frac{\lambda}{p^2+\lambda^2} \log\left(1+p^2\right) dp\right)
\\&= \exp\left(2 \log\left(\lambda+1\right)\right) = (\lambda+1)^2 = -(i\lambda+i)^2, \quad \lambda \in \R_+.
\end{align*}

This shows that the holomorphic functions \(\Out(K)\) and \(z \mapsto -(z+i)^2\) coincide on \(i\R_+\) and are therefore equal.

* By (a), (d) and lem:OuterHomo we have

\[\Out(K)(z) = \frac{-iz}{-(z+i)^2} = \frac {iz}{(z+i)^2}, \quad z \in \C_+.\]

* By lem:OuterHomo and (d), for \(z \in \C_+\), we have

\[\Out(K)(z) \cdot \Out(K)(z) = \frac 1{-(z+i)^2},\]

\[\Out(K)(z) = \pm \frac i{z+i}.\]

The statement then follows by lem:OuterSymmetric since

\[+ \frac {i}{i+i} = \frac 12 \in \R_+. \qedhere\]

§.§ Inner functions

In this section we will introduce the concept of inner functions and present some of their basic properties:

Let \(\cH\) be a complex Hilbert space.

* We call a function \(\phi \in H^\infty(\C_+,B(\cH))\) an inner fuction, if \(\phi(x) \in \U(\cH)\) for almost every \(x \in \R\).

* We write \(\gls*{Inn}\) for the set of inner functions.

* We set \(\Inn(\C_+) \coloneqq \Inn(\C_+,B(\C))\), i.e. \(\Inn(\C_+)\) is the set of functions \(\phi \in H^\infty(\C_+)\) with \(\left|\phi(x)\right| = 1\) for almost every \(x \in \R\).

The following theorem links Hardy spaces, outer functions and inner functions:
Let \(f \in H^2(\C_+) \setminus \{0\}\). Then there exists an inner function \(\phi \in \Inn(\C_+)\) and an outer function \(F \in \Out^2(\C_+)\) such that

\[f = \phi \cdot F.\]

We now consider a special class of inner functions, namely the so called Blaschke products:

For \(\omega \in \C_+\), we define the Blaschke factor \(\gls*{phio} \in \Inn(\C_+)\) by

\[\phi_\omega(z) \coloneqq \frac{z-\omega}{z-\overline{\omega}}, \quad z \in \C_+.\]

A finite Blaschke product is a function \(\phi \in \Inn(\C_+)\) of the form

\[\phi = c \cdot \prod_{n = 1}^N \phi_{\omega_n}\]

with \(c \in \T\), \(N \in \N_0\) and \(\omega_1,\dots,\omega_N \in \C_+\).

By construction the zeros of a finite Blaschke product \(\phi = c \cdot \prod_{n = 1}^N \phi_{\omega_n}\) are precisely the numbers \(\omega_1,\dots,\omega_N\) counted with multiplicity.

Let \(\omega \in \C_+\) and \(f \in H^2(\C_+)\) with \(f(\omega) = 0\). Then there exists \(g \in H^2(\C_+)\) such that

\[f = \phi_\omega \cdot g.\]

If \(f = 0\) the statement follows by choosing \(g = 0\). Therefore, we can assume that \({f \neq 0}\). Then, by thm:OuterInnerDecomp, there exists an inner function \(\phi \in \Inn(\C_+)\) and an outer function \({F \in \Out^2(\C_+)}\) such that

\[f = \phi \cdot F.\]

Since \(F(\C_+) \subeq \exp(\C) = \C^\times\), we have \(F(\omega) \neq 0\) and therefore \(\phi(\omega) = 0\). This implies that there exists a function \(m \in \cO(\C_+)\) such that \(\phi(z) = (z-\omega) \cdot m(z)\), \(z \in \C_+\). Then, defining \(h \in \cO(\C_+)\) by \(h(z) \coloneqq (z-\overline{\omega}) \cdot m(z)\), \(z \in \C_+\), we have

\[\phi = \phi_\omega \cdot h.\]

Since \(h\) is holomorphic and therefore continuous, for \(\epsilon \in (0,\Im(\omega))\), we have

\[\sup_{z \in U_{\leq \epsilon}(\omega)} |h(z)| < \infty.\]

Further, for every \(z \in \C_+ \setminus U_{\leq \epsilon}(\omega)\), one has

\begin{align*}
|h(z)| &= \left|\frac{z-\overline{\omega}}{z-\omega} \cdot \phi(z)\right| \leq \left|\frac{z-\overline{\omega}}{z-\omega}\right| \cdot \left\lVert \phi\right\rVert_\infty = \left|1 + \frac{\omega-\overline{\omega}}{z-\omega}\right| \cdot 1 \leq 1 + \frac{|\omega-\overline{\omega}|}{|z-\omega|} \leq 1 + \frac{2 \Im(\omega)}{\epsilon},
\end{align*}

so also

\[\sup_{z \in \C_+ \setminus U_{\leq \epsilon}(\omega)} |h(z)| < \infty.\]

This shows that \(h \in H^\infty(\C_+)\). Then, setting

\[g \coloneqq h \cdot F \in H^\infty(\C_+) \cdot \Out^2(\C_+) \subeq H^2(\C_+),\]

we get

\[f = \phi \cdot F = \phi_\omega \cdot h \cdot F = \phi_\omega \cdot g. \qedhere\]

§.§ Pick functions

After considering inner functions, in this section we want to take a closer look at the infinitesimal generators of one-parameter semigroups

\[\phi: \R_{\geq 0} \to \Inn(\C_+,B(\cH)).\]

We will see that these infinitesimal generators are precisely operator-valued Pick functions, defined as follows:

Let \(\cH\) be a finite dimensional complex Hilbert space.

* We call a function \({F \in \cO(\C_+,B(\cH))}\) an operator-valued Pick function if

\[\Im(F(z)) \coloneqq \frac 1{2i} (F(z)-F(z)^*) \geq 0 \quad \forall z \in \C_+.\]

* We write \(\gls*{Pick}\) for the set of operator-valued Pick functions.

* We write \(\gls*{PickR}\) for the set of Pick functions \(F \in \mathrm{Pick}(\C_+,B(\cH))\), for which the limit

\[\gls*{Fstar}(x) \coloneqq \lim_{\epsilon \downarrow 0} F(x+i\epsilon)\]

exists for almost every \(x \in \R\) and is self-adjoint. We say such a Pick function has self-adjoint boundary values.

* We set

\[\mathrm{Pick}(\C_+) \coloneqq \mathrm{Pick}(\C_+,B(\C)) \quad \text{and} \quad \mathrm{Pick}(\C_+)_\R \coloneqq \mathrm{Pick}(\C_+,B(\C))_\R\]

and in the latter case we say these functions have real boundary values.

To better understand this definition of Pick functions, we first want to see some consequences of the condition \(\Im(F(z)) \geq 0\) and what happens if \(0\) is contained in the spectrum of \(\Im(F(z))\):

Let \(\cH\) be a finite dimensional Hilbert space and \(A \in B(\cH)\) such that \(\Im(A) \geq 0\). Then the following hold:

* \(\Spec(A) \subeq \overline{\C_+}\).

* If \(x \in \Spec(A) \cap \R\) and \(v \in \cH \setminus \{0\}\) with \(Av = xv\), then \(\Im(A)v=0\) and \(A^*v = xv\).

* For every eigenvalue \(\lambda \in \C\) of \(F(z)\) and every normalized eigenvector \(v \in \cH\) to this eigenvalue \(\lambda\), we have

\[\lambda = \braket*{v}{\lambda v} = \braket*{v}{A v} = \braket*{v}{\frac 12(A+A^*) v} + i \braket*{v}{\frac 1{2i}(A-A^*) v} \in \overline{\C_+},\]

so \(\Spec(A) \subeq \overline{\C_+}\).

* Since \(x \in \R\), we have

\[0 = \braket*{v}{xv} - \braket*{xv}{v} = \braket*{v}{Av} - \braket*{Av}{v} = \braket*{v}{Av} - \braket*{v}{A^*v} = 2i \braket*{v}{\Im(A)v}.\]

Since \(\Im(A) \geq 0\), this implies \(\Im(A)v = 0\) and therefore

\[A^*v = Av = xv. \qedhere\]

Let \(f \in \mathrm{Pick}(\C_+)\). If there exists \(z_0 \in \C_+\) such that \(\Im(f(z_0)) = 0\), then \(f\) is constant.

Using the biholomorphic Cayley transform

\[h(z) \coloneqq \frac{z-i}{z+i}\]

\[h(\R) = \T \setminus \{1\} \quad \text{and} \quad h(\C_+) = \D,\]

we get

\[h(f(z_0)) \in h(\R) \subeq \T\]

\[h(f(z)) \in h(\overline{\C_+}) \subeq \overline{\D} \quad \forall z \in \C_+.\]

\[|h(f(z))| \leq 1 = h(f(z_0)) \quad \forall z \in \C_+.\]

This, by the Maximum modulus principle, implies that \(h \circ f\) and therefore also \(f\) is constant.

We now present some results about Pick functions with self-adjoint/real boundary values:
(cf. <cit.>)
Let \(\cH\) be a finite dimensional complex Hilbert space and let \(\phi: \R_{\geq 0} \to \Inn(\C_+,B(\cH))\) be a strongly continuous one-parameter semigroup of inner functions. Then there exists a Pick function \({F \in \mathrm{Pick}(\C_+,B(\cH))_\R}\) such that

\[\phi_t(z) = e^{itF(z)}, \quad z \in \C_+.\]

Let \(\cH\) be a finite dimensional complex Hilbert space and let \({F \in \mathrm{Pick}(\C_+,B(\cH))_\R}\). Then the function \(F_*\) is measurable.

Let \(v,w \in \cH\). For \(n \in \N\), we define

\[g_n: \R \to \C, \quad x \mapsto \braket*{v}{F\left(x+\frac in\right)w}.\]

Then, for every \(n \in \N\), the function \(g_n\) is continuous and therefore measurable. This implies that also the function

\[\braket*{v}{F_*(\cdot)w} = \limsup_{n \to \infty} g_n\]

is measurable. Since \(v,w \in \cH\) were chosen arbitrarily, this implies that \(F_*\) is measurable.

(Nevanlinna Representation)(cf. <cit.>, <cit.>)
Let \(F \in \cO(\C_+)\). Then \(F \in \mathrm{Pick}(\C_+)\), if and only if there exist \(C \in \R\), \(D \geq 0\) and some Borel measure \(\mu\) with

\[\int_\R \frac{d\mu(\lambda)}{1+\lambda^2} < \infty\]

such that

\[F(z) = C + Dz + \int_\R \frac 1{\lambda - z} - \frac{\lambda}{1+\lambda^2} \,d\mu(\lambda) \quad \forall z \in \C_+.\]

In this case, one has \(F \in \mathrm{Pick}(\C_+)_\R\), if and only if \(\mu\) is singular with respect to the Lebesgue measure on \(\R\). If \(\mu\) is singular, a minimal support for it is given by

\[\gls*{Smu} \coloneqq \left\{x \in \R: \lim_{\epsilon \downarrow 0} \Im(F(x+i\epsilon)) = +\infty\right\}.\]

If one takes the measure \(\mu\) from this theorem and defines a measure \(\tilde \nu\) on \(\R\) by

\[d\tilde \nu(\lambda) \coloneqq \frac{d\mu(\lambda)}{1+\lambda^2},\]

one gets that

\[F(z) = C + Dz + \int_\R \frac {1+\lambda z}{\lambda - z} \,d\tilde \nu(\lambda)\]

for a finite measure \(\tilde \nu\) on \(\R\). Notice that

\[\lim_{\lambda \to \pm\infty} \frac {1+\lambda z}{\lambda - z} = z,\]

so, if we extend our measure \(\tilde \nu\) to a measure \(\nu\) on \(\overline{\R}\coloneqq \R \cup \{\infty\}\) by \(\nu(\{\infty\}) \coloneqq D\), we get

\[F(z) = C + \int_{\overline{\R}} \frac {1+\lambda z}{\lambda - z} \,d\nu(\lambda).\]

Then, setting

\[\gls*{SF} \coloneqq \begin{cases} S_\mu & \text{if }D = 0 \\ S_\mu \cup \{\infty\} & \text{if }D > 0,\end{cases}\]

in the case that \(\mu\) was singular, the set \(S_F\) is a minimal support for the measure \(\nu\).

Given a Pick function \(F \in \mathrm{Pick}(\C_+)_\R\), we want to understand how the function \(F_*\) behaves on \(\overline{\R} \setminus S_F\). For this we need the following definition:

Let \(\overline{\R}\coloneqq \R \cup \{\infty\}\) be the one-point compactification of \(\R\). For \(a,b \in \overline{\R}\), we set

\[(a,b) \coloneqq \begin{cases} \{x \in \R: a<x<b\} & \text{if }a,b \in \R, a \leq b \\ \{x \in \R: b < x\} \cup \{\infty\} \cup \{x \in \R: x < a\} & \text{if }a,b \in \R, a > b \\ \{x \in \R: a<x\} & \text{if }a \in \R,b = \infty \\ \{x \in \R: x<b\} & \text{if }b \in \R, a = \infty \\ \eset & \text{if }a,b = \infty.\end{cases}\]

Let \(A \subeq \overline{\R}\) be a closed subset. Then there exists a countable family \(\left(a_j,b_j\right)_{j \in J}\) in \(\overline{\R} \times \overline{\R}\) such that \(\overline{\R} \setminus A = \bigsqcup_{j \in J} \left(a_j,b_j\right)\).

The maximally connected subsets of \(\overline{\R} \setminus A\) are of the form \((a,b)\) for some \(a,b \in \overline{\R}\). Therefore

\[\overline{\R} \setminus A = \bigsqcup_{j \in J} \left(a_j,b_j\right)\]

for some family \(\left(a_j,b_j\right)_{j \in J}\) in \(\overline{\R} \times \overline{\R}\) with \(\left(a_j,b_j\right) \neq \eset\) for every \(j \in J\). Since \(\left(a_j,b_j\right) \neq \eset\) implies that \(\left(a_j,b_j\right) \cap \Q \neq \eset\), for every \(j \in J\), there exists \(q_j \in \Q\) such that \(q_j \in \left(a_j,b_j\right)\). From this follows the existence of a surjective function \(f: \Q \to J\) and therefore \(J\) is countable.

Let \(F \in \mathrm{Pick}(\C_+)_\R\). Further let \(a,b \in \overline{\R}\) with \(a \neq b\) such that \((a,b) \cap S_F = \eset\) and \(a,b \in S_F\). Then the map

\[F_*\big|_{\left(a,b\right)}: (a,b) \to \R\]

is monotonically increasing, bijective and continuously differentiable.

By replacing \(F\) with \(F \circ M\), where \(M: \C_+ \to \C_+\) is a suitable Möbius transformation, we can without loss of generality assume that \(a,b \in \R\) with

\[a<0<b.\]

Now, since \((a,b) \cap S_F = \eset\), by <cit.>, there exists a function \(\tilde F \in \cO(\C_+ \cup (a,b) \cup \C_-)\) with

\[\tilde F(z) = \overline{\tilde F(\overline{z})} \quad \forall z \in \C_+ \cup (a,b) \cup \C_-\]

such that

\[\tilde F \big|_{\C_+} = F.\]

Since \(\tilde F\) is holomorphic, we have that \(F_*\big|_{(a,b)}= \tilde F\big|_{a,b}\) is continuously differentiable. Further, we have

\[\Im \tilde F(z) = \Im F(z) \geq 0 \quad \forall z \in \C_+,\]

which implies that \(\Im \tilde F(z) \leq 0\) for all \(z \in \C_-\) and therefore

\[\frac d{dy}\bigg|_{y=0} \Im \tilde F(x+iy) \geq 0 \quad \forall x \in (a,b).\]

This, by the Cauchy–Riemann equations implies that

\[0 \leq \frac d{dx} \Re \tilde F(x) = \frac d{dx} F_*(x) \quad \forall x \in (a,b),\]

so \(F_*\big|_{(a,b)}\) is monotonically increasing.
Now we assume that \(F_*\big|_{(a,b)} = \tilde F\big|_{(a,b)}\) is not injective, i.e. that there exist \(c,d \in (a,b)\) with \(c<d\) such that

\[\tilde F(c) = \tilde F(d).\]

Then by the monotony of \(F_*\big|_{(a,b)} = \tilde F\big|_{(a,b)}\), the function \(\tilde F\) would be constant on \([c,d]\) and therefore constant in contradiction to the assumption that \(a,b \in S_F\). Therefore \(F_*\big|_{\left(a,b\right)}\) is injective.

To show that the function \(F_*\big|_{\left(a,b\right)}\) is surjective, due to its monotony, it suffices to show that

\[\lim_{x \downarrow a} F_*(x) = -\infty \qquad \text{and} \qquad \lim_{x \uparrow b} F_*(x) = +\infty.\]

For \(x \in (a,b)\), \(\epsilon \in (0,1)\) and \(\lambda \in \overline{\R} \setminus (a,b)\), setting \(m \coloneqq \max\{|a|,|b|\}\), we have

\begin{align*}
\left|\frac{1+ \lambda (x+i\epsilon)}{\lambda - (x+i\epsilon)}\right| &= \left|(x+i\epsilon) + \frac{1+(x+i\epsilon)^2}{\lambda - (x+i\epsilon)}\right| \leq |(x+i\epsilon)| + \frac{1+\left|x+i\epsilon\right|^2}{|\lambda - (x+i\epsilon)|}
\\&\leq \sqrt{1+m^2} + \frac{2+m^2}{|\lambda-x|} \leq \sqrt{1+m^2} + \frac{2+m^2}{\min\{|x-a|,|x-b|\}}.
\end{align*}

Therefore, writing

\[F(z) = C + \int_{\overline{\R}} \frac {1+\lambda z}{\lambda - z} \,d\nu(\lambda), \quad z \in \C_+,\]

with \(\nu\) defined as in rem:GePick, by the Dominated Convergence Theorem, we get

\begin{align*}
F_*(x) &= \lim_{\epsilon \downarrow 0} F(x+i\epsilon) = C + \lim_{\epsilon \downarrow 0} \int_{\overline{\R}} \frac {1+\lambda (x+i\epsilon)}{\lambda - (x+i\epsilon)} \,d\nu(\lambda)
\\&= C + \lim_{\epsilon \downarrow 0} \int_{\overline{\R} \setminus (a,b)} \frac {1+\lambda (x+i\epsilon)}{\lambda - (x+i\epsilon)} \,d\nu(\lambda) = C + \int_{\overline{\R} \setminus (a,b)} \lim_{\epsilon \downarrow 0} \frac {1+\lambda (x+i\epsilon)}{\lambda - (x+i\epsilon)} \,d\nu(\lambda)
\\&= C + \int_{\overline{\R} \setminus (a,b)} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda) = C + \int_{\overline{\R}} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda).
\end{align*}

For \(x \in \left(0,b\right)\) and \(\lambda \in \overline{\R} \setminus (a,b+1)\), we have

\begin{align*}
\left|\frac{1+ \lambda x}{\lambda - x}\right| &= \left|x + \frac{1+x^2}{\lambda - x}\right| \leq |x| + \frac{1+x^2}{|\lambda - x|} \leq b + \frac{1+b^2}{\min\left\{|a|,1\right\}},
\end{align*}

so, by the Dominated Convergence Theorem, we get

\begin{align*}
\lim_{x \uparrow b} \int_{\overline{\R} \setminus (a,b+1)} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda) = \int_{\overline{\R} \setminus (a,b+1)} \frac {1+\lambda b}{\lambda - b}. \,d\nu(\lambda)
\end{align*}

\[\frac {1+\lambda x}{\lambda - x} \geq 0 \quad \forall x \in (0,b), \lambda \geq b\]

and since for every \(\lambda \geq b\) the function

\[(0,b) \ni x \mapsto \frac {1+\lambda x}{\lambda - x} = \frac{1+\lambda^2}{\lambda - x} - \lambda\]

is monotonically increasing, by the Monotone Convergence Theorem, we get

\[\lim_{x \uparrow b} \int_{[b,b+1)} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda) = \int_{[b,b+1)} \frac {1+\lambda b}{\lambda - b} \,d\nu(\lambda).\]

This yields

\begin{align*}
\lim_{x \uparrow b} F_*(x) &= C + \lim_{x \uparrow b} \int_{\overline{\R}} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda) = C + \lim_{x \uparrow b} \int_{\overline{\R} \setminus (a,b)} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda)
\\&= C + \lim_{x \uparrow b} \int_{\overline{\R} \setminus (a,b+1)} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda) + \lim_{x \uparrow b} \int_{[b,b+1)} \frac {1+\lambda x}{\lambda - x} \,d\nu(\lambda)
\\&= C + \int_{\overline{\R} \setminus (a,b+1)} \frac {1+\lambda b}{\lambda - b} \,d\nu(\lambda) + \int_{[b,b+1)} \frac {1+\lambda b}{\lambda - b} \,d\nu(\lambda) = C + \int_{\overline{\R}} \frac {1+\lambda b}{\lambda - b} \,d\nu(\lambda).
\end{align*}

If \(\lim_{x \uparrow b} F_*(x)<\infty\), this would imply that \(\left(\lambda \mapsto \frac {1+\lambda b}{\lambda - b}\right) \in L^1(\overline{\R},\C;\nu)\) and therefore also

\[\left(\lambda \mapsto \frac1{1+b^2} \cdot \left(\frac {1+\lambda b}{\lambda - b} - b\right) = \frac 1{\lambda - b}\right) \in L^1(\overline{\R},\C;\nu).\]

Then, for \(\epsilon \in (0,1)\), we have

\begin{align*}
\left|\frac{1+ \lambda (b+i\epsilon)}{\lambda - (b+i\epsilon)}\right| &= \left|(b+i\epsilon) + \frac{1+\left(b+i\epsilon\right)^2}{\lambda - (b+i\epsilon)}\right| \leq \left|b+i\epsilon\right| + \frac{1+\left|b+i\epsilon\right|^2}{\left|\lambda - (b+i\epsilon)\right|} \leq \sqrt{1+b^2} + \frac{2+b^2}{\left|\lambda - b\right|},
\end{align*}

so, by the Dominated Convergence Theorem, we would get

\begin{align*}
\lim_{\epsilon \downarrow 0} F(b+i\epsilon) &= C + \lim_{\epsilon \downarrow 0} \int_{\overline{\R}} \frac {1+\lambda (b+i\epsilon)}{\lambda - (b+i\epsilon)} \,d\nu(\lambda)
\\&= C + \int_{\overline{\R}} \lim_{\epsilon \downarrow 0} \frac {1+\lambda (b+i\epsilon)}{\lambda - (b+i\epsilon)} \,d\nu(\lambda) = C + \int_{\overline{\R}} \frac {1+\lambda b}{\lambda - b} \,d\nu(\lambda) = \lim_{x \uparrow b} F_*(x)<\infty
\end{align*}

in contradiction to the assumption \(b \in S_F\). Therefore

\[\lim_{x \uparrow b} F_*(x) = \infty.\]

Analogously one shows that

\begin{align*}
\lim_{x \downarrow a} F_*(x) = C + \int_{\overline{\R}} \frac {1+\lambda a}{\lambda - a} \,d\nu(\lambda)
\end{align*}

and sees that \(\lim_{x \downarrow a} F_*(x) > -\infty\) is in contradiction to the assumption \(a \in S_F\), which shows that

\[\lim_{x \downarrow a} F_*(x) = -\infty. \qedhere\]

§ AN APPROXIMATION THEOREM

In this chapter, we will prove the existence of a limit appearing in the proof of prop:FixedPointFromConstruction and calculate its value. Since this proof is very technical and lengthy, we decided to remove it from the proof of prop:FixedPointFromConstruction and include it here as an appendix. We start with a definition:

Let \(n \in \N\) and \(p \in \R_+\). We define the following three functions:

\begin{align*}
\gls*{gn}(x) &\coloneqq \frac{x^2}{n\left(\frac 1{n^2}+x^2\right)\left(1+\frac{x^2}{n^2}\right)}, \quad x \in \R
\\~
\\\gls*{dpn}(x) &\coloneqq g_n(x) \cdot \frac{\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}, \quad x \in \R
\\~
\\\gls*{fpn}(x) &\coloneqq d_{p,n}(x) + g_n(x) \cdot \left(\chi_{\left(-1,1\right)}(x) \cdot \frac 1{p^2} + \chi_{\left(-\infty,-1\right) \cup \left(1,\infty\right)}(x)\right), \quad x \in \R
\end{align*}

* For every \(n \in \N\) and \(x \in \R^\times\), one has

\[g_n(x) = \frac{x^2}{n\left(\frac 1{n^2}+x^2\right)\left(1+\frac{x^2}{n^2}\right)} \leq \frac {x^2}{n \cdot x^2 \cdot 1} = \frac 1n,\]

\[\lim_{n \to \infty} g_n(x) = 0.\]

* For every \(x \in \R^\times\) with \(|x| \neq p\) one has

\[\lim_{n \to \infty} d_{p,n}(x) = \lim_{n \to \infty} g_n(x) \cdot \frac{\left(x^2-p^2\right)\left(1-x^2\right)+2x^2}{\left(x^2-p^2\right)^2} = 0.\]

The goal of this section is to show that for any measurable function \(\phi: \R \to \C\) with

\[\int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty\]

one has

\[\frac{\phi(p) + \phi(-p)}2 = \lim_{n \to \infty} \frac 1\pi \int_\R f_{p,n}(x) \,\phi(x) \,dx\]

for almost every \(p \in \R_+\). We start by the following lemma:

For any measurable function \(\phi: \R \to \C\) with

\[\int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty\]

and every \(p \in \R_+\) one has

\[\lim_{n \to \infty} \int_{\left(-p+\frac 1{\sqrt n},p-\frac 1{\sqrt n}\right)} \left(d_{p,n}(x)+ \frac{g_n(x)}{p^2}\right)\phi(x) \,dx = 0.\]

For \(n \in \N\) with \(\frac 1{\sqrt n} < p\) and \(x \in \left(-p+\frac 1{\sqrt n},p-\frac 1{\sqrt n}\right)\) one has

\begin{align*}
&\left|d_{p,n}(x)+ \frac{g_n(x)}{p^2}\right| = \frac{g_n(x)}{p^2} \cdot \left|\frac{\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}} \cdot p^2 + 1\right|
\\&\qquad=\frac{g_n(x)}{p^2} \cdot \left|\frac{\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)p^2+2x^2\left(\frac 1{n^2} + 1\right)p^2 + \left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}\right|
\\&\qquad=\frac{g_n(x)}{p^2} \cdot \left|\frac{x^4\left(1-p^2\right) + x^2\left(p^2 + \frac 3{n^2}p^2 + p^4 + \frac 2{n^2}\right) + \left(\frac 1{n^4} + \frac {p^2}{n^2}\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}\right|
\\&\qquad\leq\frac{g_n(x)}{p^2} \cdot \frac{x^2\left|x^2\left(1-p^2\right) + \left(p^2 + \frac 3{n^2}p^2 + p^4 + \frac 2{n^2}\right)\right| + \left(\frac 1{n^4} + \frac {p^2}{n^2}\right)}{\left(p^2+\frac 1{n^2}-x^2\right)^2+4x^2 \frac 1{n^2}}
\\&\qquad\leq\frac{g_n(x)}{p^2} \cdot \frac{x^2\left(p^2\left(1+p^2\right) + \left(4p^2 + p^4 + 2\right)\right) + \frac 1{n^2}\left(\frac 1{n^2} + p^2\right)}{\left(p^2+\frac 1{n^2}-\left(p-\frac 1{\sqrt n}\right)^2\right)^2}
\\&\qquad\leq\frac{g_n(x)}{p^2} \cdot \frac{x^2\left(2p^4 + 5p^2 + 2\right) + \frac 1{n^2}\left(1 + p^2\right)}{\left(\frac 1{n^2}-\frac 1n + 2p \frac 1{\sqrt n}\right)^2}
\\&\qquad\leq\frac{x^2}{p^2n\left(\frac 1{n^2}+x^2\right)\left(1 + \frac{x^2}{n^2}\right)} \cdot \frac{x^2\left(2p^4 + 5p^2 + 2\right) + \frac 1{n^2}\left(1 + p^2\right)}{\left(0 - \frac p{\sqrt n} + 2p \frac 1{\sqrt n}\right)^2}
\\&\qquad\leq\frac{x^2}{p^2n\left(\frac 1{n^2}+x^2\right)} \cdot \frac{x^2\left(2p^4 + 5p^2 + 2\right) + \frac 1{n^2}\left(1 + p^2\right)}{\frac {p^2} n}
\\&\qquad=\frac{x^2}{p^2n\left(\frac 1{n^2}+x^2\right)} \cdot \frac{x^2\left(2p^4 + 5p^2 + 2\right)}{\frac {p^2} n} + \frac{x^2}{p^2n\left(\frac 1{n^2}+x^2\right)} \cdot \frac{\frac 1{n^2}\left(1 + p^2\right)}{\frac {p^2} n}
\\&\qquad=\frac{x^2}{\frac 1{n^2}+x^2} \cdot \frac{x^2\left(2p^4 + 5p^2 + 2\right)}{p^4} + \frac{x^2}{1+n^2x^2} \cdot \frac{\left(1 + p^2\right)}{p^4}
\\&\qquad\leq 1 \cdot \frac{x^2\left(2p^4 + 5p^2 + 2\right)}{p^4} + \frac{x^2}{1} \cdot \frac{\left(1 + p^2\right)}{p^4} = x^2 \cdot \frac{2p^4 + 6p^2 + 3}{p^4}
\\&\qquad\leq \frac{x^2}{\left(1+x^2\right)^2} \cdot \frac{\left(2p^4 + 6p^2 + 3\right)\left(1+p^2\right)^2}{p^4}.
\end{align*}

By assumption, we have

\[\int_{(-p,p)}\frac{x^2}{\left(1+x^2\right)^2} \cdot \frac{\left(2p^4 + 6p^2 + 3\right)\left(1+p^2\right)^2}{p^4} \left|\phi(x)\right| \,dx < \infty\]

and therefore, by the Dominated Convergence Theorem, we have

\begin{align*}
&\lim_{n \to \infty} \int_{\left(-p+\frac 1{\sqrt n},p-\frac 1{\sqrt n}\right)} \left(d_{p,n}(x)+ \frac{g_n(x)}{p^2}\right)\phi(x) \,dx
\\&\qquad= \lim_{n \to \infty} \int_{\left(-p,p\right)} \chi_{\left(-p+\frac 1{\sqrt n},p-\frac 1{\sqrt n}\right)}(x) \left(d_{p,n}(x)+ \frac{g_n(x)}{p^2}\right)\phi(x) \,dx
\\&\qquad= \int_{\left(-p,p\right)} \lim_{n \to \infty}\chi_{\left(-p+\frac 1{\sqrt n},p-\frac 1{\sqrt n}\right)}(x) \left(d_{p,n}(x)+ \frac{g_n(x)}{p^2}\right)\phi(x) \,dx
\\&\qquad= \int_{\left(-p,p\right)} 0 \cdot \phi(x) \,dx = 0. \qedhere
\end{align*}

For any measurable function \(\phi: \R \to \C\) with

\[\int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty\]

and every \(p \in \R_+\), one has

\[\lim_{n \to \infty} \int_{\left(-\infty,-p-\frac 1{\sqrt n}\right) \cup \left(p+\frac 1{\sqrt n},\infty\right)} \left(d_{p,n}(x)+g_n(x)\right)\phi(x) \,dx = 0.\]

For \(n \in \N\) with \(\frac 1{\sqrt n} < p\) and \(x \in \left(-\infty,-p-\frac 1{\sqrt n}\right) \cup \left(p+\frac 1{\sqrt n},\infty\right)\), one has

\begin{align*}
&\left|d_{p,n}(x)+g_n(x)\right| = g_n(p) \cdot \left|\frac{\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}} + 1\right|
\\&\quad\leq \frac 1n \cdot \left|\frac{\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right) + \left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}\right|
\\&\quad=\frac 1n \cdot \left|\frac{x^2\left(\frac 5{n^2} + 3 - p^2\right) - \left(p^2 + \frac 1{n^2}\right) + \left(p^2 + \frac 1{n^2}\right)^2}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}\right|
\\&\quad\leq \frac 1n \cdot \frac{x^2\left(\frac 5{n^2} + 3 + p^2\right) + \left(p^2 + \frac 1{n^2}\right)\left(p^2 + \frac 1{n^2} + 1\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2}
\\&\quad\leq \frac 1n \cdot \frac{x^2\left(8 + p^2\right) + x^2\left(p^2+2\right)}{\left(x^2 \cdot \frac{p^2 + \frac 1{n^2}}{\left(p + \frac 1{\sqrt n}\right)^2}-p^2-\frac 1{n^2} + x^2 \left(1-\frac{p^2 + \frac 1{n^2}}{\left(p + \frac 1{\sqrt n}\right)^2}\right)\right)^2} \leq \frac 1n \cdot \frac{2x^2\left(p^2+5\right)}{\left(x^2 \left(1-\frac{p^2 + \frac 1{n^2}}{\left(p + \frac 1{\sqrt n}\right)^2}\right)\right)^2}
\\&\quad= \frac 1n \cdot \frac{2x^2\left(p^2+5\right)}{\left(x^2 \cdot \frac{2p \frac 1{\sqrt n} + \frac 1n-\frac 1{n^2}}{\left(p + \frac 1{\sqrt n}\right)^2}\right)^2} = \frac 1{nx^2} \cdot \frac{2\left(p^2+5\right)\left(p + \frac 1{\sqrt n}\right)^2}{\left(2p \frac 1{\sqrt n} + \frac 1n-\frac 1{n^2}\right)^2} \leq \frac 1{nx^2} \cdot \frac{2\left(p^2+5\right)\left(p + 1\right)^2}{\left(2p \frac 1{\sqrt n}\right)^2}
\\&\quad= \frac 1{x^2} \cdot \frac{\left(p^2+5\right)\left(p + 1\right)^2}{2p^2} = \frac {x^2\left(\frac 1{x^2} + 1\right)^2}{\left(1+x^2\right)^2} \cdot \frac{\left(p^2+5\right)\left(p + 1\right)^2}{2p^2}
\\&\quad\leq \frac {x^2\left(\frac 1{p^2} + 1\right)^2}{\left(1+x^2\right)^2} \cdot \frac{\left(p^2+5\right)\left(p + 1\right)^2}{2p^2} = \frac {x^2}{\left(1+x^2\right)^2} \cdot \frac{\left(1 + p^2\right)^2\left(p^2+5\right)\left(p + 1\right)^2}{2p^6}.
\end{align*}

By assumption, we have

\[\int_{(-\infty,-p) \cup (p,\infty)}\frac {x^2}{\left(1+x^2\right)^2} \cdot \frac{\left(1 + p^2\right)^2\left(p^2+5\right)\left(p + 1\right)^2}{2p^6} \left|\phi(x)\right| \,dx < \infty\]

and therefore, by the Dominated Convergence Theorem, we have

\begin{align*}
&\lim_{n \to \infty} \int_{\left(-\infty,-p-\frac 1{\sqrt n}\right) \cup \left(p+\frac 1{\sqrt n},\infty\right)} \left(d_{p,n}(x)+g_n(x)\right)\phi(x) \,dx
\\&\qquad= \lim_{n \to \infty} \int_{\left(-\infty,-p\right) \cup \left(p,\infty\right)} \chi_{\left(-\infty,-p-\frac 1{\sqrt n}\right) \cup \left(p+\frac 1{\sqrt n},\infty\right)}(x) \left(d_{p,n}(x)+g_n(x)\right)\phi(x) \,dx
\\&\qquad= \int_{\left(-p,p\right)} \lim_{n \to \infty}\chi_{\left(-\infty,-p-\frac 1{\sqrt n}\right) \cup \left(p+\frac 1{\sqrt n},\infty\right)}(x) \left(d_{p,n}(x)+g_n(x)\right)\phi(x) \,dx
\\&\qquad= \int_{\left(-p,p\right)} 0 \cdot \phi(x) \,dx = 0. \qedhere
\end{align*}

Let \(n \in \N\) and \(p \in \R_+\). We define the following functions:

\begin{align*}
\tilde g_{p,n}(x) \coloneqq \frac{px}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)}, \quad x \in \R
\\~
\\\gls*{dpntilde}(x) \coloneqq \tilde g_{p,n}(x) \cdot \frac{2p^2\left(\frac 1{n^2} + 1\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}}, \quad x \in \R
\end{align*}

For every \(p \in \R_+\), there exist sequences \(\left(b_n\right)_{n \in \N}\) and \(\left(B_n\right)_{n \in \N}\) in \(\R\) with

\[\lim_{n \to \infty} b_{p,n} = 1 = \lim_{n \to \infty} B_{p,n}\]

such that for sufficiently large \(n \in \N\), one has

\[b_{p,n} \cdot \tilde d_{p,n}(x) \leq d_{p,n}(x) \leq B_{p,n} \cdot \tilde d_{p,n}(x)\]

for every \(x \in \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)\).

We have

\[\frac{g_n(x)}{\tilde g_{p,n}(x)} = \frac{\frac{x^2}{n\left(\frac 1{n^2}+x^2\right)\left(1+\frac{x^2}{n^2}\right)}}{\frac{px}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)}} = \frac{x\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)}{p\left(\frac 1{n^2}+x^2\right)\left(1+\frac{x^2}{n^2}\right)},\]

so for \(x \in \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)\) one has

\[\frac{g_n(x)}{\tilde g_{p,n}(x)} \leq \frac{\left(p+\frac 1{\sqrt n}\right)\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)}{p\left(\frac 1{n^2}+\left(p-\frac 1{\sqrt n}\right)^2\right)\left(1+\frac{\left(p-\frac 1{\sqrt n}\right)^2}{n^2}\right)} \eqqcolon A_{p,n} \xrightarrow{n \to \infty} 1\]

\[\frac{g_n(x)}{\tilde g_{p,n}(x)} \geq \frac{\left(p-\frac 1{\sqrt n}\right)\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)}{p\left(\frac 1{n^2}+\left(p+\frac 1{\sqrt n}\right)^2\right)\left(1+\frac{\left(p+\frac 1{\sqrt n}\right)^2}{n^2}\right)} \eqqcolon a_{p,n} \xrightarrow{n \to \infty} 1.\]

Further, we have

\begin{align*}
\frac{d_{p,n}(x)}{\tilde d_{p,n}(x)} &= \frac{g_n(x)}{\tilde g_{p,n}(x)} \cdot \frac{\frac{\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}}{\frac{2p^2\left(\frac 1{n^2} + 1\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}}}
\\&= \frac{g_n(x)}{\tilde g_{p,n}(x)} \cdot \frac{\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right)}{2p^2\left(\frac 1{n^2} + 1\right)} \cdot \left(\frac{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}}\right)^{-1}.
\end{align*}

For \(n \in \N\) with \(\frac 1{\sqrt n} \leq p\) and \(x \in \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)\) one has

\begin{align*}
&\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right) = -x^4 + x^2\left(\frac 3{n^2} + 3 + p^2\right) - \left(p^2 + \frac 1{n^2}\right)
\\&\qquad\geq -\left(p + \frac 1{\sqrt n}\right)^4 + \left(p - \frac 1{\sqrt n}\right)^2\left(\frac 3{n^2} + 3 + p^2\right) - \left(p^2 + \frac 1{n^2}\right)
\\&\qquad\eqqcolon v_{p,n} \xrightarrow{n \to \infty} -p^4 + p^2\left(3+p^2\right) - p^2 =  2p^2 > 0
\end{align*}

\begin{align*}
&\left(x^2-p^2-\frac 1{n^2}\right)\left(1-x^2\right)+2x^2\left(\frac 1{n^2} + 1\right) = -x^4 + x^2\left(\frac 3{n^2} + 3 + p^2\right) - \left(p^2 + \frac 1{n^2}\right)
\\&\qquad\leq -\left(p - \frac 1{\sqrt n}\right)^4 + \left(p + \frac 1{\sqrt n}\right)^2\left(\frac 3{n^2} + 3 + p^2\right) - \left(p^2 + \frac 1{n^2}\right)
\\&\qquad\eqqcolon V_{p,n} \xrightarrow{n \to \infty} -p^4 + p^2\left(3+p^2\right) - p^2 =  2p^2 > 0
\end{align*}

\begin{align*}
2p^2\left(\frac 1{n^2} + 1\right) \eqqcolon w_{p,n} \xrightarrow{n \to \infty} 2p^2 > 0
\end{align*}

\begin{align*}
\frac{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} &= 1 + \frac{\frac 4{n^2}\left(x^2-p^2\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} \geq 1 + \frac{\frac 4{n^2}\left(\left(p - \frac 1{\sqrt n}\right)^2-p^2\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}}
\\&\geq 1 + \frac{\frac 4{n^2}\left(\left(p - \frac 1{\sqrt n}\right)^2-p^2\right)}{4p^2 \frac 1{n^2}} = 1 + \frac {\frac 1n - \frac 2{\sqrt n}}{p^2} \eqqcolon u_{p,n} \xrightarrow{n \to \infty} 1
\end{align*}

\begin{align*}
\frac{\left(x^2-p^2-\frac 1{n^2}\right)^2+4x^2 \frac 1{n^2}}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} &= 1 + \frac{\frac 4{n^2}\left(x^2-p^2\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} \leq 1 + \frac{\frac 4{n^2}\left(\left(p + \frac 1{\sqrt n}\right)^2-p^2\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}}
\\&\leq 1 + \frac{\frac 4{n^2}\left(\left(p + \frac 1{\sqrt n}\right)^2-p^2\right)}{4p^2 \frac 1{n^2}} = 1 + \frac {\frac 1n + \frac 2{\sqrt n}}{p^2} \eqqcolon U_{p,n} \xrightarrow{n \to \infty} 1.
\end{align*}

Therefore, for sufficiently large \(n \in \N\) and \(x \in \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)\), one has

\[\frac{d_{p,n}(x)}{\tilde d_{p,n}(x)} \leq A_{p,n} \cdot \frac{V_{p,n}}{w_{p,n}} \cdot u_{p,n}^{-1} \eqqcolon B_{p,n} \xrightarrow{n \to \infty} 1 \cdot \frac{2p^2}{2p^2} \cdot 1 = 1\]

\[\frac{d_{p,n}(x)}{\tilde d_{p,n}(x)} \geq a_{p,n} \cdot \frac{v_{p,n}}{w_{p,n}} \cdot U_{p,n}^{-1} \eqqcolon b_{p,n} \xrightarrow{n \to \infty} 1 \cdot \frac{2p^2}{2p^2} \cdot 1 = 1.  \qedhere\]

Let \(f \in L^1_{\mathrm{loc}}(\R,\C)\). Then \(p \in \R\) is called a Lebesgue point of \(f\), if

\[\lim_{\epsilon \downarrow 0} \frac 1{2\epsilon} \int_{(p-\epsilon,p+\epsilon)} |f(x)-f(p)| \,dx = 0.\]

Let \(f \in L^1_{\mathrm{loc}}(\R,\C)\) and \(p \in \R\) be a Lebesgue point of \(f\). Further, let \(\left(k_n\right)_{n \in \N}\) be a sequence of non-negative functions with

\[\int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} k_n(x) \,dx = 1,\]

such that there exist constants \(r,R>0\) such that for every \(n \in \N\), one has

\[\max\left\{k_n\left(p-\frac 1{\sqrt n}\right),k_n\left(p+\frac 1{\sqrt n}\right)\right\} \leq R\sqrt{n}\]

and \(k_n \in C^1\left(\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)\right)\) with

\begin{align*}
k_n'(x) &\geq 0 &&\forall x \in \left(p-\frac 1{\sqrt n},p-\frac 1n\right)
\\\left|k_n'(x)\right| &\leq rn^2 &&\forall x \in \left(p-\frac 1n,p+\frac 1n\right)
\\k_n'(x) &\leq 0 &&\forall x \in \left(p+\frac 1n,p+\frac 1{\sqrt n}\right).
\end{align*}

\[f(p) = \lim_{n \to \infty} \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)}k_n(x)f(x) \,dx.\]

For \(\epsilon>0\) we set

\[c_\epsilon \coloneqq \frac 1{2\epsilon} \int_{(p-\epsilon,p+\epsilon)} |f(x)-f(p)| \,dx.\]

Then, since \(p\) is a Lebesgue point, we have \(\lim_{\epsilon \downarrow 0}c_\epsilon = 0\). We set

\[C_\epsilon \coloneqq \sup_{t \in (0,\epsilon]} c_t.\]

Then also \(\lim_{\epsilon \downarrow 0}C_\epsilon = 0\). We now define the function

\[m_n(x) \coloneqq \begin{cases} rn - rn^2|x-p| & \text{if } x \in \left(p-\frac 1n,p+\frac 1n\right) \\ 0 & \text{else}\end{cases}.\]

Then, for every \(x \in \left(p,p+\frac 1n\right)\), we have

\[k_n'(x) + m_n'(x) = k_n'(x) - rn^2 \leq rn^2 - rn^2 = 0\]

and for every \(x \in \left(p-\frac 1n,p\right)\), we have

\[k_n'(x) + m_n'(x) = k_n'(x) + rn^2 \geq -rn^2 + rn^2 = 0,\]

so \(k_n'(x)+m_n'(x) \leq 0\) for \(x \in \left(p,p+\frac 1{\sqrt n}\right)\) and \(k_n'(x)+m_n'(x) \geq 0\) for \(x \in \left(p-\frac 1{\sqrt n},p\right)\).
This yields

\begin{align*}
&\int_p^{p+\frac 1{\sqrt n}} \left(k_n(x) + m_n(x)\right) |f(x)-f(p)| \,dx
\\&\qquad= \int_p^{p+\frac 1{\sqrt n}} \left(k_n\left(p+\frac 1{\sqrt n}\right) - \int_x^{p+\frac 1{\sqrt n}} k_n'(t) + m_n'(t) \,dt\right) |f(x)-f(p)| \,dx
\\&\qquad= k_n\left(p+\frac 1{\sqrt n}\right) \cdot \int_p^{p+\frac 1{\sqrt n}} |f(x)-f(p)| \,dx
\\&\qquad\qquad + \int_p^{p+\frac 1{\sqrt n}} \int_x^{p+\frac 1{\sqrt n}} -\left(k_n'(t) + m_n'(t)\right) |f(x)-f(p)| \,dt \,dx
\\&\qquad\leq R\sqrt{n} \cdot 2\frac 1{\sqrt n}c_{\frac 1{\sqrt n}} + \int_p^{p+\frac 1{\sqrt n}} \int_p^t -\left(k_n'(t) + m_n'(t)\right) |f(x)-f(p)| \,dx \,dt
\\&\qquad= 2R \cdot c_{\frac 1{\sqrt n}} + \int_p^{p+\frac 1{\sqrt n}} -\left(k_n'(t) + m_n'(t)\right)\int_p^t  |f(x)-f(p)| \,dx \,dt
\\&\qquad\leq 2R \cdot c_{\frac 1{\sqrt n}} + \int_p^{p+\frac 1{\sqrt n}} -\left(k_n'(t) + m_n'(t)\right) 2\left(t-p\right)c_{t-p} \,dt
\\&\qquad\leq 2R \cdot c_{\frac 1{\sqrt n}} + 2C_{\frac 1{\sqrt n}} \int_p^{p+\frac 1{\sqrt n}} -\left(k_n'(t) + m_n'(t)\right) \left(t-p\right) \,dt
\\&\qquad= 2R \cdot c_{\frac 1{\sqrt n}} + 2C_{\frac 1{\sqrt n}} \left(\left[-\left(k_n(t) + m_n(t)\right) \left(t-p\right)\right]_p^{p+\frac 1{\sqrt n}} + \int_p^{p+\frac 1{\sqrt n}} \left(k_n(t) + m_n(t)\right) \,dt\right)
\\&\qquad= 2R \cdot c_{\frac 1{\sqrt n}} + 2C_{\frac 1{\sqrt n}} \left(-k_n\left(p+\frac 1{\sqrt n}\right)\frac 1{\sqrt n} + \int_p^{p+\frac 1{\sqrt n}} k_n(t) \,dt + \frac r2\right)
\\&\qquad\leq 2R \cdot c_{\frac 1{\sqrt n}} + 2C_{\frac 1{\sqrt n}} \left(0 + 1 + \frac r2\right) \xrightarrow{n \to \infty} 0
\end{align*}

and analogously

\[\int_{p-\frac 1{\sqrt n}}^p \left(k_n(x) + m_n(x)\right) |f(x)-f(p)| \,dx \xrightarrow{n \to \infty} 0.\]

Finally, we have

\begin{align*}
\left|\int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)}k_n(x)f(x) \,dx - f(p)\right| &= \left|\int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)}k_n(x)\left(f(x)-f(p)\right) \,dx\right|
\\&\leq \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)}k_n(x)\left|f(x)-f(p)\right| \,dx
\\&\leq \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)}\left(k_n(x) + m_n(x)\right)\left|f(x)-f(p)\right| \,dx \xrightarrow{n \to \infty} 0,
\end{align*}

\[f(p) = \lim_{n \to \infty} \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)}k_n(x)f(x) \,dx. \qedhere\]

For every \(p \in \R_+\), one has

\[\lim_{n \to \infty} \frac 1\pi \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} \tilde d_{p,n}(x) \,dx = \frac 12.\]

We define the function

\[D_{p,n}(x) \coloneqq \frac{p^2\left(1+\frac 1{n^2}\right)}{2\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \arctan\left(\frac{x^2-p^2-\frac 1{n^2}}{\frac{2p}n}\right).\]

\begin{align*}
D_{p,n}'(x) &= \frac{p^2\left(1+\frac 1{n^2}\right)}{2\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \frac{\frac{nx}p}{1 + \left(\frac{x^2-p^2-\frac 1{n^2}}{\frac{2p}n}\right)^2}
\\&= \frac{px}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \frac{2p^2\left(1+\frac 1{n^2}\right)}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} = \tilde d_{p,n}(x).
\end{align*}

\[\arctan\left(\frac{\left(p\pm\frac 1{\sqrt n}\right)^2-p^2-\frac 1{n^2}}{\frac{2p}n}\right) = \arctan\left(\frac{\frac 1n \pm \frac {2p}{\sqrt n} - \frac 1{n^2}}{\frac{2p}n}\right) = \arctan\left(\frac 1{2p} \pm \sqrt{n} - \frac 1{2pn}\right),\]

we get

\begin{align*}
&\int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} \tilde d_{p,n}(x) \,dx = \left[D_{p,n}(x)\right]_{p-\frac 1{\sqrt n}}^{p+\frac 1{\sqrt n}}
\\&\qquad= \frac{p^2\left(1+\frac 1{n^2}\right)}{2\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot  \left(\arctan\left(\frac 1{2p} + \sqrt{n} - \frac 1{2pn}\right) - \arctan\left(\frac 1{2p} - \sqrt{n} - \frac 1{2pn}\right)\right)
\\&\qquad \xrightarrow{n \to \infty} \frac 12 \cdot \frac \pi 2 - \frac 12 \cdot \left(-\frac \pi 2\right) = \frac \pi 2. \qedhere
\end{align*}

Let \(\phi: \R \to \C\) be a measurable function with

\[\int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty.\]

Further let \(p \in \R_+\) be a Lebesgue point of \(\phi\). Then

\[\lim_{n \to \infty} \frac 1\pi \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} \tilde d_{p,n}(x) \phi(x) \,dx = \frac 12 \phi(p).\]

For \(n \in \N\) with \(\frac 1{\sqrt n} < p\) set

\[D_n \coloneqq \frac 1\pi \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} \tilde d_{p,n}(x) \,dx.\]

Then, by lem:dIntegral1, we have \(\lim_{n \to \infty}D_n = \frac 12\). Further, we define

\[k_n(x) \coloneqq \frac{\tilde d_{p,n}(x)}{\pi D_n}.\]

\[\int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} k_n(x) \,dx = 1\]

and \(k_n(x) \geq 0\) for \(x \in \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)\).
Then, for sufficiently large \(n \in \N\), one has

\begin{align*}
k_n\left(p\pm\frac 1{\sqrt n}\right) &= \frac 1{\pi D_n} \frac{p\left(p\pm\frac 1{\sqrt n}\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \frac{2p^2\left(\frac 1{n^2} + 1\right)}{\left(\left(p\pm\frac 1{\sqrt n}\right)^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}}
\\&\leq \frac 1{\pi D_n} \frac{p\left(p+1\right)}{n \cdot p^2 \cdot 1} \cdot \frac{4p^2}{\left(\frac 1n \pm \frac{2p}{\sqrt n}-\frac 1{n^2}\right)^2 + 0}
\\&= \frac 1{\pi D_n} \frac{4p\left(p+1\right)}{\left(\frac 1{\sqrt n} \pm 2p -\frac 1{\sqrt{n^3}}\right)^2} \xrightarrow{n \to \infty} \frac 2\pi \frac{4p\left(p+1\right)}{4p^2} = \frac{2(p+1)}{\pi p},
\end{align*}

so for sufficiently large \(n \in \N\), we have

\[k_n\left(p\pm\frac 1{\sqrt n}\right) \leq \frac{p+1}{p} \leq \frac{p+1}{p} \sqrt{n}.\]

Further, we have

\begin{align*}
k_n'(x) &= \frac 1{\pi D_n} \frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \left(\frac{x}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}}\right)'
\\&= \frac 1{\pi D_n} \frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \left(\frac{1}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} - \frac{2x^2\left(x^2-p^2-\frac 1{n^2}\right)}{\left(\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}\right)^2}\right)
\end{align*}

For \(x \in \left(p-\frac 1n,p+\frac 1n\right)\), using

\[\frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \leq \frac{4p^3}{n \cdot p^2 \cdot 1} = \frac{4p}{n}\]

\begin{align*}
\left|x^2-p^2-\frac 1{n^2}\right| &= \left|\left(p+(x-p)\right)^2-p^2-\frac 1{n^2}\right| = \left|2p(x-p) + (x-p)^2-\frac 1{n^2}\right|
\\&\leq 2p\frac 1n + \left(\frac 1n\right)^2 + \frac 1{n^2} \leq \frac{2(p+1)}n,
\end{align*}

this yields

\begin{align*}
\left|k_n'(x)\right| &\leq\frac 1{\pi D_n} \frac{4p}{n} \cdot \left(\frac{1}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} + \frac{2x^2\left|x^2-p^2-\frac 1{n^2}\right|}{\left(\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}\right)^2}\right)
\\&\leq\frac 1{\pi D_n} \frac{4p}{n} \cdot \left(\frac{1}{4p^2 \frac 1{n^2}} + \frac{2(p+1)^2 \cdot \frac{2(p+1)}n}{\left(4p^2 \frac 1{n^2}\right)^2}\right)
\\&=\frac 1{\pi D_n} \left(\frac{n}{p} + \frac{(p+1)^3n^2}{p^3}\right) \leq \frac 1{\pi D_n} \frac{2(p+1)^3}{p^3} \cdot n^2.
\end{align*}

Then the fact that \(\lim_{n \to \infty}D_n = \frac 12\) implies that there is \(r>0\) such that \(\left|k_n'(x)\right| \leq rn^2\) for every \(x \in \left(p-\frac 1n,p+\frac 1n\right)\).

Further, for \(x \in \left(p-\frac 1{\sqrt n},p-\frac 1n\right)\), we have

\[\left(x^2-p^2-\frac 1{n^2}\right) \leq 0,\]

\begin{align*}
k_n'(x) &= \frac 1{\pi D_n} \frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \left(\frac{1}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} - \frac{2x^2\left(x^2-p^2-\frac 1{n^2}\right)}{\left(\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}\right)^2}\right)
\\&\geq \frac 1{\pi D_n} \frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \left(0 - 0\right) = 0.
\end{align*}

For \(x \in \left(p+\frac 1n,p+\frac 1{\sqrt n}\right)\), we get

\begin{align*}
&\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2} - 2x^2\left(x^2-p^2-\frac 1{n^2}\right) = \left(-x^2-p^2-\frac 1{n^2}\right)\left(x^2-p^2-\frac 1{n^2}\right)+4p^2 \frac 1{n^2}
\\&\qquad\qquad\qquad= \left(p^2+\frac 1{n^2}\right)^2-x^4+4p^2 \frac 1{n^2} \leq \left(p^2+\frac 1{n^2}\right)^2-\left(p+\frac 1n\right)^4+4p^2 \frac 1{n^2}
\\&\qquad\qquad\qquad= \left(p^2+\frac 1{n^2}\right)^2-\left(p^2+\frac 1{n^2} + 2\frac pn\right)^2+\left(2\frac pn\right)^2 \leq 0,
\end{align*}

\begin{align*}
k_n'(x) &= \frac 1{\pi D_n} \frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \left(\frac{1}{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}} - \frac{2x^2\left(x^2-p^2-\frac 1{n^2}\right)}{\left(\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}\right)^2}\right)
\\&= \frac 1{\pi D_n} \frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \left(\frac{\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2} - 2x^2\left(x^2-p^2-\frac 1{n^2}\right)}{\left(\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}\right)^2}\right)
\\&\leq \frac 1{\pi D_n} \frac{2p^3\left(\frac 1{n^2} + 1\right)}{n\left(\frac 1{n^2}+p^2\right)\left(1+\frac{p^2}{n^2}\right)} \cdot \left(\frac{0}{\left(\left(x^2-p^2-\frac 1{n^2}\right)^2+4p^2 \frac 1{n^2}\right)^2}\right) = 0.
\end{align*}

These estimates show that we can apply lem:LebesguePointApprox and get

\[\frac 1\pi \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} \tilde d_{p,n}(x) \phi(x) \,dx = D_n \cdot \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} k_n(x) \phi(x) \,dx \xrightarrow{n \to \infty} \frac 12 \cdot \phi(p). \qedhere\]

Let \(\phi: \R \to \C\) be a measurable function with

\[\int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty.\]

Further let \(p \in \R_+\) such that \(p\) and \(-p\) are Lebesgue points of \(\phi\). Then

\[\frac{\phi(p) + \phi(-p)}2 = \lim_{n \to \infty} \frac 1\pi \int_{\left(-p-\frac 1{\sqrt n},-p+\frac 1{\sqrt n}\right) \cup \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} d_{p,n}(x) \,\phi(x) \,dx.\]

By lem:localPhiIntegral and lem:dSlightChange, we have

\[\lim_{n \to \infty} \frac 1\pi \int_{\left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} d_{p,n}(x)\phi(x) \,dx = \frac 12 \phi(p).\]

Since for every \(p \in \R_+\) and every \(n \in \N\) the function \(d_{p,n}\) is symmetric, we then also get

\[\lim_{n \to \infty} \frac 1\pi \int_{\left(-p-\frac 1{\sqrt n},-p+\frac 1{\sqrt n}\right)} d_{p,n}(x)\phi(x) \,dx = \frac 12 \phi(-p),\]

\[\frac{\phi(p) + \phi(-p)}2 = \lim_{n \to \infty} \frac 1\pi \int_{\left(-p-\frac 1{\sqrt n},-p+\frac 1{\sqrt n}\right) \cup \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} d_{p,n}(x) \,\phi(x) \,dx. \qedhere\]

(Lebesgue Differentiation Theorem)(cf. [12])
Let \(f \in L^1_{\mathrm{loc}}(\R,\C)\). Then almost every \(p \in \R\) is a Lebesgue point of \(f\).

For any measurable function \(\phi: \R \to \C\) with

\[\int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty\]

almost every \(p \in \R\) is a Lebesgue point of \(\phi\).

For \(0<a<b<\infty\), one has

\begin{align*}
\int_{[-b,-a] \cup [a,b]} |\phi(x)| \,dx &\leq \int_{[-b,-a] \cup [a,b]} \frac{\left(1+b^2\right)^2}{a^2} \cdot \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx
\\&\leq \frac{\left(1+b^2\right)^2}{a^2} \cdot \int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty.
\end{align*}

The statement then follows by thm:LebesguePointAe.

For any measurable function \(\phi: \R \to \C\) with

\[\int_\R \frac{x^2}{\left(1+x^2\right)^2} \left|\phi(x)\right| \,dx < \infty\]

one has

\[\frac{\phi(p) + \phi(-p)}2 = \lim_{n \to \infty} \frac 1\pi \int_\R f_{p,n}(x) \,\phi(x) \,dx\]

for almost every \(p \in \R^\times\).

Let \(p \in \R_+\) such that \(p\) and \(-p\) are Lebesgue points of \(\phi\). For \(n \in \N\), we set

\[c_n \coloneqq \int_{\left[-p - \frac 1{\sqrt n},-1\right) \cup \left(1,p + \frac 1{\sqrt n}\right]} g_n(x) \phi(x) \,dx \quad \text{if } p + \frac 1{\sqrt n} \geq 1\]

\[c_n \coloneqq - \int_{\left[-1,-p - \frac 1{\sqrt n}\right) \cup \left(p + \frac 1{\sqrt n},1\right]} g_n(x) \phi(x) \,dx \quad \text{if } p + \frac 1{\sqrt n} < 1.\]

Then, for \(p + \frac 1{\sqrt n} \geq 1\), one has

\begin{align*}
|c_n| \leq \int_{\left[-p - 1,-1\right) \cup \left(1,p + 1\right]} \frac 1n |\phi(x)| \,dx &\leq \int_{\left[-p - 1,-1\right) \cup \left(1,p + 1\right]} \frac 1n \frac{\left(1+\left(p+1\right)^2\right)^2}{1^2} \cdot \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx
\\&\leq \frac{\left(1+\left(p+1\right)^2\right)^2}{n} \cdot \int_\R \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx \xrightarrow{n \to \infty} 0
\end{align*}

and for \(p + \frac 1{\sqrt n} < 1\), one has

\begin{align*}
|c_n| \leq \int_{\left[-1,-p - \frac 1{\sqrt n}\right) \cup \left(p + \frac 1{\sqrt n},1\right]} \frac 1n |\phi(x)| \,dx &\leq \int_{\left[-1,-p\right) \cup \left(p,1\right]} \frac 1n \frac{\left(1+\left(1\right)^2\right)^2}{p^2} \cdot \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx
\\&\leq \frac 4{np^2} \cdot \int_\R \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx \xrightarrow{n \to \infty} 0,
\end{align*}

so in every case

\[\lim_{n \to \infty} c_n = 0.\]

Further, for \(n \in \N\) with \(\frac 1{\sqrt n} \leq \frac p2\), we set

\[k_n \coloneqq \int_{\left[-p + \frac 1{\sqrt n},-1\right) \cup \left(1,p - \frac 1{\sqrt n}\right]} g_n(x) \phi(x) \,dx \quad \text{if } p - \frac 1{\sqrt n} \geq 1\]

\[k_n \coloneqq - \int_{\left[-1,-p + \frac 1{\sqrt n}\right) \cup \left(p - \frac 1{\sqrt n},1\right]} g_n(x) \phi(x) \,dx \quad \text{if } p - \frac 1{\sqrt n} < 1.\]

Then, for \(p - \frac 1{\sqrt n} \geq 1\), one has

\begin{align*}
|k_n| \leq \int_{\left[-p,-1\right) \cup \left(1,p\right]} \frac 1n |\phi(x)| \,dx &\leq \int_{\left[-p,-1\right) \cup \left(1,p\right]} \frac 1n \frac{\left(1+p^2\right)^2}{1^2} \cdot \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx
\\&\leq \frac{\left(1+p^2\right)^2}{n} \cdot \int_\R \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx \xrightarrow{n \to \infty} 0
\end{align*}

and for \(p - \frac 1{\sqrt n} < 1\), one has

\begin{align*}
|k_n| \leq \int_{\left[-1,-\frac p2\right) \cup \left(\frac p2,1\right]} \frac 1n |\phi(x)| \,dx &\leq \int_{\left[-1,-\frac p2\right) \cup \left(\frac p2,1\right]} \frac 1n \frac{\left(1+\left(1\right)^2\right)^2}{\left(\frac p2\right)^2} \cdot \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx
\\&\leq \frac {16}{np^2} \cdot \int_\R \frac{x^2}{\left(1+x^2\right)^2} |\phi(x)| \,dx \xrightarrow{n \to \infty} 0,
\end{align*}

so in every case

\[\lim_{n \to \infty} k_n = 0.\]

Then, by lem:approximationZero, lem:approximationInfinity and lem:LebesguePointKonvergence, one has

\begin{align*}
\lim_{n \to \infty} \frac 1\pi \int_\R f_{p,n}(x) \,\phi(x) \,dx &= \lim_{n \to \infty} \frac 1\pi \int_{\left(-p-\frac 1{\sqrt n},-p+\frac 1{\sqrt n}\right) \cup \left(p-\frac 1{\sqrt n},p+\frac 1{\sqrt n}\right)} d_{p,n}(x) \,\phi(x) \,dx
\\&\qquad + \lim_{n \to \infty} \frac 1\pi \int_{\left(-p+\frac 1{\sqrt n},p-\frac 1{\sqrt n}\right)} \left(d_{p,n}(x)+ \frac{g_n(x)}{p^2}\right)\phi(x) \,dx
\\&\qquad + \lim_{n \to \infty} \frac 1\pi \int_{\left(-\infty,-p-\frac 1{\sqrt n}\right) \cup \left(p+\frac 1{\sqrt n},\infty\right)} \left(d_{p,n}(x)+g_n(x)\right)\phi(x) \,dx
\\&\qquad + \lim_{n \to \infty} \frac 1\pi c_n + \lim_{n \to \infty} \frac 1\pi \frac{k_n}{p^2}
\\&= \frac{\phi(p) + \phi(-p)}2 + 0 + 0 + 0 + 0
\\&= \frac{\phi(p) + \phi(-p)}2.
\end{align*}

The assertion then follows, since by cor:LebesguePointPhi almost every \(p \in \R^\times\) is a Lebesgue point of the function \(\phi\).

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