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\begin{document}

\section{A lemma for dissipative $2 \times 2$ matrices}

The goal of these notes is to prove an analogue of Lemma 4.1 from \cite{}, i.e., 
\begin{lemma}
For every $s \in (0,1)$, there exists a number $C(s)< \infty$ for which
\begin{equation}
\int_I \left\| \left( A + x \idty \right)^{-1} \right\|^s \, dx \leq C(s) 
\end{equation} 
for every dissipative $2 \times 2$ matrix $A$ and every unit interval $I$.
\end{lemma}

The analogue corresponds to replacing $\idty$ with the matrix $J$ given by
\begin{equation}
J = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \, .
\end{equation}
\begin{conj} For every $s \in (0,1)$, there exists a number $C(s)< \infty$ for which
\begin{equation}
\int_I \left\| \left( A + x J \right)^{-1} \right\|^s \, dx \leq C(s) 
\end{equation} 
for every dissipative $2 \times 2$ matrix $A$ and every unit interval $I$.
\end{conj}

Looking at the proof of Lemma 4.1 in \cite{}, the first step in the argument is
to replace a general dissipative $A$ with a unitarily equivalent, upper triangular
matrix $\tilde{A}$. In other words, use that there exists a unitary matrix $U$ for which
\begin{equation}
UAU^* = \tilde{A}
\end{equation} 
in which case, 
\begin{equation}
\left\| \left( A + x \idty \right)^{-1} \right\| = \left\| U \left( A + x \idty \right)^{-1} U^* \right\| = \left\| \left[ U \left( A + x \idty \right) U^* \right]^{-1} \right\| = \left\| \left( \tilde{A} + x \idty \right)^{-1} \right\| \, ,
\end{equation}
and therefore it suffices to look at upper-triangular, dissipative matrices. Unfortunately, $UJU^* \neq J$, and so this
part of the argument breaks down in the proof of the analogue. Here is a proposition.
\begin{prop}
For every $s \in (0,1)$, there exists a number $C(s)< \infty$ for which
\begin{equation} \label{prop:2by2}
\int_I \left\| \left( A + x J \right)^{-1} \right\|^s \, dx \leq C(s) 
\end{equation} 
for every upper-triangular, dissipative $2 \times 2$ matrix $A$ and every unit interval $I$.
\end{prop}
\begin{proof}
Let us write
\begin{equation}
A = \left( \begin{array}{cc} a & b \\ 0 & d \end{array} \right) \, .
\end{equation}
The fact that $A$ is dissipative means that 
\begin{equation}
{\rm Im} \left[ A \right] =   \left( \begin{array}{cc} {\rm Im}[a] & \frac{b}{2i} \\ - \frac{\overline{b}}{2i} & {\rm Im}[d] \end{array} \right)
\end{equation}
is a positive matrix and hence
\begin{equation} \label{eq:det}
{\rm det} \left( {\rm Im} [A] \right) = {\rm Im}[a] {\rm Im}[d] - \frac{|b|^2}{4} \geq 0.
\end{equation}
It is easy to calculate
\begin{equation}
\left(A + x J \right)^{-1} = \left( \begin{array}{cc} a +x & b \\ 0 & d -x \end{array} \right)^{-1} = \left( \begin{array}{cc} \frac{1}{a+x} & \frac{-b}{(a+x)(d-x)} \\ 0 & \frac{1}{d-x} \end{array} \right) \, .
\end{equation}
As in the proof of Lemma 4.1 of \cite{}, the bound claimed in (\ref{prop:2by2}) follows if we prove the integral bound
for each of the entires of the above matrix.  For the diagonal entries this is clear. In fact, if $b=0$, we are done.

Otherwise, we need only consider the upper right entry.
It is easy to see that
\begin{equation}
\frac{b}{(a+x)(d-x)} = \frac{b}{a+d} \left( \frac{1}{a+x} + \frac{1}{d-x} \right)
\end{equation}
and therefore,
\begin{equation}
\int_I \left| \frac{b}{(a+x)(d-x)} \right|^s \, dx \leq \left| \frac{b}{a+d} \right|^s \int_I  \left( \frac{1}{|a+x|^s} + \frac{1}{|d-x|^s} \right) \, dx \, .
\end{equation}
Clearly,
\begin{equation}
\left| \frac{b}{a+d} \right|^2 \leq \frac{|b|^2}{( {\rm Im}[a] +{\rm Im}[d])^2} \leq \frac{|b|^2}{2 {\rm Im}[a] {\rm Im}[d]} \leq 2 \,
\end{equation}
by (\ref{eq:det}). This completes the proof of the proposition.
\end{proof}

Let us now consider the self-adjoint case.
Suppose that
\begin{equation}
A =  \left( \begin{array}{cc} a & b \\ \overline{b} & d \end{array} \right) \, 
\end{equation}
with $a,d \in \mathbb{R}$ and $b \in \mathbb{C}$. It is easy to
calculate
\begin{equation}
( A + x J)^{-1} = \left( \begin{array}{cc} a +x & b \\ \overline{b} & d -x \end{array} \right)^{-1} = \frac{1}{(a+x)(d-x) - |b|^2} \left( \begin{array}{cc} d-x & -b \\ - \overline{b} & a+x \end{array} \right) \, .
\end{equation}

The polynomial
\begin{equation}
p(x) = (a+x)(d-x) - |b|^2 = - \left( x^2 +(a-d)x + (|b|^2-ad) \right) = -(x- x_+)(x-x_-) \,
\end{equation}
with
\begin{equation}
x_{\pm} =  \frac{1}{2} \left( d-a \pm \sqrt{(a+d)^2-4|b|^2} \right) \, .
\end{equation}
In this case, the upper right off-diagonal entry is
\begin{equation}
\frac{b}{(x-x_+)(x-x_-)} = \frac{b}{x_+-x_-} \left( \frac{1}{x-x_+} - \frac{1}{x-x_-} \right) \, .
\end{equation}

Let $\alpha$ and $\beta$ be real parameters.
Define the $2 \times 2$ matrices
\begin{equation}
U_{\alpha, \beta} = \left( \begin{array}{cc} e^{i \alpha} \cosh( \beta) &  e^{-i \alpha} \sinh(\beta) \\ e^{i \alpha} \sinh( \beta) & e^{-i \alpha} \cosh(\beta) \end{array} \right) \, .
\end{equation}
Observe that ${\rm det}(U_{\alpha, \beta}) =1$ and
\begin{equation}
U_{\alpha, \beta}^* = \left( \begin{array}{cc} e^{-i \alpha} \cosh( \beta) &  e^{-i \alpha} \sinh(\beta) \\ e^{i \alpha} \sinh( \beta) & e^{i \alpha} \cosh(\beta) \end{array} \right) 
\quad \mbox{whereas} \quad U_{\alpha, \beta}^{-1} = \left( \begin{array}{cc} e^{-i \alpha} \cosh( \beta) &  -e^{-i \alpha} \sinh(\beta) \\ -e^{i \alpha} \sinh( \beta) & e^{i \alpha} \cosh(\beta) \end{array} \right) \, .
\end{equation}
Moreover, $U_{\alpha, \beta} J U_{\alpha, \beta}^* = J$.
\end{document}