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\begin{document}

\author[Harald Grosse, Christian Krattenthaler and Paul B. Slater]
{Harald Grosse, Christian Krattenthaler and Paul B. Slater}
\address{Institut f\"ur Mathematik\\
Universit\"at Wien\\
Strudlhofgasse 4\\
A-1090 Vienna, Austria\\
e-mail: {\tt kratt@pap.univie.ac.at}\\
WWW: {\tt http://radon.mat.univie.ac.at/People/kratt}}
\address{Institut f\"ur Theoretische Physik\\
Universit\"at Wien\\
Strudlhofgasse 4\\
A-1090 Vienna, Austria\\
e-mail: {\tt Harald.Grosse@univie.ac.at}}
\address{Community and Organization Research Institute\\
University of California\\
Santa Barbara\\
CA 93106-2150\\
e-mail: slater@itp.ucsb.edu}
\title[Asymptotic Redundancies II]{Asymptotic Redundancies for Universal
Quantum Coding II}

\begin{abstract}
\end{abstract}

\maketitle

The three-dimensional convex set of $2\times 2$ density matrices
that will be the focus of our study has members representable in the form,
\begin{equation} \label{eq:6}
\rho = {\frac 1  2} \begin{pmatrix}1+z&x-iy\\
                   x+iy&1-z\end{pmatrix}\quad .
\end{equation}

\begin{theorem}\label{t15}
Let $w$ be a spherically symmetric probability density on the unit
ball, i.e., $w=w(x,y,z)$ depends only on $r=\sqrt{x^2+y^2+z^2}$.
Furthermore, let $\ze_n(w)$ be the average 
$\int_{x^2+y^2+z^2\le 1} \big(\overset n  \otimes \rho \big)\,w \,dx\,dy\,dz$.
Then the eigenvalues of 
$\ze_n(w)$ are
\begin{equation}\label{e50}
\la_h=\frac {\pi} {2^{n-1}(n-2h+1)}\int_{-1}^{1}
s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,ds, 
\quad h=0,1,\dots,\fl{\frac {n} {2}},
\end{equation}
with respective multiplicities
\begin{equation}\label{e51}
\frac {(n-2h+1)^2} {(n+1)}\binom {n+1}h,
\end{equation}
and corresponding eigenspaces 
$\{v_{h,s}(P):h\le s\le n-h,\ P$ a ballot path from $(0,0)$
to $(n,n-2h)\}$ as described in \cite[Sec.~2.2]{KrSlAA}.

The relative entropy of $\overset n  \otimes \rho$ with respect to
$\ze_n(w)$ is given by 
\begin{multline}\label{e32}   
\frac {n} {2}(1-r)\log((1-r)/2)+\frac {n}
{2}(1+r)\log((1+r)/2)\\
-\sum _{h=0} ^{\fl{n/2}}\frac {(n-2h+1)} {(n+1)}\binom {n+1}h\hskip5cm\\
\cdot\frac
{1} {2^{n+1}r}\((1+r)^{n-h+1}(1-r)^h-(1+r)^h(1-r)^{n-h+1}\)\log\la_h,
\end{multline}
with $\la_h$ as given in
{\em(\ref{e50})}.
\end{theorem}

On the basis of this theorem we can prove the following asymptotic
result.
\begin{theorem}\label{tasy}
Let $w$ be a spherically symmetric probability density on the unit
ball, i.e., $w=w(x,y,z)$ depends only on $r=\sqrt{x^2+y^2+z^2}$, with
the property that the probability is not totally concentrated at $r=0$
or $r=1$
(i.e., $w(r)$ is not just a multiple of a Dirac delta at $r=0$ or $r=1$).
For a fixed $r$ with $0\le r< 1$ such that $w$ is continuous and
nonzero at $r$, the asymptotics of the relative entropy of 
$\overset n\otimes \rho$ with respect to $\ze_n(w)$ is given by
\begin{equation}\label{a1} 
\frac {3} {2}\log n  -\frac {1} {2}- \frac {3} {2} \log 2  
- \frac {3} {2} \log \pi -\log(1-r^2)
+\frac {1} {2r}\log\(\frac {1-r} {1+r}\)
-\log w(r) +o\(1\).
\end{equation}
In the case $r=0$, this means that the asymptotics is given by the
expression {\em(\ref{a1})} in the limit $r\downarrow0$, i.e., by
\begin{equation}\label{a2} 
\frac {3} {2}\log n  -\frac {3} {2}- \frac {3} {2} \log 2  
- \frac {3} {2} \log \pi 
- \log w(0) +o\(1\).
\end{equation}

For $r=1$, in the case that $w$ is continuous and
nonzero at $1$, the asymptotics is given by
\begin{equation}\label{a3}
 2\log n-3\log 2- \log\pi -\log w(1) + o\(1\).
\end{equation}
\end{theorem}

\begin{proof} We start with the case $0<r<1$.
We concentrate first on the sum in (\ref{e32}).
Because of $\la_{n+1-h}=\la_h$, 
we have
\begin{multline}\label{a4}
\frac {1} {2^{n+1}r}
\sum _{h=0} ^{\fl{n/2}}\frac {(n-2h+1)} {(n+1)}\binom {n+1}h\hskip5cm\\
\cdot\((1+r)^{n-h+1}(1-r)^h-(1+r)^h(1-r)^{n-h+1}\)\log\la_h\\
=\frac {1} {2^{n+1}r}\sum _{h=0} ^{n+1}\frac {(n-2h+1)} {(n+1)}\binom {n+1}h
(1+r)^{n-h+1}(1-r)^h\log\la_h.
\end{multline}

Our first goal is to detect where the dominating terms in the sum
(\ref{a4}) are. As we shall see, they are the terms with $h$ close to
$n(1-r)/2$. As a first step towards the verification of this claim, 
we observe that the absolute value of $\log\la_h$, $h\ne (n+1)/2$, is bounded
by a linear function in $n$. Indeed, consider the expression for $\la_h$ as given
by (\ref{e50}) in a slightly rewritten form,
\begin{equation}\label{a5}
\la_{h}=\frac {\pi} {2^{n-1}(n-2h+1)}\int_{0}^{1}
s(1-s^2)^h\((1+s)^{n-2h+1}-(1-s)^{n-2h+1}\) w(s)\,ds.
\end{equation}
Because of $\la_{n+1-h}=\la_h$, 
it suffices to consider (\ref{a5}) for $0\le h\le
n/2$ only. Clearly, the integrand is smallest for $h=n/2$ and largest for $h=0$.
Therefore we have
\begin{multline} \notag
\frac {\pi} {2^{n-1}(n+1)}\int_{0}^{1}
s(1-s^2)^{n/2}\((1+s)-(1-s)\) w(s)\,ds\\
\le \la_h
\le \frac {\pi} {2^{n-1}}\int_{0}^{1}
s\((1+s)^{n+1}-(1-s)^{n+1}\) w(s)\,ds.
\end{multline}
Since $w(r)$ is not just a multiple of 
a Dirac delta at $r=0$ or $r=1$, there must be an $a<1$
such that $\int _{a} ^{1-a}w(s)\,ds>0$. Then we have
\begin{equation} \notag
\frac {\pi} {2^{n-1}(n+1)}
2a^2(2a-a^2)^{n/2} 
\int_{a}^{1-a} w(s)\,ds
\le \la_h
\le \frac {\pi} {2^{n-1}}
2^{n+1} \int_{0}^{1} w(s)\,ds.
\end{equation}
Clearly this implies $An\le \log\la_h\le Bn$ for some $A,B$
independent of $h$ and $n$. Phrased differently,
\begin{equation} \label{b1}
\log\la_h=O(n),
\end{equation}
as we claimed.

Now it is an easy exercise, using Stirling's approximation
and (\ref{b1}), that for\break 
$\v{h-n(1-r)/2}>\sqrt n\log n$ the summand corresponding to $h$
in (\ref{a4}) is at most of order $O(n^{-\log n+1})$, and,
hence, the contribution of the summands with 
$\v{h-n(1-r)/2}>\sqrt n\log n$ is
at most $O(n^{-\log n+2})$. Therefore the dominating terms in the sum
(\ref{a4}) are those with $\vert{h-n(1-r)/2}\vert\le \sqrt n\log n$.
Consequently, the sum (\ref{a4}) equals
{\refstepcounter{equation}\label{b3}}
\alphaeqn
\begin{align} \notag
O&\(\frac {1} {n}\)+
\frac {1} {2^{n+1}r}\sum _{\v{h-n(1-r)/2}\le \sqrt n\log n} 
\frac {(n-2h+1)} {(n+1)}\binom {n+1}h
(1+r)^{n-h+1}(1-r)^h\\
\notag
&\kern4cm
\cdot\log\bigg(\frac {\pi} {2^{n-1}(n-2h+1)}\int_{-1}^{1}
s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,ds\bigg)\\
\notag
&=O\(\frac {1} {n}\)+
\frac {1} {2^{n+1}r}\sum _{\v{h-n(1-r)/2}\le \sqrt n\log n} 
\frac {(n-2h+1)} {(n+1)}\binom {n+1}h
(1+r)^{n-h+1}(1-r)^h\\
\label{b3a}
&\kern6cm \cdot
\log\bigg(\frac {8\pi\,\Ga(h+1)\,\Ga(n-h+2)} {\Ga(n+4)}\bigg)\\
\notag
&\quad 
+\frac {1} {2^{n+1}r}\sum _{\v{h-n(1-r)/2}\le \sqrt n\log n} 
\frac {(n-2h+1)} {(n+1)}\binom {n+1}h
(1+r)^{n-h+1}(1-r)^h\\
\label{b3b}
&\kern4cm\cdot\log\int_{-1}^{1}
\frac {s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}ds
\end{align}
\reseteqn
as $n$ tends to infinity.

It is easily seen, using arguments similar to those in the previous
paragraph, that we may extend the summation in (\ref{b3a}) to
all integers $h$, at the loss of making again an error of at most
$O(n^{-\log n+2})$, an error which is absorbed by the term $O(1/n)$
in (\ref{b3}).

By additivity of logarithm we have
\begin{multline*} 
\log\bigg(\frac {8\pi\,\Ga(h+1)\,\Ga(n-h+2)} {\Ga(n+4)}\bigg)\\
=3\log2 +\log\pi +\log\Ga(h+1)+\log\Ga(n-h+2)-\log\Ga(n+4).
\end{multline*}
On the right-hand side we use Stirling's formula
\begin{equation} \label{eq:Stirling}
\log\Ga(z)=\(z-\frac {1} {2}\)\log(z)-z+\frac {1} {2}\log2
+\frac {1} {2}\log\pi+O\(\frac {1} {z}\)
\end{equation}
three times, and substitute the result in (\ref{b3a}). The
according summations, over all integers $h$, 
can be carried out or estimated by means of (\ref{a8}),
(\ref{a9}), (\ref{a13}), and (\ref{a15}) respectively. Thus, after
simplification, we
obtain
\begin{multline} \label{b4}
\frac {n} {2}(1-r)\log((1-r)/2)+\frac {n} {2}(1+r)\log((1+r)/2)\\
-\frac {3} {2}\log n
+\frac {3} {2}\log 2+\frac {3} {2}\log\pi +\log(1-r^2)+\frac {1}
{2r}\log\frac {1+r} {1-r}+\frac {1} {2}+O\(\frac {1} {n}\)
\end{multline}
for the expression (\ref{b3a}) as $n$ tends to infinity.

On the other hand, by Lemma~\ref{la3} we have, 
as long as $D=\v{h-n(1-r)/2}\le \sqrt n\log n$, that
\begin{align} \notag
\log&\!\int_{-1}^{1}
\frac {s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}ds\\
\notag
&=\log\( (w({r})+o(1))
\bigg(1+A_1\frac {D} {n}+O\(\frac {D^2} {n^2}\)+A_2\frac {D^3} {n^2}
+O\(\frac {D^4} {n^3}\)+O\(\frac {1} {n}\)\bigg)\)\\
\label{log1}
&=\log w(r)+B_1\frac {D} {n}+O\(\frac {D^2} {n^2}\)+B_2\frac {D^3} {n^2}
+O\(\frac {D^4} {n^3}\)+O\(\frac {1} {n}\)+o(1),
\end{align}
where $A_1$,
$A_2$, $B_1$, $B_2$ are explicit constants independent of $h$ (and,
hence, of $D$) and
$n$. Substitution of this in
(\ref{b3b}) and subsequent use of (\ref{a8}), (\ref{a9}), 
(\ref{a10}), (\ref{a11}), and (\ref{a12}) yields the estimation 
\begin{equation} \label{b5}
\log w(r)+O(1/n)+o(1)
\end{equation} 
for the expression (\ref{b3b}) as $n$ tends to
infinity. 

A combination of the last two observations shows that the sum in
(\ref{e32}) equals the sum of (\ref{b4}) and (\ref{b5}).
This establishes (\ref{a1}).

\medskip
Now we turn to the two exceptional cases $r=0$ and $r=1$. 

In the case $r=1$, by (\ref{e32}) the relative entropy of 
$\overset n\otimes \rho$ with respect to $\ze_n(w)$ equals
$-\log\la_0$,
$\la_0$ being given by (\ref{e50}), i.e.,
\begin{multline*} 
\frac {\pi} {2^{n-1}(n+1)}\int_{-1}^{1}
s(1+s)^{n+1} w(\v{s})\,ds\\=
\frac {8\pi} {(n+2)(n+3)}\int_{-1}^{1}
\frac {s(1+s)^{n+1} w(\v{s})\,\Ga(n+4)} {2^{n+2}(n+1)\,\Ga(n+2)}ds.
\end{multline*}
Application of Lemma~\ref{la3} with $r=1$ and $h=0$, plus a simple
computation, making use of Stirling's formula, 
show immediately that $-\log\la_0$ is approximated by
(\ref{a2}).

In the case $r=0$, the relative entropy (\ref{e32}) of 
$\overset n\otimes \rho$ with respect to $\ze_n(w)$ reduces to
\begin{equation}\notag
-n\log 2
-\frac {1} {2^n}\sum _{h=0} ^{\fl{n/2}}\frac {(n-2h+1)^2} {(n+1)}\binom {n+1}h
\log\la_h.
\end{equation}
The asymptotics of that expression can be determined in a similar way
to what was done for $0<r<1$. For the sake of brevity, we omit the derivation.
The result is (\ref{a4}). 

\medskip
This completes the proof of the Theorem.
\end{proof}


\begin{remarks}
(1) If $w$ is two times differentiable in 
a neighbourhood of $r$ then,
in view of the remark after the proof of Lemma~\ref{la3}, 
in each of\/ (\ref{a1}), (\ref{a2}), respectively
(\ref{a3}),
the error term $o(1)$ can be
replaced by $O(1/n)$.

\smallskip
(2) The assertion of Theorem~\ref{tasy} is restricted to a situation
where $w$ is nonsingular at $r$. It is however not very difficult to
extend Theorem~\ref{tasy} to singular cases as well. For example, let
us suppose that $w(r)=w_0(r)(1-r)^{-u}$, $u<1$, where $w_0(r)$ is
continuous and nonzero at $r=1$. Then the relative entropy of 
$\overset n\otimes \rho$ with respect to $\ze_n(w)$ is given by
$-\log\la_0$,
$\la_0$ being given by (\ref{e50}), i.e.,
\begin{multline*} 
\frac {\pi} {2^{n-1}(n+1)}\int_{-1}^{1}
\frac {s(1+s)^{n+1}} {(1-s^2)^u} w_0(\v{s})\,(1+\v{s})^u\,ds\\=
\frac {8\pi\Ga(n+2-u)\,\Ga(1-u)} {\Ga(n+4)}\int_{-1}^{1}
\frac {s(1+s)^{n+1}} {(1-s^2)^u} 
\frac {\Ga(n+4)\,w_0(\v{s})\,(1+\v{s})^u} 
{2^{n+2}(n+1)\,\Ga(n+2-u)\,\Ga(1-u)}ds.
\end{multline*}
Now we apply the (appropriate extension of) Lemma~\ref{la3}, with $n$
replaced by $n-2u$ and with $h=-u$. A simple computation, which uses
Stirling's formula, then shows
that $-\log\la_0$, and, hence, the relative entropy of 
$\overset n\otimes \rho$ with respect to $\ze_n(w)$ at $r=1$,
is approximated by 
\begin{equation} \label{r1}
(2-u)\log n  + (u-3) \log 2  - \log \pi - \log\Ga(1-u)
- \log w_0(1) +o\(1\).
\end{equation}
\end{remarks}

\begin{lemma}\label{la1}
We have the following summations:
\begin{equation}\label{a8}
\frac {1} {2^{n+1}r}\sum _{h=0} ^{n+1}\frac {(n-2h+1)}
{(n+1)} \binom {n+1}h
(1+r)^{n+1-h}(1-r)^h=1.
\end{equation}
\begin{equation}\label{a9}
\frac {1} {2^{n+1}r}\sum _{h=0} ^{n+1}\frac {(n-2h+1)}
{(n+1)} \binom {n+1}h
(1+r)^{n+1-h}(1-r)^hh=\frac {(1-r)(nr-1)} {2r}.
\end{equation}
\begin{multline}\label{a10}
\frac {1} {2^{n+1}r}\sum _{h=0} ^{n+1}\frac {(n-2h+1)}
{(n+1)} \binom {n+1}h
(1+r)^{n+1-h}(1-r)^h\(h-n\frac {1-r} {2}\)^2\\
=\frac {-{r}^{3}n+2 r+nr-2}{4 r}.
\end{multline}
\begin{multline}\label{a11}
\frac {1} {2^{n+1}r}\sum _{h=0} ^{n+1}\frac {(n-2h+1)}
{(n+1)} \binom {n+1}h
(1+r)^{n+1-h}(1-r)^h\(h-n\frac {1-r} {2}\)^3\\
=\frac
{-2 {r}^{4}n-3 {r}^{3}n+5 n{r}^{2}+3 nr+4 r-3 n-4}{8 r}.
\end{multline}
\begin{multline}\label{a12}
\frac {1} {2^{n+1}r}\sum _{h=0} ^{n+1}\frac {(n-2h+1)}
{(n+1)} \binom {n+1}h
(1+r)^{n+1-h}(1-r)^h\(h-n\frac {1-r} {2}\)^4\\
=\frac {3 {r}^{5}{n}^{2}-6 {r}^{5}n-8 {r}^{4}n-6 {r}^{3}{n}^{2}+4
 {r}^{3}n+20 n{r}^{2}+8 r+2 nr+3 {n}^{2}r-8-12 n}{16 r}.
\end{multline}
\begin{equation}\label{a13}
\frac {1} {2^{n+1}r}\sum _{h=-1} ^{n+1}\frac {(n-2h+1)}
{(h+1)(n+1)} \binom {n+1}{h}
(1+r)^{n+1-h}(1-r)^h=\frac {2(1+2r+nr)} {(n+1)(n+2)r(1-r)}.
\end{equation}

\end{lemma}

\begin{proof} 
In all the cases, the sums can be split into several simpler
sums, each  of which can itself be summed using
 the binomial theorem.
\end{proof} 

\begin{lemma}\label{la2}
For fixed $r$ with $0< r<1$
we have the following asymptotic expansion:
\begin{multline}\label{a15} 
\frac {1} {2^{n+1}r}\sum _{h=0} ^{n+1}\frac {(n-2h+1)}
{(n+1)} \binom {n+1}h
(1+r)^{n+1-h}(1-r)^h \(h+\frac {1} {2}\)\log(1+h)\\
=\(\frac {n} {2}(1-r)+1-\frac {1} {2r}\)\big(\log
n+\log(1-r)-\log2\big)
\\+\frac {7} {4}+\frac {r} {4}-\frac {1} {2r}+O\(\frac {1} {n}\).
\end{multline}
\end{lemma}
For a proof see \cite[Lemma~11]{KrSlAA}.

\begin{lemma}\label{la3}
Let $w$ be a (generalized) function on the unit interval.
Let $r$ be fixed with $0\le r\le 1$ such that $w$ is continuous and
nonzero at $r$. For integers $h$ with $\v{h-n(1-r)/2}\le \sqrt n\log
n$ we have
\begin{multline} \label{w1}
\int_{-1}^{1}
\frac {s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}ds\\
=(w(r)+o(1))
\bigg(1+A_1\frac {D} {n}+O\(\frac {D^2} {n^2}\)+A_2\frac {D^3} {n^2}
+O\(\frac {D^4} {n^3}\)+O\(\frac {1} {n}\)\bigg),
\end{multline}
where $D=h-n(1-r)/2$,
where $A_1$ and $A_2$ are explicit constants independent of $h$ (and,
hence, of $D$) and $n$,
and where the constants in the error terms are independent of $h$ and $n$.
\end{lemma}

\begin{proof}  Roughly speaking, the assertion of the lemma is true
because of the following facts: First of all, we have
\begin{equation} \label{w2}
\int_{-1}^{1}
\frac {s(1+s)^{n-h+1}(1-s)^h \,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}ds=1.
\end{equation}
(This is the same integrand as in (\ref{w1}), except that the term
$w(\v s)$ is missing. The evaluation (\ref{w2}) follows upon the
substitution $s\to 1-2s$ from the beta integral evaluation.) 
Secondly, suppose that $h=n(1-R)/2$, for fixed
$R$ between 0 and 1. Then, as $n$ tends to infinity,
the integrand in (\ref{w2}) approaches 0 for
$s\ne R$. Both facts combined imply that, as $n$ tends to infinity,
the integrand approaches a Dirac delta at $R$, so that the integral
in (\ref{w1}) will approach $w(R)$. Since the $h$'s which are
considered in the statement of the lemma are ``close" to $n(1-r)/2$, the
$R$ to $h$ will be close to $r$, and, thus, the integral
(\ref{w1}) will be close to $w(r)$.

Of course, the above arguments do not constitute a proof of the
lemma. A detailed justification is the subject of the
following paragraphs.

Writing $h=n(1-r)/2+D$, we transform the integrand in (\ref{w1}) 
as follows,
\begin{align*} 
&\frac {s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}\\
&\kern1cm
=\frac {s(1+s)^{n(1+r)/2+1}(1-s)^{n(1-r)/2} w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(rn-2D+1)\,\Ga(n(1-r)/2+1)\,\Ga(n(1+r)/2+2)}
\(\frac {1-s} {1+s}\)^D\\
&\kern3cm
\times
\frac {\(\frac {n(1+r)} {2}-D+2\)\(\frac {n(1+r)} {2}-D+3\)\cdots 
\(\frac {n(1+r)} {2}+1\)} 
{\(\frac {n(1-r)} {2}+1\)\(\frac {n(1-r)} {2}+2\)\cdots 
\(\frac {n(1-r)} {2}+D\)}\\
&\kern1cm
=\frac {s(1+s)^{n(1+r)/2+1}(1-s)^{n(1-r)/2} w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(rn-2D+1)\,\Ga(n(1-r)/2+1)\,\Ga(n(1+r)/2+2)}
\(\frac {(1+r)} {(1-r)}\frac {(1-s)} {(1+s)}\)^D\\
&\kern3cm
\times
\frac {\(1+\frac {-D+2} {n(1+r)/2}\)\(1+\frac {-D+3} {n(1+r)/2}\)\cdots 
\(1+\frac {1} {n(1+r)/2}\)} 
{\(1+\frac {1} {n(1-r)/2}\)\(1+\frac {2} {n(1-r)/2}\)\cdots 
\(1+\frac {D} {n(1-r)/2}\)}.
\end{align*}
In this expression we use Stirling's formula (\ref{eq:Stirling}) in
all three instances of occurence of the gamma function. Moreover, we
replace each term $(1+x)$ in the last line by
$\exp(x-x^2/2+O(x^3))$. After some further manipulation we arrive at the following
expression,
\begin{multline} \label{w3}
 w(\v{s})\frac {s(1+s)} {r(1+r)}\frac {rn} 
{(rn-2D+1)}
\sqrt{\frac {n} {2\pi(1-r^2)}}
\exp\bigg(\(\frac {n(1+r)} {2}-D\)\log\(1-\frac {r-s} {1+r}\)\\
+
\(\frac {n(1-r)} {2}+D\)\log\(1+\frac {r-s} {1-r}\)+O\(\frac {1}
{n}\)\bigg)\\
\times\,
\exp\(-\frac {2D^2} {n(1-r^2)}+a_1\frac {D} {n}+a_2\frac
{D^3} {n^2}+O\(\frac {D^4} {n^3}\)\),
\end{multline}
where $a_1$ and $a_2$ are explicit constants independent of $D$ and $n$.

The next step consists of the analysis of the exponential
\begin{equation} \label{w4}
\exp\(\(\frac {n(1+r)} {2}-D\)\log\(1-\frac {r-s} {1+r}\)+
\(\frac {n(1-r)} {2}+D\)\log\(1+\frac {r-s} {1-r}\)\)
\end{equation}
in the middle of expression (\ref{w3}) as it behaves as a function of $s$.
The derivative with respect to $s$ of the exponent is
\begin{multline*} 
\frac {d} {ds}\(\(\frac {n(1+r)} {2}-D\)\log\(\frac {1+s} {1+r}\)+
\(\frac {n(1-r)} {2}+D\)\log\(\frac {1-s} {1-r}\)\)\\=
\frac {n} {1-s^2}\(r-s-2{D}/ {n}\).
\end{multline*}
So it is increasing for $s<r-2D/n$ and decreasing for $s>r-2D/n$. In
addition, we have $\v D=\v{h-n(1-r)/2}\le \sqrt n\log n$ by assumption.
Applying Taylor expansion of the logarithm in the expression
(\ref{w4}), 
\begin{equation} \label{w5}
\exp\(-2D\frac {(s-r)} {(1-r^2)}+b_1D(s-r)^2
-\frac {n} {2}\frac {(s-r)^2} {(1-r^2)}+
b_2n(s-r)^3+O\(n(s-r)^4\)\),
\end{equation}
we see that 
it is of order $O(n^{-\log n})$ for $s=r\pm 5\log n/\sqrt
n$. Hence, it must be of order $O(n^{-\log n})$ for all
$s$ with $\v{s-r}> 5\log n/\sqrt n$. Thus, the complete expression
(\ref{w3}) is of order $w(\v s)O(n^{-\log n+1/2})$ for all
$s$ with $\v{s-r}> 5\log n/\sqrt n$. This implies that we have
\begin{multline} \label{w6}
\underset{\v s\le1}{\underset{\v{s-r}> 5\log n/\sqrt n}{\int}}
\frac {s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}ds\\=
O(n^{-\log n+1/2})\underset{\v s\le1}{\underset{\v{s-r}> 5\log n/\sqrt
n}{\int}}w(\v s)\,ds=O(n^{-\log n+1/2}).
\end{multline}
Hence, the dominating part of the integral (\ref{w1}) will come from
the contribution of the range where $\v{s-r}\le 5\log n/\sqrt n$.
By virtue of (\ref{w3}) and (\ref{w5}), this is
\begin{multline} \label{w7}
\underset{\v s\le1}{\underset{\v{s-r}\le 5\log n/\sqrt n}{\int}}
\frac {s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}ds\\=
\underset{\v s\le1}{\underset{\v{s-r}\le 5\log n/\sqrt n}{\int}}
 w(\v{s})\frac {s(1+s)} {r(1+r)}
\sqrt{\frac {n} {2\pi(1-r^2)}}\hskip6cm\\
\times\,
\exp\bigg(-\frac {n} {2(1-r^2)}\(s-r+\frac {2D} {n}\)^2
+a_1\frac {D} {n}+b_1D(s-r)^2+b_2n(s-r)^3\\
+a_2\frac {D^3} {n^2}+O\(\frac {D^4} {n^3}\)+O\(n(s-r)^4\)+O\(\frac
{1} {n}\)\bigg)ds.
\end{multline}

By assumption, $w$ is continuous at $r$. Accordingly, we
may replace $w(\v{s})$ in (\ref{w7}) by $w(r)+o(1)$ as $n\to\infty$.
Furthermore, if we extend the range of integration on the right-hand side 
to $-\infty< s< \infty$ then we make an error of at most $O(1/n)$.

Now, we have 
\begin{multline*} 
\int _{-\infty} ^{\infty}s(1+s)
\exp\bigg(-\frac {n} {2(1-r^2)}\(s-r+\frac {2D} {n}\)^2\bigg)ds\\
=\sqrt{\frac {2\pi(1-r^2)} {n}}\(r(1+r)-\frac {4Dr} {n}+\frac {4D^2}
{n^2}+\frac {1-r^2} {n}\),
\end{multline*}
\begin{multline*} 
\int _{-\infty} ^{\infty}(s-r)^2s(1+s)
\exp\bigg(-\frac {n} {2(1-r^2)}\(s-r+\frac {2D} {n}\)^2\bigg)ds\\
=\sqrt{\frac {2\pi(1-r^2)} {n}}
\bigg(\frac {r(1+r)(1-r^2)} {n}-\frac {6D(1+2r)(1-r^2)} {n^2}\\
+\frac {4D^2r(1+r)} {n^2}+O\(\frac {D^3} {n^3}\)\bigg),
\end{multline*}
\begin{multline*} 
\int _{-\infty} ^{\infty}(s-r)^3s(1+s)
\exp\bigg(-\frac {n} {2(1-r^2)}\(s-r+\frac {2D} {n}\)^2\bigg)ds\\
=\sqrt{\frac {2\pi(1-r^2)} {n}}
\bigg(-\frac {6Dr(1+r)(1-r^2)} {n^2}+O\(\frac {D^2} {n^3}\)\\
-\frac {8D^3r(1+r)} {n^3}+O\(\frac {D^4} {n^4}\)+O\(\frac {1}
{n^2}\)\bigg),
\end{multline*}
\begin{multline*} 
\int _{-\infty} ^{\infty}(s-r)^4s(1+s)
\exp\bigg(-\frac {n} {2(1-r^2)}\(s-r+\frac {2D} {n}\)^2\bigg)ds\\
=\sqrt{\frac {2\pi(1-r^2)} {n}}
\(O\(\frac {1} {n^2}\)+O\(\frac {D^2} {n^3}\)+O\(\frac {D^4}
{n^4}\)\).
\end{multline*}
Thus, from (\ref{w7}) (with $w(\v{s})$ replaced by $w(r)+o(1)$ and
range of integration extended to all s with from $-\infty$ to $\infty$) 
we derive
\begin{multline} \label{w8}
\underset{\v s\le1}{\underset{\v{s-r}\le 5\log n/\sqrt n}{\int}}
\frac {s(1+s)^{n-h+1}(1-s)^h w(\v{s})\,\Ga(n+4)} 
{2^{n+2}(n-2h+1)\,\Ga(h+1)\,\Ga(n-h+2)}ds\\=
 (w({r})+o(1))
\bigg(1+A_1\frac {D} {n}+O\(\frac {D^2} {n^2}\)+A_2\frac {D^3} {n^2}
+O\(\frac {D^4} {n^3}\)+O\(\frac {1} {n}\)\bigg),
\end{multline}
where $A_1$ and $A_2$ are explicit constants independent of $h$ and $n$.
A combination of (\ref{w6}) and (\ref{w8}) gives (\ref{w1})
immediately. This completes the proof of the lemma.
\end{proof}

\begin{remark}
If $w$ is two times differentiable in 
a neighbourhood of $r$ then we could have replaced $w(\v{s})$ in (\ref{w7})
by $w(r)+w'(r)(s-r)+O((s-r)^2)$ (instead of by $w(r)+o(1)$).
This would result in the conclusion that in that case the error term
$o(1)$ in (\ref{w1}) can be replaced by $O(1/n)$.
\end{remark}

\begin{thebibliography}{10}

\bibitem{KrSlAA}
C.    Krattenthaler and P. Slater, {\it Asymptotic redundancies for 
universal quantum coding}, preprint.

\end{thebibliography}


\end{document}
