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\begin{document}
\begin{center}
Strong nonnegative linearization\\
of orthogonal polynomials\\ \bigskip
Ryszard Szwarc \\University of Wrocław, Poland
\end{center}

\begin{slide}
One of the main problems in the theory of orthogonal polynomials is to determine
whether any expansion of the product  of two orthogonal polynomials
in terms of these polynomials has nonnegative coefficients. We want to decide 
which orthogonal systems $\{p_n\}_{n=0}^N$  have the property
$$p_n(x)p_m(x)=\sum c(n,m,k)p_k(x)$$ 
with nonnegative coefficients $c(n,m,k)$ for every  $n,m$ and $k.$
Numerous classical orthogonal polynomials as well as their $q$-analogues satisfy 
nonnegative linearization property (Askey, Gasper, Rahman).
There are many criteria for nonnegative linearization given in terms of the 
coefficients of the recurrence relation the orthogonal polynomials satisfy, that can be applied to general orthogonal polynomials
systems (Askey, Sz.).
These criteria are based on the connection between the linearization property and a 
certain discrete boundary value problem of hyperbolic type.
\end{slide}
\begin{slide}
Let orthogonal polynomials $\{p_n\}_{n=0}^N,$ where $N$ may be infinite, satisfy
$$xp_n=\gamma_np_{n+1}+\beta_np_n+\alpha_np_{n-1},\quad 0\le n\le N,
$$
where $\gamma_n, \alpha_{n+1}>0,$ $\beta_n$ are real numbers and $\alpha_0=\gamma_N=p_{-1}=0.$
Assume that  for $0\le n<N$ we have \vspace{-30pt}
\begin{enumerate}\itemsep -5pt
\item[(i)] $\alpha_n\le \alpha_{n+1}.$ 
\item[(ii)] $\beta_n\le \beta_{n+1}.$
\item[(iii)] $\alpha_n+\gamma_n\le \alpha_{n+1}+\gamma_{n+1}.$  
\item[(iv)] $\alpha_n\le \gamma_{n}$ for $n<N.$
\end{enumerate}\vspace{-30pt}
Then $c(n,m,k)\ge 0$ for any $n,m,k$ (Sz1992).
\vspace{30pt}

This criterion yields nonnegative linearization
for the associated polynomials of any order.
\end{slide}
\begin{slide}
There is a wide class of orthogonal polynomials which resist any general criteria known 
so far. These are finite systems of orthogonal polynomials. The simplest family
consists of Krawtchouk polynomials, orthogonal with respect to the binomial 
distribution
$$\mu = \sum_{n=0}^N {N\choose n}p^nq^{N-n}\delta_n, \quad q=1-p,$$
where $0< p<1.$ Upon normalization $$K_n(0)=(-1)^n(p/q)^{n/2}$$ they satisfy
\begin{align*}xK_n=&\sqrt{pq}\,(N-n)K_{n+1}+[p(N-n)+qn]K_n\\
&+\sqrt{pq}\,nK_{n-1}, \quad 0\le n\le N.
\end{align*}
Eagleson (1960) has shown that $K_n$ admit nonnegative linearization if and only if 
$0< p \le {1\over 2}.$
\end{slide}
\begin{slide}
When we apply our criterion to this case the assumptions (i)--(iii) are satisfied
for $0<p\le {1\over 2}.$ The assumption (iv) is valid only for $n\le N/2.$\vspace{-20pt}

Let's modify the recurrence relation by removing the middle term. Then we obtain 
$$x\widetilde{K}_n=\sqrt{pq}\,(N-n)\widetilde{K}_{n+1} 
+\sqrt{pq}\,n\widetilde{K}_{n-1}, \ 0\le n\le N.$$
It is possible to show nonnegative linearization for these polynomials by using
the criterion given previously and some symmetry property of this new system.\vspace{-20pt}

Now the following question arises. Given a system of polynomials
satisfying
$$xp_n=\gamma_np_{n+1}+\beta_np_n+\alpha_np_{n-1}.$$
What property should this system satisfy so that for any 
nondecreasing sequence $\varepsilon_n$ of real numbers the new
system
$$xq_n= \gamma_nq_{n+1}+(\beta_n+\varepsilon_n)q_n+\alpha_nq_{n-1}$$
satisfied nonnegative linearization property ?
Clearly the system $p_n$ should  satisfy nonnegative linearization property, but
it is not sufficient.
\end{slide}
\begin{slide}
This new property will be called strong nonnegative linearization.
Unfortunately the polynomials $\widetilde{K}_n$ do not satisfy it. Indeed, let
$$\varepsilon_n=\left\{\begin{array}{ll}
0 & \quad0\le n\le N-1,\\
\varepsilon &\quad n=N.
\end{array}\right.$$
Let
$$xq_n=(N-n)q_{n+1}+\varepsilon_nq_n+nq_{n-1}.$$
Then the nonnegative linearization fails for\\ $0<\varepsilon <\sqrt{N-2}.$
Therefore the finite system case remains unsolved.
\end{slide}
\begin{slide}
Let $p_n^{(l)}(x)$ denote the associated polynomials of order $l.$
By definition they satisfy the shifted recurrence relation
$$xp_n^{(l)}=\gamma_{n+l}p_{n+1}^{(l)}+\beta_{n+l}p_n^{(l)}+
\alpha_{n+l}p_{n-1}^{(l)}.$$
The system $p_n$ satisfy the strong nonnegative linearization property
if and only if 
$$p_n^{(l)}p_m^{(l)}=\sum_{k=|n-m|}^{n+m}C_l(n,m,k)p_k^{(l)},$$
with nonnegative coefficients $C_l(n,m,k)$ for all $n,m,k$ and $l\ge 0.$\\
For convenience we will use the polynomials $p_n^{[l]}$ satisfying
\begin{align*}
xp_n^{[l]}&=\gamma_np_{n+1}^{[l]}+\beta_np_n^{[l]}+\alpha_np_{n-1}^{[l]},\quad
n\ge l+1,\\
p_0^{[l]}&=p_1^{[l]}=\ldots =p_l^{[l]}=0,\ p_{l+1}^{[l]}={1\over \gamma_l}.
\end{align*}
Then we have 
$$p_n^{[l]}=p_{n-l-1}^{(l+1)}.$$
The advantage is that the polynomials $p_n^{[l]}$ satisfy  the same recurrence 
relation for each $l.$
\end{slide}
\begin{slide}
We have
$$p_n^{[l]}p_m^{[l]}=\sum_{k=|n-m|+l+1}^{n+m-l-1}c_l(n,m,k)p^{[l]}_k.$$
The nonnegativity of the coefficients $C_l(n,m,k),$  is equivalent
to the nonnegativity of the linearization coefficients $c(n,m,k)=C_0(n,m,k)$
and that of  $c_l(n,m,k)$ because
$$c_l(n,m,k)=C_{l+1}(n-l-1,m-l-1,k-l-1).$$ 

We are going to show three equivalent conditions for each of the properties
nonnegative linearization and strong nonnegative linearization. \\
\end{slide}
\begin{slide}
Let $u(n,m)$ be a matrix of complex numbers
indexed by $0\le n,m< N.$ Define two operators $L_1$ and $L_2$
acting on such matrices by the rule
\begin{eqnarray*}
(L_1u)(n,m)&=&\gamma_{n}u(n+1,m)\\
&&+\beta_nu(n,m)+\alpha_{n}u(n-1,m),\\
(L_2u)(n,m)&=&\gamma_{m}u(n,m+1)
\\&&+\beta_mu(n,m)+\alpha_{m}u(n,m-1).
\end{eqnarray*}
Let $$H=L_1-L_2.$$
\end{slide}
\begin{slide}
\begin{pro}
\begin{enumerate}
\item[(i)]
The polynomials $p_n$ admit nonnegative product linearization if and only if
every matrix $
u=\{u(n,m)\}$ such that
$$
\left\{\begin{array}{rcl}Hu(n,m)&=&0,\ \mbox{for}\ 0\le m<n< N\\
u(n,0)&\ge&  0, \ \mbox{for}\ 0\le n\le N,
\end{array}\right.
$$
satisfies $u(n,m)\ge 0$ for $0\le m\le n\le N.$
\bigskip

\item[(ii)]
The polynomials $\{p_n\}_{n=0}^N$ admit strong nonnegative product linearization if and only if
every matrix $
u=\{u(n,m)\}$ such that
$$
\left\{\begin{array}{rcl}Hu(n,m)&\le &0,\ \mbox{for}\ 0\le m<n< N\\
u(n,0)&\ge&  0, \ \mbox{for}\ 0\le n\le N,
\end{array}\right.
$$
satisfies $u(n,m)\ge 0$ for $0\le m\le n\le  N.$
\end{enumerate}
\end{pro}
\end{slide}
\begin{slide}
The proof of both propositions follows from the lemma.
\begin{lem}
 Given a matrix $v=\{v(n,m)\}_{n> m\ge 0}$ and a sequence
$f=\{f(n)\}_{n\ge 0}.$ Let a matrix $u=\{u(n,m)\}_{n\ge m\ge 0}$ 
satisfy
\begin{align*}
Hu(n,m)&=v(n,m), &\mbox{for\ }& 0\le m<n<N,\\
u(n,0)&=f(n), &\mbox{for\ }&0\le n\le N.
\end{align*}
Then
\begin{multline*}
u(n,m)=-\sum_{k> l\ge 0}v(k,l)c_l(n,m,k)\\ +\sum_{k\ge 0}f(k)c(n,m,k).
\end{multline*}
\end{lem}
The summations are finite because $c_l(n,m,k)\neq 0$ implies
$|n-m|\le k+l+1\le n+m.$
\end{slide}
\begin{slide}
For each point $(n,m)$ with $1\le m\le n\le N,$ let $\Delta_{n,m}$ denote the set of lattice 
points in the plane defined by
$$\Delta_{n,m}=\{(i,j)\,|\,0\le j\le i\le N,\ 
|n-i|<m-j\}.$$
The set $\Delta_{n,m}$ is depicted in below for $n+m\le N.$ (the points in $\Delta_{n,m}$ 
are marked with empty circles).
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\end{slide}
\begin{slide}
In case $N$ is finite and $n+m> N$ the corresponding picture is
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\end{slide}
\begin{slide}
Let $H^*$ denote  the  adjoint operator to $H$ with respect to the inner product of matrices
$$\langle u,v\rangle=\sum_{n,m=0}^{N-1} u(n,m)\overline{v(n,m)}.$$ This operator  acts
according to
\begin{multline*}
(H^*v)(n,m)=\\ \alpha_{n+1}v(n+1,m)+\beta_nv(n,m)+\gamma_{n-1}v(n-1,m)\\-\alpha_{m+1}v(n,m+1)
-\beta_mv(n,m)-\gamma_{m-1}v(n,m-1).
\end{multline*}
\end{slide}
\begin{slide}
\begin{thm}
\begin{enumerate}
\item[(a)]
The orthogonal polynomials $\{p_n\}_{n=0}^{N}$   admit nonnegative linearization if 
 for  every $(n,m),$ with $1\le m\le n\le N,$ there exists a matrix $v(i,j)$ such 
that 
\begin{enumerate}\itemsep -5pt
\item[\rm (i)] $\supp v\subset \Delta_{n,m}.$
\item[\rm (ii)] $(H^*v)(n,m)<0.$
\item[\rm (iii)] $(H^*v)(i,j)\ge 0$ for $(i,j)\neq (n,m).$
\end{enumerate}
\item[(b)]
The orthogonal polynomials $\{p_n\}_{n=0}^{N}$   admit strong  nonnegative linearization   if for  every $(n,m),$ with $1\le m\le n\le N,$ there exists a matrix $v(i,j)$ such 
that 
\begin{enumerate}\itemsep -5pt
\item[\rm (i)] $\supp v\subset \Delta_{n,m}.$
\item[\rm (ii)] $(H^*v)(n,m)<0.$
\item[\rm (iii)] $(H^*v)(i,j)\ge 0$ for $(i,j)\neq (n,m).$
\item[\rm (iv)] $v(i,j)\ge 0$ for $(i,j)\in \Delta_{n,m}.$
\end{enumerate}
\end{enumerate}
\end{thm}
\end{slide}
\begin{slide}
Let $v_{n,m}$ denote a  matrix such that
\begin{eqnarray*}
\supp v_{n,m}&\subset &\Delta_{n,m},\\
(H^*v_{n,m})(n,m)&=&-1,\\
(H^*v_{n,m})(i,j)&=&0, \ \mbox{for}\ 0<j<m.
\end{eqnarray*}
This is a special choice of the matrix which satisfies 
conditions (i), (ii) and partially (iii), of Theorem 1.
The matrix $v_{n,m}$ is uniquely determined. Moreover
we have the following.
\bigskip

\begin{thm}\label{t1}
For any $n\ge m\ge 0$ and $k> l\ge 0$ we have
$$v_{n,m}(k,l)=c_l(n,m,k).$$
Moreover
$$H^*v_{n,m}=-\delta_{(n,m)}+\sum_{k=n-m}^{n+m}c(n,m,k)\delta_{(k,0)}.$$
\end{thm}
In this way the conditions given in Theorem 1 become equivalent to nonnegative (or strong nonnegative)
linearization.
\end{slide}
\begin{slide}
By Theorem 2 the following becomes evident.
\bigskip

\begin{thm}
\begin{enumerate}
\item[(a)]
The orthogonal polynomials $\{p_n\}_{n=0}^{N}$   admit nonnegative linearization if 
and only if for  every $(n,m),$ with $1\le m\le n\le N,$ 
the matrix $v_{n,m}$ satisfies\\
$(H^*v_{n,m})(j,0)\ge 0.$
\item[(b)]
The orthogonal polynomials $\{p_n\}_{n=0}^{N}$   admit strong nonnegative linearization if 
and only if for  every $(n,m),$ with $1\le m\le n\le N,$ 
the matrix $v_{n,m}$ satisfies
$(H^*v_{n,m})(j,0)\ge 0$ and $v_{n,m}(i,j)\ge 0$ for $(i,j)\in \Delta_{n,m}.$
\end{enumerate}
\end{thm}
\end{slide}
\begin{slide}
{\it Proof of Theorem 1(b)} \\
Let $u=\{u(n,m)\}_{n\ge m\ge 0}$ satisfy  
\begin{align*}
(Hu)(n,m)&\le 0,\ \mbox{for\ }n>m\ge 0,\\
u(n,0)&\ge 0.
\end{align*} 
  We will show that
$u(n,m)\ge 0,$ by induction on $m.$ Assume that $u(i,j)\ge 0$ for $j<m.$
Let $v$ be a matrix satisfying the assumptions  of Theorem~1(b). 
 Then
\begin{multline*}
0\ge \langle Hu,v\rangle =\langle u,H^*v\rangle\\=
u(n,m)(H^*v)(n,m)+\sum_{\scriptstyle i\ge j \ge 0 \atop j<m} u(i,j)(H^*v)(i,j)
\end{multline*}
Therefore
$$-u(n,m)(H^*v)(n,m)\ge \sum_{\scriptstyle i\ge j \ge 0 \atop j<m} u(i,j)(H^*v)(i,j),$$
and the conclusion follows.\qed

\end{slide}
\begin{slide}
\begin{thm}\label{new}
  Assume that \vspace{-15pt}
\begin{enumerate}
\item[(i)] $\beta_m\le \beta_n$ for $m\le n.$\vspace{-15pt}
\item[(ii)] $\alpha_m\le \alpha_n$ for $m<n.$\vspace{-15pt}
\item[(iii)] $\alpha_m+\gamma_m\le \alpha_n+\gamma_n$ for $m<n.$\vspace{-15pt}
\item[(iv)] $\alpha_m\le \gamma_n$ for $m\le n.$\vspace{-15pt}
\end{enumerate}
Then the system $\{p_n\}_{n=0}^\infty$ satisfies the strong nonnegative
linearization property.
\end{thm}
 
\end{slide}
\begin{slide}
It suffices to construct for every $(n,m)$ with $n\ge m,$ a  matrix $v$ 
  satisfying the  assumptions of Theorem 1. 
Fix $(n,m).$ Define the matrix $v$ according to the following.
$$v(i,j)=
\left\{ \begin{array}{ll}
c_ic_j& (i,j)\in \Delta_{n,m}, \ n+m-i-j\ \mbox{odd}\\
0 & \mbox{otherwise}
\end{array}\right.
$$
where  
$$c_0=1,\qquad c_i={\gamma_{0}\gamma_{1}\ldots\gamma_{i-1}\over 
\alpha_{1}\alpha_{2}\ldots\alpha_{i}}.$$
The points in the support of $v$  are marked by empty circles in the picture 
below.
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\end{slide}
\begin{slide}
Then  $\supp H^*v$ consists of the points marked by
$\circ , \bullet , \triangleleft ,\triangleright$ and $\diamond.$ A straightforward
computation gives the following.
 $${(H^*v)(i,j)\over c_ic_j}=
\left\{ 
\begin{array}{ll}
-\alpha_m  &(i,j)=(n,m)\\
\beta_i-\beta_j & (i,j) \ - \ \circ \\
\alpha_{i}+\gamma_i-\alpha_j-\gamma_j & (i,j)\ - \ \bullet\\
\alpha_i-\alpha_j & (i,j)\ - \ \triangleright\\
\gamma_i-\alpha_j & (i,j)\ - \ \triangleleft
\end{array}\right.$$
\end{slide}
\begin{slide}
\begin{thm}
Assume orthogonal polynomial system $\{p_n\}_{n=0}^N$ satisfies   strong
nonnegative linearization property.
 Let $\varepsilon_n$
be a nondecreasing sequence. Let $q_n$ be a sequence of polynomials satisfying
the perturbed recurrence relation
$$xq_n=\gamma_nq_{n+1}+(\beta_n+\varepsilon_n)q_n+\alpha_nq_{n-1},$$
for $n\ge 0.$ Then the system $\{q_n\}_{n=0}^N$ satisfies strong nonnegative
linearization property.
\end{thm} 

{\it Proof.}   We will make use of Theorem 1(b). 
 Let $H$ and $H_{\varepsilon}$ denote the hyperbolic operators corresponding to
the unperturbed and perturbed system, respectively. For any matrix $v(i,j)$ we have
$$
(H_{\varepsilon}^*v)(i,j)=(H^*v)(i,j)+(\varepsilon_i-\varepsilon_j)v(i,j).$$
 
By assumptions for any $n\ge m\ge 0,$ there exists a  matrix $v$ 
satisfying the assumptions of Theorem 1(b), with respect to
$H.$   
\end{slide}
\begin{slide} 
 The same matrix $v$ satisfies these assumptions with respect to
$H_\varepsilon.$ Indeed, the assumptions (i) and (iv) do not depend on the perturbation.
Since $v(n,m)=0$ the assumption (ii) is not affected, as well.
Concerning (iii), since $v\ge 0$ and $\varepsilon_n $ is nondecreasing we have
$$(H_{\varepsilon}^*v)(i,j)\ge (H^*v)(i,j)\ge 0,$$
for $i\ge j\ge 0$ and $j<m.$  Hence the perturbed system of polynomials satisfies the strong
nonnegative linearization property \qed
\medskip

This theorem is not valid for standard nonnegative linearization property
e.g. the Krawtchouk polynomials case.
\end{slide}
\begin{slide}
 

Since
$$\gamma_0p_1p_n=\gamma_np_{n+1}+(\beta_n-\beta_0)p_n+\alpha_np_{n-1},$$
nonnegative linearization requires $\beta_n\ge \beta_0.$
There are many examples of orthogonal polynomial systems satisfying
 nonnegative linearization for which
$\beta_1\ge \beta_2\ge ... \ge \beta_0.$ For example the Jacobi polynomials
with $\alpha> \beta>-1$ and $-1\le \alpha+\beta<0,$ are such. Therefore the associated polynomials
do not satisfy nonnegative linearization. Moreover the following holds.
\bigskip

\begin{pro}
Assume orthogonal polynomial system $\{p_n\}_{n=0}^N$ satisfies   strong
nonnegative linearization property and
$$xp_n=\gamma_np_{n+1}+\beta_np_n+\alpha_np_{n-1}.$$
Then $\beta_n$ is nondecreasing.
\end{pro}
\end{slide}
\begin{slide}
\begin{pro}
The Jacobi polynomials satisfy  strong nonnegative linearization property if and 
only if either $\alpha> \beta>-1$ and $\alpha+\beta\ge 0$ or $\alpha=\beta\ge- {1\over 2}.$
 \end{pro}
By Gasper's result the Jacobi polynomials satisfy nonnegative linearization property
if $\alpha\ge \beta>-1$ and $c(2,2,2)\ge 0.$ In particular the conditions
$\alpha\ge \beta>-1$ and $\alpha+\beta\ge -1$ are sufficient.
\end{slide}
\begin{slide}\setlength\textwidth{8in}
\setlength\textheight{11in}
\setlength\topmargin{-3.0cm}
\setlength\oddsidemargin{-2.2cm}
\begin{tabular}{|p{3.7in}|p{3.7in}|}
\hline
 Nonnegative \par linearization & Strong nonnegative \par linearization\\
\hline & \\
 $c(n,m,k)\ge 0$ & $c(n,m,k)\ge 0$\par
$c_l(n,m,k)\ge 0$\\ &\\
\hline &\\
 $\left.\begin{array}{r}
Hu(n,m)=0\\
n>m\ge 0\\
u(n,0)\ge 0
\end{array}\right\}\Rightarrow u\ge 0 $&  $\left.\begin{array}{r}
Hu(n,m)\le 0\\
n>m\ge 0\\
u(n,0)\ge 0
\end{array}\right\}\Rightarrow u\ge 0 $\\ &\\ \hline &\\
 For $n\ge m\ge 0$ there exists a matrix $v$ such that
\begin{enumerate}
\item[(i)] $\supp v \subset \Delta_{n,m}.$
\item[(ii)] $(H^*v)(n,m)< 0.$
\item[(iii)] $(H^*v)(i,j)\ge 0,$\par $0\le j<m.$
\end{enumerate} &For $n\ge m\ge 0$ there exists a matrix $v$ such that
\begin{enumerate}
\item[(i)] $\supp v \subset \Delta_{n,m}.$
\item[(ii)] $(H^*v)(n,m)< 0.$
\item[(iii)] $(H^*v)(i,j)\ge 0,$\par $0\le j<m.$
\item[(iv)] $v(i,j)\ge 0.$ 
\end{enumerate}\\ \hline &\\
\begin{enumerate}\item[(i)] $(H^*v_{n,m})(i,j)\ge 0,$ \par $0\le j<m.$\end{enumerate}&
\begin{enumerate}\item[(i)]$(H^*v_{n,m})(i,j)\ge 0,$ \par $0\le j<m.$
\item[(ii)]
$v_{n,m}(i,j)\ge 0.$\end{enumerate}\\ \hline
\end{tabular}\bigskip

\hspace{0.5cm}We always assume that $n\ge m\ge 0$ i $i\ge j\ge 0.$
\end{slide}

\end{document}
