%%  chapter-09.tex
%%  Approximation Theory: Chapter 9
%%  Neal L. Carothers
%%  Bowling Green State University
%%  Bowling Green, Ohio  43403
%%  carother@math.bgsu.edu
%%  http://www.bgsu.edu/~carother/


\input 680-setup.tex

\chaptertitle = {Orthogonal Polynomials}

\centerline{\hfil\tf Orthogonal Polynomials\hfil}
\vskip-\baselineskip
\line{\sc Math 680 \hfil 7/20/94}

\noindent
Given a positive (except possibly at finitely many
points), Riemann integrable {\sl weight function\/}
$w(x)$ on $[\,a,b\,]$, the expression 
$$\langle f,g\rangle \ = \ \int_a^bf(x)\,g(x)\,w(x)\,dx$$
defines an inner product on $C[\,a,b\,]$ and
$$\Vert f\Vert_2 \ = \ \left(\int_a^bf(x)^2\,dx\right)^{1/2}
	\ = \ \sqrt{\langle f,f\rangle}$$
defines a strictly convex norm on $C[\,a,b\,]$.  
Thus, given a finite dimensional subspace $E$ of
$C[\,a,b\,]$ and an element $f\in C[\,a,b\,]$, there is a 
unique $g\in E$ such that 
$$\Vert f-g\Vert_2 \ = \ \Min_{\strut h\in E}\Vert f-h\Vert_2.$$
We say that $g$ is the {\sl least-squares approximation\/}
to $f$ out of $E$.

Now if we
apply the Gram-Schmidt procedure to the sequence
$1,x,x^2,\ldots$, we will arrive at a
sequence $(Q_n)$  of {\sl orthogonal polynomials\/}
relative to the above inner product.  
In this special case, however, 
the Gram-Schmidt procedure simplifies substantially:  

\proclaim Theorem.
The following procedure defines a sequence
$(Q_n)$ of orthogonal polynomials {\rm(}relative to
$w${\rm)}.
$$\displaylines{
Q_0(x)=1,\quad Q_1(x)=x-a_0,\quad \hbox{and} \quad
	Q_{n+1}(x)=(x-a_n)Q_n(x)-b_nQ_{n-1}(x),\cr
\noalign{\hbox{where}}
a_n=\langle\,x\,Q_n,Q_n\,\rangle\big/\langle\,Q_n,Q_n\,\rangle
	\quad\hbox{and}\quad
	b_n=\langle\,x\,Q_n,Q_{n-1}\,\rangle\big/\langle\,Q_{n-1},Q_{n-1}\,\rangle\cr
}$$
{\rm(}and where $x\,Q_n$ is shorthand for the polynomial 
$x\,Q_n(x)$\/{\rm)}.

\proof
It's easy to see from these formulas that $Q_n$ is a monic 
polynomial of degree exactly $n$.  In particular, the
$Q_n$'s are linearly independent (and nonzero).

Now we checked in class that $Q_0$, $Q_1$, and $Q_2$ are
mutually 
orthogonal, so let's check that $Q_{n+1}$ is orthogonal 
to each $Q_k$, $k\le n$.  First,
$$\displaylines{
\langle\,Q_{n+1},Q_n\,\rangle
	=\langle\,x\,Q_n,Q_n\,\rangle-a_n\langle\,Q_n,Q_n\,\rangle
	-b_n\langle\,Q_{n-1},Q_n\,\rangle=0\cr
\noalign{\hbox{and}}
\langle\,Q_{n+1},Q_{n-1}\,\rangle
	=\langle\,x\,Q_n,Q_{n-1}\,\rangle-a_n\langle\,Q_n,Q_{n-1}\,\rangle
	-b_n\langle\,Q_{n-1},Q_{n-1}\,\rangle=0,\cr
}$$
since $\langle\,Q_{n-1},Q_n\,\rangle=0$.  Next, we take
$k<n-1$ and use the recurrence formula twice:
$$\openup1\jot\eqalign{
\langle\,Q_{n+1},Q_k\,\rangle
	&=\langle\,x\,Q_n,Q_k\,\rangle-a_n\langle\,Q_n,Q_k\,\rangle
	-b_n\langle\,Q_{n-1},Q_k\,\rangle\cr
	&=\langle\,x\,Q_n,Q_k\,\rangle=\langle\,Q_n,x\,Q_k\,\rangle
		\qquad\hbox{(Why?)}\cr
	&=\langle\,Q_n,Q_{k+1}+a_kQ_k+b_kQ_{k-1}\,\rangle=0,\cr
}$$
since $k+1<n$.~\qed

\noindent
{\bf Observations}

\item{1.}
Using the same trick as above, we have
$$b_n=\langle\,x\,Q_n,Q_{n-1}\,\rangle\big/\langle\,Q_{n-1},Q_{n-1}\,\rangle
	=\langle\,Q_n,Q_n\,\rangle\big/\langle\,Q_{n-1},Q_{n-1}\,\rangle>0.$$

\item{2.}
Each $p\in{\cal P}_n$ can be uniquely written 
$p=\sum_{i=0}^n\alpha_iQ_i$, where 
$\alpha_i=\langle\,p,Q_i\,\rangle\big/\langle\,Q_i,Q_i\,\rangle$.

\item{3.}
If $Q$ is any monic polynomial of degree exactly $n$, then
$Q=Q_n+\sum_{i=0}^{n-1}\alpha_iQ_i$ (why?)\ and hence
$$\Vert Q\Vert_2^2 \ = \ \Vert Q_n\Vert_2^2 \ + \ 
	\sum_{i=0}^{n-1}\alpha_i^2\Vert Q_i\Vert_2^2 \ 
	> \ \Vert Q_n\Vert_2^2,$$
unless $Q=Q_n$.  That is, $Q_n$ has the least $\Vert\cdot\Vert_2$
norm of all monic polynomials of degree $n$.

\item{4.}
The $Q_n$'s are {\sl unique\/} in the following sense: 
If $(P_n)$ is another sequence of orthogonal polynomials
such that $P_n$ has degree exactly $n$, then $P_n=\alpha_nQ_n$
for some $\alpha_n\ne0$.  (Why?)  Consequently, there's 
no harm in referring to the $Q_n$'s as {\sl the\/} sequence
of orthogonal polynomials relative to $w$.  

\noindent
{\bf Examples}

\item{\bf 1.}
On $[-1,1\,]$, the Chebyshev
polynomials of the first kind $(T_n)$ are orthogonal
relative to the weight $w(x)=1/\sqrt{1-x^2}$.  
$$\int_{-1}^1T_m(x)\,T_n(x)\,{{dx}\over{\sqrt{1-x^2}}}
	=\int_0^\pi \cos m\theta\,\cos n\theta\,d\theta 
	=\cases{0, & $m\ne n$\cr \pi, & $m=n=0$\cr \pi/2, & $m=n\ne0$.}$$
Since $T_n$ has degree exactly $n$,
this must be the right choice.
Notice, too, that ${1\over\sqrt{2}}\,T_0,T_1,T_2,\ldots$ are
{\sl orthonormal\/} relative to the weight
$2/\pi\sqrt{1-x^2}$.

\item{}
In terms of the inductive procedure given earlier, we
must have $Q_n=2^{-n+1}T_n$ and from this it follows that
$a_n=0$, $b_1=1/2$, and $b_n=1/4$ for $n\ge2$.  
(Why?)  That is,
the recurrence formula given in our first Theorem
reduces to the familar relationship 
$T_{n+1}(x)=2x\,T_n(x)-T_{n-1}(x)$.  Curiously, 
$Q_n=2^{-n+1}T_n$ minimizes both
$$\Max_{\strut-1\le x\le1}|p(x)|\qquad\hbox{and}\qquad
	\left(\int_{-1}^1p(x)^2\,{{dx}\over{\sqrt{1-x^2}}}\right)^{1/2}$$
over all monic polynomials of degree exactly $n$.

\item{}
The Chebyshev polynomials satisfy
$(1-x^2)\,T_n''(x)-x\,T_n'(x)+n^2\,T_n(x)=0$.
Since this is a polynomial identity, it suffices to check
it for all $x=\cos\theta$.  In this case,
$$T_n'(x) \ = \ {{n\sin n\theta}\over{\sin\theta}}$$
and 
$$T_n''(x) \ = \ {{n^2\cos n\theta\sin\theta-n\sin n\theta\cos\theta}
		\over{\sin^2\theta\,(-\sin\theta)}}.$$
Hence,
$$\displaylines{
\qquad(1-x^2)\,T_n''(x)-x\,T_n'(x)+n^2\,T_n(x)\hfill\cr
	\hfill= \ -n^2\cos n\theta+n\sin n\theta\cot\theta
	-n\sin n\theta\cot\theta+n^2\cos\theta=0\qquad\cr
}$$

\item{\bf2.}
On $[-1,1\,]$, the Chebyshev
polynomials of the second kind $(U_n)$ are orthogonal
relative to the weight $w(x)=\sqrt{1-x^2}$.  
$$\openup2\jot\displaylines{
\qquad\int_{-1}^1U_m(x)\,U_n(x)\,(1-x^2)\,{{dx}\over{\sqrt{1-x^2}}}\hfill\cr
\hfill=\int_0^\pi{{\sin\,(m+1)\theta}\over{\sin\theta}}\cdot
		{{\sin\,(n+1)\theta}\over{\sin\theta}}\cdot\sin^2\theta\,d\theta
	=\cases{0, & $m\ne n$\cr \pi/2, & $m=n$.}\qquad\cr
}$$
While we're at it, notice that 
$$T_n'(x) \ = \ {{n\sin n\theta}\over{\sin\theta}} \ = \ U_{n-1}(x).$$
As a rule, the derivatives of a sequence of orthogonal 
polynomials are again orthogonal polynomials, but 
relative to a different weight.

\item{\bf3.}
On $[-1,1\,]$ with weight $w(x)\equiv1$, the sequence
$(P_n)$ of {\sl Legendre polynomials\/} are orthogonal,
and are typically normalized by $P_n(1)=1$.  The first
few Legendre polynomials are
$P_0(x)=1$, $P_1(x)=x$, $P_2(x)={3\over2}\,x^2-{1\over2}$, 
and	$P_3(x)={5\over2}\,x^3-{3\over2}$.  (Check this!)
After we've
seen a few more examples, we'll come back and give
an explicit formula for $P_n$.

\item{\bf4.}
All of the examples we've seen so far are special cases of
the following: On $[-1,1\,]$, consider the weight
$w(x)=(1-x)^\alpha(1+x)^\beta$, where $\alpha,\beta>-1$.
The corresponding orthogonal polynomials
$(P_n^{\,(\alpha,\beta)})$ are called the {\sl Jacobi polynomials\/}
and are typically normalized by requiring that 
$$P_n^{\,(\alpha,\beta)}(1) \ = \ {{n+\alpha}\choose\alpha} \ = \ 
	{{(\alpha+1)(\alpha+2)\cdots(\alpha+n)}\over{n!}}.$$
It follows that $P_n^{\,(0,0)}=P_n$, 
$$\displaylines{
P_n^{\,(-1/2,-1/2)} \ = \ {{1\cdot3\cdot5\cdots(2n-1)}\over{2^nn!}}\,\,T_n,\cr
\noalign{and}
P_n^{\,(1/2,1/2)} \ = \ {{1\cdot3\cdot5\cdots(2n+1)}\over{2^n(n+1)!}}\,\,U_n.\cr
}$$
The polynomials $P_n^{\,(\alpha,\alpha)}$ are called
{\sl ultraspherical\/} polynomials.

\item{\bf5.}
There are also several classical examples of orthogonal
polynomials on unbounded intervals.  In particular,
$$\vbox{\halign{%
\hfil$#$\hfil
    &\qquad$#$\hfil&\quad\qquad#\hfil\cr
(0,\infty) & w(x)=e^{-x} & Laguerre polynomials,\cr
(0,\infty) & w(x)=x^\alpha e^{-x} & generalized Laguerre polynomials,\cr
(-\infty,\infty) & w(x)=e^{-x^2} & Hermite polynomials.\cr
}}$$


We next characterize the {\sl orthogonal complement\/}
of ${\cal P}_{n-1}$.

\proclaim Lemma 1. {\rm(Integration by-parts.)}
$$\int_a^bu^{(n)}v \ = \ 
	\sum_{k=1}^n(-1)^{k-1}u^{(n-k)}v^{(k-1)}\,\Bigr]_a^b
	\ + \ (-1)^n\int_a^buv^{(n)}.$$

Now if $v$ is a polynomial of degree $<n$, then $v^{(n)}=0$ and we get:

\proclaim Lemma 2.
$f\in C[\,a,b\,]$ satisfies $\displaystyle\int_a^bf(x)\,p(x)\,w(x)\,dx=0$ 
for all polynomials $p\in{\cal P}_{n-1}$ if and only if
there is an $n$-times differentiable function $u$ on 
$[\,a,b\,]$ satisfying $fw=u^{(n)}$ and 
$u^{(k)}(a)=u^{(k)}(b)=0$ for all $k=0,1,\ldots,n-1$.

\proof
One direction is clear from Lemma 1:  Given
$u$ as above, we would have
$\int_a^bfpw =\int_a^bu^{(n)}p= (-1)^n\int_a^bup^{(n)}=0$.

So, suppose we have that $\int_a^bfpw=0$ for all 
$p\in{\cal P}_{n-1}$.  By integrating $fw$ repeatedly, 
choosing constants appropriately,
we may define a function $u$ satisfying $fw=u^{(n)}$ and
$u^{(k)}(a)=0$ for all $k=0,1,\ldots,n-1$.  We want
to show that the hypotheses on $f$ force $u^{(k)}(b)=0$ 
for all $k=0,1,\ldots,n-1$.

Now Lemma 1 tells us that 
$$0 \ = \ \int_a^bfpw \ = \ 
	\sum_{k=1}^n(-1)^{k-1}u^{(n-k)}(b)\,p^{(k-1)}(b)$$
for all $p\in{\cal P}_{n-1}$.  But the numbers
$p(b),p'(b),\ldots,p^{(n-1)}(b)$ are completely
arbitrary; that is (again by integrating repeatedly,
choosing our constants as we please), we can find 
polynomials $p_k$ of degree $k<n$ such that 
$p_k^{(k)}(b)\ne0$ and $p_k^{(j)}(b)=0$ for
$j\ne k$.  In fact, $p_k(x)=(x-b)^k$ works just fine!
In any case, we must have $u^{(k)}(b)=0$ 
for all $k=0,1,\ldots,n-1$.~\qed

Rolle's theorem tells us a bit more about the functions
orthogonal to ${\cal P}_{n-1}$:

\proclaim Lemma 3.
If\/ $w(x)>0$ in $(a,b)$, and if\/ $f\in C[\,a,b\,]$ satisfies 
$\displaystyle\int_a^bf(x)\,p(x)\,w(x)\,dx=0$ 
for all polynomials $p\in{\cal P}_{n-1}$, then $f$ has at least $n$
distinct zeros in the open interval\/ $(a,b)$.

\proof
Write $fw=u^{(n)}$, where $u^{(k)}(a)=u^{(k)}(b)=0$ for all 
$k=0,1,\ldots,n-1$.  In particular, since $u(a)=u(b)=0$, Rolle's
theorem tells us that $u'$ would have at least one zero in $(a,b)$.
But then $u'(a)=u'(c)=u'(b)=0$, and so $u''$ must have at least
two zeros in $(a,b)$.  Continuing, we find that $fw=u^{(n)}$ must have
at least $n$ zeros in $(a,b)$.  Since $w>0$, the result follows.~\qed

\proclaim Corollary.
Let $(Q_n)$ be the sequence of orthogonal polynomials associated to
a given weight $w>0$ on $[\,a,b\,]$.  Then, the roots of $Q_n$ are
real, simple, and lie in \/ $(a,b)$.

\proclaim Lemma 4.
If\/ $p^*$ is the least-squares approximation to $f\in C[\,a,b\,]$
out of\/ ${\cal P}_{n-1}$, then $f-p^*$ has at least $n$ distinct
zeros in\/ $(a,b)$.

\proof
The least-squares approximation satisfies $\langle\,f-p^*,p\,\rangle=0$
for all $p\in{\cal P}_{n-1}$.~\qed

The sheer volume of literature on orthogonal polynomials and
other ``special functions'' is truly staggering.  We'll content
ourselves with the Legendre and the Chebyshev polynomials.
In particular, let's return to the problem of finding an 
explicit formula for the Legendre polynomials.  We could,
as Rivlin does, use induction and a few observations that
simplify the basic recurrence formula (you're encouraged
to read this; see pp.\ 53--54).  Instead we'll give a simple
(but at first sight intimidating) formula that is of use in
more general settings than ours.  

Lemma 2 (with $w\equiv1$ and $[\,a,b\,]=[-1,1\,]$\/)
says that if we want to find a {\sl polynomial\/} $f$ of
degree $n$ which is orthogonal to ${\cal P}_{n-1}$, then we'll
need to take a {\sl polynomial\/} for $u$, and this $u$ will
have to be divisible by $(x-1)^n(x+1)^n$.  (Why?)  That is,
we must have $P_n(x)=c_n\cdot D^n\bigl[(x^2-1)^n\bigr]$, where
$D$ denotes differentiation, and we'll find $c_n$ by evaluating
the right-hand side at $x=1$.

\proclaim Lemma 5. {\rm(Leibniz's formula.)}\quad
$\displaystyle D^n(f\,g) \ = \ 
	\sum_{k=0}^n{n\choose k}\,D^k(f)\,D^{n-k}(g)$.

\proof
Induction and the fact that 
${{n-1}\choose{k-1}}+{{n-1}\choose k}={n\choose k}$.~\qed

Consequently, 
$Q(x)=D^n\bigl[(x-1)^n(x+1)^n\bigr]=
	\sum_{k=0}^n{n\choose k}\,D^k(x-1)^n\,D^{n-k}(x+1)^n$
and it follows that $Q(1)=2^nn!$ and $Q(-1)=(-1)^n2^nn!$.
This, finally, gives us the formula discovered by
Rodrigues in 1814:
$$P_n(x) \ = \ {1\over{2^nn!}}\,D^n\bigl[(x^2-1)^n\bigr].$$
The Rodrigues formula is quite useful (and easily
generalizes to the Jacobi polynomials).  

\noindent
{\bf Observations}

\item{\bf1.}
By Lemma 3, the roots of $P_n$ are real, distinct, and lie in
$(-1,1)$.

\item{\bf2.}
$(x^2-1)^n=\sum_{k=0}^n(-1)^k{n\choose k}x^{2n-2k}$.  If we apply
${1\over{2^nn!}}\,D^n$ and simplify, we get another formula
for the Legendre polynomials.
$$P_n(x) \ = \ {1\over{2^n}}\,\sum_{k=0}^{[n/2]}
	(-1)^k{n\choose k}{{2n-2k}\choose n}\,x^{n-2k}.$$
In particular, if $n$ is even (odd), then $P_n$ is even (odd).
Notice, too, that if we
let $\widetilde P_n$ denote the polynomial given by the
standard construction, then we must have 
$P_n=2^{-n}{{2n}\choose n}\widetilde P_n$.

\item{\bf3.}
In terms of our standard recurrence formula, it follows
that $a_n=0$ (because $xP_n(x)^2$ is always odd).
It remains to compute $b_n$.  First, integrating by parts,
$$\int_{-1}^1P_n(x)^2\,dx \ = \ xP_n(x)^2\Bigr]_{-1}^1 \ - \ 
	\int_{-1}^1x\cdot2P_n(x)\,P_n'(x)\,dx,$$
or $\langle\,P_n,P_n\,\rangle=2-2\langle\,P_n,xP_n'\,\rangle$.
But $xP_n'=nP_n+{}$lower degree terms; hence,
$\langle\,P_n,xP_n'\,\rangle=n\langle\,P_n,P_n\,\rangle$.
Thus, $\langle\,P_n,P_n\,\rangle=2/(2n+1)$.  Using this and 
the fact that $P_n=2^{-n}{{2n}\choose n}\widetilde P_n$,
we'd find that $b_n=n^2/(4n^2-1)$.  Thus,
$$\openup1\jot\eqalign{
P_{n+1} \ = \ 2^{-n-1}{{2n+2}\choose{n+1}}\widetilde P_{n+1} \ 
	&= \ 2^{-n-1}{{2n+2}\choose{n+1}}\left[x\,\widetilde P_n \ - \ 
		{{n^2}\over{(4n^2-1)}}\,\widetilde P_{n-1}\right]\cr
	&= \ {{2n+1}\over{n+1}}\,x\,P_n \ - \ {n\over{n+1}}\,P_{n-1}.\cr
}$$
That is, the Legendre polynomials satisfy the recurrence formula
$$(n+1)\,P_{n+1}(x) \ = \ (2n+1)\,x\,P_n(x)-n\,P_{n-1}(x).$$

\item{\bf4.}
It follows from {\bf3} that the sequence 
$\widehat P_n=\sqrt{{2n+1}\over2}\,P_n$ is 
{\sl orthonormal\/} on $[-1,1\,]$.

\item{\bf5.}
The Legendre polynomials satisfy
$(1-x^2)\,P_n''(x)-2x\,P_n'(x)+n\,(n+1)\,P_n(x)=0$.
If we set $u=(x^2-1)^n$; that is, if $u^{(n)}=2^nn!P_n$, 
note that $u'(x^2-1)=2nxu$.
Now we apply $D^{n+1}$ to both sides of this last equation 
(using Leibniz's formula) and simplify:  
$$\openup2\jot\displaylines{
u^{(n+2)}(x^2-1)+(n+1)\,u^{(n+1)}\,2x+{{(n+1)n}\over2}\,u^{(n)}\,2
	=2n\,\bigl[u^{(n+1)}\,x + (n+1)\,u^{(n)}\bigr]\cr
\implies \ (1-x^2)\,u^{(n+2)}-2x\,u^{(n+1)}+n\,(n+1)\,u^{(n)}=0.\cr
}$$

\item{\bf6.}
Through a series of exercises, similar in spirit to
{\bf5}, Rivlin shows that $|P_n(x)|\le1$
on $[-1,1\,]$.  See pp.\ 63--64 for details.

Given an orthogonal sequence, it makes sense to
consider ``generalized Fourier series'' 
relative to the sequence and to
find analogues of the Dirichlet kernel,
Lebesgue's theorem, and so on.  In case of the Legendre
polynomials we have the following:

\proclaim Example.
The ``Fourier-Legendre'' series for $f\in C[-1,1\,]$ is
given by $\sum_k\langle\,f,\widehat P_k\,\rangle\,\widehat P_k$,
where 
$$\widehat P_k=\sqrt{{2k+1}\over2}\,P_k\qquad\hbox{and}\qquad
	\langle\,f,\widehat P_k\,\rangle=\int_{-1}^1f(x)\,\widehat P_k(x)\,dx.$$
The partial sum operator 
$S_n(f)=\sum_{k=0}^n\langle\,f,\widehat P_k\,\rangle\,\widehat P_k$
is a linear projection onto ${\cal P}_n$ and 
may be written as 
$$S_n(f)(x)=\int_{-1}^1f(t)\,K_n(t,x)\,dt,$$
where $K_n(t,x)=\sum_{k=0}^n\widehat P_k(t)\,\widehat P_k(x)$.

Since the $\widehat P_k$'s are orthonormal, we have
$\sum_{k=0}^n|\langle\,f,\widehat P_k\,\rangle|^2  =  
	\Vert S_n(f)\Vert_2^2  \le  \Vert f\Vert_2^2$,
and so the generalized Fourier coefficients 
$\langle\,f,\widehat P_k\,\rangle$ are square summable; 
in particular, $\langle\,f,\widehat P_k\,\rangle\to0$ as $k\to\infty$.
As we saw in the case of Fourier series, the fact that the polynomials
(i.e., the span of the $\widehat P_k$'s) 
are dense in $C[\,a,b\,]$ implies that $S_n(f)$ actually
converges to $f$ in the $\Vert\cdot\Vert_2$ norm.  These
same observations remain valid for any sequence of orthogonal
polynomials.  The real question remains, just as with
Fourier series, whether $S_n(f)$ is a good uniform
(or even pointwise) approximation to $f$.

If you're willing to swallow the fact that $|P_n(x)|\le1$,
then 
$$|K_n(t,x)| \ \le \ \sum_{k=0}^n\sqrt{{2k+1}\over2}\,\sqrt{{2k+1}\over2}
	\ = \ {1\over2}\sum_{k=0}^n(2k+1) \ = \ {{(n+1)^2}\over2}.$$
Hence, $\Vert S_n(f)\Vert\le(n+1)^2\Vert f\Vert$.  That is, the
``Lebesgue numbers'' for this process are $(n+1)^2$.  The analogue of
Lebesgue's theorem in this case would then read:
$$\Vert f-S_n(f)\Vert \ \le \ Cn^2E_n(f).$$
Thus, $S_n(f)\uniformto f$ whenever $n^2E_n(f)\to0$, and Jackson's
theorem tells us when this will happen:  
{\sl If $f$ is twice continuously differentiable,
then the Fourier-Legendre series for $f$ converges uniformly
to $f$ on $[-1,1\,]$.}

\noindent
{\bf The Christoffel-Darboux Identity}

\noindent
It would also be of interest to have a closed form for
$K_n(t,x)$.  That this is indeed always possible, for
any sequence of orthogonal polynomials, is a very
important fact.

Using our original notation, let $(Q_n)$ be the sequence
of monic orthogonal polynomials corresponding to a given
weight $w$, and let $(\widehat Q_n)$ be the {\sl orthonormal\/}
counterpart of $(Q_n)$; in other words, $Q_n=\lambda_n\widehat Q_n$,
where $\lambda_n=\sqrt{\,\langle\,Q_n,Q_n\,\rangle\,}$.  It
will help things here if you recall (from Observation 1 on page 2)
that $\lambda_n^2=b_n\lambda_{n-1}^2$.

As with the Legendre polynomials, each $f\in C[\,a,b\,]$
is represented by the generalized Fourier series 
$\sum_k\langle\,f,\widehat Q_k\,\rangle\,\widehat Q_k$,
with partial sum operator 
$$S_n(f)(x)=\int_a^bf(t)\,K_n(t,x)\,w(t)\,dt,$$
where $K_n(t,x)=\sum_{k=0}^n\widehat Q_k(t)\,\widehat Q_k(x)$.
As before, $S_n$ is a projection onto ${\cal P}_n$; in
particular, $S_n(1)=1$ for every $n$.

\proclaim Theorem. {\rm(Christoffel-Darboux)} \ 
The kernel $K_n(t,x)$ can be written
$$\sum_{k=0}^n\widehat Q_k(t)\,\widehat Q_k(x) \ = \ 
	\lambda_{n+1}\lambda_n^{-1}\,
{{\widehat Q_{n+1}(t)\,\widehat Q_n(x)
	-\widehat Q_n(t)\,\widehat Q_{n+1}(x)}\over{t-x}}.$$

\proof
We begin with the standard recurrence formulas
$$\displaylines{
Q_{n+1}(t) = (t-a_n)\,Q_n(t) - b_nQ_{n-1}(t)\cr
Q_{n+1}(x) = (x-a_n)\,Q_n(x) - b_nQ_{n-1}(x)\cr
}$$
(where $b_0=0$\/).  
Multiplying the first by $Q_n(x)$, the second
by $Q_n(t)$, and subtracting:
$$\displaylines{
\qquad Q_{n+1}(t)\,Q_n(x)-Q_n(t)\,Q_{n+1}(x)\hfill\cr 
	\hfill= \ (t-x)\,Q_n(t)\,Q_n(x) \ + \ 
	b_n\bigl[\,Q_n(t)\,Q_{n-1}(x) \ - \ 
		Q_n(x)\,Q_{n-1}(t)\,\bigr]\qquad\cr
}$$
(and again, $b_0=0$\/).
If we divide both sides of this equation by $\lambda_n^2$
we get 
$$\displaylines{
\qquad \lambda_n^{-2}\bigl[\,
		Q_{n+1}(t)\,Q_n(x)-Q_n(t)\,Q_{n+1}(x)\,\bigr]\hfill\cr 
	\hfill= \ (t-x)\,\widehat Q_n(t)\,\widehat Q_n(x) \ + \ 
	\lambda_{n-1}^{-2}\bigl[\,Q_n(t)\,Q_{n-1}(x) \ - \ 
		Q_n(x)\,Q_{n-1}(t)\,\bigr].\qquad\cr
}$$
Thus, we may repeat the process; arriving finally at 
$$\lambda_n^{-2}\bigl[\,Q_{n+1}(t)\,Q_n(x)-Q_n(t)\,Q_{n+1}(x)\,\bigr] \ 
	= \ (t-x)\,\sum_{k=0}^n\widehat Q_n(t)\,\widehat Q_n(x).$$
The Christoffel-Darboux identity now follows by writing 
$Q_n=\lambda_n\widehat Q_n$, etc.~\qed

And we now have a version of the Dini-Lipschitz theorem:

\proclaim Theorem.
Let $f\in C[\,a,b\,]$ and suppose that at some point $x_0$ in
$[\,a,b\,]$ we have 
\item{\rm(i)}
$f$ is Lipschitz at $x_0$; that is, $|f(x_0)-f(x)|\le K|x_0-x|$
for some constant $K$ and all $x$ in $[\,a,b\,]$; and
\item{\rm(ii)}
the sequence $(\widehat Q_n(x_0))$ is bounded.
\vskip0pt\noindent
Then, the series $\sum_k\langle\,f,\widehat Q_k\,\rangle\,\widehat Q_k(x_0)$
converges to $f(x_0)$.

\proof
First note that the sequence $\lambda_{n+1}\lambda_n^{-1}$ is bounded:
Indeed, by Cauchy-Schwarz, 
$$\eqalign{
\lambda_{n+1}^2 \ &= \ \langle\,Q_{n+1},Q_{n+1}\,\rangle \ = \ 
		\langle\,Q_{n+1},x\,Q_n\,\rangle\cr
	&\le \ \Vert Q_{n+1}\Vert_2\cdot\Vert\,x\,\Vert\cdot\Vert Q_n\Vert_2 \ 
	= \ \Max\{|a|,|b|\}\,\lambda_{n+1}\lambda_n.\cr
}$$
Thus,  $\lambda_{n+1}\lambda_n^{-1}\le c=\Max\{|a|,|b|\}$.
Now, using the Christoffel-Darboux identity,
$$\openup2\jot\eqalignno{
S_n(f)(x_0)&-f(x_0) \ 
	= \ \int_a^b\bigl[\,f(t)-f(x_0)\,\bigr]\,K_n(t,x_0)\,w(t)\,dt\cr
	&= \ \lambda_{n+1}\lambda_n^{-1}\,\int_a^b{{f(t)-f(x_0)}\over{t-x_0}}\,
		\bigl[\,\widehat Q_{n+1}(t)\,\widehat Q_n(x_0)
		-\widehat Q_n(t)\,\widehat Q_{n+1}(x_0)\,\bigr]\,w(t)\,dt\cr
	&= \ \lambda_{n+1}\lambda_n^{-1}\,
		\bigl[\,\langle\,h,\widehat Q_{n+1}\,\rangle\,\widehat Q_n(x_0)
		-\langle\,h,\widehat Q_n\,\rangle\,\widehat Q_{n+1}(x_0)\,\bigr],\cr
}$$
where $h(t)=(f(t)-f(x_0))/(t-x_0)$.  But $h$ is bounded
(and continuous everywhere except, possibly, at $x_0$) by hypothesis (i), 
$\lambda_{n+1}\lambda_n^{-1}$ is bounded, and $\widehat Q_n(x_0)$ is 
bounded by hypothesis (ii).  All that remains is to notice that the numbers 
$\langle\,h,\widehat Q_n\,\rangle$ are the generalized Fourier
coefficients of the bounded, Riemann integrable function $h$, and so 
must tend to zero (since, in fact, they're even square summable).~\qed

We end this section with a negative result, due to Nikolaev:

\proclaim Theorem.
There is\/ {\bf no} weight $w$ such that\/ {\bf every} 
$f\in C[\,a,b\,]$ has a uniformly convergent 
expansion in terms of orthogonal polynomials.  
In fact, given any $w$, there is always some
$f$ for which $\Vert f-S_n(f)\Vert$ is unbounded.




\bye


%%  end of chapter-09.tex



