%%  chapter-02.tex
%%  Approximation Theory: Chapter 2
%%  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 = {Algebraic Polynomials}

\centerline{\hfil\tf Approximation by Algebraic Polynomials\hfil}
\vskip-\baselineskip
\line{\sc Math 680 \hfil 6/29/94}

\noindent{\bf Introduction}

\noindent
Let's begin with some notation.  Throughout, we're concerned
with the problem of best (uniform) approximation of a given
function $f\in C[\,a,b\,]$ by elements from ${\cal P}_n$,
the subspace of algebraic polynomials of degree at most $n$ 
in $C[\,a,b\,]$.  We know that the problem has a solution 
(possibly more than one), which we've chosen to write as $p_n^*$.  
We set
$$E_n(f)=\Min_{p\in{\cal P}_n}\Vert f-p\Vert=\Vert f-p_n^*\Vert.$$
Since ${\cal P}_n\subset{\cal P}_{n+1}$ for each $n$, it's
clear that $E_n(f)\ge E_{n+1}(f)$ for each $n$.  Our goal in this 
chapter is to prove that $E_n(f)\to0$.  We'll accomplish this by 
proving:

\proclaim Theorem. \ 
{\rm(The Weierstrass Approximation Theorem, 1885):} \ 
Let $f\in C[\,a,b\,]$.  Then, for every $\eps>0$, there is
a polynomial $p$ such that $\Vert f-p\Vert<\eps$.

It follows from the Weierstrass theorem that $p_n^*\uniformto f$
for each $f\in C[\,a,b\,]$.  (Why?)  This is an important first
step in determining the exact nature of $E_n(f)$ as a function
of $f$ and $n$.  We'll look for much more precise information
in later sections.  

Now there are many proofs of the Weierstrass theorem (a mere
three are outlined in the exercises, but there are hundreds!),
but all of them start with one simplification: The underlying
interval $[\,a,b\,]$ is of no consequence here.  

\proclaim Lemma.
If the Weierstrass theorem holds for $C[\,0,1\,]$, then it also
holds for $C[\,a,b\,]$.  In fact, $C[\,0,1\,]$ and $C[\,a,b\,]$ are, 
for all practical purposes, identical:  They are linearly isometric 
as normed spaces, order isomorphic as lattices, and isomorphic 
as algebras {\rm(}rings\/{\rm)}.

\proof
We'll settle for proving only the first assertion; the second
is outlined in the exercises (and uses a similar argument).

Given $f\in C[\,a,b\,]$, notice that the function
$$g(x)=f\bigl(a+(b-a)x\bigr), \quad 0\le x\le 1,$$
defines an element of $C[\,0,1\,]$.  Now, given $\eps>0$,
suppose that we can find a polynomial $p$ such that
$\Vert g-p\Vert<\eps$; in other words, suppose that 
$$\Max_{0\le x\le1}\bigl|f\bigl(a+(b-a)x\bigr)-p(x)\bigr|<\eps.$$
Then, 
$$\Max_{a\le t\le b}\left|f(t)-p\left({{t-a}\over{b-a}}\right)\right|<\eps.$$
(Why?)  But if $p(x)$ is a polynomial in $x$, then 
$q(t)=p\left({{t-a}\over{b-a}}\right)$ is a polynomial in $t$
(again, why?)\ and $\Vert f-q\Vert<\eps$.~\qed

The point to our first result is that it suffices to prove the
Weierstrass theorem for any interval we like; $[\,0,1\,]$ and
$[-1,1\,]$ are popular choices, but it hardly matters which 
interval we use.  

\noindent{\bf Bernstein's Proof}

\noindent
The proof of the Weierstrass theorem we present here is due to the
great Russian mathematician S.\ N.\ Bernstein in 1912.  Bernstein's
proof is of interest to us for a variety of reasons; perhaps most
important is that Bernstein actually {\sl displays\/} a sequence
of polynomials that approximate a given $f\in C[\,0,1\,]$.  Moreover,
as we'll see later, Bernstein's proof generalizes to yield a powerful,
unifying theorem, the Bohman-Korovkin theorem.  

If $f$ is any {\sl bounded\/} function on $[\,0,1\,]$, we define
the sequence of {\sl Bernstein polynomials\/} for $f$ by 
$$\bigl(B_n(f)\bigr)(x) = \sum_{k=0}^n f\left({k\over n}\right)
	\cdot {n\choose k} x^k(1-x)^{n-k},\qquad 0\le x\le1.$$
Please note that $B_n(f)$ is a polynomial of degree
at most $n$.  Also, it's easy to see that 
$\bigl(B_n(f)\bigr)(0)=f(0)$, and
$\bigl(B_n(f)\bigr)(1)=f(1)$.  In general, 
$\bigl(B_n(f)\bigr)(x)$ is an {\sl average\/}
of the numbers $f(k/n)$, $k=0,\ldots,n$.
Bernstein's theorem states that $B_n(f)\uniformto f$
for each $f\in C[\,0,1\,]$.  Surprisingly, the proof 
actually only requires that we check three easy cases:
$$f_0(x)=1,\quad f_1(x)=x,\quad \hbox{and}\quad f_2(x)=x^2.$$
This, and more, is the content of the following lemma.

{\baselineskip = 32 true pt
\proclaim Lemma.  {\rm(i)}~$B_n(f_0)=f_0$ \ 
and \ $B_n(f_1)=f_1$.
\item{\rm(ii)} $B_n(f_2)=
	\displaystyle{\Bigl(1-{1\over n}\Bigr)f_2 + {1\over n}f_1}$, \ 
and hence \ $B_n(f_2)\uniformto f_2$.
\item{\rm(iii)} $\displaystyle{%
	\sum_{k=0}^n \Bigl({k\over n}-x\Bigr)^2{n\choose k}x^k(1-x)^{n-k}
	= {{x(1-x)}\over n} \le {1\over{4n}}}$, \ if \ $0\le x\le 1$.
\item{\rm(iv)} Given $\delta>0$ and $0\le x\le 1$, let
$F$ denote the set of $k$'s in $\{0,\ldots,n\}$ for which 
$\displaystyle{\Bigl|{k\over n}-x\Bigr|}\ge \delta$.  Then
$\displaystyle{\sum_{k\in F}{n\choose k}x^k(1-x)^{n-k}
	\le {1\over{4n\delta^2}}}$.\bigskip\par
}

\proof
That $B_n(f_0)=f_0$ follows from the binomial formula:
$$\sum_{k=0}^n {n\choose k}x^k(1-x)^{n-k} = [x+(1-x)]^n = 1.$$
To see that $B_n(f_1)=f_1$, first notice that for $k\ge 1$
we have
$${k\over n}{n\choose k} = 
	{{(n-1)\,!}\over{(k-1)\,!\,(n-k)\,!}} = {{n-1}\choose{k-1}}.$$
Consequently,
$$\openup 1 \jot\eqalign{%
\sum_{k=0}^n {k\over n}{n\choose k}x^k(1-x)^{n-k} \ 
	&= \ x\sum_{k=1}^n {{n-1}\choose{k-1}}x^{k-1}(1-x)^{n-k}\cr
	&= \ x\sum_{j=0}^{n-1} {{n-1}\choose j}x^j(1-x)^{(n-1)-j}
	 \ = \ x.\cr
}$$

Next, to compute $B_n(f_2)$, we rewrite twice:
$$\openup 1 \jot\eqalign{%
\left({k\over n}\right)^2{n\choose k}  
	=  {k\over n}{{n-1}\choose{k-1}}  
	&=  {{n-1}\over n}\cdot{{k-1}\over{n-1}}{{n-1}\choose{k-1}}
		  +  {1\over n}{{n-1}\choose{k-1}}, \hbox{ if } k\ge 1\cr
	&=  \left(1-{1\over n}\right){{n-2}\choose{k-2}}
		  +  {1\over n}{{n-1}\choose{k-1}}, \hbox{ if } k\ge 2.\cr
}$$
Thus,
$$\openup 1 \jot\eqalign{%
&\sum_{k=0}^n \left({k\over n}\right)^2{n\choose k}x^k(1-x)^{n-k}\cr
&\qquad= 
\left(1-{1\over n}\right)\sum_{k=2}^n{{n-2}\choose{k-2}}x^k(1-x)^{n-k}
 \ + \ {1\over n}\sum_{k=1}^n{{n-1}\choose{k-1}}x^k(1-x)^{n-k}\cr
&\qquad= \left(1-{1\over n}\right)x^2 + {1\over n}\,x,\cr
}$$
which establishes (ii) since 
$\Vert B_n(f_2)-f_2\Vert 
	= {1\over n}\Vert f_1-f_2\Vert\to 0$ as $n\to\infty$.

To prove (iii) we combine the results in (i) and (ii)
and simplify.  Since $((k/n)-x)^2 = (k/n)^2 - 2x(k/n) + x^2$,
we get 
$$\eqalign{%
\sum_{k=0}^n \left({k\over n}-x\right)^2{n\choose k}x^k(1-x)^{n-k} \ 
&= \ \left(1-{1\over n}\right)x^2 + {1\over n}x-2x^2+x^2\cr
& = \ {1\over n}\,x(1-x) \ \le \ {1\over {4n}},\cr
}$$
for $0\le x\le 1$.  

Finally, to prove (iv), note that
$1\le ((k/n)-x)^2/\delta^2$ for $k\in F$, and hence 
$$\openup 1 \jot\eqalign{%
\sum_{k\in F}{n\choose k}x^k(1-x)^{n-k} \ 
&\le \ {1\over{\delta^2}}\sum_{k\in F}\Bigl({k\over n}-x\Bigr)^2
	{n\choose k}x^k(1-x)^{n-k}\cr
&\le \ {1\over{\delta^2}}\sum_{k=0}^n\Bigl({k\over n}-x\Bigr)^2
	{n\choose k}x^k(1-x)^{n-k}\cr
&\le \ {1\over{4n\delta^2}}, \ \ \hbox{from (iii).~\qed}\cr
}$$

Now we're ready for {\sl the proof of Bernstein's theorem\/}:

\proof
Let $f\in C[\,0,1\,]$ and let $\eps>0$.  Then, since $f$ is
uniformly continuous, there is a $\delta>0$ such that
$|f(x)-f(y)|<\eps/2$ whenever $|x-y|<\delta$.  Now we use
the previous lemma to estimate $\Vert f-B_n(f)\Vert$.  First
notice that since the numbers ${n\choose k} x^k(1-x)^{n-k}$ 
are nonnegative and sum to $1$, we have  
$$\openup 1\jot\eqalign{%
&\left| f(x)-\sum_{k=0}^n {n\choose k} 
	f\left({k\over n}\right) x^k(1-x)^{n-k}\right|\qquad\cr
&\qquad\qquad\qquad= \ 
	\left|\sum_{k=0}^n\left(f(x) - f\left({k\over n}\right)\right)
	{n\choose k} x^k(1-x)^{n-k}\right|\cr
&\qquad\qquad\qquad\le \ 
	\sum_{k=0}^n\left|f(x) - f\left({k\over n}\right)\right|
	{n\choose k} x^k(1-x)^{n-k},\cr
}$$
Now fix $n$ (to be specified in a moment)
and let $F$ denote the set of $k$'s in $\{0,\ldots,n\}$
for which $|(k/n)-x|\ge \delta$.  Then 
$|f(x)-f(k/n)|<\eps/2$ for $k\notin F$, while 
$|f(x)-f(k/n)|\le 2\Vert f\Vert$ for $k\in F$.
Thus, 
$$\openup 1\jot\eqalign{%
&\left|f(x) - \bigl(B_n(f)\bigr)(x)\right|\cr
&\quad\qquad\le \ 
{\eps\over 2}\sum_{k\notin F} {n\choose k} x^k(1-x)^{n-k}
+ 2\Vert f\Vert \sum_{k\in F} {n\choose k} x^k(1-x)^{n-k}\cr
&\quad\qquad< \ {\eps\over 2}\cdot 1 \ 
+ \ 2\Vert f\Vert\cdot {1\over{4n\delta^2}}, 
	\ \ \hbox{from (iv) of the Lemma,}\cr
&\quad\qquad< \ \eps, \ \ \hbox{provided that} \ 
	n>\Vert f\Vert/\eps\delta^2.~\qed\cr
}$$

\noindent{\bf Landau's Proof}

\noindent
Just because it's good for us, let's give {\sl a second proof
of Weierstrass's theorem}.  This one is due to Landau in 1908.
First, given $f\in C[\,0,1\,]$, notice that it suffices to 
approximate $f-p$, where $p$ is any polynomial.  (Why?)
In particular, by subtracting the {\sl linear\/} function 
$f(0) + x(f(1)-f(0))$, we may suppose that $f(0)= f(1) = 0$
and, hence, that $f\equiv0$ outside $[\,0,1\,]$.
That is, we may suppose that $f$ is defined and uniformly
continuous on all of $\R$.  

Again we will display a sequence of polynomials
that converge uniformly to $f$; this time we define 
$$L_n(x) =Êc_n\int_{-1}^1Êf(x+t)\,(1-t^2)^n\,dt,$$
where $c_n$ is chosen so that 
$$c_n\int_{-1}^1(1-t^2)^n\,dt=1.$$
Note that by our assumptions on $f$, we may rewrite this
expression as
$$L_n(x)=c_n\int_{-x}^{1-x}f(x+t)\,(1-t^2)^n\,dt
	=c_n\int_0^1f(t)\,(1-(t-x)^2)^n\,dt.$$
Written this way, it's clear that 
$L_n$ is a polynomial of degree at most $n$.

We first need to estimate $c_n$.  Since
$(1-x^2)^n\ge1-nx^2$, we get
$$\int_{-1}^1(1-x^2)^n\,dx\ge2\int_0^{1/\sqrt{n}}(1-nx^2)\,dx
	={4\over{3\sqrt{n}}}>{1\over\sqrt{n}},$$
and so $c_n<\sqrt{n}$.  In particular, 
$$c_n\int_\delta^1(1-x^2)^n\,dx<\sqrt{n}\,(1-\delta^2)^n\to0
	\qquad(n\to\infty)$$
for any $0<\delta<1$, which is the inequality we'll need.

Next, let $\eps>0$ be given, and choose $0<\delta<1$ such that
$$|f(x)-f(y)|\le\eps/2\ \hbox{whenever}\ |x-y|\le\delta.$$
Then, since $c_n(1-x^2)^n\ge0$ and integrates to $1$, we get
$$\openup1\jot\eqalign{%
|L_n(x)-f(x)|
	&=\left|c_n\int_0^1\bigl[f(x+t)-f(x)\bigr](1-t^2)^n\,dt \right|\cr
	&\le c_n\int_0^1|f(x+t)-f(x)|(1-t^2)^n\,dt\cr
	&\le {\eps\over2}\,c_n\int_0^\delta(1-t^2)^n\,dt
		+2\Vert f\Vert\,c_n\int_\delta^1(1-t^2)^n\,dt\cr
	&\le{\eps\over2}+6\Vert f\Vert\,\sqrt{n}\,(1-\delta^2)^n<\eps,
}$$
provided that $n$ is sufficiently large.~\qed

A third proof of the Weierstrass theorem, due to Lebesgue
in 1898, is outlined in the exercises.  Lebesgue's proof
is of particular interest since, according to Cheney, it
inpsired Stone's version of the Weierstrass theorem (in part).
We'll talk about the Stone-Weierstrass theorem a bit later
in the course.

Before we go on, let's stop and make an observation or two:
While the Bernstein polynomials $B_n(f)$ offer a convenient and 
explicit polynomial approximation to $f$, they are by no means
the best approximations.  Indeed, recall that if $f_1(x)=x$
and $f_2(x)=x^2$, then $B_n(f_2)=(1-{1\over n})f_2+{1\over n}f_1\ne f_2$.
Clearly, the best approximation to $f_2$ out of ${\cal P}_n$ should
be $f_2$ itself whenever $n\ge2$.  On the other hand, since
we always have 
$$E_n(f)\le\Vert f-B_n(f)\Vert,$$
a detailed understanding of Bernstein's proof will lend
insight into the general problem of polynomial approximation.
Our next project, then, is to improve upon our estimate 
of the error $\Vert f-B_n(f)\Vert$.

\filbreak

\noindent{\bf Improved Estimates}

\noindent
To begin, we will need a bit more notation.  The
{\sl modulus of continuity\/} of a bounded function
$f$ on the interval $[\,a,b\,]$ is defined by
$$\omega_f(\delta)=\omega_f([\,a,b\,];\delta)
	=\sup\bigl\{|f(x)-f(y)|:x,y\in[\,a,b\,], \ |x-y|\le\delta\bigr\}$$
for any $\delta>0$.  Note that $\omega_f(\delta)$ is a
measure of the ``$\eps$'' (in the definition of uniform continuity)
that goes along with $\delta$; literally, we have written
$\eps=\omega_f(\delta)$ as a function of $\delta$.

Here are a few easy facts about the modulus of continuity:

\noindent{\bf Exercises}

\item{1.}
We always have 
$|f(x)-f(y)|\le\omega_f(\,|x-y|\,)$ for any $x\ne y\in[\,a,b\,]$.

\item{2.}
If $0<\delta'\le\delta$, then $\omega_f(\delta')\le\omega_f(\delta)$.

\item{3.}
$f$ is {\sl uniformly continuous\/} if and only if
$\omega_f(\delta)\to0$ as $\delta\to0^+$.

\item{4.}
If $f\,'$ exists and is bounded on $[\,a,b\,]$,
then $\omega_f(\delta)\le K\delta$ for some constant $K$.

\item{5.}
More generally, we say that $f$ satisfies a {\sl Lipschitz condition\/}
of order $\alpha$ with constant $K$, where $0<\alpha\le1$ and $0\le K<\infty$,
if $|f(x)-f(y)|\le K|x-y|^\alpha$ for all $x$, $y$.  We abbreviate
this statement by the symbols: $f\in{\rm lip}_K\alpha$.   Check that
if $f\in{\rm lip}_K\alpha$, then $\omega_f(\delta)\le K\delta^\alpha$
for all $\delta>0$.

For the time being, we actually only need one simple fact about $\omega_f(\delta)$:

\proclaim Lemma.
Let $f$ be a bounded function on $[\,a,b\,]$ and let $\delta>0$.
Then, $\omega_f(n\delta)\le n\,\omega_f(\delta)$ for $n=1,2,\ldots$.
Consequently, $\omega_f(\lambda\delta)\le(1+\lambda)\,\omega_f(\delta)$
for any $\lambda>0$.

\proof
Given $x<y$ with $|x-y|\le n\,\delta$, split the interval
$[\,x,y\,]$ into $n$ pieces, each of length $\delta$.
Specifically, if we set $z_k=x+k(y-x)/n$, for $k=0,1,\ldots,n$,
then $|z_k-z_{k-1}|\le\delta$ for any $k\ge1$, and so
$$\eqalign{%
|f(x)-f(y)| \ &= \ \left| \sum_{k=1}^n f(z_k)-f(z_{k-1})\right|\cr
	&\le \ \sum_{k=1}^n |f(z_k)-f(z_{k-1})|\cr
	&\le \ n\,\omega_f(\delta).\cr
}$$
Thus, $\omega_f(n\delta)\le n\,\omega_f(\delta)$.

The second assertion follows from the first (and one of our
exercises).  Given $\lambda>0$, choose an integer $n$ so
that $n-1<\lambda\le n$.  Then,
$$\omega_f(\lambda\delta)\le \omega_f(n\,\delta)
	\le n\,\omega_f(\delta)\le (1+\lambda)\,\omega_f(\delta).
	\eqno\qed$$

We next repeat the proof of Bernstein's theorem, making a few
minor adjustments here and there.

\proclaim Theorem.
For any bounded function $f$ on $[\,0,1\,]$ we have
$$\Vert f-B_n(f)\Vert\le
	{3\over2}\,\omega_f\!\left({1\over\sqrt{n}}\right).$$
In particular, if $f\in C[\,0,1\,]$, then
$E_n(f)\le{3\over2}\,\omega_f({1\over\sqrt{n}})\to0$
as $n\to\infty$.

\proof
We first do some term juggling:
$$\openup 3\jot\eqalign{%
|f(x)-B_n(f)(x)| \ 
	&= \ \left|\sum_{k=0}^n\left(f(x) - f\left({k\over n}\right)\right)
	{n\choose k} x^k(1-x)^{n-k}\right|\cr
	&\le \ \sum_{k=0}^n\left|f(x) - f\left({k\over n}\right)\right|
	{n\choose k} x^k(1-x)^{n-k}\cr
	&\le \ \sum_{k=0}^n \,\omega_f\!\left(\,\left|x-{k\over n}\right|\,\right)
	{n\choose k} x^k(1-x)^{n-k}\cr
	&\le \ \omega_f\!\left({1\over\sqrt{n}}\right)\,
	\sum_{k=0}^n \,\left[\,1+\sqrt{n}\,\left|x-{k\over n}\right|\,\right]
	{n\choose k} x^k(1-x)^{n-k}\cr
	&= \ \omega_f\!\left({1\over\sqrt{n}}\right)\,
	\left[\,1 \ + \ \sqrt{n}\,\sum_{k=0}^n \left|x-{k\over n}\right|\,
	{n\choose k} x^k(1-x)^{n-k}\right],\cr
}$$
where the third inequality follows from our previous Lemma
(where we've taken $\lambda=\sqrt{n}\,\left|x-{k\over n}\right|$ and
$\delta={1\over\sqrt{n}}\,$).  All that remains is to estimate the sum, 
and for this we'll use Cauchy-Schwarz (and our earlier observations
about Bernstein polynomials).
$$\openup 1\jot\eqalign{%
&\sum_{k=0}^n \left|x-{k\over n}\right|\,{n\choose k} x^k(1-x)^{n-k}\cr
	&\qquad\qquad\le\left[\sum_{k=0}^n \left|x-{k\over n}\right|^2\,
	{n\choose k} x^k(1-x)^{n-k}\right]^{1/2}\cdot
	\left[\sum_{k=0}^n {n\choose k} x^k(1-x)^{n-k}\right]^{1/2}\cr
	&\qquad\qquad\le\left[{1\over{4n}}\right]^{1/2}={1\over{2\sqrt{n}}}.\cr
}$$
Finally, 
$$|f(x)-B_n(f)(x)|\le\omega_f\!\left({1\over\sqrt{n}}\right)
	\left[1 \ + \ \sqrt{n}\cdot{1\over{2\sqrt{n}}}\right]
	={3\over2}\,\omega_f\!\left({1\over\sqrt{n}}\right).
	\eqno\qed$$

\noindent{\bf Examples}

\item{1.}
If $f\in{\rm lip}_K\alpha$, it follows that
$\Vert f-B_n(f)\Vert\le {3\over2}Kn^{-\alpha/2}$ and
hence $E_n(f)\le {3\over2}Kn^{-\alpha/2}$.

\item{2.}
As a particular case of the first example, consider
$f(x)=\left|x-{1\over2}\right|$ on $[\,0,1\,]$.
Then $f\in{\rm lip}_11$, and so 
$\Vert f-B_n(f)\Vert\le{3\over2}\,n^{-1/2}$.  But,
as Rivlin points out (see the Remark on p.\ 16), 
$\Vert f-B_n(f)\Vert>{1\over2}\,n^{-1/2}$.
Thus, we can't hope to improve on the power of $n$
in this estimate.  Nevertheless, we will see an
improvement in our estimate of $E_n(f)$.

\noindent{\bf The Bohman-Korovkin Theorem}

\noindent
The real value to us in Bernstein's approach is that the
map $f\mapsto B_n(f)$, while providing a simple formula for
an approximating polynomial, is also {\sl linear\/} and
{\sl positive}.  In other words,
$$\displaylines{%
B_n(f+g)=B_n(f)+B_n(g),\cr
B_n(\alpha f)=\alpha B_n(f),\quad\alpha\in\R,\cr
\noalign{\hbox{and}}
B_n(f)\ge0\quad\hbox{whenever}\quad f\ge0.\cr
}$$
As it happens, any positive, linear map $T:C[\,0,1\,]\to C[\,0,1\,]$
is necessarily also continuous!  Here's why: 
$$-f,\,f\le|f|\implies -T(f),\,T(f)\le T(|f|);\ \hbox{i.e.,}\ 
	|T(f)|\le T(|f|).$$
But $|f|\le\Vert f\Vert\cdot1$, where $1$ denotes the constant
$1$ function, and so 
$$|T(f)|\le T(|f|)\le\Vert f\Vert\,T(1).$$
Thus,
$$\Vert T(f)\Vert\le\Vert f\Vert\,\Vert T(1)\Vert$$
for any $f\in C[\,0,1\,]$.  It follows that $T$ is Lipschitz
with constant $\Vert T(1)\Vert$:
$$\Vert T(f)-T(g)\Vert=\Vert T(f-g)\Vert\le\Vert T(1)\Vert\,\Vert f-g\Vert.$$

Now positive, linear maps abound in analysis, so this is a
fortunate turn of events.  What's more, Bernstein's theorem 
generalizes very nicely when placed in this new setting.
The following elegant theorem was proved (independently) by
Bohman and Korovkin in, roughly, 1952.

\proclaim Theorem.
Let $T_n:C[\,0,1\,]\to C[\,0,1\,]$ be a sequence of positive,
linear maps, and suppose that $T_n(f)\to f$ uniformly in each
of the three cases
$$f_0(x)=1,\quad f_1(x)=x,\quad \hbox{and}\quad f_2(x)=x^2.$$
Then, $T_n(f)\to f$ uniformly for every $f\in C[\,0,1\,]$.

The proof of the Bohman-Korovkin theorem is essentially identical
to the proof of Bernstein's theorem except, of course, we write 
$T_n(f)$ in place of $B_n(f)$.  For full details, see Cheney's
book {\it An Introduction to Approximation Theory}, Chelsea, 1982.  
Rather than proving the theorem, let's settle for a quick application.

\noindent{\bf Example}

\noindent
Let $f\in C[\,0,1\,]$ and, for each $n$, let $L_n(f)$ be the 
``polygonal'' approximation to $f$ with nodes at $k/n$, $k=0,1,\ldots,n$.
That is, $L_n(f)$ is linear on each subinterval $[\,(k-1)/n,k/n\,]$ and
agrees with $f$ at each of the endpoints $L_n(f)(k/n)=f(k/n)$.
Then, $L_n(f)\to f$ uniformly for each $f\in C[\,0,1\,]$.  This is
actually an easy calculation all by itself, but let's see why the
Bohman-Korovkin theorem makes short work of it.

That $L_n(f)$ is positive and linear is (nearly) obvious; that
$L_n(f_0)=f_0$ and $L_n(f_1)=f_1$ are really easy since, in fact,
$L_n(f)=f$ for any linear function $f$.  We just need to show
that $L_n(f_2)\uniformto f_2$.  But a picture will convince you
that the maximum distance between $L_n(f_2)$ and $f_2$ on the
interval $[\,(k-1)/n,k/n\,]$ is at most
$$\left({k\over n}\right)^2-\left({{k-1}\over n}\right)^2
	={{2k-1}\over{n^2}}\le{2\over n}.$$
That is, $\Vert f_2-L_n(f_2)\Vert\le2/n\to0$ as $n\to\infty$.~\qed



\bye


%%  end of chapter-02.tex


