%%  chapter-08.tex
%%  Approximation Theory: Chapter 8
%%  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 = {Jackson's Theorems}

\centerline{\hfil\tf Jackson's Theorems\hfil}
\vskip-\baselineskip
\line{\sc Math 680 \hfil 7/18/94}

\noindent
We continue our investigations of the ``middle ground''
between algebraic and trigonometric approximation by
presenting several results due to the great American
mathematician Dunham Jackson (from roughly 1911--1912).
The first of these results will give us the best possible
estimate of $E_n(f)$ in terms of $\omega_f$ and $n$.

\proclaim Jackson's Theorem 1.
If\/ $f\in C^{2\pi}$, then\/ 
$E_n^T(f)\le6\,\omega_f([-\pi,\pi\,];{1\over n}\,)$.

Theorem 1 should be viewed as an improvement over
Bernstein's Theorem, which stated that 
$E_n(f)\le{3\over2}\omega_f({1\over\sqrt{n}})$
for $f\in C[-1,1\,]$.  As we'll see, the proof of Theorem 1
not only mimics the proof of Bernstein's result, but also
uses some of the ideas we talked about in the last section.
In particular, the proof we'll give involves integration against
an ``improved'' Dirichlet kernel.

Before we dive into the proof, let's list several immediate and
important Corollaries:

\proclaim Corollary.  Weierstrass's second theorem
{\rm(since $\omega_f({1\over n})\to0$ for any $f\in C^{2\pi}\,$).}

\proclaim Corollary.  The Dini-Lipschitz theorem:  If\/
$\omega_f({1\over n})\log n\to0$ as $n\to\infty$, then the Fourier
series for $f$ converges uniformly to $f$.

\proof
From Lebesgue's theorem,
$$\Vert f-s_n(f)\Vert \ \le \ (4+\log n)\,E_n^T(f) \ 
	\le \ 6\,(4+\log n)\,\omega_f\left({1\over n}\right)\to0.
	\eqno\qed$$

\proclaim Jackson's Theorem 2.
If\/ $f\in C[-1,1\,]$, then\/
$E_n(f)\le6\,\omega_f([-1,1\,];{1\over n}\,)$.

\proof
Let $\varphi(\theta)=f(\cos\theta)$.  Then, as we've seen,
$$E_n(f)  \ = \ E_n^T(\varphi) \ \le \ 
	6\,\omega_\varphi\left([-\pi,\pi\,];{1\over n}\right) \ 
	\le 6\,\omega_f\left([-1,1\,];{1\over n}\right),$$
where the last inequality follows from the fact that
$$|\varphi(\alpha)-\varphi(\beta)| \ = \ 
	|f(\cos\alpha)-f(\cos\beta)| \ \le \ 
	\omega_f(|\cos\alpha-\cos\beta\,|) \ \le \ 
	\omega_f(|\alpha-\beta\,|).
	\eqno\qed$$

\proclaim Corollary.
If\/ $f\in{\rm lip}_K\alpha$, then\/
$E_n(f)\le 6Kn^{-\alpha}$.
{\rm(Bernstein's theorem gives $n^{-\alpha/2}$.)}

\proclaim Corollary.
If\/ $f\in C[-1,1\,]$ has a bounded derivative, then\/
$E_n(f)\le {6\over n}\,\Vert f\,'\,\Vert$.

\proclaim Corollary.
If\/ $f\in C[-1,1\,]$ has a continuous derivative, then\/
$E_n(f)\le {6\over n}\,E_{n-1}(f\,'\,)$.

\proof
If $p^*\in{\cal P}_{n-1}$ is the best approximation to $f\,'$,
consider $p(x)=\int_{-1}^xp^*(t)\,dt\in{\cal P}_n$.
From the previous Corollary,
$$\openup2\jot\eqalignno{
E_n(f) \ &= \ E_n(f-p)\qquad\hbox{(Why?)}\cr
	&\le \ {6\over n}\,\Vert f\,'-p^*\Vert \ 
	= \ {6\over n}\,E_{n-1}(f\,'\,).
	&\qed\cr
}$$

Iterating this last inequality will give the following result:

\proclaim Corollary.
If\/ $f\in C[-1,1\,]$ is $k$-times continuously differentiable,
then\/
$$E_n(f) \ \le \ {{6^{k+1}}\over{n(n-1)\cdots(n-k+1)}}\,\,
	\omega_k\left({1\over{n-k}}\right),$$
where $\omega_k$ is the modulus of continuity of $f^{(k)}$.

Well, enough corollaries.  It's time we proved 
Jackson's Theorem 1.  Now Jackson's approach was to show that
$${1\over\pi}\int_{-\pi}^\pi f(x+t)\,\cdot c_n
	\left({{\sin nt}\over{\sin t}}\right)^4dt \ 
	\uniformto \ f(x),$$
where $J_n(t)=c_n(\sin nt/\sin t)^4$ is the ``improved'' kernel
we alluded to earlier (it's essentially the square of Fej\'er's
kernel).  The approach we'll take, due to Korovkin, proves the
existence of a suitable kernel without giving a tidy formula
for it.  On the other hand, it's relatively easy to outline
the idea.  The key here is that $J_n(t)$ should be an even,
nonnegative, trig polynomial of degree $n$ with
${1\over\pi}\int_{-\pi}^\pi J_n(t)\,dt=1$.  In other words, 
$$J_n(t) \ = \ {1\over2}+\sum_{k=1}^n\rho_{k,n}\cos kt,$$
where $\rho_{1,n},\ldots,\rho_{n,n}$ must be chosen so that
$J_n(t)\ge0$.  Assuming we can find such $\rho_{k,n}$'s,
here's what we get:

\proclaim Lemma.
If\/ $f\in C^{2\pi}$, then
$$\left|\,f(x)-{1\over\pi}\int_{-\pi}^\pi f(x+t)\,J_n(t)\,dt\,\right|
	\ \ \le \ \ \omega_f\left({1\over n}\right)\cdot
	\left[1 + n\pi\sqrt{\,{{1-\rho_{1,n}}\over2}}\,\,\right].$$

\proof
We already know how the first several lines of the proof will go:
$$\openup2\jot\eqalign{
\left|\,f(x)-{1\over\pi}\int_{-\pi}^\pi f(x+t)\,J_n(t)\,dt\,\right| \ 
	&= \ {1\over\pi}\left|\int_{-\pi}^\pi \bigl[f(x)-f(x+t)\bigr]\,J_n(t)\,dt\,\right|\cr
	&\le \ {1\over\pi}\int_{-\pi}^\pi |f(x)-f(x+t)|\,J_n(t)\,dt\cr
	&\le \ {1\over\pi}\int_{-\pi}^\pi \omega_f(\,|t|\,)\,J_n(t)\,dt.\cr
}$$
Next we borrow a trick from Bernstein.  We replace $\omega_f(\,|t|\,)$ by 
$$\omega_f(\,|t|\,) \ = \ \omega_f\left(n|t|\cdot{1\over n}\right) \ 
	\le \ \bigl(1+n|t|\,\bigr)\,\omega_f\left({1\over n}\right),$$
and so the last integral on the right-hand side, above, is dominated by
$$\omega_f\left({1\over n}\right)\cdot
	{1\over\pi}\int_{-\pi}^\pi \bigl(1+n|t|\,\bigr)\,J_n(t)\,dt \ 
	= \ \omega_f\left({1\over n}\right)\cdot
	\left[1+{n\over\pi}\int_{-\pi}^\pi |t|\,J_n(t)\,dt\,\right].$$
All that remains is to estimate $\int_{-\pi}^\pi |t|\,J_n(t)\,dt$,
and for this we'll appeal to the Cauchy-Schwarz inequality (again,
compare this to the proof of Bernstein's theorem).
$$\openup2\jot\eqalign{
{1\over\pi}\int_{-\pi}^\pi |t|\,J_n(t)\,dt \ 
	&= \ {1\over\pi}\int_{-\pi}^\pi |t|\,J_n(t)^{1/2}J_n(t)^{1/2}\,dt\cr
	&\le \ \left({1\over\pi}\int_{-\pi}^\pi |t|^2\,J_n(t)\,dt\right)^{1/2}
		\left({1\over\pi}\int_{-\pi}^\pi J_n(t)\,dt\right)^{1/2}\cr
	&= \ \left({1\over\pi}\int_{-\pi}^\pi |t|^2\,J_n(t)\,dt\right)^{1/2}.\cr
}$$
But,
$$|t|^2 \ \le \ \left[\pi\sin\left({t\over2}\right)\right]^2 \ 
	= \ {{\pi^2}\over2}\,(1-\cos t\,).$$
So,
$${1\over\pi}\int_{-\pi}^\pi |t|\,J_n(t)\,dt \ 
	\le \ \left({{\pi^2}\over2}\cdot{1\over\pi}\int_{-\pi}^\pi 
		(1-\cos t)\,J_n(t)\,dt\right)^{1/2} \ 
	= \ \left({{\pi^2}\over2}(1-\rho_{1,n})\right)^{1/2}.
	\eqno\qed$$

Now we still have to prove that we can actually find a suitable
choice of scalars $\rho_{1,n},\ldots,\rho_{n,n}$.  We already know
that we need to choose the $\rho_{k,n}$'s so that $J_n(t)$ will be
nonnegative, but now it's clear that we also want $\rho_{1,n}$ 
to be very close to $1$.  To get us started, let's first see
why it's easy to generate nonnegative cosine polynomials.
Given {\sl real numbers\/} $c_0,\ldots,c_n$, note that 
$$\eqalignno{
0 \ \le \ \left|\,\sum_{k=0}^nc_ke^{ikx}\,\right|^2 \ 
	&= \ \left(\sum_{k=0}^nc_ke^{ikx}\right)\left(\sum_{j=0}^nc_je^{-ijx}\right) \ 
	= \ \sum_{k,j}c_kc_je^{i(k-j)x}\cr
	&= \ \sum_{k=0}^nc_k^2 \ + \ \sum_{k>j}c_kc_j\bigl(e^{i(k-j)x}+e^{i(j-k)x}\bigr)\cr
	&= \ \sum_{k=0}^nc_k^2 \ + \ 2\sum_{k>j}c_kc_j\cos(k-j)x\cr
	&= \ \sum_{k=0}^nc_k^2 \ + \ 2\sum_{k=0}^{n-1}c_kc_{k+1}\cos x
		+\cdots+2c_0c_n\cos nx.&(\ast)\cr
}$$
In particular, we need to find $c_0,\ldots,c_n$ with 
$$\sum_{k=0}^nc_k^2 \ = \ {1\over2}\qquad\hbox{and}\qquad
	2\sum_{k=0}^{n-1}c_kc_{k+1} \ \approx \ 1.$$
What we'll do is find $c_k$'s with
$\sum_{k=0}^{n-1}c_kc_{k+1}\approx\sum_{k=0}^nc_k^2$, and
then normalize.  But, in fact, we won't actually {\sl find\/}
anything---we'll simply write down a choice of $c_k$'s that
happen to work!  Consider:
$$\openup1\jot\eqalign{
\sum_{k=0}^n\sin\left({{k+1}\over{n+2}}\,\pi\right)&
	\sin\left({{k+2}\over{n+2}}\,\pi\right) \ 
= \ \sum_{k=0}^n\sin\left({{k+1}\over{n+2}}\,\pi\right)
	\sin\left({k\over{n+2}}\,\pi\right)\cr
&= \ {1\over2}\sum_{k=0}^n\left[\sin\left({k\over{n+2}}\,\pi\right)
	+\sin\left({{k+2}\over{n+2}}\,\pi\right)\right]
	\sin\left({{k+1}\over{n+2}}\,\pi\right).\cr
}$$
By changing the index of summation, it's easy to see that
first two sums are equal and, hence, each is equal to the
average of the two.  Next we re-write this last sum, using the 
trig identity
${1\over2}\bigl(\sin A+\sin B\bigr)=\cos\left({{A-B}\over2}\right)
	\sin\left({{A+B}\over2}\right)$, to get
$$\sum_{k=0}^n\sin\left({{k+1}\over{n+2}}\,\pi\right)
	\sin\left({{k+2}\over{n+2}}\,\pi\right) \ 
	= \ \cos\left({\pi\over{n+2}}\right)
	\sum_{k=0}^n\sin^2\left({{k+1}\over{n+2}}\,\pi\right).$$
Since $\cos\left({\pi\over{n+2}}\right)\approx1$ for large $n$, 
we've done it!  If we define  
$c_k=c\cdot\sin\left({{k+1}\over{n+2}}\,\pi\right)$,
where $c$ is chosen so that $\sum_{k=0}^nc_k^2=1/2$,
and if we define $J_n(x)$ using $(\ast)$, then $J_n(x)\ge0$
and $\rho_{1,n}= \cos\left({\pi\over{n+2}}\right)$.  The
estimate needed in our Lemma becomes
$$\sqrt{\,{{1-\rho_{1,n}}\over2}} \ 
	= \ \sqrt{\,{{1-\cos\left({\pi\over{n+2}}\right)}\over2}} \ 
	= \ \sin\left({\pi\over{2n+4}}\right) \ \le \ {\pi\over{2n}},$$
and so we have
$$E_n^T(f) \ \le \ 
	\left(1+{{\pi^2}\over2}\right)\,\omega_f\left({1\over n}\right)
	\ < \ 6\,\omega_f\left({1\over n}\right).
	\eqno\qed$$

Jackson's theorems are what we might call {\sl direct theorems}.  
If we know something about $f$, then we can say something about
$E_n(f)$.  There is also the notion of an {\sl inverse theorem},
meaning that if we know something about $E_n(f)$, we should be
able to say something about $f$.  In other words, we would expect
an inverse theorem to be, more or less, the converse of some
direct theorem.  Now inverse theorems are typically much harder
to prove than direct theorems, but in order to have some idea of 
what such theorems might tell us (and to see some of the
techniques used in their proofs), we present one of the easier 
inverse theorems, due to Bernstein.  This result gives the converse
to one of our corollaries to Jackson's theorem (see the top of page 2).

\proclaim Theorem.
If\/ $f\in C^{2\pi}$ satisfies\/ $E_n^T(f)\le A\,n^{-\alpha}$, for
some constants $A$ and $0<\alpha<1$, then $f\in{\rm lip}_K\alpha$
for some constant $K$.

\proof
For each $n$, choose $U_n\in{\cal T}_n$ so that
$\Vert f-U_n\Vert\le A\,n^{-\alpha}$.  Now if we
set $V_0=U_1$ and $V_n=U_{2^n}-U_{2^{n-1}}$ for
$n\ge1$, then $V_n\in{\cal T}_{2^n}$ and
$f=\sum_{n=0}^\infty V_n$.  Indeed, 
$$\Vert V_n\Vert \ \le \ 
	\Vert U_{2^n}-f\Vert + \Vert U_{2^{n-1}}-f\Vert \ 
	\le \ A\,(2^n)^{-\alpha}+A\,(2^{n-1})^{-\alpha} \ 
	= \ B\cdot2^{-n\alpha},$$
which is summable; thus, the (telescoping) series 
$\sum_{n=0}^\infty V_n$ converges uniformly to $f$.

Next we estimate $|f(x)-f(y)|$ using finitely many of
the $V_n$'s, the precise number to be specified later.
Using the mean value theorem and Bernstein's inequality
we get 
$$\eqalign{
|f(x)-f(y)| \ &\le \ \sum_{n=0}^\infty|V_n(x)-V_n(y)|\cr
	&\le \ \sum_{n=0}^{m-1}|V_n(x)-V_n(y)|
		\ + \ 2\sum_{n=m}^\infty\Vert V_n\Vert\cr
	&= \ \sum_{n=0}^{m-1}|V_n'(\xi_n)|\,|x-y|
		\ + \ 2\sum_{n=m}^\infty\Vert V_n\Vert\cr
	&\le \ |x-y|\sum_{n=0}^{m-1}2^n\Vert V_n\Vert
		\ + \ 2\sum_{n=m}^\infty\Vert V_n\Vert\cr
	&\le \ |x-y| \sum_{n=0}^{m-1}B\,2^{n(1-\alpha)}
		\ + \ 2\sum_{n=m}^\infty B\,2^{-n\alpha}\cr
	&\le \ C\,\left[\,|x-y|\cdot2^{m(1-\alpha)} 
		\ + \ 2^{-m\alpha}\,\right],\cr
}$$
where we've used, in the fourth line, the fact that 
$V_n\in{\cal T}_{2^n}$ and, in the last line, 
standard estimates for geometric series.  Now we
want the right-hand side to be dominated by a constant
times $|x-y|^\alpha$.  In other words, if we set
$|x-y|=\delta$, then we want 
$$\delta\cdot2^{m(1-\alpha)} \ + \ 2^{-m\alpha} 
	\ \le \ D\cdot\delta^\alpha$$
or, equivalently, 
$$(2^m\delta)^{(1-\alpha)} \ + \ (2^m\delta)^{-\alpha}
	\ \le \ D.$$
Thus, we should choose $m$ so that $2^m\delta$ is both
bounded above and bounded away from zero.  For example,
if $0<\delta<1$, then we could choose $m$ so that 
$1\le2^m\delta<2$.~\qed

In order to better explain the phrase ``more or less
the converse of some direct theorem,'' let's see how
the previous result falls apart when $\alpha=1$.  
Although we might hope that $E_n^T(f)\le A/n$ would
imply that $f\in{\rm lip}_K1$, it happens not to be true. 
The best result in this regard is due to Zygmund, who 
gave necessary and sufficient conditions on $f$ so
that $E_n^T(f)\le A/n$ (and these conditions do not 
characterize ${\rm lip}_K1$ functions).  Instead of
pursuing Zygmund's result, we'll
settle for simple ``surgery'' on our previous result,
keeping an eye out for what goes wrong.
This result is again due to Bernstein.

\proclaim Theorem.
If\/ $f\in C^{2\pi}$ satisfies\/ $E_n^T(f)\le A/n$, 
then $\omega_f(\delta)\le K\delta|\log\delta\,|$
for some constant $K$ and all $\delta$ sufficiently small.

\proof
If we repeat the previous proof, setting $\alpha=1$,
only a few lines change.  In particular, $2^n\Vert V_n\Vert$
is now equal to $1$ and so the conclusion of that long string 
of inequalites would read
$$|f(x)-f(y)| \ \le \ C\,\bigl[\,|x-y|\cdot m \ + \ 2^{-m}\,\bigr]
	\ = \ C\,\bigl[\,m\delta \ + \ \ 2^{-m}\,\bigr].$$
Clearly, we can't hope for the right-hand side to be dominated
by a constant times $\delta$, as we might have hoped, for this
would force $m$ to be bounded (independent of $\delta$\/), which
in turn bounds $2^{-m}$ away from zero.  But, if we again think of 
$2^m\delta$ as the ``variable'' in this inequality, then 
the term $m\delta$ suggests that the correct order of magnitude
of the right-hand side is $\delta|\log\delta\,|$.  
Thus, we would try to find a constant $D$ so that 
$$\displaylines{
m\delta \ + \ \ 2^{-m} \ \le \ D\cdot\delta|\log\delta\,|\cr
\noalign{\hbox{or}}
m(2^m\delta) \ + \ 1 \ \le \ D\cdot(2^m\delta)|\log\delta\,|.\cr
}$$
Now if we take $0<\delta<1/2$, then $\log 2<-\log\delta=|\log\delta\,|$.
Hence, if we again choose $m\ge1$ so that $1\le2^m\delta<2$, we'll get
$$m\log 2+\log\delta < \log 2\quad\implies\quad
	m<{{\log 2-\log\delta}\over{\log 2}}<{2\over{\log 2}}\,|\log\delta\,|$$
and, finally,
$$m(2^m\delta)+1 \ \le \ 2m+1 \ \le \ 3m \ \le \ 
	{6\over{\log 2}}\,|\log\delta\,| \ \le \ 
	{6\over{\log 2}}\,(2^m\delta)\,|\log\delta\,|.
	\eqno\qed$$


\bye


%%  end of chapter-08.tex





