%%  chapter-07.tex
%%  Approximation Theory: Chapter 7
%%  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 = {Fourier Series}

\centerline{\hfil\tf A Brief Introduction to Fourier Series\hfil}
\vskip-\baselineskip
\line{\sc Math 680 \hfil 7/18/94}

\noindent
The {\sl Fourier series\/} of a $2\pi$-periodic, integrable 
function $f$ is 
$${{a_0}\over2}+\sum_{k=1}^\infty\bigl(a_k\cos kx+b_k\sin kx\bigr),$$
where the coefficients are defined by
$$a_k={1\over\pi}\int_{-\pi}^\pi f(t)\,\cos kt\,dt
	\qquad\hbox{and}\qquad
	b_k={1\over\pi}\int_{-\pi}^\pi f(t)\,\sin kt\,dt.$$
Please note that each of these integrals is well-defined and finite;
indeed, 
$$|a_k|\le{1\over\pi}\int_{-\pi}^\pi |f(t)|\,dt$$
and so, for example, we would have 
$|a_k|\le2\Vert f\Vert$ for $f\in C^{2\pi}$.

We write the partial sums of the series as
$$s_n(f)(x)={{a_0}\over2}+\sum_{k=1}^n\bigl(a_k\cos kx+b_k\sin kx\bigr).$$
Now while $s_n(f)$ need not converge pointwise to $f$
(in fact, it may even diverge at a given point), 
and while $s_n(f)$ is not typically a good uniform
approximation to $f$, it is still a very natural
choice for an approximation to $f$ in the ``least-squares''
sense (which we'll make precise shortly).  Said in other
words, the Fourier series for $f$ provides a useful
{\sl representation\/} for $f$ even if it fails to
converge pointwise to $f$.

\noindent{\bf Observations}

\item{\bf1.}
The functions $1,\cos x,\cos 2x,\ldots$, $\sin x,\sin 2x,\ldots$,
are {\sl orthogonal\/} on $[-\pi,\pi\,]$.  That is,
$$\int_{-\pi}^\pi \cos mx\,\cos nx\,dx=
	\int_{-\pi}^\pi \sin mx\,\sin nx\,dx=
		\int_{-\pi}^\pi \cos mx\,\sin nx\,dx=0$$
for any $m\ne n$ (and the last equation even holds for $m=n$),
$$\int_{-\pi}^\pi \cos^2 mx\,dx=\int_{-\pi}^\pi \sin^2 mx\,dx=\pi$$
for any $m\ne0$, and, of course, $\int_{-\pi}^\pi 1\,dx=2\pi$.

\item{\bf2.}
What this means is that if 
$T(x)={{a_0}\over2}+\sum_{k=1}^n\bigl(a_k\cos kx+b_k\sin kx\bigr)$,
then, 
$$\int_{-\pi}^\pi T(x)\,\cos mx\,dx=a_m\int_{-\pi}^\pi \cos^2 mx\,dx
	=\cases{\pi a_m &$(m\ne0)$\cr 2\pi a_0 &$(m=0)$.}$$
That is, if $T\in{\cal T}_n$,
then $T$ is actually equal to its own Fourier series.

\item{\bf3.}
The partial sum operator $s_n(f)$ is a {\sl linear projection\/} from 
$C^{2\pi}$ onto ${\cal T}_n$.

\item{\bf4.}
If some $f\in C^{2\pi}$ has $a_k=b_k=0$ for all $k$, then $f\equiv0$.

\item{}
Indeed, by linearity, this means that 
$$\int_{-\pi}^\pi f(x)\,T(x)\,dx=0$$
for any trig polynomial $T$.  But from Weierstrass's second theorem
we know that $f$ is the uniform limit of some sequence of trig
polynomials $(T_n)$.  Thus,
$$\int_{-\pi}^\pi f(x)^2\,dx=\Limonn\int_{-\pi}^\pi f(x)\,T_n(x)\,dx=0.$$
Since $f$ is continuous, this easily implies that $f\equiv0$.

\item{\bf5.}
If $f$, $g\in C^{2\pi}$ have the same Fourier series, then $f\equiv g$.

\item{\bf6.}
The coefficients $a_k$ and $b_k$ minimize the expression
$$\varphi(a_0,a_1,\ldots,b_n)
	=\int_{-\pi}^\pi \bigl[f(x)-s_n(f)(x)\bigr]^2\,dx.$$

\item{}
It's not hard to see, for example, that 
$${{\partial\,\varphi}\over{\partial\,a_k}}
	=\int_{-\pi}^\pi 2\bigl[f(x)-s_n(f)(x)\bigr]\cos kx\,dx=0$$
precisely when $a_k$ satisfies
$$\int_{-\pi}^\pi f(x)\,\cos kx\,dx=a_k\int_{-\pi}^\pi \cos^2 kx\,dx.$$

\item{\bf7.}
The partial sum $s_n(f)$ is the best approximation to $f$ out of ${\cal T}_n$
{\sl relative to the\/} $L_2$ {\sl norm}
$$\Vert f\Vert_2=\left({1\over\pi}\int_{-\pi}^\pi f(x)^2\,dx\right)^{1/2}.$$
(Some authors prefer $1/2\pi$ in place of $1/\pi$.)
Moreover,
$$\Vert f-s_n(f)\Vert_2^2={1\over\pi}\int_{-\pi}^\pi f(x)^2\,dx
	-{\textstyle{1\over4}}a_0^2-\sum_{k=1}^n\bigl(a_k^2+b_k^2\bigr)
	=\Vert f\Vert_2^2-\Vert s_n(f)\Vert_2^2.$$

\item{\bf8.}
It follows from {\bf7} that 
$${1\over\pi}\int_{-\pi}^\pi s_n(f)(x)^2\,dx
	={\textstyle{1\over4}}a_0^2+\sum_{k=1}^n\bigl(a_k^2+b_k^2\bigr)
	\le{1\over\pi}\int_{-\pi}^\pi f(x)^2\,dx.$$
In other words, $\Vert s_n(f)\Vert_2\le\Vert f\Vert_2$.  
In particular, the Fourier coefficients of any $f\in C^{2\pi}$ are
square summable.  (Why?)

\item{\bf9.}
It follows from {\bf8} and Weierstrass's second theorem that 
$s_n(f)\to f$ {\sl in the\/} $L_2$ {\sl norm\/} whenever $f\in C^{2\pi}$.

\item{}
Indeed, given $\eps>0$, choose a trig polynomial $T$ such that 
$\Vert f-T\Vert<\eps$.  Then, since $s_n(T)=T$ for large enough $n$,
we have 
$$\eqalign{
\Vert f-s_n(f)\Vert_2&\le \Vert f-T\Vert_2+\Vert s_n(T-f)\Vert_2\cr
	&\le2\Vert f-T\Vert_2\le2\sqrt{2}\,\Vert f-T\Vert<2\sqrt{2}\,\eps.
}$$

By way of comparison, let's give a simple class of functions whose 
Fourier partial sums provide good {\sl uniform\/} approximations.

\proclaim Theorem.
If\/ $f\,''\in C^{2\pi}$, then the Fourier series for $f$ converges
uniformly to $f$.

\proof
First notice that integration by-parts gives us an estimate
on the order of growth of the Fourier coefficients:
$$\pi a_k = \int_{-\pi}^\pi f(x)\,\cos kx\,dx
	=\int_{-\pi}^\pi f(x)\,d\left({{\sin kx}\over k}\right)
	=-{1\over k}\int_{-\pi}^\pi f\,'(x)\,\sin kx\,dx$$
(because $f$ is $2\pi$-periodic).  Thus, 
$|a_k|\le2\Vert f\,'\,\Vert/k\to0$ as $k\to\infty$.
Now we integrate by-parts again:
$$-\pi ka_k=\int_{-\pi}^\pi f\,'(x)\,\sin kx\,dx
	=\int_{-\pi}^\pi f\,'(x)\,d\left({{\cos kx}\over k}\right)
	={1\over k}\int_{-\pi}^\pi f\,''(x)\,\cos kx\,dx$$
(because $f\,'$ is $2\pi$-periodic).  Thus, 
$|a_k|\le2\Vert f\,''\,\Vert/k^2\to0$ as $k\to\infty$.
More importantly, this inequality (along with the Weierstrass 
$M$-test) implies that the Fourier 
series for $f$ is both uniformly and absolutely convergent:
$$\left|{{a_0}\over2}+\sum_{k=1}^\infty\bigl(a_k\cos kx+b_k\sin kx\bigr)\right|
	\le \left|{{a_0}\over2}\right|+\sum_{k=1}^\infty\bigl(|a_k|+|b_k|\bigr)
	\le C\sum_{k=1}^\infty{1\over{k^2}}.$$
But why should the series actually converge to $f\,$?
Well, if we call the sum 
$$g(x)={{a_0}\over2}+\sum_{k=1}^\infty\bigl(a_k\cos kx+b_k\sin kx\bigr),$$
then $g\in C^{2\pi}$ (why?)\ and $g$ has the same Fourier coefficients
as $f\,$!  Hence, $g=f$.~\qed

Our next chore is to find a closed expression for 
$s_n(f)$.  For this we'll need a couple of trig identities;
the first two need no explanation.
$$\displaylines{
\cos kt\cos kx + \sin kt\sin kx = \cos k(t-x)\cr
2\cos\alpha\sin\beta = \sin(\alpha+\beta)-\sin(\alpha-\beta)\cr
{\textstyle{1\over2}}+\cos\theta+\cos2\theta+\cdots+\cos n\theta=
	{{\sin\,(n+{1\over2})\,\theta}\over{2\sin{1\over2}\theta}}\cr
}$$
Here's a short proof for the third:
$$\sin{\textstyle{1\over2}}\theta + 
	\sum_{k=1}^n2\cos k\theta\,\sin{\textstyle{1\over2}}\theta
	=\sin{\textstyle{1\over2}}\theta+\sum_{k=1}^n\bigl[\,
		\sin\,(k+{\textstyle{1\over2}})\,\theta
			-\sin\,(k-{\textstyle{1\over2}})\,\theta\,\bigr]
	=\sin\,(n+{\textstyle{1\over2}})\,\theta.$$
The function 
$$D_n(t)={{\sin\,(n+{1\over2})\,t}\over{2\sin{1\over2}\,t}}$$
is called {\sl Dirichlet's kernel}.  It plays an important role
in our next calculation.

Now we're ready to re-write our formula for $s_n(f)$. 
$$\openup2\jot\eqalign{
s_n(f)(x)&={\textstyle{1\over2}}a_0+
		\sum_{k=1}^n\bigl(a_k\cos kx+b_k\sin kx\bigr)\cr
	&={1\over\pi}\int_{-\pi}^\pi f(t)\left[{\textstyle{1\over2}}+
		\sum_{k=1}^n\cos kt\cos kx + \sin kt\sin kx\right]dt\cr
	&={1\over\pi}\int_{-\pi}^\pi f(t)\left[{\textstyle{1\over2}}+
		\sum_{k=1}^n\cos k(t-x)\right]dt\cr
	&={1\over\pi}\int_{-\pi}^\pi f(t)\cdot
		{{\sin\,(n+{1\over2})\,(t-x)}\over{2\sin{1\over2}\,(t-x)}}\,dt\cr
	&={1\over\pi}\int_{-\pi}^\pi f(t)\,D_n(t-x)\,dt
	={1\over\pi}\int_{-\pi}^\pi f(x+t)\,D_n(t)\,dt.\cr
}$$
It now follows easily that $s_n(f)$ is linear in $f$ (because 
integration against $D_n$ is linear), that $s_n(f)\in{\cal T}_n$ 
(because $D_n\in{\cal T}_n$), and, in fact, that 
$s_n({\cal T}_m)={\cal T}_{\min(m,n)}$.  In other words, $s_n$
is indeed a linear projection onto ${\cal T}_n$.

While we know that $s_n(f)$ is a good approximation to $f$ in
the $L_2$ norm, a better understanding of its effectiveness as 
a uniform approximation will require a better understanding
of the Dirichlet kernel $D_n$.  Here are a few pertinent facts:

{\baselineskip=24 true pt
\proclaim Lemma. \ 
{\rm(a)} \ $D_n$ is even,
\item{\rm(b)} \ 
$\displaystyle{1\over\pi}\int_{-\pi}^\pi D_n(t)\,dt
	={2\over\pi}\int_0^\pi D_n(t)\,dt=1$,
\item{\rm(c)} \ 
$|D_n(t)|\le n+{1\over2}$ \ and \ $D_n(0)=n+{1\over2}$,
\item{\rm(d)} \ 
$\displaystyle{{{|\sin\,(n+{1\over2})\,t\,|}\over t}}
	\ \le \ |D_n(t)|\le {\pi\over{2t}}$ \ for \ $0<t<\pi$,
\item{\rm(e)} \ 
If \ $\lambda_n=\displaystyle{1\over\pi}\int_{-\pi}^\pi |D_n(t)|\,dt$, \ 
then \ $\displaystyle{{4\over{\pi^2}}\log n \ \le \ \lambda_n \ \le \ 3+\log n}$.
\par}

\proof
(a), (b), and (c) are relatively clear from the fact that
$$D_n(t)={\textstyle{1\over2}}+\cos t+\cos2t+\cdots+\cos nt.$$
(Notice, too, that (b) follows from the fact that $s_n(1)=1$.)
For (d) we use a more delicate estimate: Since
$2\theta/\pi\le\sin\theta\le\theta$ for $0<\theta<\pi/2$,
it follows that $2t/\pi\le2\sin(t/2)\le t$ for $0<t<\pi$.
Hence, 
$${\pi\over{2t}} \ \ge \ 
	{{|\sin\,(n+{1\over2})\,t\,|}\over{2\sin{1\over2}t}}
	\ \ge \ {{|\sin\,(n+{1\over2})\,t\,|}\over t}$$
for $0<t<\pi$.  Next, the upper estimate in (e) is easy:
$$\openup2\jot\eqalign{
{2\over\pi}\int_0^\pi|D_n(t)|\,dt \ 
	&= \ {2\over\pi}\int_0^\pi{{|\sin\,(n+{1\over2})\,t\,|}\over{2\sin{1\over2}t}}\,dt\cr
	&\le \ {2\over\pi}\int_0^{1/n}(n+{\textstyle{1\over2}})\,dt
		+{2\over\pi}\int_{1/n}^\pi{\pi\over{2t}}\,dt\cr
	&= \ {{2n+1}\over{\pi n}}+\log\pi +\log n<3+\log n.\cr
}$$
The lower estimate takes some work:
$$\openup2\jot\eqalign{
{2\over\pi}\int_0^\pi|D_n(t)|\,dt \ 
	&= \ {2\over\pi}\int_0^\pi{{|\sin\,(n+{1\over2})\,t\,|}\over{2\sin{1\over2}t}}\,dt
		\ \ge \ {2\over\pi}\int_0^\pi{{|\sin\,(n+{1\over2})\,t\,|}\over t}\,dt\cr
	&= \ {2\over\pi}\int_0^{(n+{1\over2})\pi}{{|\sin x\,|}\over x}\,dx
		\ \ge \ {2\over\pi}\int_0^{n\pi}{{|\sin x\,|}\over x}\,dx\cr
	&= \ {2\over\pi}\,\sum_{k=1}^n\int_{(k-1)\pi}^{k\pi}{{|\sin x\,|}\over x}\,dx\cr
	&\ge \ {2\over\pi}\,\sum_{k=1}^n{1\over{k\pi}}\int_{(k-1)\pi}^{k\pi}|\sin x\,|\,dx
		\ = \ {4\over{\pi^2}}\,\sum_{k=1}^n{1\over k}\cr
	&\ge \ {4\over{\pi^2}}\,\log n,\cr
}$$
because $\sum_{k=1}^n{1\over k}\ge\log n$.~\qed

The numbers $\lambda_n=\Vert D_n\Vert_1$ are called the 
{\sl Lebesgue numbers\/} associated to this process
(compare this to the terminology we used for interpolation).
The point here is that $\lambda_n=\Vert D_n\Vert_1$
gives the norm of the partial sum operator (projection)
on $C^{2\pi}$ and, as before, $\lambda_n\to\infty$
as $n\to\infty$.  (As a matter of no small curiosity, notice 
that, from Observation {\bf8}, the norm
of $s_n$ as an operator on $L_2$ is $1$.)

\proclaim Corollary.
If $f\in C^{2\pi}$, then
$$|s_n(f)(x)|\le{1\over\pi}\int_{-\pi}^\pi|f(x+t)|\,|D_n(t)|\,dt
	\le\lambda_n\Vert f\Vert.\eqno(\ast)$$
In particular,
$\Vert s_n(f)\Vert\le\lambda_n\Vert f\Vert\le(3+\log n)\Vert f\Vert$.

If we approximate the function ${\rm sgn}\,D_n$ by a
continuous function $f$ of norm one, then 
$$s_n(f)(0)\approx{1\over\pi}\int_{-\pi}^\pi|D_n(t)|\,dt=\lambda_n.$$
Thus, $\lambda_n$ is the smallest constant that works in $(\ast)$.
The fact that the partial sum operators are not uniformly
bounded on $C^{2\pi}$, along with the Baire category theorem,
tells us that there must be some $f\in C^{2\pi}$ for which
$\Vert s_n(f)\Vert$ is unbounded.  But, as in the case of
projections onto ${\cal P}_n$, the algebraic polynomials, 
this has more to do with projections than it does with
Fourier series:

\proclaim Theorem. \ {\rm(Kharshiladze, Lozinski)} \ 
For each $n$, let $L_n$ be a continuous, linear projection
from $C^{2\pi}$ onto ${\cal T}_n$.  Then, there is some
$f\in C^{2\pi}$ for which $\Vert L_n(f)-f\Vert$ is unbounded.

Although our last Corollary may not look very useful, it
does give us some information about the effectiveness of
$s_n(f)$ as a uniform approximation to $f$.  Specifically, 
we have {\sl Lebesgue's theorem\/}:

\proclaim Theorem.
If\/ $f\in C^{2\pi}$, and if we set \ 
$E_n^T(f)=\Min_{\strut T\in{\cal T}_n}\Vert f-T\Vert$, \ then
$$E_n^T(f) \ \le \ \Vert f-s_n(f)\Vert \ \le \ (4+\log n)\,E_n^T(f).$$

\proof
Let $T^*$ be the best approximation to $f$ out of ${\cal T}_n$.
Then, since $s_n(T^*)=T^*$, we get
$$\Vert f-s_n(f)\Vert\le\Vert f-T^*\Vert+\Vert s_n(T^*-f)\Vert
	\le (4+\log n)\,\Vert f-T^*\Vert.
	\eqno\qed$$

As an application of Lebesgue's theorem, let's speak briefly
about ``Chebyshev series,'' a notion that fits neatly in between
our discussions of approximation by algebraic polynomials and
by trig polynomials.

\proclaim Theorem.
Suppose that $f\in C[-1,1\,]$ is twice continously differentiable.  
Then, $f$ may be written
as a uniformly and absolutely convergent Chebyshev series; that is,
$f(x)=\sum_{k=0}^\infty a_kT_k(x)$, where $\sum_{k=0}^\infty|a_k|<\infty$.

\proof
As usual, consider $\varphi(\theta)=f(\cos\theta)\in C^{2\pi}$.  
Since $\varphi$
is {\sl even\/} and twice differentiable, its Fourier series is an
absolutely and uniformly convergent cosine series:
$$f(\cos\theta)=\sum_{k=0}^\infty a_k\cos kx=\sum_{k=0}^\infty a_kT_k(\cos x),$$
where $|a_k|\le 2\Vert \varphi\,''\,\Vert/k^2$.  Thus, 
$f(x)=\sum_{k=0}^\infty a_kT_k(x)$.~\qed

If we write $S_n(f)(x)=\sum_{k=0}^n a_kT_k(x)$, we get an
interesting consequence of this Theorem.  First, notice that
$$S_n(f)(\cos\theta)=s_n(\varphi)(\theta).$$
Thus, from Lebesgue's theorem,
$$\eqalign{
E_n(f) \ \le \ \Vert f-S_n(f)\Vert_{C[-1,1\,]} \ 
	&= \ \Vert \varphi-s_n(\varphi)\Vert_{C^{2\pi}}\cr
	&\le \ (4+\log n)\,E_n^T(\varphi) \ 
	= \ (4+\log n)\,E_n(f).\cr
}$$
For $n<400$, this reads
$$E_n(f) \ \le \ \Vert f-S_n(f)\Vert \ \le \ 10\,E_n(f).$$
That is, for numerical purposes, the error incurred by using
$\sum_{k=0}^n a_kT_k(x)$ to approximate $f$ is 
{\sl within one decimal place accuracy\/} of the best
approximation!  Notice, too, that $E_n(f)$ would be very
easy to estimate in this case, since
$$E_n(f) \ \le \ \Vert f-S_n(f)\Vert \
	= \ \left\Vert\,\sum_{k>n} a_kT_k\right\Vert \ 
	\le \ \sum_{k>n}|a_k| \ 
	\le \ 2\,\Vert \varphi\,''\,\Vert\,\sum_{k>n}{1\over{k^2}}.$$

Lebesgue's theorem should remind you of our ``fancy'' version of Bernstein's
theorem; if we knew that $E_n^T(f)\log n\to0$ as $n\to\infty$,
then we'd know that $s_n(f)$ converged
uniformly to $f$.  Our goal, then, is to improve our estimates
on $E_n^T(f)$, and the idea behind these improvements is
to {\sl replace $D_n$ by a better kernel\/} (with regard to uniform
approximation).  Before we pursue anything quite so delicate
as an estimate on $E_n^T(f)$, though, let's investigate a
simple (and useful) replacement for $D_n$.

Since the sequence of partial sums $(s_n)$ need not 
converge to $f$, we might try looking at their 
arithmetic means (or Ces\`aro sums):
$$\sigma_n={{s_0+s_1+\cdots+s_{n-1}}\over n}.$$
(These averages typically have better convergence properties
than the partial sums themselves.  Just consider $\sigma_n$
in the (scalar) case $s_n=(-1)^n$, for example.)  Specifically,
we set
$$\eqalign{
\sigma_n(f)(x)
	&={1\over n}(s_0(f)+\cdots+s_{n-1}(f))\cr
	&={1\over\pi}\int_{-\pi}^\pi f(x+t)\,
		\left[{1\over n}\sum_{k=0}^{n-1}D_k(t)\right]\,dt
	={1\over\pi}\int_{-\pi}^\pi f(x+t)\,K_n(t)\,dt,
}$$
where $K_n=(D_0+D_1+\cdots+D_{n-1})/n$ is called
{\sl Fej\'er's kernel}.  
The same techniques
we used earlier can be applied to find a closed form for
$\sigma_n(f)$ which, of course, reduces to simplifying
$(D_0+D_1+\cdots+D_{n-1})/n$.  As before, we begin
with a trig identity:
$$\eqalign{
2\sin\theta\,\sum_{k=0}^{n-1}\sin\,(2k+1)\theta
	&=\sum_{k=0}^{n-1}\bigl[\,\cos2k\theta-\cos\,(2k+2)\theta\,\bigr]\cr
	&=1-\cos2n\theta=2\sin^2n\theta.\cr
}$$
Thus,
$$K_n(t) \ = \ {1\over n}\sum_{k=0}^{n-1}
	{{\sin\,(2k+1)\,t/2}\over{2\sin\,(t/2)}}
	\ = \ {{\sin^2(nt/2)}\over{2n\sin^2(t/2)}}.$$
Please note that $K_n$ is even, {\sl nonnegative},
and ${1\over\pi}\int_{-\pi}^\pi K_n(t)\,dt=1$.  Thus,
$\sigma_n(f)$ is a positive, linear map from $C^{2\pi}$
onto ${\cal T}_n$ (but it's not a projection---why?),
satisfying $\Vert\sigma_n(f)\Vert_2\le\Vert f\Vert_2$ (why?).

Now the arithmetic mean operator $\sigma_n(f)$ is still
a good approximation $f$ in $L_2$ norm.  Indeed,
$$\Vert f-\sigma_n(f)\Vert_2
	={1\over n}\left\Vert\,\sum_{k=0}^{n-1}(f-s_k(f))\,\right\Vert_2
	\le{1\over n}\sum_{k=0}^{n-1}\Vert f-s_k(f)\Vert_2
	\to0$$
as $n\to\infty$ (since $\Vert f-s_k(f)\Vert_2\to0$\/). 
But, more to the point, $\sigma_n(f)$ is
actually a good {\sl uniform\/} approximation to $f$,
a fact that we'll call {\sl Fej\'er's theorem\/}:

\proclaim Theorem.
If\/ $f\in C^{2\pi}$, then\/ $\sigma_n(f)$ converges uniformly
to $f$ as $n\to\infty$.

Note that since $\sigma_n(f)\in{\cal T}_n$, Fej\'er's theorem implies
Weierstrass's (second) theorem.  Curiously, Fej\'er was only 
19 years old when he proved this result (about 1900)
while Weierstrass was 
75 at the time he proved his approximation theorems.  

We'll give
two proofs of Fej\'er's theorem; one with details, one without.
But both follow from quite general considerations.  First:

{\baselineskip=24 true pt
\proclaim Theorem.
Suppose that $k_n\in C^{2\pi}$ satisfies
\item{\rm(a)} \ 
$k_n\ge0$,
\item{\rm(b)} \ 
$\displaystyle{{1\over\pi}\int_{-\pi}^\pi k_n(t)\,dt=1}$, and
\item{\rm(c)} \ 
$\displaystyle\int_{\delta\le|t|\le\pi}k_n(t)\,dt\to0$ for every $\delta>0$.
\vskip0pt\noindent
Then, \ $\displaystyle{1\over\pi}\int_{-\pi}^\pi f(x+t)\,k_n(t)\,dt\uniformto f(x)$ \ 
for each $f\in C^{2\pi}$.
\par}

\proof
Let $\eps>0$.  Since $f$ is uniformly continuous, we may choose
$\delta>0$ so that $|f(x)-f(x+t)|<\eps$, for any $x$, whenever $|t|<\delta$.
Next, we use the fact that 
$k_n$ is nonnegative and integrates to $1$ to write
$$\openup2\jot\eqalign{
\left|f(x)-{1\over\pi}\int_{-\pi}^\pi f(x+t)\,k_n(t)\,dt\,\right| \ 
	&= \ {1\over\pi}\left|\int_{-\pi}^\pi \bigl[f(x)-f(x+t)\bigr]\,k_n(t)\,dt\,\right|\cr
	&\le \ {1\over\pi}\int_{-\pi}^\pi \bigl|f(x)-f(x+t)\bigr|\,k_n(t)\,dt\cr
	&\le \ {\eps\over\pi}\int_{|t|<\delta}k_n(t)\,dt \ 
		+ \ {{2\Vert f\Vert}\over\pi}\int_{\delta\le|t|\le\pi}k_n(t)\,dt\cr
	&<\eps+\eps=2\eps,
}$$
for $n$ sufficiently large.~\qed

To see that Fej\'er's kernel satisfies the conditions of the
Theorem is easy: In particular, (c) follows from the fact that
$K_n(t)\uniformto0$ on the set $\delta\le|t|\le\pi$.  Indeed,
since $\sin(t/2)$ increases on $\delta\le t\le\pi$ we have
$$K_n(t) \ = \ {{\sin^2(nt/2)}\over{2n\sin^2(t/2)}} \ 
	\le \ {1\over{2n\sin^2(\delta/2)}} \ \to0.$$

Our second proof, or sketch, really, is based on a variant
of the Bohman-Korovkin theorem for $C^{2\pi}$, due to Korovkin.  
In this
setting, the three ``test cases'' are 
$$f_0(x)=1,\qquad f_1(x)=\cos x,\qquad\hbox{and}\qquad
	f_2(x)=\sin x.$$

\proclaim Theorem.
Let $(L_n)$ be a sequence of positive, linear maps on $C^{2\pi}$.
If\/ $L_n(f)\uniformto f$ for each of the three functions  
$f_0(x)=1$, $f_1(x)=\cos x$, and $f_2(x)=\sin x$, then
$L_n(f)\uniformto f$ for every $f\in C^{2\pi}$.

We won't prove this theorem; rather, we'll check
that $\sigma_n(f)\uniformto f$ in each of the three test
cases.  Since $s_n$ is a projection, this is painfully simple!
$$\eqalign{
\sigma_n(f_0)&=\textstyle{1\over n}\,(f_0+f_0+\cdots+f_0)=f_0,\cr
\sigma_n(f_1)&=\textstyle{1\over n}\,(0+f_1+\cdots+f_1)
	={{n-1}\over n}\cdot f_1\uniformto f_1,\cr
\sigma_n(f_2)&=\textstyle{1\over n}\,(0+f_2+\cdots+f_2)
	={{n-1}\over n}\cdot f_2\uniformto f_2.\cr
}$$

Kernel operators abound in analysis; for example,
Landau's proof of the Weierstrass theorem uses the
kernel $L_n(x)=c_n(1-x^2)^n$.  And, in the next section,
we'll encounter {\sl Jackson's kernel\/}
$J_n(t)=\sin^4nt/n^3\sin^2t$, which is essentially
the square of Fej\'er's kernel.  While we will have
no need for a general theory of such operators, please
note that the key to their utility is the fact that they're
nonnegative!

Lastly, a word or two about Fourier series involving
complex coefficients.  Most modern textbooks consider
the case of a $2\pi$-periodic, integrable function
$f:\R\to\C$ and define the Fourier series of $f$ by
$$\sum_{k=-\infty}^\infty c_ke^{ikt},$$
where now we have only one formula for the $c_k$'s:
$$c_k={1\over{2\pi}}\int_{-\pi}^\pi f(t)\,e^{-ikt}\,dt,$$
but, of course, the $c_k$'s may well be complex.
This somewhat simpler approach has other advantages; for one,
the exponentials $e^{ikt}$ are now an orthonormal set
(relative to the normalizing constant $1/2\pi$).  And,
if we remain consistent with this choice and define
the $L_2$ norm by 
$$\Vert f\Vert_2
	=\left({1\over{2\pi}}\int_{-\pi}^\pi |f(t)|^2\,dt\right)^{1/2},$$
then we have the simpler estimate $\Vert f\Vert_2\le\Vert f\Vert$ 
for $f\in C^{2\pi}$.

The Dirichlet and Fejer kernels are essentially the same
in this case, too, except that we would now write
$s_n(f)(x)=\sum_{k=-n}^nc_ke^{ikx}$.  Given this, the
Dirichlet and Fej\'er kernels can be written 
$$\openup2\jot\eqalign{
D_n(x)=\sum_{k=-n}^ne^{ikx}&=1+\sum_{k=1}^n(e^{ikx}+e^{-ikx})\cr
	&=1+2\sum_{k=1}^n\cos kx\cr
	&={{\sin\,(n+{1\over2})\,x}\over{\sin{1\over2}\,x}}\cr
}$$
and 
$$\openup2\jot\eqalign{
K_n(x)&={1\over n}\sum_{m=0}^{n-1}\sum_{k=-m}^me^{ikx}
	=\sum_{k=-n}^n\left(1-{|k|\over n}\right)\,e^{ikx}\cr
	&={1\over n}\sum_{m=0}^{n-1}{{\sin\,(m+{1\over2})\,x}\over{\sin{1\over2}\,x}}\cr
	&={{\sin^2(nt/2)}\over{n\sin^2(t/2)}}.\cr
}$$
In other words, each is twice its real coefficient counterpart.
Since the choice of normalizing constant ($1/\pi$ versus
$1/2\pi$, and sometimes even $1/\sqrt{\pi}$ or $1/\sqrt{2\pi}\,$)
has a (small) effect on these formulas, you may find some
variation in other textbooks.




\bye


%%  end of chapter-07.tex


