# 3.9 Uniform convergence  (Page 3/3)

 Page 3 / 3

The next theorem shows that continuous, real-valued functions on closed bounded intervals are uniform limits of step functions.Step functions have not been mentioned lately, since they aren't continuous functions, but this next theorem will be crucial for us when we study integration in [link] .

Let $f$ be a continuous real-valued function on the closed and bounded interval $\left[a,b\right].$ Then there exists a sequence $\left\{{h}_{n}\right\}$ of step functions on $\left[a,b\right]$ that converges uniformly to $f.$

We use the fact that a continuous function on a compact set is uniformly continuous ( [link] ).

For each positive integer $n,$ let ${\delta }_{n}$ be a positive number satisfying $|f\left(x\right)-f\left(y\right)|<1/n$ if $|x-y|<{\delta }_{n}.$ Such a ${\delta }_{n}$ exists by the uniform continuity of $f$ on $\left[a,b\right].$ Let ${P}_{n}=\left\{{x}_{0}<{x}_{1}<...<{x}_{{m}_{n}}\right\}$ be a partition of $\left[a,b\right]$ for which ${x}_{i}-{x}_{i-1}<{\delta }_{n}$ for all $1\le i\le {m}_{n}.$ Define a step function ${h}_{n}$ on $\left[a,b\right]$ as follows:

If ${x}_{i-1}\le x<{x}_{i},$ then ${h}_{n}\left(x\right)=f\left({x}_{i-1}\right).$ This defines ${h}_{n}\left(x\right)$ for every $x\in \left[a,b\right),$ and we complete the definition of ${h}_{n}$ by setting ${h}_{n}\left(b\right)=f\left(b\right).$ It follows immediately that ${h}_{n}$ is a step function.

Now, we claim that $|f\left(x\right)-{h}_{n}\left(x\right)|<1/n$ for all $x\in \left[a,b\right].$ This is clearly the case for $x=b,$ since $f\left(b\right)={h}_{n}\left(b\right)$ for all $n.$ For any other $x,$ let $i$ be the unique index such that ${x}_{i-1}\le x<{x}_{i}.$ Then

$|f\left(x\right)-{h}_{n}\left(x\right)|=|f\left(x\right)-f\left({x}_{i-1}\right)|<1/n$

because $|x-{x}_{i-1}|<{\delta }_{n}.$

So, we have defined a sequence $\left\{{h}_{n}\right\}$ of step functions, and the sequence $\left\{{h}_{n}\right\}$ converges uniformly to $f$ by [link] .

We close this chapter with a famous theorem of Abel concerning the behavior of a power series function on the boundary of its disk of convergence.See the comments following [link] .

## Abel

Suppose $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_{n}{z}^{n}$ is a power series function having finite radius of convergence $r>0,$ and suppose there exists a point ${z}_{0}$ on the boundary of ${B}_{r}\left(0\right)$ that is in the domain of $f;$ i.e., $\sum {a}_{n}{z}_{0}^{n}$ converges to $f\left({z}_{0}\right).$ Suppose $g$ is a continuous function whose domain contains the open disk ${B}_{r}\left(0\right)$ as well as the point ${z}_{0},$ and assume that $f\left(z\right)=g\left(z\right)$ for all $z$ in the open disk ${B}_{r}\left(0\right).$ Then $f\left({z}_{0}\right)$ must equal $g\left({z}_{0}\right).$

For simplicity, assume that $r=1$ and that ${z}_{0}=1.$ See the exercise that follows this proof. Write ${S}_{n}$ for the partial sum of the ${a}_{n}$ 's: ${S}_{n}={\sum }_{n=0}^{n}{a}_{n}.$ In the following computation, we will use the Abel Summation Formula in the form

$\sum _{n=0}^{N}{a}_{n}{z}^{n}={S}_{N}{z}^{N}+\sum _{n=0}^{N-1}{S}_{n}\left({z}^{n}-{z}^{n+1}\right).$

See [link] . Let $ϵ$ be a positive number. Then, for any $0 and any positive integer $N,$ we have

$\begin{array}{ccc}\hfill |g\left(1\right)-f\left(1\right)|& =& |g\left(1\right)-f\left(t\right)+f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}+\sum _{n=0}^{N}{a}_{n}{t}^{n}-f\left(1\right)|\hfill \\ & \le & |g\left(1\right)-g\left(t\right)|+|f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}|+|\sum _{n=0}^{N}{a}_{n}{t}^{n}-f\left(1\right)|\hfill \\ & \le & |g\left(1\right)-g\left(t\right)|+|f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}|\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|{S}_{N}{t}^{N}+\sum _{n=0}^{N-1}{S}_{n}\left({t}^{n}-{t}^{n+1}\right)-f\left(1\right)|\hfill \\ & =& |g\left(1\right)-g\left(t\right)|+|f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|{S}_{N}{t}^{N}+\sum _{n=0}^{N-1}\left({S}_{n}-{S}_{N}\right)\left({t}^{n}-{t}^{n+1}\right)+{S}_{N}\sum _{n=0}^{N-1}\left({t}^{n}-{t}^{n+1}\right)-f\left(1\right)|\hfill \\ & =& |g\left(1\right)-g\left(t\right)|+|f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|\sum _{n=0}^{N-1}\left({S}_{n}-{S}_{N}\right)\left({t}^{n}-{t}^{n+1}\right)+{S}_{N}\left({t}^{N}+\sum _{n=0}^{N-1}\left({t}^{n}-{t}^{n+1}\right)\right)-f\left(1\right)|\hfill \\ & \le & |g\left(1\right)-g\left(t\right)|+|f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}|\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|\sum _{n=0}^{N-1}\left({S}_{n}-{S}_{N}\right)\left({t}^{n}-{t}^{n+1}\right)|+|{S}_{N}-f\left(1\right)|\hfill \\ & \le & |g\left(1\right)-g\left(t\right)|+|f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}|\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|\sum _{n=0}^{P}\left({S}_{n}-{S}_{N}\right)\left({t}^{n}-{t}^{n+1}\right)|+|\sum _{n=P+1}^{N-1}\left({S}_{n}-{S}_{N}\right)\left({t}^{n}-{t}^{n+1}\right)|+|{S}_{N}-f\left(1\right)|\hfill \\ & \le & |g\left(1\right)-g\left(t\right)|+|f\left(t\right)-\sum _{n=0}^{N}{a}_{n}{t}^{n}|\hfill \\ & & \phantom{\rule{1.em}{0ex}}\phantom{\rule{1.em}{0ex}}+|\sum _{n=0}^{P}\left({S}_{n}-{S}_{N}\right)\left({t}^{n}-{t}^{n+1}\right)|+\sum _{n=P+1}^{N-1}|{S}_{n}-{S}_{N}|\left({t}^{n}-{t}^{n+1}\right)+|{S}_{N}-f\left(1\right)|\hfill \\ & =& {t}_{1}+{t}_{2}+{t}_{3}+{t}_{4}+{t}_{5}.\hfill \end{array}$

First, choose an integer ${M}_{1}$ so that if $P$ and $N$ are both larger than ${M}_{1},$ then ${t}_{4}<ϵ.$ (The sequence $\left\{{S}_{k}\right\}$ is a Cauchy sequence, and $\sum \left({t}^{k}-{t}^{k+1}$ is telescoping.)

Fix such a $P>{M}_{1}.$ Then choose a $\delta >0$ so that if $1>t>1-\delta ,$ then both ${t}_{1}$ and ${t}_{3}<ϵ.$ How?

Fix such a $t.$ Finally, choose a $N,$ greater than ${M}_{1},$ and also large enough so that both ${t}_{2}$ and ${t}_{5}$ are less than $ϵ.$ (How?)

Now, $|g\left(1\right)-f\left(1\right)|<5ϵ.$ Since this is true for every $ϵ>0,$ it follows that $f\left(1\right)=g\left(1\right),$ and the theorem is proved.

Let $f,g,r,$ and ${z}_{0}$ be as in the statement of the preceding theorem. Define $\stackrel{^}{f}\left(z\right)=f\left({z}_{0}z\right)$ and $\stackrel{^}{g}\left(z\right)=g\left({z}_{0}z\right).$

1. Prove that $\stackrel{^}{f}$ is a power series function $\stackrel{^}{f}\left(z\right)={\sum }_{n=0}^{\infty }{b}_{n}{z}^{n},$ with radius of convergence equal to 1, and such that ${\sum }_{n=0}^{\infty }{b}_{n}$ converges to $\stackrel{^}{f}\left(1\right);$ i.e., 1 is in the domain of $\stackrel{^}{f}.$
2. Show that $\stackrel{^}{g}$ is a continuous function whose domain contains the open disk ${B}_{1}\left(0\right)$ and the point $z=1.$
3. Show that, if $\stackrel{^}{f}\left(1\right)=\stackrel{^}{g}\left(1\right),$ then $f\left({z}_{0}\right)=g\left({z}_{0}\right).$ Deduce that the simplification in the preceding proof is justified.
4. State and prove the generalization of Abel's Theorem to a function $f$ that is expandable in a Taylor series around a point $c.$

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!