3.9 Uniform convergence  (Page 3/3)

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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 $c.$

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