3.9 Uniform convergence  (Page 2/3)

The uniform limit of continuous functions is continuous.

Suppose $\left\{f_n\right\}$ is a sequence of continuous functions on a set $S\subseteq C,$ and assume that the sequence $\left\{f_n\right\}$ converges uniformly to a function $f.$ Then $f$ is continuous on $S.$

This proof is an example of what is called by mathematicians a “ $3ϵ$ argument.”

Fix an $x\in S$ and an $ϵ>0.$ We wish to find a $\delta >0$ such that if $y\in S$ and $|y-x|<\delta$ then $\left|f\right(y)-f(x\left)\right|<ϵ.$

We use first the hypothesis that the sequence converges uniformly. Thus, given this $ϵ>0,$ there exists a natural number $N$ such that if $n\ge N$ then $|f\left(z\right)-f_n\left(z\right)|<ϵ/3$ for all $z\in S.$ Now, because $f_N$ is continuous at $x,$ there exists a $\delta >0$ such that if $y\in S$ and $|y-x|<\delta$ then $|f_N\left(y\right)-f_N\left(x\right)|<ϵ/3.$ So, if $y\in S$ and $|y-x|<\delta ,$ then

This completes the proof.

REMARK Many properties of functions are preserved under the taking of uniform limits, e.g., continuity, as we have just seen.However, not all properties are preserved under this limit process. Differentiability is not, integrability is sometimes, being a power series function is, and so on.We must be alert to be aware of when it works and when it does not.

Weierstrass m-test

Let $\left\{u_n\right\}$ be a sequence of complex-valued functions defined on a set $S\subseteq C.$ Write $S_N$ for the partial sum $S_N\left(x\right)={\sum }_{n=0}^N{u}_n\left(x\right).$ Suppose that, for each $n,$ there exists an $M_n>0$ for which $|u_n\left(x\right)|\le M_n$ for all $x\in S.$ Then

1. If $\sum M_n$ converges, then the sequence $\left\{S_N\right\}$ converges uniformly to a function $S.$ That is, the infinite series $\sum u_n$ converges uniformly.
2. If each function $u_n$ is continuous, and $\sum M_n$ converges, then the function $S$ of part (1) is continuous.

Because $\sum M_n$ is convergent, it follows from the Comparison Test that for each $x\in S$ the infinite series ${\sum }_{n=0}^{\infty }{u}_n\left(x\right)$ is absolutely convergent, hence convergent. Define afunction $S$ by $S\left(x\right)={\sum }_{n=0}^{\infty }{u}_n\left(x\right)=lim{S}_N\left(x\right).$

To show that $\left\{S_N\right\}$ converges uniformly to $S,$ let $ϵ>0$ be given, and choose a natural number $N$ such that ${\sum }_{n=N+1}^{\infty }{M}_n<ϵ.$ This can be done because $\sum M_n$ converges. Now, for any $x\in S$ and any $m\ge N,$ we have

This proves part (1).

Part (2) now follows from part (1) and [link] , since the $S_N$ 's are continuous.

Let $f\left(z\right)={\sum }_{n=0}^{\infty }{a}_n{z}^n$ be a power series function with radius of convergence $r>0,$ and let $\left\{S_N\left(z\right)\right\}$ denote the sequence of partial sums of this series:

If $0<r^{\text{'}}<r,$ then the sequence $\left\{S_N\right\}$ converges uniformly to $f$ on the disk $B_{r^{\text{'}}}\left(0\right).$

Define a power series function $g$ by $g\left(z\right)={\sum }_{n=0}^{\infty }\left|a_n\right|z^n,$ and note that the radius of convergence for $g$ is the same as that for $f,$ i.e., $r.$ Choose $t$ so that $r^{\text{'}}<t<r.$ Then, since $t$ belongs to the disk of convergence of the power series function $g,$ we know that ${\sum }_{n=0}^{\infty }\left|a_n\right|t^n$ converges. Set $m_n=\left|a_n\right|t^n,$ and note that $\sum m_n$ converges. Now, for each $z\in B_{r^{\text{'}}}\left(0\right),$ we have that

so that the infinite series $\sum a_n{z}^n$ converges uniformly on $B_{r^{\text{'}}}\left(0\right)$ by the Weierstrass M-Test.