Uniform Convergence of Series of Functions

Recall that a sequence of functions $(f_{n}(x))_{n=1}^{\infty }$ with common domain $X$ is said to be pointwise convergent if for all $x\in X$ and for all $\varepsilon >0$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ then:

${\begin{aligned}\quad \left|f_{n}(x)-f(x)\right|<\varepsilon \end{aligned}}$

Also recall that a sequence of functions $(f_{n}(x))_{n=1}^{\infty }$ with common domain $X$ is said to be uniformly convergent if for all $\varepsilon >0$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ then for all $x\in X$ we have that:

${\begin{aligned}\quad \left|f_{n}(x)-f(x)\right|<\varepsilon \end{aligned}}$

We will now extend the concept of pointwise convergence and uniform convergence to series of functions.

Definition: Let $(f_{n}(x))_{n=1}^{\infty }$ be a sequence of functions with common domain $X$. The corresponding series $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ is said to be Pointwise Convergent to the sum function $f(x)$ if the corresponding sequence of partial sums $(s_{n}(x))_{n=1}^{\infty }$ (where $\displaystyle {s_{n}(x)=\sum _{k=1}^{n}f_{n}(x)=f_{1}(x)+f_{2}(x)+...+f_{n}(x)}$) is pointwise convergent to $f(x)$.

For example, consider the following sequence of functions defined on the interval $(-1,1)$:

${\begin{aligned}\quad (f_{n}(x))_{n=1}^{\infty }=(x^{n-1})_{n=1}^{\infty }=(1,x,x^{2},x^{3},...)\end{aligned}}$

We now that this series converges pointwise for all $x\in (0,1)$ since the result series $\sum _{n=1}^{\infty }x^{n-1}$ is simply a geometric series to the sum function $\displaystyle {f(x)={\frac {1}{1-x}}}$.

Definition: Let $(f_{n}(x))_{n=1}^{\infty }$ be a sequence of functions with common domain $X$. The corresponding series $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ is said to be Uniformly Convergent to the sum function $f(x)$ if the corresponding sequence of partial sums is uniformly convergent to $f(x)$.

The geometric series given above actually converges uniformly on $(-1,1)$, though, showing this with the current definition of uniform convergence of series of functions is laborious. We will soon develop methods to determine whether a series of functions converges uniformly or not without having to brute-force apply the definition for uniform convergence for the sequence of partial sums.

Cauchy's Uniform Convergence Criterion for Series of Functions

If we have a sequence of functions $(f_{n}(x))_{n=1}^{\infty }$ with common domain $X$ then the corresponding series of functions $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ is said to be uniformly convergent if the corresponding sequence of partial sums $(s_{n}(x))_{n=1}^{\infty }$ is a uniformly convergent sequence of functions.

We will now look at a nice theorem known as Cauchy's uniform convergence criterion for series of functions.

Theorem 1: Let $(f_{n}(x))_{n=1}^{\infty }$ be a sequence of real-valued functions with common domain $X$. Then $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ is uniformly convergent on $X$ if and only if for all $\varepsilon >0$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ and for all $x\in X$ we have that $\displaystyle {\left|\sum _{k=n+1}^{n+p}f_{k}(x)\right|<\varepsilon }$ for all $p\in \mathbb {N}$.

• Proof: $\Rightarrow$ Suppose that $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ is uniformly convergent to some limit function $f(x)$ on $X$. Let $(s_{n}(x))_{n=1}^{\infty }$ denote the sequence of partial sums for this series. Then we must have that $\displaystyle {\lim _{n\to \infty }s_{n}(x)=f(x)}$ uniformly on $X$. So, for $\varepsilon _{1}={\frac {\varepsilon }{2}}$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ and for all $x\in X$ we have that:

${\begin{aligned}\quad \left|s_{n}(x)-f(x)\right|<\varepsilon \end{aligned}}$

• For any $p\in \mathbb {N}$ let $m=n+p$. Then $m\geq N$ and so:

${\begin{aligned}\quad \quad \left|\sum _{k=1}^{m}f_{k}(x)-\sum _{k=1}^{n}f_{k}(x)\right|=\left|\sum _{k=1}^{n+p}f_{k}(x)-\sum _{k=1}^{n}f_{k}(x)\right|=\left|\sum _{k=n+1}^{n+p}f_{k}(x)\right|=\left|s_{m}(x)-s_{n}(x)\right|\leq \left|s_{m}(x)-f(x)\right|+\left|f(x)-s_{n}(x)\right|<\varepsilon _{1}+\varepsilon _{1}={\frac {\varepsilon }{2}}+{\frac {\varepsilon }{2}}=\varepsilon \end{aligned}}$

• $\Leftarrow$ Suppose that for all $\varepsilon >0$ there exists an $N\in \mathbb {N}$ such that if $n\geq N$ and for all $x\in X$ we have that:

${\begin{aligned}\quad \left|\sum _{k=n+1}^{n+p}f_{k}(x)\right|<\varepsilon \end{aligned}}$

• Let $m,n\geq N$. Assume without loss of generality that $m>n$ and that $m=n+p$ for some $p\in \mathbb {N}$. Then from above we see that for all $x\in X$:

${\begin{aligned}\quad \left|s_{m}(x)-s_{n}(x)\right|=\left|\sum _{k=1}^{n+p}f_{k}(x)-\sum _{k=1}^{n}f_{k}(x)\right|=\left|\sum _{k=n+1}^{n+p}f_{k}(x)\right|<\varepsilon \end{aligned}}$

• So $(s_{n}(x))_{n=1}^{\infty }$ converges uniformly by the Cauchy uniform convergence criterion for sequences of functions. So $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ converges uniformly on $X$. $\blacksquare$

The Weierstrass M-Test for Uniform Convergence of Series of Functions

Recall that if $(f_{n}(x))_{n=1}^{\infty }$ is a sequence of real-valued functions with common domain $X$, then we say that the corresponding series of functions $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ is uniformly convergent if the sequence of partial sums $(s_{n}(x))_{n=1}^{\infty }$ is a uniformly convergent sequence.

We will now look at a very nice and relatively simply test to determine uniform convergence of a series of real-valued functions called the Weierstrass M-test.

Theorem 1: Let $(f_{n}(x))_{n=1}^{\infty }$ be a sequence of real-valued functions with common domain $X$, and let $(M_{n})_{n=1}^{\infty }$ be a sequence of nonnegative real numbers such that $\left|f_{n}(x)\right|\leq M_{n}$ for each $n\in \mathbb {N}$ and for all $x\in X$. If $\displaystyle {\sum _{n=1}^{\infty }M_{n}}$ converges then $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ uniformly converges on $X$.

• Proof: Suppose that there exists a sequence of nonnegative real numbers $(M_{n})_{n=1}^{\infty }$ such that for all $n\in \mathbb {N}$ and for all $x\in X$ we have that:

${\begin{aligned}\quad \left|f_{n}(x)\right|\leq M_{n}\end{aligned}}$

• Furthermore, suppose that $\displaystyle {\sum _{n=1}^{\infty }M_{n}}$ converges to some $M\in \mathbb {R}$, $M\geq 0$. Then we have that for all $x\in X$:

${\begin{aligned}\quad \left|\sum _{n=1}^{\infty }f_{n}(x)\right|\leq \sum _{n=1}^{\infty }\left|f_{n}(x)\right|\leq \sum _{n=1}^{\infty }M_{n}=M\end{aligned}}$

• So the $\displaystyle {\sum _{n=1}^{\infty }f_{n}(x)}$ converges for each $x\in X$ by the comparison test. $\blacksquare$

Weierstrass Example

The series expansion of the exponential function can be shown to be uniformly convergent on any bounded subset $S\subset \mathbb {C}$ using the Weierstrass M-test.

Theorem (Weierstrass M-test). Let $(f_{n})$ be a sequence of functions $f_{n}:E\to \mathbb {C}$ and let $M_{n}$ be a sequence of positive real numbers such that $|f_{n}(x)|\leq M_{n}$ for all $x\in E$ and $n=1,2,3,\ldots$ If ${\textstyle \sum _{n}M_{n}}$ converges, then ${\textstyle \sum _{n}f_{n}}$ converges uniformly on $E$.

The complex exponential function can be expressed as the series:

$\sum _{n=0}^{\infty }{\frac {z^{n}}{n!}}.$

Any bounded subset is a subset of some disc $D_{R}$ of radius $R,$ centered on the origin in the complex plane. The Weierstrass M-test requires us to find an upper bound $M_{n}$ on the terms of the series, with $M_{n}$ independent of the position in the disc:

$\left|{\frac {z^{n}}{n!}}\right|\leq M_{n},\forall z\in D_{R}.$

To do this, we notice

$\left|{\frac {z^{n}}{n!}}\right|\leq {\frac {|z|^{n}}{n!}}\leq {\frac {R^{n}}{n!}}$

and take $M_{n}={\tfrac {R^{n}}{n!}}.$

If $\sum _{n=0}^{\infty }M_{n}$ is convergent, then the M-test asserts that the original series is uniformly convergent.

The ratio test can be used here:

$\lim _{n\to \infty }{\frac {M_{n+1}}{M_{n}}}=\lim _{n\to \infty }{\frac {R^{n+1}}{R^{n}}}{\frac {n!}{(n+1)!}}=\lim _{n\to \infty }{\frac {R}{n+1}}=0$

which means the series over $M_{n}$ is convergent. Thus the original series converges uniformly for all $z\in D_{R},$ and since $S\subset D_{R}$, the series is also uniformly convergent on $S.$

Licensing

Content obtained and/or adapted from:

• Pointwise Convergent and Uniformly Convergent Series of Functions, mathonline.wikidot.com under a CC BY-SA license
• Uniform convergence, Wikipedia under a CC BY-SA license