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. $\blacksquare$

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Content obtained and/or adapted from:

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