Uniform Convergence of Sequences of Functions

Definition: An Infinite Sequence of Functions $(f_{n}(x))_{n=1}^{\infty }=(f_{1}(x),f_{2}(x),...,f_{n}(x),...)$ is a sequence of functions with a common domain. The $n^{\mathrm {th} }$ Term of the sequence is the function $f_{n}(x)$.

We can define a finite sequence of functions analogously. A finite sequence of functions is denoted $(f_{n})_{n=1}^{\infty }$.

We can also denote an infinite sequence of functions as simply $(f_{n}(x))$. We can also use curly brackets to denote a sequence of functions such as $\{f_{n}(x)\}_{n=1}^{\infty }$ or simply $\{f_{n}(x)\}$.

For example, consider the following sequence of functions:

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

This is a sequence of diagonal straight lines that pass through the origin and whose slope is increasing. The following illustrates a few of the functions in this sequence:

<img src="http://mathonline.wdfiles.com/local--files/sequences-of-functions/Screen%20Shot%202015-10-19%20at%205.44.29%20PM.png" alt="Screen%20Shot%202015-10-19%20at%205.44.29%20PM.png" class="image" />

For another example, consider the following sequence of functions:

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

This is a sequence of the simplest $n^{\mathrm {th} }$ degree polynomials whose exponent is increasing. The following illustrates a few of the functions in this sequence:

<img src="http://mathonline.wdfiles.com/local--files/sequences-of-functions/Screen%20Shot%202015-10-19%20at%205.48.10%20PM.png" alt="Screen%20Shot%202015-10-19%20at%205.48.10%20PM.png" class="image" />

Definition: Let $(f_{n}(x))_{n=1}^{\infty }$ be a sequence of functions with common domain $X$. Then $(f_{n})_{n=1}^{\infty }$ is said to be Pointwise Convergent to the the function $f$ written $\lim _{n\to \infty }f_{n}(x)=f(x)$ if for all $x\in X$ and for all $\varepsilon >0$ there exists a $N\in \mathbb {N}$ such that if $n\geq N$ then $\mid f_{n}(x)-f(x)\mid <\varepsilon$.

For example, consider the following sequence of functions defined on $[0,1]$:

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

We claim that $(f_{n}(x))_{n=1}^{\infty }$ is pointwise convergent to $f(x)=0$. The following image shows the first six functions in the sequence given above. It should be intuitively clear that the sequence $(f_{n}(x))_{n=1}^{\infty }$ converges to the limit function $f(x)=0$.

To show this, fix $x\in [0,1]$ and assume that $x\neq 0$ and let $\varepsilon >0$ be given. Then since $n,x>0$ we have that:

Failed to parse (unknown function "\lvert"): {\displaystyle \begin{align} \quad |f_n(x) - f(x)| = \lvert \frac{1}{n} x - 0 \rvert = \lvert \frac{x}{n} \rvert = \frac{x}{n} \end{align}}

Choose $N\in \mathbb {N}$ such that $N>{\frac {x}{\varepsilon }}$ which can be done by the Archimedean property. Then ${\frac {1}{N}}<{\frac {\varepsilon }{x}}$ and so for $n\geq N$ we have that:

${\begin{aligned}\quad {\frac {x}{n}}\leq {\frac {x}{N}}<{\frac {\varepsilon x}{x}}=\varepsilon \end{aligned}}$

Therefore $\lim _{n\to \infty }f_{n}(x)=f(x)$ for $x\in (0,1]$. Now, for $x=0$, notice that:

${\begin{aligned}\quad (f_{n}(0))_{n=1}^{\infty }=(0)_{n=1}^{\infty }=(0,0,...)\end{aligned}}$

This sequence clearly converges to $f(0)=0$. So, we conclude that $\displaystyle {\lim _{n\to \infty }f_{n}(x)=f(x)}$ for all $x\in [0,1]$. Hence the sequence $(f_{n}(x))_{n=1}^{\infty }$ is pointwise convergent on all of $[0,1]$.

Uniform Convergence of Sequences of Functions

Recall from the <a href="/pointwise-convergence-of-sequences-of-functions">Pointwise Convergence of Sequences of Functions</a> page that we say the sequence of functions $(f_{n}(x))_{n=1}^{\infty }$ with common domain $X$ is convergent to the limit function $f(x)$ 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 $\mid f_{n}(x)-f(x)\mid <\varepsilon$.

Another somewhat stronger type of convergence of a sequence of functions is called uniform convergence which we define below. Note the subtle but very important difference in the definition below!

Definition: Let $(f_{n}(x))_{n=1}^{\infty }$ be a sequence of functions with common domain $X$. Then $(f_{n}(x))_{n=1}^{\infty }$ is said to be Uniformly Convergent to the the limit function $f$ written Failed to parse (syntax error): {\displaystyle \lim_{n \to \infty} f_n(x) = f(x) \: \mathit{uniformly \: on} \: X} or Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f_n \to f \: \mathit{uniformly \: on} \: X} if for all $\varepsilon >0$ there exists a $N\in \mathbb {N}$ such that if $n\geq N$ then $\mid f_{n}(x)-f(x)\mid <\varepsilon$ for all $x\in X$.

Graphically, if the sequence of functions $(f_{n}(x))_{n=1}^{\infty }$ are all real-valued and uniformly converge to the limit function $f$, then from the definition above, we see that for all $\varepsilon >0$ there exists an $N\in \mathbb {N}$ such that for all $n\geq N$ we have that the following inequality holds for all $x\in X$:

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

The following graphic illustrates the concept of uniform convergence of a sequence of functions <math>(f_n(x))_{n=1}^{\infty}<math>:

<img src="http://mathonline.wdfiles.com/local--files/uniform-convergence-of-sequences-of-functions/Screen%20Shot%202015-10-19%20at%209.25.38%20PM.png" alt="Screen%20Shot%202015-10-19%20at%209.25.38%20PM.png" class="image" />

Licensing

Content obtained and/or adapted from:

• Sequences of Functions, mathonline.wikidot.com under a CC BY-SA license
• [1] under a CC BY-SA license
• [2] under a CC BY-SA license