Editing Uniform convergence (section)

== Examples ==
For <math>x \in [0,1)</math>, a basic example of uniform convergence can be illustrated as follows: the sequence <math>(1/2)^{x+n}</math> converges uniformly, while <math>x^n</math> does not.  Specifically, assume <math>\varepsilon=1/4</math>. Each function <math>(1/2)^{x+n}</math> is less than or equal to <math>1/4</math> when <math>n \geq 2</math>, regardless of the value of <math>x</math>. On the other hand, <math>x^n</math> is only less than or equal to <math>1/4</math> at ever increasing values of <math>n</math> when values of <math>x</math> are selected closer and closer to 1 (explained more in depth further below).

Given a [[topological space]] ''X'', we can equip the space of [[bounded function|bounded]] [[real number|real]] or [[complex number|complex]]-valued functions over ''X'' with the [[uniform norm]] topology, with the uniform metric defined by 

:<math>d(f,g)=\|f-g\|_{\infty}=\sup_{x\in X} |f(x)-g(x)|.</math>

Then uniform convergence simply means [[Limit of a sequence|convergence]] in the [[uniform norm]] topology: 

:<math>\lim_{n\to\infty}\|f_n-f\|_{\infty}=0</math>.

The sequence of functions <math>(f_n)</math> 

:<math>\begin{cases} f_n:[0,1]\to [0,1] \\ f_n(x)=x^n \end{cases}</math> 

is a classic example of a sequence of functions that converges to a function <math>f</math> pointwise but not uniformly. To show this, we first observe that the pointwise limit of <math>(f_n)</math> as <math>n\to\infty</math> is the function <math>f</math>, given by

: <math>f(x) = \lim_{n\to \infty} f_n(x) = \begin{cases} 0, & x \in [0,1); \\ 1, & x=1. \end{cases} </math>

''Pointwise convergence:'' Convergence is trivial for <math>x=0</math> and <math>x=1</math>, since <math>f_n(0)=f(0)=0</math> and <math>f_n(1)=f(1)=1</math>, for all <math>n</math>. For <math>x \in (0,1)</math> and given <math>\varepsilon>0</math>, we can ensure that <math>|f_n(x)-f(x)|<\varepsilon</math> whenever <math>n\geq N</math> by choosing <math>N = \lceil\log\varepsilon/\log x\rceil</math>, which is the minimum integer exponent of <math>x</math> that allows it to reach or dip below <math>\varepsilon</math> (here the upper square brackets indicate rounding up, see [[Floor and ceiling functions|ceiling function]]). Hence, <math>f_n\to f</math> pointwise for all <math>x\in[0,1]</math>. Note that the choice of <math>N</math> depends on the value of <math>\varepsilon</math> and <math>x</math>. Moreover, for a fixed choice of <math>\varepsilon</math>, <math>N</math> (which cannot be defined to be smaller) grows without bound as <math>x</math> approaches 1. These observations preclude the possibility of uniform convergence.

''Non-uniformity of convergence:'' The convergence is not uniform, because we can find an <math>\varepsilon>0</math> so that no matter how large we choose <math>N,</math> there will be values of <math>x \in [0,1]</math> and <math>n \geq N</math> such that <math>|f_n(x)-f(x)|\geq\varepsilon.</math> To see this, first observe that regardless of how large <math>n</math> becomes, there is always an <math>x_0 \in [0,1)</math> such that <math>f_n(x_0)=1/2.</math> Thus, if we choose <math>\varepsilon = 1/4,</math> we can never find an <math>N</math> such that <math>|f_n(x)-f(x)|<\varepsilon</math> for all <math>x\in[0,1]</math> and <math>n\geq N</math>. Explicitly, whatever candidate we choose for <math>N</math>, consider the value of <math>f_N</math> at <math>x_0 = (1/2)^{1/N}</math>. Since 

:<math>\left|f_N(x_0) - f(x_0)\right| = \left| \left[ \left(\frac{1}{2}\right)^{\frac{1}{N}} \right]^N - 0 \right| = \frac{1}{2} > \frac{1}{4} = \varepsilon,</math>
the candidate fails because we have found an example of an <math>x\in[0,1]</math> that "escaped" our attempt to "confine" each <math>f_n\ (n\geq N)</math> to within <math>\varepsilon</math> of <math>f </math> for all <math>x\in[0,1]</math>. In fact, it is easy to see that
:<math>\lim_{n\to\infty}\|f_n-f\|_{\infty}=1,</math>
contrary to the requirement that <math>\|f_n-f\|_{\infty}\to 0</math> if <math>f_n \rightrightarrows f</math>.

In this example one can easily see that pointwise convergence does not preserve differentiability or continuity. While each function of the sequence is smooth, that is to say that for all ''n'', <math>f_n\in C^{\infty}([0,1])</math>, the limit <math>\lim_{n\to \infty}f_n</math> is not even continuous.

=== Exponential function ===
The series expansion of the [[exponential function]] can be shown to be uniformly convergent on any bounded subset <math>S \subset \C</math> using the [[Weierstrass M-test]].

'''Theorem (Weierstrass M-test).''' ''Let <math>(f_n)</math> be a sequence of functions <math>f_n:E\to \C</math> and let <math>M_n </math> be a sequence of positive real numbers such that <math>|f_n(x)|\le M_n</math> for all <math>x\in E</math> and <math>n=1,2, 3, \ldots</math> If <math display="inline">\sum_n M_n</math> converges, then <math display="inline">\sum_n f_n</math> converges absolutely and uniformly on <math>E</math>.''

The complex exponential function can be expressed as the series:

:<math>\sum_{n=0}^{\infty}\frac{z^n}{n!}.</math>

Any bounded subset is a subset of some disc <math>D_R</math> of radius <math>R,</math> centered on the origin in the [[complex plane]]. The Weierstrass M-test requires us to find an upper bound <math>M_n</math> on the terms of the series, with <math>M_n</math> independent of the position in the disc:

:<math>\left| \frac{z^n}{n!} \right|\le M_n, \forall z\in D_R.</math>

To do this, we notice

:<math>\left| \frac{z^n}{n!}\right| \le \frac{|z|^n}{n!} \le \frac{R^n}{n!}</math>

and take <math>M_n=\tfrac{R^n}{n!}.</math>

If <math>\sum_{n=0}^{\infty}M_n</math> is convergent, then the M-test asserts that the original series is uniformly convergent.

The [[ratio test]] can be used here:

:<math>\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</math>

which means the series over <math>M_n</math> is convergent. Thus the original series converges uniformly for all <math>z\in D_R,</math> and since <math>S\subset D_R</math>, the series is also uniformly convergent on <math>S.</math>