Editing Uniform convergence (section)

== Applications ==

===To continuity===
{{Main|Uniform limit theorem}}
[[Image:Drini nonuniformconvergence SVG.svg|thumb|350px|right|Counterexample to a strengthening of the uniform convergence theorem, in which pointwise convergence, rather than uniform convergence, is assumed. The continuous green functions <math>\sin^n(x)</math> converge to the non-continuous red function. This can happen only if convergence is not uniform.]]

If <math>E</math> and <math>M</math> are [[topological space|topological spaces]], then it  makes sense to talk about the [[continuous function (topology)|continuity]] of the functions <math>f_n,f:E\to M</math>.  If we further assume that <math>M</math> is a [[metric space]], then (uniform) convergence of the <math>f_n</math> to <math>f</math> is also well defined.  The following result states that continuity is preserved by uniform convergence:

{{math theorem | name = Uniform limit theorem | math_statement = Suppose <math>E</math> is a  topological space, <math>M</math> is a metric space, and <math>(f_n)</math> is a sequence of continuous functions <math>f_n:E\to M</math>.  If <math>f_n \rightrightarrows f</math> on <math>E</math>,  then <math>f</math> is also continuous.}}

This theorem is proved by the "{{math|&epsilon;/3}} trick", and is the archetypal example of this trick: to prove a given inequality ({{math|&epsilon;}}), one uses the definitions of continuity and uniform convergence to produce 3 inequalities ({{math|&epsilon;/3}}), and then combines them via the [[triangle inequality]] to produce the desired inequality.

{{Proof|Let <math> x_0\in E </math> be an arbitrary point. We will prove that <math> f</math> is continuous at <math>x_0</math>. Let <math>\varepsilon >0 </math>. By uniform convergence, there exists a natural number <math>N</math> such that

<math>\forall x \in E\quad d(f_N(x),f(x))\leq \tfrac{\varepsilon}{3}</math>

(uniform convergence shows that the above statement is true for all <math>n\geq N</math>, but we will only use it for one function of the sequence, namely <math>f_N</math>).

It follows from the continuity of <math> f_N</math> at <math> x_0\in E</math> that there exists an [[open set]] <math> U</math> containing <math> x_0</math> such that

<math>\forall x\in U\quad d(f_N(x),f_N(x_0))\leq\tfrac{\varepsilon}{3}</math>.

Hence, using the [[triangle inequality]],

<math>\forall x\in U\quad d(f(x), f(x_0))\leq d(f(x),f_N(x))+d(f_N(x),f_N(x_0))+d(f_N(x_0),f(x_0))\leq\varepsilon</math>,

which gives us the continuity of <math> f</math> at <math> x_0</math>.<math>\quad\square</math>}}

This theorem is an important one in the history of real and Fourier analysis, since many 18th century mathematicians had the intuitive understanding that a sequence of continuous functions always converges to a continuous function.  The image above shows a counterexample, and many discontinuous functions could, in fact, be written as a [[Fourier series]] of continuous functions.  The erroneous claim that the pointwise limit of a sequence of continuous functions is continuous (originally stated in terms of convergent series of continuous functions) is infamously known as "Cauchy's wrong theorem".  The uniform limit theorem shows that a stronger form of convergence, uniform convergence, is needed to ensure the preservation of continuity in the limit function.

More precisely, this theorem states that the uniform limit of ''[[uniformly continuous]]'' functions is uniformly continuous; for a [[locally compact]] space, continuity is equivalent to local uniform continuity, and thus the uniform limit of continuous functions is continuous.
===To differentiability===
If <math>S</math> is an interval and all the functions <math>f_n</math> are [[derivative|differentiable]] and converge to a limit <math>f</math>, it is often desirable to determine the derivative function <math>f'</math> by taking the limit of the sequence <math>f'_n</math>. This is however in general not possible: even if the convergence is uniform, the limit function need not be differentiable (not even if the sequence consists of everywhere-[[analytic function|analytic]] functions, see [[Weierstrass function]]), and even if it is differentiable, the derivative of the limit function need not be equal to the limit of the derivatives. Consider for instance <math>f_n(x) = n^{-1/2}{\sin(nx)}</math> with uniform limit <math>f_n\rightrightarrows f\equiv 0</math>. Clearly, <math>f'</math> is also identically zero.  However, the derivatives of the sequence of functions are given by <math>f'_n(x)=n^{1/2}\cos nx,</math> and the sequence <math>f'_n</math> does not converge to <math>f',</math> or even to any function at all.  In order to ensure a connection between the limit of a sequence of differentiable functions and the limit of the sequence of derivatives, the uniform convergence of the sequence of derivatives plus the convergence of the sequence of functions at at least one point is required:<ref>Rudin, Walter (1976). ''[[iarchive:PrinciplesOfMathematicalAnalysis|Principles of Mathematical Analysis]]'' 3rd edition, Theorem 7.17.  McGraw-Hill: New York.</ref>

: ''If <math>(f_n)</math> is a sequence of differentiable functions on <math>[a,b]</math> such that <math>\lim_{n\to\infty} f_n(x_0)</math> exists (and is finite) for some <math>x_0\in[a,b]</math> and the sequence <math>(f'_n)</math> converges uniformly on <math>[a,b]</math>, then <math>f_n</math> converges uniformly to a function <math>f</math> on <math>[a,b]</math>, and <math> f'(x) = \lim_{n\to \infty} f'_n(x)</math> for <math>x \in [a, b]</math>.''

===To integrability===
Similarly, one often wants to exchange integrals and limit processes. For the [[Riemann integral]], this can be done if uniform convergence is assumed:
: ''If <math>(f_n)_{n=1}^\infty</math> is a sequence of Riemann integrable functions defined on a [[compact space|compact]] interval <math>I</math>  which uniformly converge with limit <math> f</math>, then <math> f</math> is Riemann integrable and its integral can be computed as the limit of the integrals of the <math> f_n</math>:'' <math display="block">\int_I f = \lim_{n\to\infty}\int_I f_n.</math>
In fact, for a uniformly convergent family of bounded functions on an interval, the upper and lower Riemann integrals converge to the upper and lower Riemann integrals of the limit function.  This follows because, for ''n'' sufficiently large, the graph of <math>f_n</math> is within {{math|&epsilon;}} of the graph of ''f'', and so the upper sum and lower sum of <math>f_n</math> are each within <math>\varepsilon |I|</math> of the value of the upper and lower sums of <math>f</math>, respectively.

Much stronger theorems in this respect, which require not much more than pointwise convergence, can be obtained if one abandons the Riemann integral and uses the [[Lebesgue integration|Lebesgue integral]] instead.

===To analyticity===
Using [[Morera's Theorem]], one can show that if a sequence of [[Analytic function|analytic]] functions converges uniformly in a region S of the complex plane, then the limit is analytic in S. This example demonstrates that complex functions are more well-behaved than real functions, since the uniform limit of analytic functions on a real interval need not even be differentiable (see [[Weierstrass function]]).

===To series===
We say that <math display="inline">\sum_{n=1}^\infty f_n</math> converges:

{{ordered list | list-style-type=lower-roman
 | pointwise on ''E'' if and only if the sequence of partial sums <math>s_n(x)=\sum_{j=1}^{n} f_j(x)</math> converges for every <math>x\in E</math>.
 | uniformly on ''E'' if and only if ''s''<sub>''n''</sub> converges uniformly as <math>n\to\infty</math>.
 | absolutely on ''E'' if and only if <math display="inline">\sum_{n=1}^\infty |f_n|</math> converges for every <math>x \in E</math>.
}}

With this definition comes the following result:
<blockquote>Let ''x''<sub>0</sub> be contained in the set ''E'' and each ''f''<sub>''n''</sub> be continuous at ''x''<sub>0</sub>. If <math display="inline"> f = \sum_{n=1}^\infty f_n</math> converges uniformly on ''E'' then ''f'' is continuous at ''x''<sub>0</sub> in ''E''. Suppose that <math>E = [a, b]</math> and each ''f''<sub>''n''</sub> is integrable on ''E''. If <math display="inline">\sum_{n=1}^\infty f_n</math> converges uniformly on ''E'' then ''f'' is integrable on ''E'' and the series of integrals of ''f''<sub>''n''</sub> is equal to integral of the series of f<sub>n</sub>.</blockquote>