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Uniform convergence: Difference between revisions

← Older edit
asy code does nothing? Also, informal definition needed. Also, more examples would be nice.
+ asymptote
 
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A [[sequence]] of [[function]]s <math>\{f_n\},\ f_n: X \to \mathbb{R}</math> is said to '''uniformly converge''' to a function <math>f: X \to \mathbb{R}</math> if for every positive real number <math>\varepsilon > 0</math>, then there exists <math>N</math> such that for all positive integers <math>n \ge N</math>, we have <math>|f_n(x) - f(x)| < \varepsilon</math>.  (More generally, we can replace <math>\mathbb{R}</math> with any [[metric space]] <math>Y</math>.)
A [[sequence]] of [[function]]s <math>\{f_n\},\ f_n: X \to \mathbb{R}</math> is said to '''uniformly converge''' to a function <math>f: X \to \mathbb{R}</math> if for every positive real number <math>\varepsilon > 0</math>, then there exists <math>N</math> such that for all positive integers <math>n \ge N</math>, we have <math>|f_n(x) - f(x)| < \varepsilon</math>.  (More generally, we can replace <math>\mathbb{R}</math> with any [[metric space]] <math>Y</math>.)


<!--<center><asy>
This is different from [[pointwise convergence]], where a sequence of functions converge pointwise if at every point in the domain, the functions converge. Uniform convergence is a stronger condition, because (speaking informally) the function has to converge at a similar rate everywhere on its [[domain]].
import graph;
function f(real r){ return r/(1 + 5 * r^2); }
graph(f,0,1); xaxis(); yaxis();


</asy></center>-->
<center><asy> size(150);
import graph; real epsilon = 0.1; pen d = linewidth(0.7); defaultpen(d);
  // implement [an apparently non-functional] version of Lagrange.
real symproduct(real a, real x[]){
real product = 1;
for(int i = 0; i < x.length; ++i)
  if(a != x[i]) 
  product = product * (a - x[i]);
return product;
}
 
real baryweight(int i, real x[]){
real product = 1;
for(int j = 0; j < x.length; ++j)
  if(j != i) 
  product = product * (x[j] - x[i]);
return 1/product;
}
 
real lagrange(real a, real x[], real y[]){
real sum = 0;
for(int i = 0; i < x.length; ++i)
  if(a != x[i])
  sum = sum + baryweight(i,x)*y[i]/(a - x[i]);
return symproduct(a, x) * sum;
}
 
real func(real x){  // note there is a sign error. use odd # of points
real LISTx[] = {-1.5, -0.5 , 0 ,0.4, 0.8, 1, 4};
real LISTy[] = {-4, -0.5 ,  0.2 , 0.5, 0.35, 0.6, 1.2};
return lagrange(x,LISTx,LISTy);
}
 
real func_plus(real x) { return (func(x) + epsilon); }
real func_minus(real x) { return (func(x) - epsilon); }
 
draw(graph(func,-0.00001,1.00001),linewidth(1.2));
draw(graph(func_plus,-0.00001,1.00001),dashed+linewidth(0.7));
draw(graph(func_minus,-0.00001,1.00001),dashed+linewidth(0.7));
 
  // convergents
real func_conv1(real x) { return (func(x) + 1.5*epsilon*sin(12*x+pi/3)); }
real func_conv2(real x) { return (func(x) + 0.9*epsilon*sin(12*x+pi/4)); }
real func_conv3(real x) { return (func(x) + 0.5*epsilon*sin(12*x+pi/5)); }
draw(graph(func_conv1,-0.00001,1.00001),red+d);
draw(graph(func_conv2,-0.00001,1.00001),green+d);
draw(graph(func_conv3,-0.00001,1.00001),blue+d);
 
  // draw axes
Label f; f.p=fontsize(8);
xaxis(0, 1, Ticks(f, 1, 1));
yaxis(0, 1, Ticks(f, 1, 1));
 
pair a = (-0.1,func(0.00001));
draw(a--a+(0,epsilon), Arrows(6));
draw(a--a-(0,epsilon), Arrows(6));
label("$\varepsilon$",a+(0,epsilon/2),W,f.p);
label("$\varepsilon$",a-(0,epsilon/2),W,f.p);
 
</asy><br />If the functions <math>\color{red}{f_1}, \color{green}{f_2}, \color{blue}{f_3}, \color{black} \ldots</math> continue to approach <math>f</math> as shown, they converge uniformly.</center>


Every uniformly convergent sequence converges [[pointwise convergence | pointwise]], but the [[converse]] is not necessarily true.  For example, the sequence of functions defined by <math>f_n(x) = x^n</math> for <math>x \in [0, 1]</math> converges pointwise to the function <math>f(x) =\begin{cases} 1, & x = 1 \\0, & \text{otherwise}\end{cases}</math>, but this convergence is ''not'' uniform.
Every uniformly convergent sequence converges [[pointwise convergence | pointwise]], but the [[converse]] is not necessarily true.  For example, the sequence of functions defined by <math>f_n(x) = x^n</math> for <math>x \in [0, 1]</math> converges pointwise to the function <math>f(x) =\begin{cases} 1, & x = 1 \\0, & \text{otherwise}\end{cases}</math>, but this convergence is ''not'' uniform.


Uniformly convergent sequences have a number of nice properties that pointwise convergent sequences do not necessarily have.  A uniformly convergent sequence of [[continuous]] functions converges to a continuous function. A uniformly convergent sequence of differentiable functions defined on a [[closed set|closed interval]] converges to a differentiable function, and a sequence of [[Riemann-Stieltjes Integral|Stieltjes-integrable]] functions converges to a Stieltjes-integrable function.  
== Properties ==
An equivalent definition is that if <math>\lim_{n \to \infty} f_n(x) = f(x)</math> for all <math>x \in X</math> and
 
<center><cmath>\lim_{n \to \infty} \sup_{x \in X} |f(x) - f_n(x)| = 0,</cmath></center>
 
then <math>\{f_n\}</math> converges uniformly.
 
Uniformly convergent sequences have a number of nice properties that pointwise convergent sequences do not necessarily have.  A uniformly convergent sequence of [[continuous]] functions converges to a continuous function. A uniformly convergent sequence of differentiable functions defined on a [[closed set|closed interval]] converges to a differentiable function, and a sequence of [[Riemann-Stieltjes Integral|Stieltjes-integrable]] functions converges to a Stieltjes-integrable function. It is possible to show by example that these properties do not have to hold for pointwise convergent functions.  


{{stub}}
{{stub}}


[[Category:Analysis]]
[[Category:Analysis]]

Latest revision as of 18:20, 3 March 2010

A sequence of functions $\{f_n\},\ f_n: X \to \mathbb{R}$ is said to uniformly converge to a function $f: X \to \mathbb{R}$ if for every positive real number $\varepsilon > 0$, then there exists $N$ such that for all positive integers $n \ge N$, we have $|f_n(x) - f(x)| < \varepsilon$. (More generally, we can replace $\mathbb{R}$ with any metric space $Y$.)

This is different from pointwise convergence, where a sequence of functions converge pointwise if at every point in the domain, the functions converge. Uniform convergence is a stronger condition, because (speaking informally) the function has to converge at a similar rate everywhere on its domain.

[asy] size(150); import graph; real epsilon = 0.1; pen d = linewidth(0.7); defaultpen(d); // implement [an apparently non-functional] version of Lagrange. real symproduct(real a, real x[]){ real product = 1; for(int i = 0; i < x.length; ++i) if(a != x[i]) product = product * (a - x[i]); return product; } real baryweight(int i, real x[]){ real product = 1; for(int j = 0; j < x.length; ++j) if(j != i) product = product * (x[j] - x[i]); return 1/product; } real lagrange(real a, real x[], real y[]){ real sum = 0; for(int i = 0; i < x.length; ++i) if(a != x[i]) sum = sum + baryweight(i,x)*y[i]/(a - x[i]); return symproduct(a, x) * sum; } real func(real x){ // note there is a sign error. use odd # of points real LISTx[] = {-1.5, -0.5 , 0 ,0.4, 0.8, 1, 4}; real LISTy[] = {-4, -0.5 , 0.2 , 0.5, 0.35, 0.6, 1.2}; return lagrange(x,LISTx,LISTy); } real func_plus(real x) { return (func(x) + epsilon); } real func_minus(real x) { return (func(x) - epsilon); } draw(graph(func,-0.00001,1.00001),linewidth(1.2)); draw(graph(func_plus,-0.00001,1.00001),dashed+linewidth(0.7)); draw(graph(func_minus,-0.00001,1.00001),dashed+linewidth(0.7)); // convergents real func_conv1(real x) { return (func(x) + 1.5*epsilon*sin(12*x+pi/3)); } real func_conv2(real x) { return (func(x) + 0.9*epsilon*sin(12*x+pi/4)); } real func_conv3(real x) { return (func(x) + 0.5*epsilon*sin(12*x+pi/5)); } draw(graph(func_conv1,-0.00001,1.00001),red+d); draw(graph(func_conv2,-0.00001,1.00001),green+d); draw(graph(func_conv3,-0.00001,1.00001),blue+d); // draw axes Label f; f.p=fontsize(8); xaxis(0, 1, Ticks(f, 1, 1)); yaxis(0, 1, Ticks(f, 1, 1)); pair a = (-0.1,func(0.00001)); draw(a--a+(0,epsilon), Arrows(6)); draw(a--a-(0,epsilon), Arrows(6)); label("$\varepsilon$",a+(0,epsilon/2),W,f.p); label("$\varepsilon$",a-(0,epsilon/2),W,f.p); [/asy]
If the functions $\color{red}{f_1}, \color{green}{f_2}, \color{blue}{f_3}, \color{black} \ldots$ continue to approach $f$ as shown, they converge uniformly.

Every uniformly convergent sequence converges pointwise, but the converse is not necessarily true. For example, the sequence of functions defined by $f_n(x) = x^n$ for $x \in [0, 1]$ converges pointwise to the function $f(x) =\begin{cases} 1, & x = 1 \\0, & \text{otherwise}\end{cases}$, but this convergence is not uniform.

Properties

An equivalent definition is that if $\lim_{n \to \infty} f_n(x) = f(x)$ for all $x \in X$ and

\[\lim_{n \to \infty} \sup_{x \in X} |f(x) - f_n(x)| = 0,\]

then $\{f_n\}$ converges uniformly.

Uniformly convergent sequences have a number of nice properties that pointwise convergent sequences do not necessarily have. A uniformly convergent sequence of continuous functions converges to a continuous function. A uniformly convergent sequence of differentiable functions defined on a closed interval converges to a differentiable function, and a sequence of Stieltjes-integrable functions converges to a Stieltjes-integrable function. It is possible to show by example that these properties do not have to hold for pointwise convergent functions.

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