Home

About

Advanced Search

Browse by Discipline

Scientific Societies

E-print Alerts

Add E-prints

E-print Network
FAQHELPSITE MAPCONTACT US


  Advanced Search  

 
1. Uniform convergence. Suppose X is a set and (Y, ) is a metric space. We let
 

Summary: 1. Uniform convergence.
Suppose X is a set and (Y, ) is a metric space. We let
B(X, Y )
be the set of bounded functions from X to Y ; that is, f B(X, Y ) if f : X Y
and diam rng f < . For each f, g B(X, Y ) we set
(f, g) = sup{(f(x), g(x)) : x X}.
Proposition 1.1. is metric on B(X, Y ).
Proof. Suppose f, g B(X, Y ) and a X. Then
(f(x), g(x)) (f(x), f(a)) + (f(a), g(a)) + (g(a), g(x))
diam rng f + (f(a), g(a)) + diam rng g
for any x X. Thus (f, g) < . It is evident that (g, f) = (f, g) and that if
(f, g) = 0 then f = g.
Suppose f, g, h B(X, Y ). Then
(f(x), h(x)) (f(x), g(x)) + (g(x), h(x)) (f, g) + (g, h)
for any x X from which we conclude that (f, g) (f, g) + (g, h).
Example 1.1. Suppose Y is a vector space normed by || and is the corresponding
metric. Note that
B(X, Y )
is then the set of functions f : X Y such that
sup{|f(x)| : x A} < .

  

Source: Allard, William K. - Department of Mathematics, Duke University

 

Collections: Mathematics