MT2002 Analysis
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Properties of uniform convergence

We recall that uniform convergence is convergence in the metric dinfinity on a space of functions.

Some of the examples we considered in the last section show that it is possible for a sequence of continuous functions to be pointwise convergent to a non-continuous function.
Some of the other metrics we have considered give some similar unpleasant results.

For example, consider the sequence of functions (fn) where fn is:




Then the pointwise limit of (fn) is the function f given by f(x) = 0 if x < 1/2 and f(x) = 1 otherwise. Its graph is:


Then it is easy to see that d1(fn, f) = 1/2n rarrow 0 as n rarrow infinity. Unfortunately, f does not lie in C[0, 1] and so we cannot quite talk about convergence in this space. However, the sequence (fn) is a Cauchy sequence in C[0, 1] with the metric d1.

This motivates:

Definition

A space in which every Cauchy sequence is convergent is called a complete space.

Remarks

  1. What we saw earlier about the Completeness axiom can now be phrased as:
    The real numbers R form a complete space under the usual metric.

  2. The last example shows that C[0, 1] under the metric d1 is not a complete space. We saw earlier that the set of rationals Q under the usual metric is not a complete space.

The nice thing about the metric dinfinity is:

Theorem

The space C[0, 1] under uniform convergence (the metric dinfinity) is a complete space.


Proof
We need to show that if (fn) is a Cauchy sequence under dinfinity (that is: Given epsilon > 0 there exists N such that if m, n > N then dinfinity< epsilon) then the sequence has a limit in C[0, 1].

First we show that the sequence converges to its pointwise limit in the metric dinfinity.

Proof of that
Take a point x belongs [0, 1]. Then the sequence (fn(x)) is a real sequence which, from the fact that (fn) is a Cauchy sequence, is a Cauchy sequence in R and hence has a limit which we will call f(x). Thus the pointwise limit f is defined.

Since at each point fn(x) is close to f(x) we have the lub{ |fn(x) - f(x)| for x belongs [0, 1] } is small and so (fn) rarrow f in dinfinity.

The last thing we need to prove is that the pointwise limit f is continuous.

Proof of that
To prove that f is continuous at p belongs [0, 1], take some epsilon > 0. Choose N such that dinfinity(fn, f) < epsilon.
Then since fn is continuous, there exists delta such that if x is in the interval (p - delta, p + delta) then |fn(x) - fn(p)| < epsilon. Then |f(x) - f(p)| = |f(x) - fn(x) + fn(x) - fn(p) + fn(p) - f(p)| < |f(x) - fn(x)| + |fn(x) - fn(p)| + |fn(p) - f(p)| < epsilon + epsilon + epsilon and so can be made arbitrarily small.


Previous page
(Convergence in infinite dimensional spaces) 		Contents Next page
(The Weierstrass approximation theorem)
JOC September 2002