MT2002 Analysis

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## Properties of uniform convergence

We recall that uniform convergence is convergence in the metric d 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→ 0 as n→ ∞. 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 d is:

Theorem

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

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

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

Proof of that
Take a point x ∈ [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 ∈ [0, 1] } is small and so (fn)→ f in d.

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 ∈ [0, 1], take some ε > 0. Choose N such that d(fn, f) < ε.
Then since fn is continuous, there exists δ such that if x is in the interval (p - δ, p + δ) then |fn(x) - fn(p)| < ε. 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)| < ε + ε + ε and so can be made arbitrarily small.

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JOC September 2001