Uniform Convergence

Consider the sequence of functions , on the unit interval . For each fixed , the sequence is convergent:

as .

For each fixed , is continuous, but the limiting function is not. This is an unfortunate state of affairs which is remedied by the introduction of . This is a stronger notion of convergence for a sequence of functions, sufficient to guarantee, in particular, that continuity is preserved in the passage to limits.

Let be any set (you may think that it is a subset of or ) and let be sequence of functions defined in .

Definition 1
The sequence  converges uniformly to the function  if: given , there exists  (depending upon ) such that 
 


Remark 2

An equivalent formulation of Uniform convergence is

 


If and is any function, we say that is continuous at in if given , there exists such that

Then is said to be continuous in if it is continuous at every point of .

Now we show that 'the uniform limit of a sequence of continuous functions is again continuous'.


Theorem 1

Let be continuous for each and suppose that uniformly in as . Then is continuous in .

 


Proof

Pick . We need to estimate

  for all  close to .  For this, use

Given , by uniform convergence, there exists so that
Substituting this into Uniform convergence , we obtain
Now , and find, by continuity of at , a such that
Combining Uniform convergence and Uniform convergence we see that for this same ,
which proves that is continuous.

 


Example 3

We can see that the example we started with fails to be uniformly convergent. Indeed, with and as in Uniform convergence , if , we have and this tends to as . Thus and so does not converge uniformly to in this case.

 

Uniformly convergent series[edit | edit source]

As usual, we pass from uniformly convergent sequences to uniformly convergent series via partial sums. That is, if is a sequence of functions, then we say that the series is uniformly convergent in if the corresponding sequence of partial sums

is uniformly convergent in . There is a for uniform convergence of sums, with the following statement:

Theorem 2

Let be a sequence of functions. The series converges uniformly in if and only if:

Given any , there exists depending upon , such that

for all , we have

 


Finally, we state the Weierstrass -test, which is a very effective way of checking uniform convergence of many of the series which arise in complex analysis.


Theorem 3

Let be a sequence of functions. Suppose that there exist such that

and also
Then the series converges absolutely and uniformly in . In particular, if each of the is continuous, then so is .

 
Proof

Use the Cauchy criterion. By the triangle inequality,

for every . The Cauchy convergence criterion for the convergent series states that, given , there exists such that implies . Substituting this into Uniformly convergent series we have
as required for the use of Theorem~ Uniformly convergent series to get absolute and uniform convergence of . The continuity statement follows from Theorem Uniform convergence .

 
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