In my complex analysis class I was grading a problem of the following type:

given where is compact and given a sequence of continuous functions which converges to 0 pointwise on and if for all show that the convergence is uniform.

The proof is easy enough to do; my favorite way is to pick for a given and such that find a “delta disk” about so that for all in that disk, also. Then cover by these open “delta disks” and then one can select a finite number of such disks, each with an associated and then let be the maximum of this finite collection of .

But we used the fact that is continuous in our proof.

Here is what can happen if the in question are NOT continuous:

Let’s work on the real interval . Define if in lowest terms, and let if is irrational.

Now let . Clearly converges to 0 pointwise and the have the decreasing function property. Nevertheless, it is easy to see that the convergence is far from uniform; in fact for each is unbounded!

Of course, we can also come up with a sequence of bounded functions that converge to 0 pointwise but fail to converge uniformly.

For this example, choose as our domain and let if in lowest terms, and let if is irrational. Now let our sequence . Clearly converges to zero pointwise. To see that this convergence is not uniform: let be given and if and the right hand side of the inequality varies with and is, in fact, unbounded. Given a fixed and one can always find a to exceed the fixed . Hence varies with

[…] In a previous post, I talked about why it was important that each be continuous. […]

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