# College Math Teaching

## April 12, 2012

### Pointwise vs. Uniform convergence for functions: Importance of being continuous

In my complex analysis class I was grading a problem of the following type:
given $K \subset C$ where $K$ is compact and given a sequence of continuous functions $f_n$ which converges to 0 pointwise on $K$ and if $|f_1(z)| > |f_2(z)|>...|f_k(z)|...$ for all $z \in K$ show that the convergence is uniform.

The proof is easy enough to do; my favorite way is to pick $\epsilon > 0$ for a given $z \in K$ and $n$ such that $|f_n(z)| < \epsilon$ find a “delta disk” about $z$ so that for all $w$ in that disk, $|f_n(w)| < \epsilon$ also. Then cover $K$ by these open “delta disks” and then one can select a finite number of such disks, each with an associated $n$ and then let $M$ be the maximum of this finite collection of $n$.

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

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

Let’s work on the real interval $[0,1]$. Define $g(x) = q$ if $x = \frac{p}{q}$ in lowest terms, and let $g(x) = 0$ if $x$ is irrational.

Now let $f_n(x) = \frac{g(x)}{n}$. Clearly $f_n$ converges to 0 pointwise and the $f_n$ have the decreasing function property. Nevertheless, it is easy to see that the convergence is far from uniform; in fact for each $n, f_n$ 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 $[0,1]$ and let $h(x) = \frac{q-1}{q}$ if $x = \frac{p}{q}$ in lowest terms, and let $h(x) = 0$ if $x$ is irrational. Now let our sequence $f_n(x) = h(x)^n$. Clearly $f_n$ converges to zero pointwise. To see that this convergence is not uniform: let $\epsilon > 0$ be given and if $(\frac{q-1}{q})^n < \epsilon, n > \frac{ln(\epsilon)}{ln(\frac{q-1}{q})}$ and the right hand side of the inequality varies with $q$ and is, in fact, unbounded. Given a fixed $n$ and $\epsilon$ one can always find a $q$ to exceed the fixed $n$. Hence $n$ varies with $q$