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


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

    Pingback by Pointwise versus Uniform convergence of sequences of continuous functions: Part II « College Math Teaching — April 17, 2012 @ 12:48 am

  2. […] the case of pointwise convergence of functions (studied earlier in this blog here and here. […]

    Pingback by Convergence of functions and nets (from advanced calculus) | College Math Teaching — January 17, 2015 @ 4:57 am

  3. […] now, let and we will show that converges uniformly on some open disk containing and . That will justify “term by term” integration that […]

    Pingback by Cauchy integral formula; power series for analytic functions, etc. | Bradley University Complex Variables Class — March 14, 2018 @ 9:28 pm

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