# College Math Teaching

## August 7, 2014

### Engineers need to know this stuff part II

This is a 50 minute lecture in a engineering class; one can easily see the mathematical demands put on the students. Many of the seemingly abstract facts from calculus (differentiability, continuity, convergence of a sequence of functions) are heavily used. Of particular interest to me is the remarks from 45 to 50 minutes into the video:

Here is what is going on: if we have a sequence of functions $f_n$ defined on some interval $[a,b]$ and if $f$ is defined on $[a,b]$, $lim_{n \rightarrow \infty} \int^b_a (f_n(x) - f(x))^2 dx =0$ then we say that $f_n \rightarrow f$ “in mean” (or “in the $L^2$ norm”). Basically, as $n$ grows, the area between the graphs of $f_n$ and $f$ gets arbitrarily small.

However this does NOT mean that $f_n$ converges to $f$ point wise!

If that seems strange: remember that the distance between the graphs can say fixed over a set of decreasing measure.

Here is an example that illustrates this: consider the intervals $[0, \frac{1}{2}], [\frac{1}{2}, \frac{5}{6}], [\frac{3}{4}, 1], [\frac{11}{20}, \frac{3}{4}],...$ The intervals have length $\frac{1}{2}, \frac{1}{3}, \frac{1}{4},...$ and start by moving left to right on $[0,1]$ and then moving right to left and so on. They “dance” on [0,1]. Let $f_n$ the the function that is 1 on the interval and 0 off of it. Then clearly $lim_{n \rightarrow \infty} \int^b_a (f_n(x) - 0)^2 dx =0$ as the interval over which we are integrating is shrinking to zero, but this sequence of functions doesn’t converge point wise ANYWHERE on $[0,1]$. Of course, a subsequence of functions converges pointwise.

## April 17, 2012

### Pointwise versus Uniform convergence of sequences of continuous functions: Part II

Filed under: analysis, calculus, complex variables, uniform convergence — collegemathteaching @ 12:48 am

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.

Now what about the $|f_1(z)| > |f_2(z)|>...|f_k(z)|...$ hypothesis? Can it be dispensed with?

Let’s look at an example in real variables:

Let $f_n(x) = sin(\frac{e \pi}{2}e^{-nx})sin(\frac{n \pi}{2} x)$ with $x \in [0,1]$. $f_n(0) = 0$ for all $n$. To see that $f_n$ converges to zero pointwise, note that $lim_{n \rightarrow \infty}e^{-nx} = 0$ for all $x > 0$, hence $lim_{n \rightarrow \infty}sin(\frac{e \pi}{2}e^{-nx}) = 0$ which implies that $f_n \rightarrow 0$ by the squeeze theorem. But $f_n$ does not converge to 0 uniformly as for $t = \frac{1}{n}$ we have $f_n(t) = 1$

Here is a graph of the functions for $n = 5, 10, 20, 40$

## 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 $q$