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.

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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.

In a previous post, I talked about why it was important that each f_k(z) be continuous.

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

Answer: well, no.

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

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