An introduction to the Weierstrass M-test: Part II

In this part, we continue from where we left off in Part I of this series on the Weierstrass M-test.

First, let's recall our standing assumptions and notation.

Throughout, I is a non-degenerate interval in \R, and (f_n)_{n=0}^\infty is a sequence of functions from I to \R

We denote the set  \N \cup \{0\} of non-negative integers by \N_0.

For each m \in \N_0 we define the function S_m: I\to \R by S_m = \sum_{n=0}^m f_n\,. So, for each x \in I, we have

\Jdisplay S_m(x) = \sum_{n=0}^m f_n (x)\,.

We also have a sequence (M_n)_{n=0}^\infty of non-negative real numbers such that

 \Jdisplay\sum_{n=0}^\infty M_n < \infty\,.

For each m \in \N_0, we set R_m=\sum_{n=m+1}^\infty M_n\,.

Recall that R_m \to 0 as m \to \infty, because the "tails of a convergent series" always tend to 0.

Our first task is to prove the following theorem that we stated last time.

Theorem 1 (Part of the Weierstrass M-test)

Suppose that, for all n \in \N_0 and all x \in I, we have

\Jdisplay |f_n(x)| \leq M_n\,.

Then, for all x \in I, the series

 \Jdisplay\sum_{n=0}^\infty f_n(x)

is absolutely convergent, and so we may define f:I \to \R by

 \Jdisplay f(x)= \sum_{n=0}^\infty f_n(x)\,.

Then for all m \in \N_0 and all x \in I we have

\Jdisplay |S_m(x) - f(x)| \leq R_m\,

(where R_m is as defined above). 

Proof of Theorem 1

We first prove the claim about absolute convergence. 

Let x \in I. Then we have

\Jdisplay \sum_{n=0}^\infty |f_n(x)| \leq \sum_{n=0}^\infty M_n < \infty\,,

and so the series 

 \Jdisplay\sum_{n=0}^\infty f_n(x)

is absolutely convergent (and hence convergent).

We set f(x)= \sum_{n=0}^\infty f_n(x), giving us a function f:I \to \R.

It remains to check our "error estimate" for the partial sum functions S_m.

Let  m \in \N_0, and let x \in I. Then

\Jdisplay |S_m(x) - f(x)| = \left |\sum_{n=m+1}^\infty f_n (x) \right | \leq \sum_{n=m+1}^\infty |f_n (x)| \leq  \sum_{n=m+1}^\infty M_n =  R_m\,,

as claimed. \quad\square

We now prove another part of the Weierstrass M-test. In the setting of Theorem 1, if we also know that the functions f_n are continuous, then so is the function f. Now one approach here is to prove the more general result that a uniform limit of continuous functions is always continuous. But here we give a direct proof in the setting of the Weierstrass M-test.

Theorem 2 (Weierstrass M-test continued)

Under the conditions of Theorem 1

Let x_0 \in I, and suppose that all of the functions f_n (n \in \N_0) are continuous at x_0. Then the function f defined in Theorem 1 is also continuous at x_0.

In particular, if all of the functions f_n are continuous on I, then so is the function f.

Comments

In other words, we are claiming that, under these conditions, the function f:I \to \R defined by

 \Jdisplay f(x) = \sum_{n=0}^\infty f_n(x)

is continuous at any point of I at which all of the functions f_n are continuous (n \in \N_0).

We will have to make use of the conditions/conclusions of Theorem 1: without some assumptions on f_n, we don't even know whether the series defining f(x) is convergent.

Proof of Theorem 2 (with comments)

The last part of the theorem follows from the version at a single point, so we just prove that part. We are given all of the conditions of Theorem 1, and that all of the functions f_n are continuous at x_0.

Let \ve>0.

We need to show that, for all x \in I sufficiently close to x_0, we can guarantee that |f(x)-f(x_0)|<\ve.

More formally, our target (not yet proven!) is to find a \delta>0 with the following property:

(*)\Jdisplay for all x \in I with |x-x_0| < \delta, we have |f(x)-f(x_0)|<\ve\,.

We find such a \delta>0 using an "\ve/3 trick".

First, since R_m \to 0 as m \to \infty, we can choose an N \in \N_0 such that R_N < \ve/3

This means that the tail of the series contributes very little to the value of f, and we can look at the function S_N as a very good approximation to f.

This N will remain constant for the rest of the argument. (It depends on \ve, but that is allowed!)

Consider the function S_N = \sum_{n=0}^N f_n\,. This is a sum of finitely many real-valued functions on I which are all continuous at x_0, and so S_N itself is also continuous at x_0.

We apply the definition of continuity for S_N, but using \ve/3 instead of \ve.

Since S_N is continuous at x_0, we can find a \delta>0 such that, for all x \in I with |x-x_0|<\delta, we have |S_N(x)-S_N(x_0)| < \ve/3.

We claim that this \delta is good enough to satisfy (*) above.

Let x \in I with |x-x_0|<\delta. Then

\Jdisplay \begin{align*} |f(x)-f(x_0)| &= |f(x)-S_N(x)+S_N(x)-S_N(x_0) + S_N(x_0)-f(x_0)|\\&\leq |f(x)-S_N(x)|+|S_N(x)-S_N(x_0)| + |S_N(x_0)-f(x_0)| \\&< \ve/3 + \ve/3 + \ve/3 = \ve\,.\end{align*}

This proves (*), and so f is continuous at x_0, as required.\quad\square

Comments

Here, the terms |f(x)-S_N(x)| and |S_N(x_0)-f(x_0| are both \leq R_N < \ve/3, using Theorem 1. This estimate didn't use the continuity of the functions f_n at all. We only used the continuity of the functions f_n at x_0 when claiming that the partial sum function S_N was continuous at x_0. This then allowed us to choose a suitable \delta>0 guaranteeing our estimate |S_N(x)-S_N(x_0)|<\ve/3 whenever x \in I with |x-x_0|<\delta.

In the next part we move on to look at differentiability of the function f, and when we can differentiate the series for f term by term.

In particular we want to see what we can say about differentiating functions given by power series.


Comments

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