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 \(S_N = \sum_{n=0}^m f_n\,.\) 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.