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.