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.
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