An introduction to the Weierstrass M-test: Part III

In this post, following on from the first two posts in the series, we look at differentiating suitable series of functions term by term. We make as much use as possible of the Weierstrass M-test (in the spirit of this series of posts), instead of discussing uniform convergence in general. Of course, uniform convergence is an important topic, and I'll come back to it at some point.

First we recall a characterization of differentiability in terms of "chord functions".

Note, for functions defined on intervals, when we discuss differentiability at endpoints (if any) of the interval, we use one-sided derivatives.

Proposition 3 (Differentiability in terms of chord functions)

Let $I$ be a non-degenerate interval in $\R$, let $f:I\to \R$, and let $x_0 \in I.$ Then the following statements are equivalent.

(a) The function $f$ is differentiable at $x_0$.

(b)  There is a function $g:I \to \R$ such that $g$ is continuous at $x_0$ and, for all $x \in I$,
\[f(x) = f(x_0) + (x-x_0) g(x)\,.\]

Comments

• Under these conditions, we then have $f'(x_0)=g(x_0)$ and, for all $x \in I\setminus\{x_0\},$ we have
  \[g(x)= \frac{f(x)-f(x_0)}{x-x_0}\,.\]
• For such $x,$ $g(x)$ is the gradient of the chord joining $(x_0,f(x_0))$ to $(x,f(x)).$ 
• I have seen $g$ called the chord function for $f$, but this may not be standard. Of course if we change $x_0$ we need a different function. 
• In my recent analysis lectures I called this function $g$ the chord function for $f$ at $x_0.$ But possibly "centred at $x_0$" would have been better.
• In the lecture notes I was following, written by my colleague Matthias Kurzke, the function $g$ was denoted by $\hat f.$ 

We omit the easy proof of Proposition 3 (but the details were covered in the live class and in the printed notes). 

Notation: Recall that we denote the set  \(\N \cup \{0\}\) of non-negative integers by \(\N_0.\)

For the rest of this post, $I$ will be a non-degenerate interval, and $M_n (n\in\N_0)$ will be non-negative real numbers with $\sum_{n=0}^\infty M_n < \infty\,.$

Next, for convenience, we combine Theorems 1 and 2 from the previous post, in terms of functions $g_n.$ We'll call this Proposition 4. See the previous post for details of the proof.

Proposition 4 (Selected portions of the Weierstrass M-test)

Let $g_n$ $(n \in \N_0)$ be functions from $I$ to $\R$, and suppose that, for all $n \in \N_0$ and all $x \in I$, we have $|g_n(x)| \leq M_n.$

(a) For each $x \in I$, the series $\sum_{n=0}^\infty g_n(x)$ is absolutely convergent (and hence convergent).

(b) Define a function $g:I \to \R$ by

\[ g(x)=\sum_{n=0}^\infty g_n(x)\,.\]

Let $x_0 \in I$, and suppose that, for all $n \in \N_0$, $g_n$ is continuous at $x_0.$ Then $g$ is also continuous at $x_0$.

Combining Propositions 3 and 4 allows us to prove a result about differentiability of sums of series.

Theorem 5 (Term by term differentiation of series)

Let $f_n$ ($n \in \N_0$) be differentiable functions from $I$ to $\R,$ and suppose that, for all $n \in \N_0$ and all $x \in I,$ we have $|f_n'(x)|\leq M_n.$

Suppose further that, for  all $x \in I$, the series $\sum_{n=0}^\infty f_n(x)$ converges.

Define $f:I \to \R$ by

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

Then $f$ is differentiable on $I$, and, for each $x \in I$,

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

Comments

• It is enough to assume that $\sum_{n=0}^\infty f_n(a)$ for at least one point $a \in I$, as (given the other assumptions) it then follows that $\sum_{n=0}^\infty f_n(x)$ converges for all $x \in I.$ We do not require this to be absolute convergence.
• We do know that, for each $x \in I$, the series $\sum_{n=0}^\infty f_n'(x)$ converges absolutely.

Proof of Theorem 5

Let $x_0 \in I$. Since each function $f_n$ is differentiable at $x_0,$ by Proposition 3, we can find functions $g_n:I \to \R$ $(n \in \N_0)$ such that each $g_n$ is continuous at $x_0$, and, for all $x \in I$,

\[f_n(x) = f_n(x_0) + (x-x_0) g_n(x)\,.\]

We then have $f_n'(x_0)=g_n(x_0)$ and, for all $x \in I\setminus\{x_0\},$ we have

\[g_n(x)= \frac{f_n(x)-f_n(x_0)}{x-x_0}\,.\]

Since $|f_n'(t)| \leq M_n$ for all $t \in I,$ we can apply the Mean Value Theorem to see that $|g_n(x)| \leq M_n$ for all $x \in I\setminus\{x_0\}$, and we also have $|g_n(x_0)|=|f_n'(x_0)| \leq M_n$. So we can apply Theorem 4 to define $g:I \to \R$ by

\[g(x)=\sum_{n=0}^\infty g_n(x)\,,\]

and this function $g$ is continuous at $x_0$, with 

\[g(x_0)=\sum_{n=0}^\infty g_n(x_0)=\sum_{n=0}^\infty f_n'(x_0)\,.\]

Let $x \in I$. Then 

\[f(x)=\sum_{n=0}^\infty f_n(x) = \sum_{n=0}^\infty (f_n(x_0)+(x-x_0)g_n(x))\]

\[= \sum_{n=0}^\infty f_n(x_0) + (x-x_0) \sum_{n=0}^\infty g_n(x) = f(x_0) + (x-x_0) g(x)\,. \]

By Proposition 3 again, since $g$ is continuous at $x_0$, it follows that $f$ is differentiable at $x_0$ with \[f'(x_0)=g(x_0)= \sum_{n=0}^\infty f_n'(x_0)\,,\] as required.$\qquad\square$

In the next part, we will look at the special case of power series.