I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. I would like someone to verify if my proof for the below exercise problem on a continuous nowhere differentiable function is correct.
[Abbott 6.4.3] (a) Show that:
\begin{equation*} g( x) =\sum _{n=0}^{\infty }\frac{\cos\left( 2^{n} x\right)}{2^{n}} \end{equation*} is continuous on all of $\displaystyle \mathbf{R}$.
Proof.
Since, $\displaystyle ( \exists M_{n})$ such that
\begin{equation*} |g_{n}( x) |=\left| \frac{\cos\left( 2^{n} x\right)}{2^{n}}\right| \leq \frac{1}{2^{n}} =M_{n} \end{equation*} and $\displaystyle \sum _{n=1}^{\infty } M_{n}$ converges, by the Weierstrass M-Test, $\displaystyle \sum _{n=1}^{\infty } g_{n}$ converges uniformly on $\displaystyle \mathbf{R}$.
Since each $\displaystyle g_{n}( x)$ is continuous on $\displaystyle \mathbf{R}$, by the term-by-term Continuity theorem, $\displaystyle \sum _{n=1}^{\infty } g_{n}( x)$ is continuous on $\displaystyle \mathbf{R}$.
(b) The function $\displaystyle g$ was cited in section 5.4 as an example of a continuous nowhere differentiable function. What happens if we try to use the theorem 6.4.3 to explore whether $\displaystyle g$ is differentiable?
Proof.
Let
\begin{equation*} g_{n}( x) =\frac{\cos\left( 2^{n} x\right)}{2^{n}} \end{equation*} So,
\begin{equation*} g_{n} '( x) =-\sin\left( 2^{n} x\right) \end{equation*} Since $\displaystyle g$ is nowhere differentiable, by the contrapositive of the term-by-term differentiability theorem, we have that:
If $\displaystyle g$ is not differentiable on $\displaystyle A$, then either $\displaystyle \sum g_{n}( x)$ converges NOT pointwise for all $\displaystyle x\in A$, or $\displaystyle \sum g_{n} '( x)$ converges NOT uniformly on $\displaystyle A$.
Since $\displaystyle \sum g_{n}$ converges uniformly on $\displaystyle \mathbf{R}$, the only possibility is that $\displaystyle \sum _{n=1}^{\infty } g_{n} '=\sum -\sin\left( 2^{n} x\right)$ does not converge uniformly on $\displaystyle \mathbf{R}$.
The question I have is, are we actually required prove this result? And how to go about proving it anyway? Does the below check out?
If a function $\sum h_n$ is uniformly convergent on $A$, it is uniformly convergent $(\forall S) \subseteq A$.
If $(\exists S \subseteq A)$ where $\sum h_n$ converges NOT uniformly, then $\sum h_n$ converges NOT uniformly on $A$.
Consider the point $x_0 = \frac{\pi}{3}$. The sequence $g_n'(x_0)=(-1)^n \frac{\sqrt{3}}{2}$. Thus, $\lim g_n'(x_0) \neq 0$. Consequently, by the $n$th term test, $\sum g_n'(x_0)$ does not converge pointwise on any interval containing $x_0 = \frac{\pi}{3}$.
Therefore, $\sum g_n'$ converges NOT uniformly on $\mathbf{R}$.
