Exercise 5.4.6

Question

\begin{enumerate}[label=(\alph*)] 
\item Modify the previous argument to show that $g^{\prime}(1)$ does not exist. Show that $g^{\prime}(1 / 2)$ does not exist.
\item Show that $g^{\prime}(x)$ does not exist for any rational number of the form $x=$ $p / 2^{k}$ where $p \in \mathbf{Z}$ and $k \in \mathbf{N} \cup\{0\}$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item Let $x_m = 1 + 1/2^m$ with $m \geq 0$. Then $h_0(x_m) = 1 - 1/2^m$, $h_n(x_m) = 1/2^m$ for $1 \leq n \leq m$, and $h_n(x_m) = 0$ for $n > m$. $g(1) = 1$, so
    $$\frac{g(x_m) - g(1)}{x_m - 1} = \frac{1 - 1/2^m + m/2^m - 1}{1/2^m} = m-1$$
    and for the same reason as in Exercise 5.4.5, $g'(1)$ does not exist.

$h_0(x)$ is differentiable at $1/2$, so we can instead consider whether $(g - h_0)(x)$ is differentiable at $1/2$. But since $h_n(x) = 2h_{n+1}(x/2)$, $g(x) = 2(g-h_0)(x/2)$. Since $g'(1)$ does not exist, both $g-h_0$ and $g$ are not differentiable at $1/2$.

\item Note that $h_n(x)$ is only non-differentiable at points of the form $q/2^n$ where $q \in \mathbf{Z}$. Express $x$ in lowest form, so that $p$ is odd, and consider
    $$i(x) = g(x) - \sum^{k-1}_{n=0}h_n(x) = \sum^\infty_{n=k}h_k(x)$$
    which is differentiable at $x$ if and only iff $g$ is as well. Since $h_{a+b}(x) = h_a(2^b x) / 2^b$,
    $$i(x) = \frac{1}{2^k} \sum^\infty_{n=0} h_0(2^k x) = \frac{1}{2^k}g(2^k x) = \frac{1}{2^k}g(p)$$
    We've shown that $g$ is not differentiable at 0 or 1, and since $g$ is periodic it's easy to show it's not differentiable at any $p \in \mathbf{Z}$, completing the proof.
 \end{enumerate}

Exercise 5.4.6

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Exercise 5.4.6

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