Exercise 7.6.16

Question

\begin{enumerate}[label=(\alph*)] 
    \item Explain why $f'(x)$ exists for all $x \notin C$.
    \item If $c \in C$, argue that $|f(x)| \leq (x-c)^2$ for all $x \in [0,1]$. Show how this implies $f'(c) = 0$.
    \item Give a careful argument for while $f'(x)$ fails to be continuous on $C$. Remember that $C$ contains many points besides the endpoints of the intervals that make up $C_1, C_2, C_3, \dots$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item As noted in Exercise 7.6.15 (b), if $x \notin C$ then $f_n(x) = f_N(x)$ for some $N$ and all $n \geq N$. Since $[0,1]\backslash C$ is open, there must be some $[a,b] \subseteq [0,1] \backslash C$ with $x \in [a,b]$; clearly $(f_n)$ converges uniformly in $[a,b]$ so by the Differentiable Limit Theorem $f'(x)$ exists and is equal to $f_N'(x)$. \item Note that for $c \in C_n$, we have $\left|f_n(x)\right| \leq (x-c)^2$. Given $c \in C$, for all $n$, $c \in C_n$ and so $\left|f_n(x)\right| \leq (x-c)^2$. Thus by the Algebraic Limit Theorem $\left|f(x)\right| \leq (x-c)^2$. Now, $$f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} = \frac{f(x)}{x-c}$$ And given $\left|f(x)\right| \leq (x-c)^2$, we can use the Squeeze Theorem to find $f'(c) = 0$. \item Although $C$ contains many points other than the interval endpoints of $C_n$, every point in $C$ is ``close'' to some interval endpoint. More formally, if $c \in C$, $\forall \epsilon > 0 \ \exists x$ satisfying $|x-c| < \epsilon$, where $x$ is an endpoint of one of the intervals in $C_n$. To see this, simply take $n$ large enough that $3^{-n}$ (the length of each interval in $C_n$) is less than $\epsilon$; then since $c \in C_n$ (and specifically, is in one of the intervals that make up $C_n$) there must be an endpoint of an interval in $C_n$ which is less than $\epsilon$ away from $c$. To show that $f'$ is discontinuous at any $c \in C$, we need to show that for some fixed $\epsilon > 0$, $\forall \delta > 0 \ \exists x \in V_\delta(c)$ where $\left|f'(x) - f'(c)\right| = \left|f'(x)\right| > \epsilon$. Take $\epsilon = 1/2$, and let $p$ be an interval endpoint in $C_n$ where $|c - p| < \delta/2$. We also have, using Exercise 7.6.14 (c), that over $V_{\delta/2}(p)$, $\left|f_n'(p)\right| > \epsilon$ at some $x$, and it's clear that $x \notin C_n$. Therefore $\left|f'(x)\right| = \left|f_n'(x)\right| > \epsilon$, and by the Triangle Inequality $x \in V_\delta(c)$. \end{enumerate}

Exercise 7.6.16

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Exercise 7.6.16

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