Exercise 5.4.4

Question

As the graph in Figure 5.7 suggests, the structure of $g(x)$ is quite intricate. Answer the following questions, assuming that $g(x)$ is indeed continuous.
\begin{enumerate}[label=(\alph*)] 
\item How do we know $g$ attains a maximum value $M$ on $[0,2]$? What is this value?
\item Let $D$ be the set of points in $[0,2]$ where $g$ attains its maximum. That is $D=\{x \in[0,2]: g(x)=M\}$. Find one point in $D$.
\item Is $D$ finite, countable, or uncountable?
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item $[0,2]$ is a compact set, so by the Extreme Value Theorem $g$ must have a maximum. Since the infinite series defining $g$ converges (and converges absolutely, for that matter), we are free to use associativity to analyze it. Group the terms in pairs, so that $f_0(x) = h_0(x) + h_1(x)$ and in general $f_n(x) = h_{2n}(x) + h_{2n + 1}(x)$. Note that over $[0,2]$,
$$
    f_0(x) = \begin{cases}
        2x & x \leq 1/2 \
        1 & 1/2 \leq x \leq 3/2 \
        -2x + 4 & 3/2 \leq x
    \end{cases}
     $$
and in particular, $f_0(x)$ reaches a maximum of 1 over $[1/2, 3/2]$, an interval of length 1. Now, $f_1(x)$ looks like a repeated $f_0(x)$ scaled down a factor of 4, and therefore has a period of $2 / 4 = 1/2$ and a maximum value of $1/4$ over an interval of length $1/4$. Since the period of $f_1(x)$ is less than half the length of the interval $[1/2, 3/2]$, there must be one cycle of $f_1(x)$ fully within $[1/2, 3/2]$, and therefore the maximum value of $f_0(x) + f_1(x)$ is $1 + 1/4$. (One cycle is when the function starts at 0, goes to a maximum and plateaus, then comes back down to 0.)

A scaling argument between $f_n$ and $f_{n+1}$ can then be used to show that $\max g(x) > \sum^n_{k=0} \frac{1}{4^k}$ for all $n \in \mathbf{N}$. However,
$$g(x) = \sum^\infty_{k=0} f_k(x) \leq \sum^\infty_{k=0} \frac{1}{4^k} = 4/3$$
and therefore $\max g(x) = 4/3$.

\item For this we'll need to track the intervals where $f_n(x)$ reaches its maximum more carefully. Note that the endpoints of each cycle in $f_n(x)$ are also endpoints of cycles in $f_{n+1}(x)$.

Some computation shows that if one of the cycles of $f_n(x)$ reaches its maximum over the interval $[a_n, a_n + b_n]$, then there will be two cycles in $f_{n+1}(x)$ which cover $[a_n, a_n + b_n / 2]$ and $[a_n + b_n/2, a_n + b_n]$, leading to $f_{n+1}(x)$ having maximum intervals in $[a_n + b_n/8, a_n + 3b_n/8]$ and $[a_n + 5b_n/8, a_n + 7b_n/8]$.

To find a point in $D$ we can repeatedly only consider the lower maximum interval at each iteration of $f_n(x)$. Defining $a_0 = 1/2,\ b_0 = 1$, and $a_{n+1} =a_n + b_n/8,\ b_{n+1} = b_n/4$, clearly $b_n = 1/4^n$. $a_n$ is (nearly) a geometric series, with
$$a_n = \frac{1}{2} + \frac{1}{6} \frac{4^n-1}{4^n},\ a_n + b_n = \frac{1}{2} + \frac{4^n + 5}{6}$$
Both $a_n$ and $a_n + b_n$ approach $1/2 + 1/6 = 2/3$, so we might conjecture $2/3 \in D$. Indeed, $2/3$ is in every $[a_n, a_n + b_n]$, and therefore $2/3$ is a point in $D$.

\item $D$ is uncountable; we will use an argument similar to showing that the Cantor set is uncountable - by mapping all sequences $x_n$ of infinite 0s and 1s to a unique point in $D$. Construct a sequence of intervals $I_n$ as such: If $x_n = 0$ then take the first cycle of $f_n(x)$ which is in $I_{n-1}$ and define $I_n$ to be the maximum interval of $f_n(x)$ in that cycle; if $x_n = 1$ then take the second cycle. (For completeness, define $I_0 = [1/2, 3/2]$.) By the Nested Interval Property the infinite intersection of these intervals yields a point in $D$, while clearly each unique sequence $x_n$ will map to a unique point in $D$. Since the set of all $x_n$ is uncountable, so too is $D$.

\end{enumerate}

Exercise 5.4.4

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Exercise 5.4.4

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