Exercise 7.14

Question

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Exercise 14: 
    Let $f$ be a continuous real function on $\mathbf R$ with the following properties: $0\le f(t)\le 1$, $f(t+2)=f(t)$ for every $t$, and
    $$f(t)=
    \begin{cases}
        0 & \bigl(0\le t\le\frac{1}{3}\bigr) \
        1 & \bigl(\frac{2}{3}\le t\le 1\bigr).
    \end{cases}$$
    Put $\Phi(t)=\bigl(x(t),y(t)\bigr)$, where $$x(t)=\sum_{n=1}^\infty 2^{-n}f(3^{2n-1}t),\quad\quad y(t)=\sum_{n=1}^\infty 2^{-n}f(3^{2n}t).$$ Prove that $\Phi$ is {\it continuous} and that
    $\Phi$ maps $I=[0,1]$ {\it onto} the unit square $I^2\subset\mathbf R^2$. In fact, show that $\Phi$ maps the Cantor set onto $I^2$.

analambanomenos · Accepted Answer

\def\where{\,|\,}

Since for each $n$ we have $\big|2^{-n}f(3^{2n-1}t)\big|\le2^{-n}$ and $\big|2^{-n}f(3^{2n}t)\big|\le2^{-n}$, the series defining $x(t)$ and $y(t)$ converge uniformly since
    $\sum 2^{-n}$ converges, by Theorem 7.10. And since the partial sums are continuous functions, $x(t)$ and $y(t)$ are continuous functions, by Theorem 7.12. Hence $\Phi$ is continuous.

Following the hint, let $(x_0,y_0)\in I^2$, and let $$x_0=\sum_{n=1}^\infty 2^{-n}a_{2n-1},\quad\quad y_0=\sum_{n=1}^\infty 2^{-n}a_{2n}$$ be the binary expansions of $x_0$ and $y_0$, where
    each $a_i$ is 0 or 1. Let $$t_0=\sum_{i=1}^\infty 3^{-i-1}(2a_i).$$ By Exercise 3.19, the set of all such $t_0$ is precisely the Cantor set.

We have for $k=1,2,3,\ldots,$
    \begin{align*}
        3^kt_0 &= \sum_{i=1}^\infty 3^{k-i-1}(2a_i) \
        &= 2\sum_{n=0}^{k-2}3^na_{k-1-n}+\frac{2}{3}\,a_k+\frac{2}{3}\sum_{n=1}^\infty 3^{-n}a_{k+n}
    \end{align*}
    Note that the last term lies between 0 and $$\frac{2}{3}\sum_{n=1}^\infty 3^{-n} = \frac{2}{3}\biggl(\frac{1/3}{1-(1/3)}\biggr)=\frac{1}{3}.$$ Also, the first term is an even integer, so
    that, since $f(t+2)=f(t)$, we have $$f(3^kt_0)=f\Biggl(\frac{2}{3}\,a_k+\frac{2}{3}\sum_{n=1}^\infty 3^{-n}a_{k+n}\Biggr).$$ If $a_k=0$, then the expression in the argument of $f$ lies between
    0 and $\frac{1}{3}$, so that $f(3^kt_0)=0$. And if $a_k=1$, then the expression in the argument of $f$ lies between $\frac{2}{3}$ and 1, so that $f(3^kt_0)=1$. In either case, we have
    $f(3^kt_0)=a_k$. Hence $\Phi(t_0)=\bigl(x(t_0),y(t_0)\bigr)=(x_0,y_0)$.

Exercise 7.14

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Exercise 7.14

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