Let $f:[0,1]\rightarrow C$ be defined by $f(\sum_{n=1}^{\infty}\frac{a_n}{2^n}) = \sum_{n=1}^{\infty}\frac{2a_n}{3^n}.$ Prove or disprove that $f$ is monotonic and continuous on $[0,1]$.
To show that it is monotone on $[0,1]$, suppose $x < y$. Let $x = \sum_{n=1}^{\infty}\frac{a_n}{2^n}$ and $y = \sum_{n=1}^{\infty}\frac{b_n}{2^n}$. Let $M$ be the smallest integer such that $a_M\neq b_M$. Since $x<y$, $a_M<b_M$ and hence $$f(x) = \sum_{n=1}^{\infty}\frac{2a_n}{3^n}< \sum_{n=1}^{\infty}\frac{2b_n}{3^n}=f(y)$$ as required.
Am I correct about this ? Also, I don't actually have a clue on how to show it is not continuous on $[0,1]$. Is there any hint about this ? (The hint for my textbook is this function is discontinuous at dyadic rationals. )
Formally $f$ is ill-defined. For example, if we take $a_1 = 0$, $a_n = 1$ if $n > 1$, and $b_1 = 1$, $b_n = 0$ if $n > 1$, we have $\sum\frac{a_n}{2^n} = \sum\frac{b_n}{2^n}$, but $\sum\frac{2 a_n}{3^n} = \frac{1}{3}$ while $\sum\frac{2 b_n}{3^n} = \frac{2}{3}$. I'll assume that we choose sequences with finite number of $1$ in such cases.
Then $f$ is discontinuous. One way to see it is note that $0$ and $\frac{2}{3}$ are in image, but $\frac{4}{9}$ isn't, and image of continuous function is connected.
Or we can take sequence $x_1 = 0.01_2$, $x_2 = 0.011_2$, ..., $x_n = \frac{1}{2} - \frac{1}{2^{n+1}}$. Then $x_n \to \frac{1}{2}$. But $f(x_n) = \frac{1}{3} - \frac{1}{3^{n + 1}}$ so $f(x_n) \to \frac{1}{3}$, while $f(\frac{1}{2}) = \frac{2}{3}$.