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. )

  • $\begingroup$ your proof seems totally right, I dont see any mistake. $\endgroup$ – Masacroso Apr 15 at 10:32
  • $\begingroup$ How about the part to show it is discontinuous ? I do not know how to start it. $\endgroup$ – Ling Min Hao Apr 15 at 10:35
  • $\begingroup$ try to prove or disprove that if $(a_n)\to a$ for some $a\in[0,1]$ then $\lim_{n\to\infty} f(a_n)=f(a)$ $\endgroup$ – Masacroso Apr 15 at 10:37
  • $\begingroup$ I tried to use the negation of the definition of continuity to prove it is discontinuous at $(0.1)_2$ but I fail to find such $\epsilon$. $\endgroup$ – Ling Min Hao Apr 15 at 10:42

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 $f(\frac{1}{2}) = \frac{2}{3}$.

  • $\begingroup$ I get what you mean for taking sequence to prove discontinuity there. But is it possible to use epsilon-delta definition to show it is discontinuous ? $\endgroup$ – Ling Min Hao Apr 15 at 14:36
  • $\begingroup$ Yes, any proof using sequences can be transformed to epsilon-delta: take epsilon less then gap ($\frac{1}{3}$ in this case), then for any $\delta$ take $n$ such that $|x_n - x| < \delta$ ($x = \frac{1}{2}$ in this case). Then we have $|x_n - x| < \delta$ but $|f(x) - f(x_n)| > \varepsilon$. $\endgroup$ – mihaild Apr 15 at 14:40
  • $\begingroup$ can I know how you get $f(x_n) = \frac{1}{3} - \frac{1}{3^{n+1}}$? Shouldn't it is $f(x_n) = \frac{2}{3} - \frac{2}{3^{n+1}}$ $\endgroup$ – Ling Min Hao Apr 15 at 15:49
  • $\begingroup$ For example, $f(x_2) = f(0.011_2) = 0.022_3 = \frac{2}{9} + \frac{2}{27} = \frac{8}{27} = \frac{1}{3} - \frac{1}{27}$. $\endgroup$ – mihaild Apr 15 at 15:53
  • $\begingroup$ But we do not allow 1 in ternary expansion because the function is mapped from $[0,1]$ to $C$, which is a Cantor set, isn't ? $\endgroup$ – Ling Min Hao Apr 15 at 16:01

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