2
$\begingroup$

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. )

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.