# Monotonicity and Continuity for a function.

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 $$x, $$a_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. )

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

## 1 Answer

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

• 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 ? – Ling Min Hao Apr 15 at 14:36
• 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 $|f(x) - f(x_n)| > \varepsilon$. – mihaild Apr 15 at 14:40
• can I know how you get $f(x_n) = \frac{1}{3} - \frac{1}{3^{n+1}}$? Shouldn't it is $f(x_n) = \frac{2}{3} - \frac{2}{3^{n+1}}$ – Ling Min Hao Apr 15 at 15:49
• For example, $f(x_2) = f(0.011_2) = 0.022_3 = \frac{2}{9} + \frac{2}{27} = \frac{8}{27} = \frac{1}{3} - \frac{1}{27}$. – mihaild Apr 15 at 15:53
• 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 ? – Ling Min Hao Apr 15 at 16:01