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I am trying to derive the fact that the function $F$ defined below is monotonically increasing. The only thing I can use is that the any member of the Cantor set has a ternary expansion involving only $0$'s and $2$'s. The function is defined as follows: Let $x\in[0,1]$ have ternary expansion $0.a_1a_2\cdots$. Define $N$ as the first index $n$ for which $a_n=1$ and set $N=\infty$ if none of the $a_n$ are $1$. Now let $F(x)=\sum_{n=1}^{N-1}\frac{a_n}{2^{n+1}}+\frac{1}{2^N}$.

I have shown that $F$ is constant while on a particular middle third removed in the construction of the Cantor set $C$ so I am really interested in showing the increasing nature on $C$. Essentially therefore I wish to show that given $x=\sum \frac{a_n}{3^n}<\sum\frac{b_n}{3^n}=y$ we have $\sum\frac{a_n}{2^{n+1}}<\sum\frac{b_n}{2^{n+1}}$ on $C$. How do I establish that?

Thanks

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Since on $C$ your $a_n$, $b_n$ are all zeroes and twos, try thinking of the relation you want to show as describing some binary expansions. –  Kevin Carlson Sep 20 '12 at 13:05

1 Answer 1

up vote 1 down vote accepted

Remember that $a_n-b_n\in\{-2,0,2\}$.

Suppose that $$ \sum_{n=1}^\infty\frac{a_n-b_n}{3^n}\gtrless0\tag{1} $$ $(1)$ happens if and only if, for the smallest $n_0$ so that $a_{n_0}\not=b_{n_0}$, it is the case that $a_{n_0}\gtrless b_{n_0}$.

The claim above is true because $$ \text{If }a_{n_0}>b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{3^n}\ge\frac2{3^{n_0}}-\sum_{n>n_0}\frac2{3^n}=\frac1{3^{n_0}}\tag{2} $$ $$ \text{If }a_{n_0}<b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{3^n}\le-\frac2{3^{n_0}}+\sum_{n>n_0}\frac2{3^n}=-\frac1{3^{n_0}}\tag{3} $$ If for the smallest $n_0$ so that $a_{n_0}\not=b_{n_0}$, it is the case that $a_{n_0}\gtrless b_{n_0}$, then $$ \sum_{n=1}^\infty\frac{a_n-b_n}{2^{n+1}}\gtreqless0\tag{4} $$ This is because $$ \text{If }a_{n_0}>b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{2^{n+1}}\ge\frac1{2^{n_0}}-\sum_{n>n_0}\frac1{2^n}=0\tag{5} $$ $$ \text{If }a_{n_0}<b_{n_0}\text{ then }\sum_{n=1}^\infty\frac{a_n-b_n}{2^{n+1}}\le-\frac1{2^{n_0}}+\sum_{n>n_0}\frac1{2^n}=0\tag{6} $$ So, if $x\gtrless y$, $F(x)\gtreqless F(y)$. We can't get strict increase since $F$ is constant on all the "middle-thirds" intervals.

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Thanks. One small question though. What is the need of establishing the inequality involving $\sum\frac{a_n-b_n}{3^n}$? –  Shahab Sep 21 '12 at 3:08
    
@Shahab: since that is $x-y$. –  robjohn Sep 21 '12 at 5:07
    
To show the increasing nature we start with $x<y$ in $C$ and attempt to prove $F(x)\le F(y)$. But if $x<y$ then it is obvious from the definition of ternary that $a_{n_0}<b_{n_0}$ and I don't understand why we need to consider all but the last inequality. –  Shahab Sep 21 '12 at 7:34
    
@Shahab: $(2)$ and $(3)$ are proof of the claim that $x>y$ if and only if the first differing base-three digit of $x$ is greater than the corresponding digit of $y$. Obviously, if all of the $a_n=b_n$ then $x=y$. However, if the first differing digit of $x$ is greater than the corresponding digit of $y$, then $x>y$ (the later digits cannot catch up). Likewise, if the first digit of $x$ is less than the corresponding digit of $y$, then $x< y$. –  robjohn Sep 21 '12 at 11:20
    
@Shahab: What $(2)$, $(3)$, and self-evidence show is $$ \begin{align} \text{first differing digit of $x$ is smaller}&\Rightarrow x< y\\ \text{all digits are equal}&\Rightarrow x=y\\ \text{first differing digit of $x$ is bigger}&\Rightarrow x>y \end{align} $$ Since all sequences fall into one of the categories on the left and the categories on the right do not overlap, we actually have bijections above (e.g. if $x< y$, then none of the other conditions on the left can hold because they imply $x=y$ or $x>y$), and this justifies the previous "if and only if" claim. –  robjohn Sep 21 '12 at 11:20

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