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This is an exercise from Problems from the Book by Andreescu and Dospinescu. When it was posted on AoPS a year ago I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me.

Problem: Prove that the function $f : [0, 1) \to \mathbb{R}$ defined by

$\displaystyle f(x) = \log_2 (1 - x) + x + x^2 + x^4 + x^8 + ...$

is bounded.

A preliminary observation is that $f$ satisfies $f(x^2) = f(x) + \log_2 (1 + x) - x$. I played around with using this functional equation for awhile, but couldn't quite make it work.

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if you differentiate your functional equation you get that limx→1f′(X)lim_{x\to 1} f'(X) exists and is finite. doesn't that do it for you? –  Eric O. Korman Aug 3 '10 at 0:46
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Hmm. Maybe. If you wrote that up together with a proof that f' actually exists, I'll accept it. That seems too easy, somehow. –  Qiaochu Yuan Aug 3 '10 at 1:11
    
A lot of solutions people claim are "from the book" are short and sweet :) –  BlueRaja - Danny Pflughoeft Aug 3 '10 at 1:23
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Plotting the function in Mathematica seems to indicate otherwise... –  Mariano Suárez-Alvarez Aug 3 '10 at 1:56
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@Danny: it is only being claimed that the problems, not their solutions, are from the Book! The authors, by their own admission, don't know how to solve some of the exercises... –  Qiaochu Yuan Aug 3 '10 at 2:02

3 Answers 3

up vote 21 down vote accepted

OK, a second trick is needed (but it actually finishes the problem). It is nice and simple enough that it's probably what the authors intended by a "Book" solution.

Let $f(x) = x \log(2) - \log(1+x)$. We want to show that $S(x) = f(x) + f(x^2) + f(x^4) + \dots$ is bounded. Because $f(0)=f(1)=0$ and $f$ is differentiable, we can find a constant $A$ such that $|f(x)| \leq Ax(1-x) = Ax - Ax^2$. The sum of this bound over the powers $x^{2^k}$ is telescopic.

Notice that the role of $\log(2)$ was to ensure that $f(1)=0$.

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Very nice! (and extra characters to satisfy the silly 15 character limit) –  Mariano Suárez-Alvarez Aug 3 '10 at 19:21
    
"Because f(0) = f(1) = 0 and f is differentiable, we can find a constant A..." ??? could you clarify? I'm missing something, is that the extreme value theorem? en.wikipedia.org/wiki/Extreme_value_theorem –  Jason S Aug 3 '10 at 22:25
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g(x) = f(x)/(x(1-x)) is defined on (0,1) and extends to a continous function on [0,1]. The extension exists because the limits of g(x) needed at the endpoints are f'(0) and f'(1). So differentiability of f(x) at 0 and 1 implies continuity of g(x), and for A just take any upper bound on the values of |g(x)| in [0,1]. –  T.. Aug 3 '10 at 22:39

Starting from (the natural logarithm of) $(1-x)^{-1} = (1+x)(1+x^2)(1+x^4) \dots$, it becomes clearer where the $\log(2)$ factor comes from.

One has to show that $\Sigma (x^{2^k} - C\log(1 + x^{2^k}))$ is bounded sum of positive terms. The sum of the first $n$ terms approaches $n - Cn\log(2)$ as $x \to 1-$, so we need $C = 1/\log(2)$ if there is to be boundedness.

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Nice observation. Can you finish the argument from here? –  Qiaochu Yuan Aug 3 '10 at 3:07
    
Yes and no. I posted a completion of the proof (see other answer) but it requires an additional, independent observation. –  T.. Aug 3 '10 at 19:04

If anyone's interested, here's another approach.

First, observe that $$ \log{2} = \int_{2^r}^{2^{r+1}} \frac{1}{x}\;dx \le \sum_{k=2^r}^{2^{r+1}-1}\frac{1}{k} \le \int_{2^r-1}^{2^{r+1}-1} \frac{1}{x}\;dx = \log{2} + \log\left(1+\frac{1}{2^{r+1}-2}\right) $$ for $r\ge1$, so using $\log(1+x) \le x$ (for $x>-1$) and $2^{r+1} - 2 \ge 2^r$, we get $$ \log{2} \le \sum_{k=2^r}^{2^{r+1}-1} \frac{1}{k} \le \log{2} + \frac{1}{2^r} $$ for all $r\ge0$ (check directly for $r=0$).

Thus for $x\in[0,1)$, we have $$\begin{align*} \lvert f(x)\log{2}\rvert = \lvert\sum_{r\ge0} x^{2^r}\log{2} - \sum_{n\ge1} \frac{x^n}{n}\rvert &= \lvert\sum_{r\ge0} x^{2^r}(\log{2} - \sum_{k=2^r}^{2^{r+1}-1}\frac{1}{k}) + \sum_{r\ge0}\sum_{k=2^r}^{2^{r+1}-1}\frac{x^{2^r} - x^k}{k} \rvert \\ &\le \sum_{r\ge0}\frac{x^{2^r}}{2^r} + \sum_{r\ge0}\sum_{k=2^r}^{2^{r+1}-1}\frac{x^{2^r} - x^k}{k} \\ &< 2 + \sum_{r\ge0}x^{2^r}(1-x)\sum_{k=2^r}^{2^{r+1}-1}\frac{1+x+\cdots+x^{k-2^r-1}}{k} \\ &\le 2 + (1-x)\sum_{r\ge0}x^{2^r}\sum_{k=2^r}^{2^{r+1}-1}\frac{k-2^r}{k} \\ &\le 2 + (1-x)\sum_{r\ge0}x^{2^r}(2^r\cdot 1) \\ &\le 2 + (1-x)(x + 2\sum_{r\ge1}\sum_{k=2^{r-1}+1}^{2^r}x^k) \\ &= 2 + x(1-x) + 2(1-x)\sum_{k\ge2} x^k \\ &= 2 + x + x^2, \end{align*}$$ so we're done.

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