Prove that this function is bounded

Question

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.

Answer 1

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 $f(1)=0$.

Answer 2

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.