Some time ago I saw a problem somewhere (I cannot remember where) that goes something like this. Let $f:[0,+\infty)\to\mathbb{R}$ be a real function with $f(x)=f(\lfloor x\rfloor)$ for all $x\geq0$, $f(0)=0$ and such that $\exists\lim\limits_{x\to+\infty}f(x)\neq0$. Can the following limit converge (in $\mathbb{R}$) for some constant $1\neq\lambda\in\mathbb{R}$?
$$\boxed{\lim_{x\to+\infty}\sum_{k=0}^\infty \left[\lambda f\left(\frac{x}{2^k}\right)-f\left(\frac{x}{3\cdot 2^k}\right)\right]}$$
Note that the sum on the limit is actually finite as, given $x\geq0$, there exists $k_0\in\mathbb{N}$ such that $\forall k\geq k_0:f(x/2^k)=f(x/(3\cdot2^k))=f(0)=0$. I have noticed that, if $f:[0,+\infty)\to\mathbb{R}$ satisfies the above-mentioned hypotheses, then the limit $L = \lim\limits_{x\to+\infty}\sum_{k=0}^\infty f\left(\frac{x}{2^k}\right)$ cannot exist since
$$\Rightarrow L = \lim_{x\to+\infty}\sum_{k=0}^\infty f\left(\frac{x/2}{2^k}\right) =\lim_{x\to+\infty}\sum_{k=0}^\infty f\left(\frac{x}{2^{k+1}}\right) =L-\lim_{x\to+\infty}f(x)\neq L$$
But it could so happen that the former limit exists (?).
In fact, if we denote $F(x)=\sum_{k=0}^\infty f\left(\frac{x}{2^{k+1}}\right)$, we can restate the troublesome condition as $\exists\lambda\neq1:\exists\lim\limits_{x\to+\infty}\lambda F(x)-F(x/3)$. Note that this condition of $F(x)$ alone can't imply the existence of $\lim\limits_{x\to+\infty}F(x)$ (and, thus, it can't lead us to a contradiction by itself) since for example, considering the 3-adic valuation, we have for $\lambda\neq0$ that $F(x)=\lambda^{-\nu_3(\lfloor x\rfloor)}$ (with $F(x):=0$ for $\lfloor x\rfloor = 0$) is divergent while $\lambda F(3n)-F(n)=F(n)-F(n)=0$ which I think implies $\lambda F(x)-F(x/3)\to0$ (?). So it could still happen that $\lim\limits_{x\to+\infty}\lambda F(x)-F(x/3)$ exists.
Best Answer
No, the limit cannot exist. Say $\lim_{x\to +\infty} f(x) = L$. Name the sum $g: [0,\infty) \to \mathbb{R}$:
$$ g(x) = \sum_{k=0}^\infty \left[\lambda f\!\left(\frac{x}{2^k}\right) - f\!\left(\frac{x}{3 \cdot 2^k}\right)\right] $$
and suppose $\lim_{x \to +\infty} g(x) = M$.
Then for every positive $x$,
$$ g(2x) = \lambda f(2x) - f\!\left(\frac{2x}{3}\right) + g(x) $$
And then
$$ 0 = \lim_{x \to +\infty} \left[g(2x) - g(x) - \lambda f(2x) + f\!\left(\frac{2x}{3}\right)\right] = M - M + (\lambda - 1) L $$
So either $L=0$ or $\lambda=1$. It is not possible that $L \neq 0$ and $\lambda \neq 1$ and the limit of $g$ exists.