Rotation number: for every $i$ there is a $x$, such that sequence terms are equal to the limit

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Suppose $\varphi\colon S^1 \to S^1$ is an orientation preserving circle homeomorphism and $f\colon\mathbb{R} \to \mathbb{R}$ a lift of $\varphi$, one can then define the lift of the Poincaré rotation number as $\tilde \rho(f)$ and a function $r_i(x) := f^i(x) - x - i \tilde \rho(f)$, which is quite helpful in other proofs about orbits of $f$. I could already show, that

  • $r_i(x+1) = r_i(x)$ is periodic (obvious since $f(x+1) = f(x) + 1$)
  • $|r_i(x)| \leq 1$ in the process of showing that the sequence $z_i = \frac{f^i(x)-x}{i}$ is Cauchy

Now I am trying to show that $r_i(x)$ has a zero. For a periodic $f$ with $f^q(x) = x+p$ for $q>0$, we have $\tilde\rho(f) = p/q$ and

$f^i(x) = f^k(x) + n\cdot p$,

where $i = n\cdot q +k$. For multiples of $q$ we thus have $z_{n\cdot q} = \frac{x-x +n\cdot p}{n\cdot q} = p/q$ and $r_{n\cdot q}(x) = 0$ for all $x$. For arbitrary $i$ I am a bit stuck and even more so for irrational $\tilde \rho$. Any hints on how to prove it?