I am struggling with exercise 3.3 in Silverman-Tate Rational Points on Elliptic Curves. Here is the paraphrased problem with necessary background:
Let $C:y^2 = x^3 + a x + b$ be a nonsingular cubic and let $P=(x,y)\in C(\mathbb{Q})$ be a rational point on the curve. We define the height $h (P)$ of $P$ to be $h (P) = \log\max\left\{ \left\vert m \right\vert , \left\vert n \right\vert \right\}$ where $x = m/n$ is in lowest terms. Prove that the limit $$\hat{h} (P) = \lim_{n\rightarrow \infty} \frac{h(2^n P)}{4^n}$$ exists (this is the canonical height of $P$).
My attempt is to prove the sequence $a_n = 4^{-n} h(2^n P)$ is Cauchy, so take $m,n\in\mathbb{N}$ and WLOG assume $m\leq n$. We proved in the text that for any point $P_0 \in C(\mathbb{Q})$ there exists a constant $\kappa_0 = \kappa_0 (P_0, a, b)$ such that for all points $Q\in C(\mathbb{Q})$ we have a bound $h(Q+P_0) \leq 2 h(Q) + \kappa_0$. Using this idea we can bound the height of $2^n P$ in terms of $2^m P$ by splitting up $2^n P$ as a sum of terms of the form $2^r P$, where $r$ decreases from $n$ to $m$. This introduces new constants $\kappa_r$ but since there are only finitely many of them we can take a maximum and not be too hindered:
$$h(2^n P)=h(2^{n-1}P + 2^{n-1} P)\leq 2 h(2^{n-1} P) + \kappa_0 = 2 h(2^{n-2}P + 2^{n-2} P) + \kappa_0$$ $$\leq 4 h(2^{n-2} P) + 2 \kappa_1 + \kappa_0 \leq \dots$$
$$\dots\leq 2^{n-m} h (2^m P) + \sum_{r=0}^{n-m-1} 2^r \kappa_r$$
Set $\kappa^* = \max\left\{\kappa_r : 0\leq r \leq n-m-1\right\}$. (Note that $\kappa^*$ depends on $m$ and $n$.) Then we have
$$h (2^n P)\leq 2^{n-m} h(2^m P) + (2^{n-m}-1)\kappa^*$$
Therefore
$$ a_n - a_m \leq \frac{2^{-m} - 2^{- n}}{2^n} (h(2^m P)+\kappa^*)$$
We also proved in the chapter that there exists a constant $\mu=\mu(a,b)$ (not depending on $P$) such that $h(2 P)\geq 4 h(P) - \mu$. If we then apply this to bound $h(2^n P)$ from below in terms of $h(2^m P)$ and $\mu$, we obtain a similar recurrence as above, which (assuming I have calculated correctly) gives
$$h(2^n P)\geq 4^{n-m} h (2^m P) - \frac{4^{n-m}-1}{3} \mu$$
Therefore we have the bounds
$$-\frac{4^{-m}-4^{-n}}{3}\mu \leq a_n - a_m \leq \frac{2^{-m} - 2^{ -n}}{2^n} (h(2^m P)+ \kappa^*)$$
So $$\left\vert a_n - a_m \right\vert \leq \max\left\{ \left\vert \frac{4^{-m}-4^{-n}}{3} \mu \right\vert , \left\vert \frac{2^{-m} - 2^{ -n}}{2^n} (h(2^m P)+ \kappa^*) \right\vert \right\}$$
$$ \leq (4^{-m} - 4^{-n}) \max\left\{ \frac{\left\vert \mu\right\vert }{3} , \left\vert h(2^m P)+ \kappa^* \right\vert \right\}$$
But I don't know where to go from here since I don't know what happens to $h(2^m P) + \kappa^*$ as $m,n\rightarrow\infty$, because $\kappa^*$ depends on $n$. We could of course apply the same procedure to $h(2^m P)$ to bound it above in terms of $h(P)$ and terms depending on $m$ and $n$, but I have not found this to be particularly helpful.
Can anyone help me on this? Thanks in advance!