This comes from the proof of the first theorem in this blog article.
The paths $x:[0,T]\to\mathbb{R}^d$ and $y:[0,T]\to\mathbb{R}^{e\times d}$ are of bounded total variation. The real numbers $p,q>1$ are such that $\frac{1}{p}+\frac{1}{q}>1$. Finally, $0\leq s\le t\leq T$.
We have the following definitions: $$\theta := \frac{1}{p}+\frac{1}{q}.$$ $$\omega(s,t):=\|x\|_{p\text{-var}[s,t]}^{1/\theta}\|y\|_{q\text{-var}[s,t]}^{1/\theta}.$$ $$\omega_{\epsilon}(s,t) := \omega(s,t)+\epsilon\left(\|x\|_{1\text{-var}[s,t]}+\|y\|_{1\text{-var}[s,t]}\right).$$ $$\Gamma_{s,t}=\int_s^t(y(u)-y(s))\mathrm{d}x(u).$$ $$\Psi(r) := \sup_{s,u,\omega_{\epsilon}(s,u)\leq r}\|\Gamma_{s,u}\|.$$
It is then claimed that
$$2^n \Psi\left(\frac{r}{2^n}\right)\xrightarrow{n\to\infty}0.$$
I can see that for any fixed $r$, $\Psi\left(\frac{r}{2^n}\right)$ goes to $0$ as $n$ goes to infinity, as $\omega$ is continuous and $\omega(s,s)=0$. But $2^n$ goes to infinity, so how do we know that $\Psi\left(\frac{r}{2^n}\right)$ goes to zero faster than $2^n$ goes to infinity?
