Let $T>1$ and $W$ be a Banach space. Assume $X:= [0,T]\to W$ is a path. For $[s,t]\subset[0,T]$, let $\| X\|_{\alpha,[s,t]}:=$$\sup_{s\le u<v\leq t} \frac{|X_v-X_u|}{|v-u|^\alpha}$ be the well-known $\alpha$-Hölder seminorm. In "A Course on Rough Paths" (Friz and Hairer, 2013), the authors write:
As a consequence of Young's inequality [You36], one has the bound
$$ \left|\int_0^1\left(Y_r-Y_0\right) d X_r\right| \left(= \left|\int_0^1 Y_r\: d X_r - Y_0 X_{0,1}\right|\:\right)\leq C\|Y\|_{\beta ;[0,1]}\|X\|_{\alpha ;[0,1]}, $$ with $C$ depending on $\alpha+\beta>1$. Given paths $X, Y$ defined on $[s, t]$ rather than $[0,1]$ it is an easy consequence of the scaling properties of Hölder seminorms, that $$ \left|\int_s^t Y_r d X_r-Y_s X_{s, t}\right| \leq C\|Y\|_\beta\|X\|_\alpha|t-s|^{\alpha+\beta} $$
It seems that the scaling property of Hölder seminorms that the authors use should be such that ($\forall s,t\in[0,T]$ with $s\le t$):
$$ \| X\|_{\alpha,[0,1]}\le \| X\|_{\alpha,[s,t]}|t-s|^\alpha . $$
I fail to see how this is true.
NOTE: I understand how to derive the second inequality for $[s,t]$ rather than $[0,1]$ from "first principles"; however, I struggle to see how it is an easy consequence of the other inequality through a "scaling property".
Let $X$ be defined on $[0,1]$ by $X=X'\circ\varphi$ with $X'$ defined on $[s,t]$ and $\varphi:[0,1]\to[s,t],\;x\mapsto s+(t-s)x$. Then, $\varphi$ is increasing and onto, hence $$\begin{align}\|X'\|_{\alpha,[s,t]}&=\sup_{s\le u<v\le t}\frac{|X'_v-X'_u|}{|v-u|^\alpha}\\&=\sup_{0\le x<y\le 1}\frac{|X'_{\varphi(y)}-X'_{\varphi(x)}|}{|\varphi(y)-\varphi(x)|^\alpha}\\&=\sup_{0\le x<y\le 1}\frac{|X_y-X_x|}{|(t-s)(y-x)|^\alpha}\\&=\frac1{|t-s|^\alpha}\|X\|_{\alpha,[0,1]}.\end{align}$$