We will follow the Lemma 8.10. (Rough Gronwall) and Proposition 8.12 from "a course in rough paths" but modify them for this particular setting of studying

$$Y_{t}=Y_{0}+\int^{t}_{0} Y dX,$$

for $Y\in C^{\alpha},X\in C^{\beta}$ where $\alpha+\beta>1$ but $\beta<\frac{1}{2}$ and $\alpha>\frac{1}{2}$.

The Lemma 8.10. (Rough Gronwall) comes verbatim with no changes

Assume $Y\in C([0,1])$, $r<1$ and $$|Y_{s,t}|\leq M ||Y||_{\infty;[s,t]}|t-s|^{r},(*)$$ for $0\leq s< t\leq 1$. Then there exists $c=c_{r}<\infty$ such that $$||Y||_{\infty;[0,1]}\leq cexp(cM^{1/r})|Y_{0}|.$$

So it suffices to prove (*). From a slight variation of the proof of Young's inequality we have the bound

$$|Y_{s,t}|=|\int_{s}^{t}Y_{r}dX_{r}|\leq c||Y||_{\infty;[s,t]}||X||_{\beta;[s,t]}|t-s|^{\beta}\leq c||Y||_{\infty;[s,t]}|t-s|^{\beta}(**)$$ and so the result follows from here. This (**)-estimate is a variation of prop.2.2 from "Stochastic differential equations driven by fractional Brownian motion". In particular, we use the representation

$$\int_{s}^{t} Y dX=Y_{s}(X_{t}-X_{s})+\sum_{k\geq 1}\sum_{i=0}^{2^{k-1}-1}(Y_{s_{2i+1}^{k}}-Y_{s_{2i}^{k}})(X_{s_{2i+1}^{k}}-X_{s_{2i}^{k}}),$$

for $s^n_i = s + (t − s) \frac{i}{2n}$.

We will follow the Lemma 8.10. (Rough Gronwall) and Proposition 8.12 from "a course in rough paths" but modify them for this particular setting of studying

$$Y_{t}=Y_{0}+\int^{t}_{0} Y dX,$$

for $Y\in C^{\alpha},X\in C^{\beta}$ where $\alpha+\beta>1$ but $\beta<\frac{1}{2}$ and $\alpha>\frac{1}{2}$.

The Lemma 8.10. (Rough Gronwall) comes verbatim with no changes

Assume $Y\in C([0,1])$, $r<1$ and $$|Y_{s,t}|\leq M ||Y||_{\infty;[s,t]}|t-s|^{r},(*)$$ for $0\leq s< t\leq 1$. Then there exists $c=c_{r}<\infty$ such that $$||Y||_{\infty;[0,1]}\leq cexp(cM^{1/r})|Y_{0}|.$$

So it suffices to prove (*). From a slight variation of the proof of Proposition 8.12 we have the bound

$$|Y_{s,t}|=|\int_{s}^{t}Y_{r}dX_{r}|\leq c||Y||_{\infty;[s,t]}||X||_{\beta;[s,t]}|t-s|^{\beta}\leq c||Y||_{\infty;[s,t]}|t-s|^{\beta},$$ where we used that $\beta<\frac{1}{2}\leq \alpha$ and so $||Y||_{\alpha;[s,t]}\leq ||Y||_{\beta;[s,t]}$. So the result follows from here.