Let $f\in\mathcal C^{\lambda}([0,T])$ and $g\in\mathcal C^{\gamma}([0,T])$ (i.e. they are Holder continous), where $\lambda+\gamma>1$.
Consider now the following estimate: $$ \left|\int_0^tf\,dg\right|\le\|f\|_{\infty}\|g\|_{\gamma}t^{\gamma}+C\|f\|_{\lambda}\|g\|_{\gamma}t^{\lambda+\gamma}\;. $$ The integral is the Young integral a generalization of the Riemann-Stiltjes integral, but don't mind about this; just consider the inequality.
Now, I saw a modification of this inequality: let $\beta\in[0,1[$ such that $\mu_{\beta}:=\gamma+\lambda(1-\beta)>1$; now $\lambda(1-\beta)\le\lambda$ thus $f\in\mathcal C^{\lambda(1-\beta)}$ and rewriting the previous inequality we have $$ \left|\int_0^tf\,dg\right|\le\|f\|_{\infty}\|g\|_{\gamma}t^{\gamma}+C\|f\|_{\lambda(1-\beta)}\|g\|_{\gamma}t^{\mu_{\beta}}\;, $$ and noting that \begin{align*} \|f\|_{\lambda(1-\beta)} &=\sup_{0\le s<t\le T}\frac{|f(t)-f(s)|}{|t-s|^{\lambda(1-\beta)}}\\ &=\sup_{0\le s<t\le T}\frac{|f(t)-f(s)|^{\beta}|f(t)-f(s)|^{1-\beta}}{|t-s|^{\lambda(1-\beta)}}\\ &\le2^{\beta}\|f\|_{\infty}^{\beta}\|f\|_{\lambda}^{1-\beta} \end{align*} the inequality reads as follows $$ \left|\int_0^tf\,dg\right|\le\|f\|_{\infty}\|g\|_{\gamma}t^{\gamma}+C2^{\beta}\|f\|_{\infty}^{\beta}\|f\|_{\lambda(1-\beta)}^{1-\beta}\|g\|_{\gamma}t^{\mu_{\beta}}\;. $$ The key point for me is that now, the $\sup$ norm of $f$ is taken into account in both summands, which near to what I want.
Is it possible to find a similar inequality but with $\|f\|_{\infty}$ at second summand raised exactly to $1?$
EDIT: The function $g$ will be a trajectory of the fractional Brownian Motion with Hurst parameter $H>1/2$; it is well known that these trajectories are $\alpha$-Holder continuous for every $\alpha<H$, but I don't know if they are also of bounded variation (I know that the trajectories of the standard BM has not finite bounded variation, but I don't know what about for the fBM with $H>1/2$). However this is the "only" restriction.