Some preliminaries: let $V$ be a topological vector space (think $V=\Bbb R^N$), $k\ge1$ integer and $T>0$.
- We define $\mathcal C_k([0,T];V)$ as the space of continous functions $f:[0,T]^k\to V$ such that $f_{t_1t_2\cdots t_k}=0$ whenever $t_i=t_{i+1}$, for some $1\le i\le k-1$.
- Let's define the operators $\delta_k$: $$ \delta_k:\mathcal C_k([0,T];V)\to\mathcal C_{k+1}([0,T];V)\\ \delta_k(f)_{t_1t_2\cdots t_{k+1}}:=\sum_{j=1}^{k+1}(-1)^{k-j}f_{t_1\cdots \widehat {t_j}\cdots t_{k+1}}\;\; $$
where the $\widehat {t_j}$ means that this argument is omitted.
- Fix $\mu>0$ and denote then by $\mathcal C_1^{\mu}([0,T];V)$ the space of $f\in\mathcal C_1([0,T];V)$ such that $$ ||f||_{\mu,[0,T]} :=\sup_{0\le r<s\le T}\frac{|(\delta_1f)_{rs}|}{|r-s|^{\mu}} =\sup_{0\le r<s\le T}\frac{|f(r)-f(s)|}{|r-s|^{\mu}} <+\infty. $$
Thus $\mathcal C_1^{\mu}([0,T];V)$ is the space of $\mu$-Hölder $V$-valued continous functions on $[0,T]$.
- Ok, taken $0\le s<t\le T$ we can define the
Young integral of $f\in\mathcal C_1^{\kappa}([0,T];\Bbb R^{n\times d})$ with respect to $g\in\mathcal C_1^{\mu}([0,T];\Bbb R^{d})$ from $s$ to $t$: \begin{align*} \mathcal I_{st}(f\,dg):=:\int_s^tf_u\,dg_u:&= f_s(\delta_1g)_{st}+\int_s^t(\delta_1f)_{su}\,dg_u\\ &=f_s(\delta_1g)_{st}+\mathcal I_{st}((\delta_1f)_{s\bullet}\,dg_{\bullet}). \end{align*}
Let's state the problem: how can I prove that
$$ \delta_2(\mathcal I_{ab}((\delta_1f)_{a\bullet}\,dg_{\bullet}))_{sut}=(\delta_1 f)_{su}(\delta_1 g)_{ut} $$
holds for every $s,u,t\in[0,T]$ ?
Here I've denoted with $a,b$ the mute variables and with the $\bullet$ the variable the integration will be performed with respect to.
It is important to prove the identity using only the given definitions.
My try was the following: \begin{align*} \delta_2(\mathcal I_{ab}((\delta_1f)_{a\bullet}\,dg_{\bullet}))_{sut} &=-\mathcal I_{ut}((\delta_1f)_{u\bullet}) +\mathcal I_{st}((\delta_1f)_{s\bullet}) -\mathcal I_{su}((\delta_1f)_{s\bullet})\\ &=-\int_u^t(f_r-f_u)\,dg_r+\int_s^t(f_r-f_s)\,dg_r-\int_s^u(f_r-f_s)\,dg_r \end{align*}
but here I'm stuck; I would able to conclude if only I could use usual integration rules (two lines more and we would have done), but I can't go on using ONLY the definitions written above.
The rules I was referring to are the following: \begin{align*} \mathcal I_{su}(f\,dg)+\mathcal I_{ut}(f\,dg)&=\mathcal I_{st}(f\,dg)\\ \mathcal I_{st}((f+h)\,dg)&=\mathcal I_{st}(f\,dg)+\mathcal I_{st}(h\,dg)\\ \mathcal I_{st}(\alpha\,dg)=\alpha\mathcal I_{st}(dg)&= \alpha(g_t-g_s)=\alpha(\delta_1g)_{st}\;,\;\;\;\alpha\in\Bbb R^{n\times d}\;\mbox{is a constant} \end{align*} which are well known in the Riemann, Lebesgue and Stiltjes integrals; in this context I'm not able to prove them directly.