In "Differential equations driven by rough paths" (Terry Lyons, et al) section 1.4.2 it's claimed that the symmetric part of the tensor:

$\int_{0 \le u_1 \le \cdots \le u_j \le t} \mathrm{d}X_{u_1} \otimes \cdots \otimes \mathrm{d}X_{u_j}$

is exactly $\frac{1}{j!}(X_t - X_0)^{\otimes j}$. It is assumed that $X:[0,T] \to V$ is a Lipschitz path in some finite dimensional vector space $V$.

Is there a simple way to establish this fact? Also I would be grateful for any reference on this type of integrals involving tensors since I've never encountered them before.

**Further background**

Integration with respect to $X$ is defined for any continuous $f: [0,T] \to L(V,W)$ ($W$ a finite dimensional vector space and $L(V,W)$ the space of linear maps from $V$ to $W$) as:

$\int f_s \mathrm{d}X_s = \lim \sum_{i = 0}^{r -1} f_{t_i}(X_{t_{i+1}} - X_{t_i})$

where the limit is taken over partitions $t_0 = 0 \le t_1 \le \cdots \le t_r = T$ whose largest interval is decreasing to $0$ in length. The result belongs to the space $W$.

Hence, to make the above iterated integral fit into the definition (and give a result in $V^{\otimes j}$) it seems to me that one must consider each element $w_1 \otimes \cdots \otimes w_i \in V^{\otimes i}$ as an element of $L(V, V^{\otimes i+1})$ via the map $v \mapsto w_1 \otimes \cdots \otimes w_i \otimes v$. However after doing this I still have trouble establishing the identity.

As an example, with the above interpretation I get:

$\int_{0 \le s \le t \le T} \mathrm{d}X_s \otimes \mathrm{d}X_t = \frac{1}{2}T^2 v_1\otimes v_1 + \frac{2}{3}T^3 v_1\otimes v_2 + \frac{1}{3}T^3 v_2 \otimes v_1 + \frac{1}{2}T^4 v_2 \otimes v_2$

for $X_t = t v_1 + t^2 v_2$ in a $2$-dimensional vector space with basis $(v_1,v_2)$. This example shows that the result of the integral itself may be non-symmetric.