Proving that the Moment Tensor is super-symmetric The Carathéodory theorem in the image bellow is the one about convex hull, isn't it? Would you please explain why can the tensor F be rewritten as that sum?
From that representation the author concludes that F is super-symmetric, but then I could the same argument to prove that everything is super-symmetric.
(Image taken from Jiang et al, 2013. Moments Tensors, Hilbert’s Identity, and k-wise Uncorrelated Random Variables)
Thank you.

 A: The decomposition in the OP is non trivial. Let us begin by fixing notation.
Let $u=(u_1,\dots,u_n)\in\mathbb R^n$. We introduce the notation $V:=\mathbb R^n$. All computations are done w.r.t. to a given basis $\{e_i\}$ of $V$, i.e. $u=u_ie_i.$
Then
$$\mathcal F_{i_1\dots i_d}=\int_\mathbb {R^n}\prod_{k=1}^d u_{i_k} p(u)du, $$
and
$$\mathcal F= \mathcal F_{i_1\dots i_d}e_{i_1}\otimes \dots\otimes e_{i_d}\in V^{\otimes d}, $$
or
$$\mathcal F= \int_\mathbb {R^n}(u\otimes\dots\otimes u)~ p(u)du\in V^{\otimes d}. $$
The non trivial observation here is that $\mathcal F$ is a symmetric tensor of order $k$ (on the finite dim. space $V$). This follows from definition of $\mathcal F$. We can write $\mathcal F\in \operatorname{Sym}^d(V)\subset V^{\otimes d}$.
The decomposition of $\mathcal F$ in (1) in the OP is called symmetric outer product decompsition (SOPD) of the given symmetric tensor. It generalizes the concept of diagonalization of symmetric matrices. 
The number $r$ there appearing (i.e. the smallest integer s.t. the  SOPD exists) is called the symmetric rank of the given symmetric tensor. The existence of SOPD's and the algorithms to find $r$ are non trivial (I will not write any proof here!).
I personally do not know any Caratheordory theorem on symmetric tensors decompositions.
For more information on the topic, please refer to here and here.
