How to prove that functional ANOVA components are orthogonal? Given a multivariate real function $f(x_1, x_2, \ldots, x_d)$, let's define:
$$
L_j f= E_{x_j \sim X_j}f(\cdot, \cdot, \ldots, x_j, \ldots, \cdot) \quad (j \in S = \{ 1,2,\ldots, d\}) \\
L_u f = \left(\prod_{j \in u} L_j \right)f \quad (u \subseteq S)
$$
The functional component of $f$ acting only on the variables whose indices are in $u \subseteq S$ is defined as:
$$
f_u = L_{S \setminus u}f- \sum_{v \subset u}f_v
$$
How can we prove that $\boldsymbol{f_u}$ and $\boldsymbol{f_v}$ are orthogonal for $\boldsymbol{u \neq v}$? This comes up in many papers on functional ANOVA, but I couldn't prove it.
I've managed to prove these results:
$$
(\forall u \subseteq S)(\forall j \in u) L_j f_u = 0 \\
(\forall u \subseteq S)(\forall j \not\in u) L_j f_u = f_u \\
(\forall u \subseteq S, u \neq \emptyset)L_u f_u = 0 \\
f = \sum_{u \subseteq S} f_u \\
$$
The orthogonality of $f_u$ and $f_v$ in the case $u \neq v$ should flow from there. I've got a feeling that it's pretty obvious, but I'm stuck. Any help?

I think I got it.
For $u \neq v$, suppose w.l.o.g. that $w = v \setminus u \neq  \emptyset$. Then, we have:
$$
\langle f_u, f_v \rangle = L_S f_uf_v = L_{(S \setminus w) \cup w} f_uf_v = L_{S \setminus w} L_w f_u f_v
$$
Since $w \cap v = \emptyset$, $f_v$ is a constant w.r.t. $L_w$, so:
$$
\langle f_u, f_v \rangle = L_{S\setminus w} (f_v (L_w f_u)) = L_{S \setminus w} (f_v0) = L_{S \setminus w} 0 = 0$$
