Independent product inputs have an orthogonal Hoeffding decomposition

Interchange the two finite sums in ${\sum }_{u\subseteq [d]}{f}_u$. The coefficient of $P_{v}f$ is ${\sum }_{u\supseteq v}(-1{)}^{\mathrm{\mid }u\mathrm{\mid }-\mathrm{\mid }v\mathrm{\mid }}=(1-1{)}^{d-\mathrm{\mid }v\mathrm{\mid }}$, which is zero unless $v=[d]$, when it is one. Since $P_{[d]}f=f$, this proves reconstruction.

Fix nonempty $u$ and $i\in u$. Apply $P_{u\setminus \{i\}}$ to the finite sum defining $f_u$. For each $w\subseteq u\setminus \{i\}$, exactly the terms $v=w$ and $v=w\cup \{i\}$ both become $P_{w}f$ by [F2]; their signs are opposite. All terms cancel, so $P_{u\setminus \{i\}}{f}_u=0$. Also $P_u{f}_u=f_u$, since every $v\subseteq u$ satisfies $P_u{P}_v=P_v$. Thus $f_u$ depends only on $u$, and for nonempty $u$ its mean $P_{\mathrm{\varnothing }}{f}_u$ is zero by applying [F2] to the preceding identity.

Let $u\mathrm{\ne }w$. Choose $i$ in their symmetric difference; after interchanging their names if needed, take $i\in u\setminus w$, and put $Q=P_{[d]\setminus \{i\}}$. Since $P_u{f}_u=f_u$, the projection composition rule [F2] and step 2.1 give $Q{f}_u=Q{P}_u{f}_u=P_{u\setminus \{i\}}{f}_u=0$. Since $P_w{f}_w=f_w$ and $i\mathrm{\notin }w$, the same rule gives $Q{f}_w=Q{P}_w{f}_w=P_w{f}_w=f_w$. Self-adjointness of $Q$ therefore gives $⟨f_u,f_w⟩=⟨f_u,Q{f}_w⟩=⟨Q{f}_u,f_w⟩=0$.

By step 1.1, $f-\mathbb{E}f={\sum }_{u\mathrm{\ne }\mathrm{\varnothing }}{f}_u$. Step 3.1 makes this a finite orthogonal sum, so its squared $L^2$ norm is the sum of the squared component norms. Apply [F4] to identify that norm with $\mathrm{Var}(f(X))$.