Absolute-value Rademacher contraction

Enlarge the evaluation-vector class by adjoining the zero vector. Since ${\mathrm{\Phi }}_i(0)=0$, for each sign vector both $A:={\sup }_{f\in \mathcal{F}\cup \{0\}}{\sum }_i{\sigma }_i{\mathrm{\Phi }}_i(f(z_i))$ and $B:={\sup }_{f\in \mathcal{F}\cup \{0\}}-{\sum }_i{\sigma }_i{\mathrm{\Phi }}_i(f(z_i))$ are nonnegative. The supremum of the absolute value is $\max (A,B)\le A+B$.

The map $\sigma \mapsto -\sigma$ preserves the uniform sign law, so ${\mathbb{E}}_{\sigma }A={\mathbb{E}}_{\sigma }B$. Therefore the left side is at most $2{\mathbb{E}}_{\sigma }A$. Applying [F1] to the enlarged class gives ${\mathbb{E}}_{\sigma }A\le L{\mathbb{E}}_{\sigma }{\sup }_{f\in \mathcal{F}\cup \{0\}}{\sum }_i{\sigma }_{i}f(z_i)\le L{\mathbb{E}}_{\sigma }{\sup }_{f\in \mathcal{F}}\mathrm{\mid }{\sum }_i{\sigma }_{i}f(z_i)\mathrm{\mid }$. Combining proves the claimed factor $2L$.

For a countable class, enumerate its members and apply the finite proof to increasing finite initial subclasses. For each sign vector, the finite suprema increase to the corresponding countable supremum; averaging over the finite sign cube passes the inequalities to the limit.