plancherel]

2026-08-26

For all $𝑓, 𝑔 : {\pm 1}^{𝑛} \to ℝ$ ,

⟨ 𝑓, 𝑔 ⟩ = 𝐄 [𝑓 𝑔] = \sum_{𝑆 \subseteq [𝑛]} \hat{𝑓} (𝑆) \hat{𝑔} (𝑆) .

Note that taking $𝑔 = 𝑓$ recovers Parseval’s identity.

Proof. [#]

Expand both functions in the Fourier basis and apply orthonormality of the characters (see the expectation inner product):

⟨ 𝑓, 𝑔 ⟩ = ⟨ \sum_{𝑆} \hat{𝑓} (𝑆) 𝜒_{𝑆}, \sum_{𝑇} \hat{𝑔} (𝑇) 𝜒_{𝑇} ⟩ = \sum_{𝑆, 𝑇} \hat{𝑓} (𝑆) \hat{𝑔} (𝑇) ⟨ 𝜒_{𝑆}, 𝜒_{𝑇} ⟩ = \sum_{𝑆} \hat{𝑓} (𝑆) \hat{𝑔} (𝑆) .