U^{3}

norm dominates the absolute mean of a function:

∣ E f ∣ \leq ∥ f ∥_{U^{3}} .

Taking

f = 1_{A}

and using

E 1_{A} = α

gives

∥ 1_{A} ∥_{U^{3}} \geq α .

Solved by gpt-5.6-sol high.

ii

Solution

For

h (x) = e_{p} (q (x))

, the product in the

U^{3}

e_{p} (\sum_{ϵ \in {0, 1}^{3}} (- 1)^{∣ ϵ ∣} q (x + ϵ \cdot h)) .

The expression in parentheses is the third additive derivative of the quadratic form

q

, so it vanishes. Every cube contributes one and therefore the quadratic phase satisfies

∥ h ∥_{U^{3}} = 1.

Solved by gpt-5.6-sol high.

Apply the Cauchy-Schwarz inequality successively in the three shift variables to the correlation with the quadratic phase

h

. At the third step the third additive derivative of its phase vanishes, leaving

∣ ⟨ f, h ⟩ ∣^{8} \leq ∥ f ∥_{U^{3}}^{8} .

Thus

∣ ⟨ f, h ⟩ ∣ \geq δ

implies

∥ f ∥_{U^{3}} \geq δ .

Solved by gpt-5.6-sol high.

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