F_{τ} (x) = \sum_{θ \in Θ (R) θ \subset τ} f_{θ} (x) w_{r^{2/3}} (x) .

Freeze

x_{3}

. In the

(x_{1}, x_{2})

Fourier variables, the functions

F_{τ} (\cdot, \cdot, x_{3})

are supported in caps of tangential length

r^{- 1/3}

and normal width

r^{- 2/3}

along the parabola. Apply part a with decoupling scale

R^{'} = r^{2/3}

and critical exponent

p = 6

. Since

(R^{'})^{1/2 - 3/6} = 1

, for each fixed

x_{3}

∥ \sum_{τ} F_{τ} (\cdot, \cdot, x_{3}) ∥_{L^{6} (R^{2})}^{6} ≲_{ϵ} r^{ϵ} (\sum_{τ} ∥ F_{τ} (\cdot, \cdot, x_{3}) ∥_{L^{6} (R^{2})}^{2})^{3} .

Integrate in

x_{3}

. Minkowski's inequality in

L^{3} (d x_{3})

gives

\int_{R^{3}} ∣ \sum_{θ} f_{θ} w_{r^{2/3}} ∣^{6} ≲_{ϵ} r^{ϵ} (\sum_{τ \in Θ (r)} (\int_{R^{3}} ∣ F_{τ} ∣^{6})^{2/6})^{6/2},

which is exactly the claimed inequality after absorbing a change in

ϵ

Solved by gpt-5.6-sol high.

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