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D. Brydges, A. Ta<strong>la</strong>rczyk / Journal of Functional Analysis 236 (2006) 682–711 693<br />

The resulting integrand is jointly mea<strong>sur</strong>ab<strong>le</strong> and non-negative so integrals and sums can be put<br />

in any or<strong>de</strong>r. Therefore both terms give the same contribution and we bound by<br />

<br />

<br />

dx dy1Uy∩Δ=∅(Fx, 1ΔFx) L2. Δ<br />

Let |Uɛ|= dy1Uy∩Δ=∅. For the future, note that |Uɛ| tends as ɛ → 0tothevolume|U| of U.<br />

Then the preceding expression equals<br />

|Uɛ| <br />

<br />

<br />

dx(Fx, 1ΔFx) L2 =|Uɛ| dx(Fx,Fx) L2 and the <strong>le</strong>mma follows by taking ɛ → 0. ✷<br />

Lemma 3.5.<br />

Δ<br />

<br />

dx (pUx ϕ,ψ)+<br />

<br />

|U|ϕ+ψ+.<br />

Proof. It is sufficient to prove case ψ = ϕ because<br />

<br />

dx (pUx ϕ,ψ)+<br />

<br />

<br />

= dx <br />

<br />

(pUx ϕ,pUx ψ)+ dxpUx ϕ+pUx ψ+<br />

<br />

<br />

<br />

pUx ϕ2 + dx<br />

1/2 <br />

pUy ψ2 + dy<br />

1/2 .<br />

Now we consi<strong>de</strong>r case ϕ = ψ for which there are no absolute values because (pUx ϕ,ϕ)+ 0.<br />

By the Cauchy–Schwartz inequality,<br />

<br />

<br />

<br />

<br />

<br />

λi(pUx ϕ,ϕ)+ = λipUx ϕ,ϕ λipUx ϕ,<br />

i i i<br />

i<br />

i<br />

+ i<br />

<br />

1/2 λj pUx ϕ ϕ+<br />

j<br />

j<br />

+<br />

<br />

<br />

1/2 = ¯λiλj (pUx ϕ,pUx ϕ)+ ϕ+.<br />

i j<br />

By consi<strong>de</strong>ring Riemann sums, this implies<br />

<br />

dx(pUxϕ,ϕ)+ <br />

<br />

<br />

dx<br />

By Lemma 3.4,<br />

<br />

dx<br />

i,j<br />

dy(pUxϕ,pUy ϕ)+<br />

<br />

=<br />

<br />

dx<br />

<br />

=|U|<br />

1/2 dy(pUxϕ,pUy ϕ)+ ϕ+. (3.3)<br />

<br />

dy(Fx,Fy) L2 |U|<br />

dx(pUxϕ,pUx ϕ)+<br />

<br />

=|U|<br />

Putting this into (3.3), we obtain the <strong>le</strong>mma for the case ψ = ϕ. ✷<br />

dx(Fx,Fx) L 2<br />

dx(pUx ϕ,ϕ)+.

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