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Too, of K 2 , K 3 , K 4 . Altogether, we get course. 23) which is much better than required. There remains to estimate I2 (which is the only place where we have to increase T to 2T in the estimate). Even if we drop all ε restrictions, Ptε (x) is evidently finite for x = 0, and has L1 -norm Pt0 1 = c/t. Despite the fact that this is divergent for t → 0, it is nevertheless true that for t ∼ 0, is Ptε (x) is essentially concentrated close to 0, and therefore there is not much difference between ps ∗ Ptε ∗ gu and (ps ∗ gu ) × Ptε 1 .

We give some details for the first part where the contribution coming ˜ cancels with κ . We then give a sketch how the rest is done the from R 1 cancellations with κ2 are done. We use i as a generic function (0, ∞) → (0, ∞) which is integrable near 0, not necessarily the same at different occurrences. 12. 1−ε dsE(e−R0,1 δ(ωs − ωs+ε )Ψ ) + κ1 (ε) (ε) ≥ −i(ε). g. ) Proof. We set δ(ωu − ωv )dudv. Ys = Ysε = u≤s≤v≤1−ε v−u≤ε Then E(e−R0,1 δ(ωs − ωs+ε )Ψ ) = E(e−R0,1 δ(ωs − ωs+ε )Ψ0,s Ψs+ε,1 (1 + O(ε))).

However pε (0)εκ1 (ε) is not integrable, a fact with which we are pleased as it will cancel the contribution coming from Ys . As Ys ≥ 0, we get e−βYs ≥ 1 − βYs , and therefore 1−ε ds E(e−R0,1 δ(ωs − ωs+ε )Ψ ) ˜s 0 1−ε ≥ pε (0)(1 + βεκ1 ((ε)) 0 ds E(e−R0,1−ε Ψ ) − βE(Ys e−R0,1−ε Ψ ) − i(ε). 36 1 On the construction of the three-dimensional polymer measure It looks obvious that E(e−R0,1−ε Ψ ) = E(e−R0,1 Ψ ) + O(ε) = (ε) + O(ε), but I don’t know how to prove this. We would need something like a bound for d E(e−R0,v Ψ ).

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