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By El-Fallah O., Kellay K., Mashreghi J., Ransford T.

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Next, here is a simple upper bound for capacity in terms of diameter. 4 If F is a compact subset of X, then cK (F) ≤ 1/K(diam(F)). 1 Potentials, energy and capacity Proof 17 If μ ∈ P(F), then IK (μ) = ≥ K(d(x, y)) dμ(x) dμ(y) K(diam(F)) dμ(x) dμ(y) = K(diam(F)). It follows that cK (F) ≤ 1/K(diam(F)). 5 For every compact set F we have cK (F) < ∞. Proof By assumption K 0, so there exists d0 > 0 such that K(d0 ) > 0. A compact set F can be covered by finitely many compact sets F1 , . . , Fn of diameter at most d0 , so cK (F) ≤ cK (F1 ) + · · · + cK (Fn ) ≤ n/K(d0 ) < ∞.

The second result that we shall need is a formula for the energy of a measure μ in terms of its Fourier coefficients μ(k), where μ(k) := T e−ikt dμ(eit ) (k ∈ Z). 4 Let X = T with the chordal metric d(z, w) := |z − w| and let K(t) := log+ (2/t). If μ is a finite positive Borel measure on T, then IK (μ) = k≥1 Proof |μ(k)|2 + μ(T)2 log 2. k We need to prove that log |eit 1 dμ(eis ) dμ(eit ) = − eis | k≥1 |μ(k)|2 . 4) Let 0 < r < 1. Then log |eit 1 dμ(eis ) dμ(eit ) = Re − reis | − log(1 − rei(s−t) ) dμ(eis ) dμ(eit ) = Re k≥1 = Re k≥1 k = k≥1 rk eik(s−t) dμ(eis ) dμ(eit ) k rk μ(k)μ(k) k r |μ(k)|2 .

And as this holds for each compact F in {Cg > t}, we get cK (Cg > t) ≤ 2 g 2 2 L2 (A) /t . It just remains to relate cK to c. For this, note that the difference between their respective kernels is log B − log 2 = log(B/2), and so 1/cK (F) − 1/c(F) = log(B/2) for all compact sets F. Hence c(F)/cK (F) = 1 + log(B/2)c(F) ≤ 1 + log(B/2)c(T). 7 2 /t2 , L2 (A) where A := 2(1 + log(B/2)c(T)). If g ∈ L2 (A), then Cg < ∞ quasi-everywhere on T. 3 Weak-type and strong-type inequalities 35 Proof Since Cg is lower semicontinuous, the set {ζ ∈ T : Cg(ζ) > t} is open in T for each t > 0.

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