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

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Then exp(| f ∗ |2 ) ∈ L1 (T). 2. | f ∗ (eiθ )| − 1) dθ = T = ∞ t=0 ∞ = t=0 ∞ ≤ 2 2tet dt dθ 0 2 {| f ∗ |>t} 2tet dθ dt 2tet |{| f ∗ | > t}| dt 2 2Atet e−Bt 2 2 / f 2 D dt t=0 < ∞. e. on T, whence e| f T ∗ (eiθ )|2 dθ ≤ e2 p ∗ 2 ∞ iθ 2 e2|g (e T )| dθ < ∞. 2 D < B/2. 3 Weak-type and strong-type inequalities 37 This result is close to being optimal. 3). 1, can be strengthened a little further. 2. Let f ∈ D. 6) where A is an absolute constant. The strategy of the proof is the same as for the weak-type inequality.

R−1 If we put all these inequalities together, then we obtain 1 π A 3 dA(w) ≤ 10 + 2 log . |w − ζ1 ||w − ζ2 | |ζ2 − ζ1 | This gives the required inequality with B = 3e5 . 6. At the end of the proof we shall make the necessary adjustments to obtain the standard logarithmic capacity c. Let t > 0 and let F be a compact subset of {ζ ∈ T : Cg(ζ) > t}. Let μ be a Borel probability measure on F. Then, clearly, Cg(ζ) dμ(ζ) ≥ t. 6, we have Cg(ζ) dμ(ζ) = F 1 π A F ≤ g L2 (A) = g L2 (A) ≤ g L2 (A) |g(w)| dμ(ζ) dA(w) |w − ζ| 2 1/2 1 1 dμ(ζ) dA(w) π A F |w − ζ| 1 dA(w) dμ(ζ1 ) dμ(ζ2 ) π |w − ζ1 ||w − ζ2 | F F A 1/2 B dμ(ζ1 ) dμ(ζ2 ) .

1 Let F be a compact subset of X such that cK (F) > 0. An equilibrium measure for F is a measure ν ∈ P(F) such that IK (μ) ≥ IK (ν) for all μ ∈ P(F). Clearly, if ν is an equilibrium measure for F, then cK (F) = 1/IK (ν). 2 Let F be a compact subset of X with cK (F) > 0. Then F has an equilibrium measure. Proof For each n ≥ 1, choose μn ∈ P(F) such that IK (μn ) < 1/cK (F) + 1/n. A subsequence of the μn converges weak* to some ν ∈ P(F). 7, this ν must satisfy IK (ν) ≤ 1/cK (F), so it is an equilibrium measure for F.