By Jarnicki M., Pflug P.

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**Example text**

There are two cases: 2 2 (a) ϕ(b) = 0: Then dmin P (p, z0 ) = |ϕ(z0 )| ≤ [mP ({a, b}, z0 )] < dP (p, z0 ). min 2 (b) ϕ(b) = 0: Then dP (p, z0 ) = |ϕ(z0 )| mE (ϕ(b), ϕ(z0 )) ≤ [mP (a, z0 )]2 mP (b, z0 ) = dP (p, z0 ). The equality dmin P (p, z0 ) = dP (p, z0 ) would imply that ϕ is simultaneously extremal for mP (a, z0 ) and mP (b, z0 ). 6), we know that such extremal functions are uniquely determined up to rotations. Hence f (1/a, −e−iϕ0 z) f (1/b, −e−iϕ0 z) ϕ(z) = eiθa f (a, z) = eiθb f (b, z); Rz Rz contradiction (both sides have different zeros).

GE (A, z) ≥ RE (A, z). We may assume that z1 · · · zd = 0. First consider the case d = k = n. Put f (ζ) := qnqn 1 n 2pj ,ζ∈ j=1 ζj (2pj ) 2pj qn n j=1 |ζj | < 1, qd E. ζ ∈ E 31 . Observe that |f (z)| = RE (A, z) and |f (ζ)| ≤ qnqn Thus gE (A, z) ≥ mE (A, z) ≥ RE (A, z). Now assume that d < n. ,pn ) . Observe that we only need to find a logarithmically plurisubharmonic function v : E −→ [0, 1), v ≡ 0, such that • v(ζ ) ≤ |ζj |, ζ = (ζd+1 , . . , ζn ) ∈ E , j = d + 1, . . , k 32 , • the mapping E ζ −→ v(ζ )rdqd (ζ ) ∈ R+ attains its maximum for ζ = (zd+1 , .

For k > 1 we first apply the case k = 1 and get u(ζ1 , . . , ζn ) ≤ |ζ1 |, ζ ∈ E n . Next we apply the inductive assumption to the functions u(ζ1 , ·)/|ζ1 |, ζ1 ∈ E∗ . Step 2. Consider the mapping Ed ι z (ζ1 , . . , ζd ) −→ ζ1 cd (z) 2p1 1 2p1 , . . , ζd cd (z) 2pd 1 2pd , zd+1 , . . , zn ∈ E. Using the holomorphic contractivity and Step 1, we get mE (A, z) ≤ gE (A, z) ≤ RE (A, z). It remains to prove that gE (A, z) ≥ RE (A, z). 29 30 In particular, if p1 = · · · = pk , then the condition simply means that |z1 | ≤ · · · ≤ |zk |.