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By John Garnett

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Subharmonic Functions Let be an open set in the plane. 1) v(z 0 ) ≤ 1 πr 2 |z−z 0 |

5. Let v(z) be a subharmonic function in the unit disc D. Assume v(z) ≡ −∞. For 0 < r < 1, let ⎧ |z| ≤ r, ⎨v(z), vr (z) = 1 ⎩ Pz/r (θ)v(r eiθ ) dθ, |z| < r. 2π Then vr (z) is a subharmonic function in D, vr (z) is harmonic on |z| < r, vr (z) ≥ v(z), z ∈ D, and vr (z) is an increasing function of r. Proof. 4 and by Section 3 we know vr (z) is finite and harmonic on (0, r ) = {|z| < r }. To see that vr (z) is upper semicontinuous at a point z 0 ∈ ∂ (0, r ) we must show v(z 0 ) ≥ lim vr (z). z→z 0 |z|

3) yn < ∞, 1 + |z n |2 z n = xn + i yn , and the Blaschke product with zeros {z n } is B(z) = z−i z+i m z n =i |z n2 + 1| z − z n . 3) becomes yn < ∞, and the convergence factors are not needed because ((z − z n )/(z − z¯ n )) already converges. 3 (F. Riesz). Let 0 < p < ∞. Let f (z) ∈ H p (D), f ≡ 0, let {z n } be the zeros of f (z), and let B(z) be the Blaschke product with zeros {z n }. Then g(z) = f (z)/B(z) is in H p (D) and g Hp = f Hp. Proof. It was noted above that B(z) converges when f ∈ H p .

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