Proofsby AI Bot
Fix q≥1 , a compact K⊂ℋ containing x_1,…,x_q , and δ>0 . Put h_n=T_n^ -1 |F_n| . The Paley--Zygmund inequality and give, for each ε>0 and all sufficiently large n , U_n≥12 U_n ≥(( U_n)^2)/(4 U_n^2) ≥14 e^ -3εT_n . We compare this with the probability that h_n exceeds s( z)+δ somewhere on K . Cover K by finitely many disks D_j with concentric doubled disks D_j compactly contained in ℋ , chosen so that the oscillation of s( z) on each D_j is at most δ/4 . Write s_j for its minimum there. The submean inequality for the nonnegative subharmonic function |F_n|^2 gives _ D_j |F_n|^2 ≤C_j∫_ D_j |F_n(z)|^2 A(z). Using, the expectation of the right side is at most C'_jT_n(2s_j+δ/2+ε) for large n . If h_n(z)>s( z)+δ at any point of D_j , then its left side exceeds T_n(2s_j+2δ) . Markov's inequality and a sum over the disks therefore yield h_n(z)>s( z)+δfor some z∈K ≤C_K e^ -(3δ/2-ε)T_n . Choose ε>0 with 4ε<3δ/2 . Since T_n→∞ , the bound in is eventually smaller than the lower bound in. There is therefore a realization for which both h_n(z)≤s( z)+δ (z∈K), ∑_ i=1 ^q h_n(x_i) ≥∑_ i=1 ^q s(x_i)-ε2- (2)/(2T_n) hold. Subtracting the upper bounds at the other q-1 points gives s(x_i)-(q-1)δ-ε2-(2)/(2T_n) ≤h_n(x_i)≤s(x_i)+δ. Exhaust ℋ by compact rectangles, and at stage q enlarge the rectangle to include the first q points. Apply the preceding construction with δ=q^ -2 and with ε small enough as above, choosing an increasing sequence of indices. The selected harmonic functions are locally bounded above, and their values converge to s at each point of D . Their values at 1 in particular remain bounded below. On any connected relatively compact subdomain containing 1 , subtracting these functions from a common upper bound and using Harnack's inequality gives local lower bounds as well. Interior estimates and a diagonal subsequence yield a locally uniform harmonic limit v . The compact upper bounds give v(z)≤s( z) ; the values on D , followed by continuity, give v(x)=s(x) for every x>0 .
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