By Luṭf Allāh Yār Muḥammadī

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Since OL < QJ < P, each hj is continuous. But if a sequence of continuous functions converges monotonically to a continuous limit, on a compact set, then the convergence is uniform. Thus /i/z) -> a uniformly on K. Now the subharmonicity of Vj, combined with the obvious inequality Vj < Qj, shows that oc < Vj < hj. Hence Vj -^ a uniformly on K, and since Uj < Vj, the lemma is proved. 6. Theorem. Suppose (a) Q is an open set in C", (b) for s = 0,1,2,... ,Fsis a homogeneous polynomial of degree s, and (c) sups I Fs(z) I < 00 for every zeQ.

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