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On the growth of universal functions. (English) Zbl 0973.30002
Let $$f=\sum_{k=0}^\infty a_kz^k$$ be holomorphic in the unit disc $$D$$. Put $$S_n(f):=\sum_{k=0}^na_kz^k$$. Assume $$f$$ is universal, i.e. for any compact set $$K\subset\mathbb C\setminus D$$ for which $$\mathbb C\setminus K$$ is connected, and for any function $$h\in\mathcal C(K)\cap\mathcal O(\text{int}K)$$, there exists a sequence $$(n_k)_{k=1}^\infty$$ such that $$S_{n_k}(f)\longrightarrow h$$ uniformly on $$K$$. The author proves that the growth of $$f$$ cannot be of the type $$|f(z)|\leq C\exp(\varPhi(\frac 1{1-|z|}))$$, where $$\varPhi:[1,+\infty)\longrightarrow[1,+\infty)$$ is a continuous increasing function with $$\int_1^\infty\log\varPhi(t)\frac{dt}{t^2}<+\infty$$. On the other hand, he constructs a universal function $$F$$ of the growth $$|F(z)|\leq C\exp(\exp(\frac{M}{1-|z|}\log\log\frac 4{1-|z|}))$$, where $$M$$ is a positive constant. The author also studies the value distribution of $$f$$. In particular, he shows that for every $$a\in\mathbb C$$ except at most one value, the equation $$f(z)=a$$ has an infinite number of distinct solutions $$(z_j^\ast(a))_{j=1}^\infty$$ such that $$\sum_{j=1}^\infty h(1-|z_j^\ast(a)|)=+\infty$$ for any increasing continuous function $$h:(0,1]\rightarrow(0,+\infty)$$ with $$\int_0^1\log\frac 1{h(s)}ds<+\infty$$.

##### MSC:
 30B10 Power series (including lacunary series) in one complex variable 30D35 Value distribution of meromorphic functions of one complex variable, Nevanlinna theory
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