# approximation of holomorphic functions on a halfplane.

Let $\mathbb {C} _ +$ denote the right halfplane and $A$ the algebra $$A = \{ f \in H^\infty({\mathbb C} _ +) \cap C(\overline{{\mathbb C} _ +}): \; |f(z)| \le M (1+|z|)^{-\epsilon} \text{ for some } \epsilon > 0, M > 0 \}$$ Assume $f \in A$ and $(f_n) \subset A$ an approximating sequence with respect to sup norm, i.e. $\sum_{\Re(z)>0} | f_n(z) - f(z) | \to 0$.

Is there a chance that the decay rates of the approximating functions $f_n$ are uniform, i.e. $\exists \epsilon, M>0 \forall n:\; |f_n(z)| \le M (1+|z|)^{-\epsilon}$?

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I seem unable to repair a Tex issue. In dispair I put the question again. Let $C_+$ denote the right halfplane and $A$ the algebra $$A = \{ f \in H^\infty(C_+) \cap C(\bar{C_+}): \; |f(z)| \le M (1+|z|)^{-\epsilon} \text{ for some } \epsilon>0, M>0\}$$ Assume $f \in A$ and $(f_n) \subset A$ an approximating sequence with respect to sup norm, i.e. $\sum_{\Re(z)>0} | f_n(z) - f(z) | \to 0$.
Is there a chance that the decay rates of the approximating functions $f_n$ are uniform, i.e. $\exists \epsilon, M>0 \forall n:\; |f_n(z)| \le M (1+|z|)^{-\epsilon}$?
@ Eric: To fix the TeX problem type $\epsilon$ instead of $\eps$. I also think that instead of {\mathbb C} you should write \mathbb{C} between dollar signs $\mathbb{C}$. In any case I had difficulties with TeX this morning. – Liviu Nicolaescu Mar 30 '12 at 13:39
No, and a counterexample is given by $f_n(z) = \frac 1 n (1+z)^{-1/n}$. Indeed, $f_n$ converges to $0$ in sup norm, and the best $\epsilon$ in the decay rate of $f_n$ is $1/n$, which goes to $0$.