Question

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} $?

Answer 1

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} $?

Answer 2

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 $0$.