I apologize if this is obvious to be true or to be false.
Let $g\in BMOA(\mathbb{C}_0)$, where $\mathbb{C}_0$ denote the open right half plane. For $t\in \mathbb{R}$, define vertical translate $g_t(z)=g(z+it), z\in \mathbb{C}_0$. Suppose there is a sequence $\{t_k\}$ such that $g_{t_k}$ converges to $f$ on each compact subset of $\mathbb{C}_0$. Prove (or show by counter example) that $f\in BMOA$ and $\|f\|_{BMO}=\|g\|_{BMO}$.
I can prove $f\in BMOA$, by use the Carlson measure condition, namely: $g\in BMOA(\mathbb{C}_0)$ if and only if $$\mu_g=|g'(\sigma+it)|^2\sigma d\sigma\frac{dt}{1+t^2}$$ is a Carlson measure for $H^2_i(\mathbb{C}_0)$, i.e. there is a constant $C$, such that $$\int |h|^2|g'(\sigma+it)|^2\sigma d\sigma \frac{dt}{1+t^2}\leq C\|h\|_{H^2_i(\mathbb{C}_0)}^2.$$ Because $g_{t_k}$ converges to $f$ uniformly on compact sets, so do $g'_{t_k}$. Use Fatou's lemma, I can prove $f\in BMOA$. But how can we prove $\|f\|_{BMO}=\|g\|_{BMO}$?
This question arise when I read this paper, lemma 2.1(ii). In that paper, $g$ is taken to be a Dirichlet series, and $f$ is its vertical normal limit. But it seems this should hold for general $g$ and its normal limit.