Firstly, let us recall that given a complex Banach space $X$ and $\mathbb{D}$ the open complex unit disc, a holomorphic mapping $f: \mathbb{D} \to X$ is said to be Bloch if $$ p(f)=\sup_{z \in \mathbb{D}} (1-|z|^2)\|f'(z)\| < \infty. $$ The space of all Bloch mappings from $\mathbb{D}$ to $X$ is denoted by $\mathcal{B}(\mathbb{D},X)$ and it is a Banach space endowed with the norm $p(\cdot)$.
In the same way, $f:\mathbb{D} \to X$ is said to be little Bloch if $$ \lim_{|z| \to 1^-} (1-|z|^2)\|f'(z)\| =0. $$ This space is denoted by $\mathcal{B}_0(\mathbb{D},X)$ and it is easy to show that $\mathcal{B}_0(\mathbb{D},X) \subsetneq \mathcal{B}(\mathbb{D},X)$. Due to little Bloch space is a closed subspace of Bloch space it follows that $\mathcal{B}_0(\mathbb{D},X)$ is a Banach space endowed with the norm $p(\cdot)$. On the other hand we denote by $\mathcal{H}^\infty (\mathbb{D},X)$ the space of all bounded holomorphic mappings $f: \mathbb{D} \to X$ endowed with the supremum norm $$ \|f\|_\infty = \sup_{z \in \mathbb{D}}\|f(z)\|. $$ $(\mathcal{H}^\infty (\mathbb{D},X),\|\cdot\|_\infty)$ is a Banach space and classical theory shows that $\mathcal{H}^\infty (\mathbb{D},X) \subsetneq \mathcal{B}(\mathbb{D},X)$ with $p(f) \leq \|f\|_\infty$.
Nevertheless, I don't know what is the relationship between $\mathcal{B}_0(\mathbb{D},X)$ and $\mathcal{H}^\infty (\mathbb{D},X)$. It seems to me that the intersection is a Banach algebra but I am not sure.
In any case, is every little Bloch mapping a bounded holomorphic mapping? If not, are there any classical counterexamples?
Is every bounded holomorphic mapping a little Bloch mapping? If not, are there any classical counterexamples?
Note that $\log (1-z)$ is the prototypical Bloch that is not bounded and has image the strip $-\pi/2 < \Im w < \pi/2$ so $g(z)=(\log (1-z)+ \pi i)^{1/2}$ is well defined and clearly is unbounded and is in little Bloch
(in the above we take the principal value of $\log$ with the argument of $1-z$ being in $(-\pi/2, \pi/2)$)
Conversely the prototypical singular inner function $f(z)=\exp \frac{z+1}{z-1}$ is bounded but not in little Bloch (also the little Bloch space is separable, while $H^{\infty}$ and the Bloch space are not)