Interpolation sequences and open mapping theorem I'm using Garnett's "Bounded Analytic Functions" as a course text and looking at interpolation sequences. $z_n$ is a sequence of interpolation if for each sequence $a_n \in l^{\infty}$ there exists $f \in \mathbb{H}^{\infty}$ such that $f(z_n) = a_n$ for all n. 
We want to show $f$ has norm control so Garnett considers the operator $T: \mathbb{H}^{\infty} \to l^{\infty}$ defined by $T(f) = (f(z_n))$ 
Garnett then uses the Open Mapping Theorem to show $f$ has norm control, i.e.
\begin{align}
\lVert f \rVert_{\mathbb{H^\infty}} \le C\lVert a_n \rVert_{l^{\infty}}. 
\end{align}
My question is that I can't see why the Open Mapping Theorem can be used. I know we need to show the operator is bounded, linear and bijective. By Open Mapping Theorem it then has a  bounded inverse and we are done. That $T$ is bounded and linear is trivial, and surjectivity comes from $z_n$ being an interpolation sequence. I've no idea how to show injectivity though. 
I have $T(f) = T(g) \implies (f(z_n)) = (g(z_n)) \implies f(z_n) = g(z_n) \forall n$ but this still isn't injectivity. Is there a way I can go from here or is there another was to show $T$ is injective? Other books also just say "by Open Mapping Theorem" without showing the conditions hold. 
For reference the bounded inverse theorem (a consequence of open mapping theorem) says that if $X$ and $Y$ are Banach spaces and $T: X \to Y$ bounded, bijective, linear operator then $T^{-1}: Y \to X$ is also bounded. 
 A: Following on from the comments we do not needed to use the bounded inverse theorem as I had originally thought. 
As $T$ is continuous and surjective by the open mapping theorem $T$ is an open map. This means that for some $r >0$ we have $r (B_{l_\infty}) \subset T(B_{\mathbb{H}^{\infty}})$ where the $B$ are the open unit balls in the respective spaces. 
To show that we have the norm control suppose $a \in l_{\infty}$ is such that $||a||_{l^{\infty}} = r/2$. By the inclusion above we can find $f \in \mathbb{H}^{\infty}$ such that $T(f) = a$ and $||f||_{\mathbb{H}^{\infty}} < 1$. Putting $c=2/r$ we have that 
\begin{equation*}
||a|| = \frac{r}{2} = \frac{2/c}{2} = \frac{1}{c}
\end{equation*}
and so $||f|| < 1 = c||a||$ as required. 
Now by a ususal scaling argument this holds for any $a \in l_\infty$ with arbitrary norm. In full, define $\hat{a} = \frac{r/2}{||a||} a \in l_\infty$. Now $||\hat{a}|| = r/2$ so there is $\hat{f} \in B_{\mathbb{H}^{\infty}}$ such that $T(\hat{f}) = \hat{a}$ and $||\hat{f}|| \leq c||\hat{a}||$ by the above. 
Define $f = \frac{||a||}{r/2} \hat{f} \in \mathbb{H}^{\infty}$. Then 
\begin{equation*}
T(f) = \frac{||a||}{r/2}T(\hat{f}) = \frac{||a||}{r/2} \hat{a} = a
\end{equation*}
and
\begin{equation*}
||f|| = ||\hat{f}||\frac{||a||}{r/2} \leq  c||\hat{a}|| \frac{||a||}{r/2} = c||a||
\end{equation*}
and we are done. 
