I am reading Agler and McCarthy's Pick Interpolation and Hilbert Function spaces. In the Chapter 9, titled "Interpolating Sequences", the authors say that every interpolating sequence is strongly separated. Here's what it means to say a sequence is interpolating sequence for $H^{\infty}({\mathbb D})$ and for it to be strongly separated:

Definition (interpolating sequence). A sequence of points $\{ \lambda _i \}_{i=1}^{\infty}$ in the open unit disc $\mathbb D$ is called interpolating sequence for $H^{\infty}({\mathbb D})$ if for any bounded sequence $\{ w_i \}_{i=1}^{\infty}$ there is a function $\phi \in H^{\infty}({\mathbb D})$ such that $\phi (\lambda_i) = w_i$ for each $i$.

Definition (strong separated sequence). A sequence of points $\{ \lambda _i \}_{i=1}^{\infty}$ in the open unit disc $\mathbb D$ is called strongly separated sequence for $H^{\infty}({\mathbb D})$ if there is some constant $\epsilon > 0$ such that for each $i$ there is a function $\phi_i \in H_1^{\infty}({\mathbb D})$, the closed unit ball of $H^\infty (\mathbb D),$ such that $\phi_i (\lambda_i) = \epsilon$ and $\phi_i (\lambda_j) = 0$ for all $j \ne i$.

The authors say that it can be shown that every interpolating sequence is strongly separated by closed graph theorem but I do not exactly see how. Here's my attempt, nonetheless:

Let $\{ \lambda _i \}_{i=1}^{\infty}$ in the open unit disc $\mathbb D$ be an interpolating sequence for $H^{\infty}({\mathbb D})$. Define a map $T : H^\infty (\mathbb D ) \to \ell ^\infty$ by $T(\phi ) = \{ \varphi (\lambda _i) \}_{i=1}^{\infty}$. This map is surjective by definition of interpolating sequences and I managed to show that $T$ is bounded by the closed graph theorem.

Now we can take $\epsilon = \lVert T \rVert$ and consider the sequence $\{ \epsilon e_i \}_{i=1}^{\infty}$ in $\ell ^\infty$ . Since $T$ is surjective, there is some $\varphi _i \in H^{\infty} (\mathbb D)$ such that $T(\varphi_i) = \{ \epsilon e_i \}_{i=1}^{\infty}$. Now $\lVert T(\varphi _i) \rVert = \varepsilon \lVert e_i \rVert =\varepsilon$. But from here, I can only conclude that $\lVert \varphi_i \rVert \ge 1$ whereas I am looking for the reverse inequality.

I am reading Agler and McCarthy's Pick Interpolation and Hilbert Function spaces. In Chapter 9, titled "Interpolating Sequences", the authors say that every interpolating sequence is strongly separated. Here's what it means to say a sequence is interpolating sequence for $H^{\infty}({\mathbb D})$ and for it to be strongly separated:

Definition (interpolating sequence). A sequence of points $\{ \lambda _i \}_{i=1}^{\infty}$ in the open unit disc $\mathbb D$ is called interpolating sequence for $H^{\infty}({\mathbb D})$ if for any bounded sequence $\{ w_i \}_{i=1}^{\infty}$ there is a function $\phi \in H^{\infty}({\mathbb D})$ such that $\phi (\lambda_i) = w_i$ for each $i$.

Definition (strong separated sequence). A sequence of points $\{ \lambda _i \}_{i=1}^{\infty}$ in the open unit disc $\mathbb D$ is called strongly separated sequence for $H^{\infty}({\mathbb D})$ if there is some constant $\epsilon > 0$ such that for each $i$ there is a function $\phi_i \in H_1^{\infty}({\mathbb D})$, the closed unit ball of $H^\infty (\mathbb D),$ such that $\phi_i (\lambda_i) = \epsilon$ and $\phi_i (\lambda_j) = 0$ for all $j \ne i$.

The authors say that it can be shown that every interpolating sequence is strongly separated by closed graph theorem but I do not exactly see how. Here's my attempt, nonetheless:

Let $\{ \lambda _i \}_{i=1}^{\infty}$ in the open unit disc $\mathbb D$ be an interpolating sequence for $H^{\infty}({\mathbb D})$. Define a map $T : H^\infty (\mathbb D ) \to \ell ^\infty$ by $T(\phi ) = \{ \varphi (\lambda _i) \}_{i=1}^{\infty}$. This map is surjective by definition of interpolating sequences and I managed to show that $T$ is bounded by the closed graph theorem.

Now we can take $\epsilon = \lVert T \rVert$ and consider the sequence $\{ \epsilon e_i \}_{i=1}^{\infty}$ in $\ell ^\infty$ . Since $T$ is surjective, there is some $\varphi _i \in H^{\infty} (\mathbb D)$ such that $T(\varphi_i) = \{ \epsilon e_i \}_{i=1}^{\infty}$. Now $\lVert T(\varphi _i) \rVert = \varepsilon \lVert e_i \rVert =\varepsilon$. But from here, I can only conclude that $\lVert \varphi_i \rVert \ge 1$ whereas I am looking for the reverse inequality.