# weakly separated sequences in RKHS are separated by Gleason metric

I am reading Agler and McCarthy's Pick Interpolation and Hilbert Function spaces. In Chapter 9, the authors ask to observe that weakly separated in a Reproducing kernel hilbert space implies separated in the Gleason metric. This is not immediate to me however.

Here are the definitions:

Fix a Hilbert function space $$\mathcal H_k$$ on $$X$$. The sequence $$\Lambda = \{ \lambda_i \}_{i=1}^{\infty}$$ in $$X$$ is weakly separated if there exist a constant $$\varepsilon > 0$$ such that whenever $$i\ne j$$, there is some function $$\varphi _{ij}$$ in the closed unit ball of the multiplier algebra $$\mathcal H _k$$ such that $$\varphi _{ij} (\lambda_i) = \varepsilon$$ and $$\varphi _{ij} (\lambda _j) = 0$$.

Also, the sequence $$\Lambda$$ in $$X$$ is said to be $$d$$-separated if there is some constant $$\varepsilon > 0$$ such that for $$i\ne j$$ we have that $$d (\lambda _i , \lambda _j) > \varepsilon$$. Note that the (pseudo)metric $$d$$ on $$X$$ is given by \begin{align*} d(\lambda _1 , \lambda_2) = \sqrt{ 1- \frac{\lvert k(\lambda_1 , \lambda _2)\rvert ^2}{k(\lambda_1 , \lambda_1 )k(\lambda_2 , \lambda_2 )}} \end{align*} where $$\lambda_1 , \lambda_2 \in X$$.

Here's my attempt: Let $$\Lambda = \{ \lambda_i \}_{i=1}^{\infty}$$ in $$X$$ be weakly separated. It suffices to show that \begin{align*} \sup \left\{ \frac{\lvert k(\lambda_i , \lambda _j)\rvert ^2}{k(\lambda_i , \lambda_i )k(\lambda_j , \lambda_j )} : \lambda_i \ne \lambda_j \right\} <1 \end{align*} in order to show that $$\Lambda$$ is $$d$$-separated. Since $$\Lambda$$ is weakly separated, there is some $$\varepsilon_0 > 0$$ and for each $$i\ne j$$, there is a function $$\varphi_{ij}$$ such that in the closed unit ball of the multiplier algebra $$\mathcal H _k$$ such that $$\varphi _{ij} (\lambda_i) = \varepsilon_0$$ and $$\varphi _{ij} (\lambda _j) = 0$$. Note that $$\varepsilon_0 \le 1$$ since $$\varepsilon_0 =\lvert \varphi_{ij} (\lambda_i)\rvert \le \lVert \varphi_{ij} \rVert_{\infty} \le \lVert \varphi \rVert _{\mathcal M (\mathcal H)} \le 1$$.

Now, note that for $$i \ne j$$, we have that \begin{align*} \varepsilon_0 ^2 \lvert k(\lambda_i , \lambda_j) \rvert &=\lvert{ \langle \epsilon_0 k_{ \lambda_j}, \epsilon_0 k_{ \lambda_i} \rangle}\rvert \\ &= \lvert \langle \varphi_{ji}(\lambda_j ) k_{\lambda_j}, \varphi_{ij}(\lambda_i ) k_{\lambda_i} \rangle \rvert \\ &= \lvert \langle {M_{\varphi_{ji}}}^* k_{\lambda_j}, {M_{\varphi_{ij}}}^* k_{\lambda_i} \rangle \rvert \\ &\le \lVert k_{\lambda _j} \rVert \lVert k_{\lambda _i} \rVert. \end{align*} This implies that \begin{align*} \frac{\lvert k(\lambda_i , \lambda _j)\rvert ^2}{k(\lambda_i , \lambda_i )k(\lambda_j , \lambda_j )} \le \frac{1}{\varepsilon_0 ^4}. \end{align*}

But $$\frac{1}{\varepsilon _0 ^4} \ge 1$$ so I do not get what I desire. I also realise that I have failed to use that fact that $$\varphi_{ij} (\lambda_j) = 0$$. Directions on proving this will be appreciated! $$\ddot \smile$$

If a sequence is weakly separated, i.e. there exists a multiplier $$\varphi_{ij}$$ of multiplier norm at most one such that $$\varphi_{ij}(\lambda_i)=\varepsilon, \varphi_{ij}(\lambda_j)=0$$, then necessarily the $$2\times2$$ Pick matrix associated to this $$2$$-point interpolation problem is positive semidefinite; $$\begin{equation*} \begin{bmatrix} (1-\varepsilon^2)k(\lambda_i,\lambda_i) & k(\lambda_i,\lambda_j)\\ k(\lambda_i,\lambda_j) & k(\lambda_j,\lambda_j) \end{bmatrix} \geq0. \end{equation*}$$ In particular its determinant has to be non negative, that is $$\begin{equation*} (1-\varepsilon^2)k(\lambda_i,\lambda_i)k(\lambda_j,\lambda_j)-|k(\lambda_i,\lambda_j)|^2 \geq0 \end{equation*}$$ which after rearranging the terms gives $$d$$-separation.