Let $u$ be an inner function and denote by $H^2$ the Hardy space on the open unit disc $\mathbb{D}$.
A model space $K_u$ associated to $u$ is a Hilbert space of the form \begin{equation} K_u = (uH^2)^{\perp} \end{equation} where $\perp$ denotes the orthogonal complement in $H^2$.
Let $\lambda_1 \in \mathbb{D}$. We consider $u = B_{\lambda_1}$ the Blaschke factor at $\lambda_1$ defined on $\mathbb{D}$ by $B_{\lambda_1}: z \mapsto \dfrac{\lambda_1 - z}{1 - \overline{\lambda}_1z}$. I am looking to prove that \begin{equation} K_u = \text{span}\{k_{\lambda_1} \} \end{equation} where $k_{\lambda_1}: z \mapsto \dfrac{1}{1-\overline{\lambda}_1 z}$ is the reproducing kernel of $H^2$ at $\lambda_1$. I managed to prove that $$\text{span} \{k_{\lambda_1} \} \subset K_u$$ by using the reproducing property of $k_{\lambda_1}$ to obtain that for all $f \in H^2$, $$ \langle uf, k_{\lambda_1} \rangle = u(\lambda_1)f(\lambda_1) = 0$$ (since $u(\lambda_1) = 0$). I struggle to prove the second inclusion (or to find another argument to prove the equality).
I am a beginner in this subject, so all hints and details are welcome.
I was dealing with the same stuff recently (a little more general). Let me just copy my lemma here. $\mathfrak I$ is the set of inner functions.
Lemma: Let $(z_j)_{j\in\mathbb N}\subset\mathbb D$ be a Blaschke sequence such that $z_i\neq z_j$ for $i\neq j$ and let $h$ be the Blaschke product with respect to $(z_j)_{j\in\mathbb N}$. Then $$ \mathcal H_h := (hH^2)^\perp = \overline{\operatorname{span}}\{k_{z_j} : j\in\mathbb N\}. $$ Proof. Let $f\in H^2$. Then $\langle hf,k_{z_j}\rangle = h(z_j)f(z_j) = 0$ for all $j\in\mathbb N$. Hence, $k_{z_j}\in\mathcal H_h$. Now, let $f\in\mathcal H_h$ such that $\langle f,k_{z_j}\rangle = 0$, i.e., $f(z_j) = 0$, for all $j\in\mathbb N$. Factorize $f$ as $f = uv$ with $u\in\mathfrak I$ and an outer function $v$. As $v$ has no zeros in $\mathbb D$, we have $u(z_j) = 0$ for all $j\in\mathbb N$. Hence, the Blaschke product $h$ must be a divisor of $u$, i.e., $u = hu_1$ with some $u_1\in\mathfrak I$. But this implies $f = h(u_1v)\in hH^2$ and thus $f=0$.