An inner product and projection property in RKHS

Question

I'm reading an article regarding the condition for the existence of a solution to a constrained Pick interpolation problem, and the author wrote the following equality - for background, we have the following spaces: $$H^2_{\alpha,\beta}\subset M_k(H^2)$$ where $M_k(H^2)$ is the set of $k\times k$ matrices with entries being functions in $H^2$, and inner product given by: $$\langle f,g\rangle=\frac{1}{2\pi}\int_{\mathbb{T}}tr[f(z)g(z)^*]|dz|$$

Now given points $z_1,\ldots,z_n$, we let: $$B_FM_k(H^2)=\{f\in M_k(H^2):f(z_i)=0,\forall i=1,\ldots,n\}$$

To the question: the author defines the subspace: $$M_{\alpha,\beta}=H^2_{\alpha,\beta}\cap(H^2_{\alpha,\beta}\cap B_FM_k(H^2))^{\perp}\subset H^2_{\alpha,\beta}$$ Now assuming that $Q\in M_k(H^{\infty})$ is such that $M_Q$ (the multiplier operator by $Q$) is invariant for $H^2_{\alpha,\beta}$, $g\in H^2_{\alpha,\beta}$ and $f\in (H^2_{\alpha,\beta}\cap B_FM_k(H^2))^{\perp}$, the author claims that: $$\langle Qg,f\rangle=\langle P_{H^2_{\alpha,\beta}}Qg,P_{(H^2_{\alpha,\beta}\cap B_FM_k(H^2))^{\perp}}f\rangle=\langle P_{M_{\alpha,\beta}}Qg,P_{M_{\alpha,\beta}}f\rangle$$

I understand the first equality, but cannot understand the second equality. Any ideas or explanations on why this is true would be appreciated. (Clarification: $P_W$ is the orthogonal projection on the subspace $W$).

Answer 1

I don't fully understand the context here, but it seems to me what you need is the following: Let $A = H^2_{\alpha, \beta}$, $B = (H^2_{\alpha, \beta} \cap B_FM_k(H^2))^\perp$, then $P_BP_A = P_{A \cap B}$. We can show that this would happen if $A \cap (A \cap B)^\perp \perp B$. (Indeed, we observe that $P_BP_A$ and $P_{A \cap B}$ both reduce to 1 on $A \cap B$, so it suffices to show $P_BP_A((A \cap B)^\perp) = 0$. Fix any $h \in (A \cap B)^\perp$, write $h = P_Ah + P_{A^\perp}h$. Since $A^\perp \subseteq (A \cap B)^\perp$, we have, $P_Ah = h - P_{A^\perp}h \in (A \cap B)^\perp$, so $P_Ah \in A \cap (A \cap B)^\perp$. As $A \cap (A \cap B)^\perp \perp B$, we have $P_BP_Ah = 0$.) Note that in this case, $(A \cap B)^\perp = \overline{A^\perp + B^\perp} = \overline{(H^2_{\alpha, \beta})^\perp + (H^2_{\alpha, \beta} \cap B_FM_k(H^2))} = (H^2_{\alpha, \beta})^\perp + (H^2_{\alpha, \beta} \cap B_FM_k(H^2))$, as the two summands are orthogonal to each other, so $A \cap (A \cap B)^\perp = H^2_{\alpha, \beta} \cap ((H^2_{\alpha, \beta})^\perp + (H^2_{\alpha, \beta} \cap B_FM_k(H^2))) = H^2_{\alpha, \beta} \cap B_FM_k(H^2) = B^\perp$. This proved the claim.