# An inner product and projection property in RKHS

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 $$W$$).

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.
• Is there a general theorem for when $P_BP_A=P_{A\cap B}$? Commented Aug 28, 2023 at 9:07
• @GBA The condition I wrote down is an equivalent condition, as should be relatively easy to verify. By taking adjoints you can see that it’s also equivalent to $P_AP_B = P_{A \cap B}$ and therefore also equivalent to $B \cap (A \cap B)^\perp \perp A$. If you’re asking for equivalent conditions, that’s all I’m aware of. Commented Aug 28, 2023 at 16:01
• A generalization of the argument I used will show that $B^\perp \subseteq A$ (or equivalently $A^\perp \subseteq B$) is a sufficient condition, which applies to your question, though this is not a necessary condition. Commented Aug 28, 2023 at 16:06