How to show that $RH^2$ is closed in $RL^2$? I have faced difficulty when I try to learn functional analysis myself and to do problem 2.14 in N. Young's "An introduction to Hilbert space".
The question is asking you to show that $RH^2$ is closed in $RL^2$ where $RL^2$ denotes the space of rational functions which are analytic on the unit circle
$$\partial D=\{z\in C:|z|=1\}$$
with the inner product
$$<f,g>=\frac{1}{2\pi i}\int_{\partial D}f(z)\overline{g(z)}\frac{dz}{z}$$
the integral being taken anti-clockwise around $\partial D$. and the $RH^2$ is the subspace of $RL^2$ consisting rational functions which are analytic on the closed unit disc clos $D$, where
$$D=\{z \in C:|z|<1\}$$ with the inner product in $RL^2$.
Could anyone give me a guide on how to do it? I try to start by checking whether the complement of $RH^2$ is open but I'm not sure what is the complement of $RH^2$.
 A: This problem is a bit weird in the sense that neither spaces above ($RH^2, RL^2$) are Hilbert so sequences from $RH^2$ can converge to more general analytic functions in the $L^2$ norm on the unit circle (leading to Hardy spaces etc).
This being said it is easy to see using Cauchy formula that for a compact set $K$ included in the open unit disc, there is $C_K$ st $\sup_{z \in K} |f(z)| \le C_K ||f||_2$ for any analytic function on the unit disc that extends continuously say to the boundary
(using the Cauchy Schwarz inequality and the fact that $|z-\zeta| >A_K>0, z \in K, \zeta \in \partial D$).
This implies that any $||.||_2$ convergent sequence converges locally uniformly hence to an analytic function $F$ on the open unit disc; if now the sequence also converges to a rational function in the $||.||_2$ norm on the unit circle, it immediately follows that $F$ is rational on the unit circle hence inside the unit disc by analytic continuation (two meromorphic functions equal on the unit circle, are equal everywhere), so $F$ is also in $RH^2$.
However I want to emphasize again that in general if $f_n$ is in $RH^2$ and $f_n$ converges into the $||.||_2$ norm it doesn't, in general, follow that the limit is in $RL^2$ (hence in $RH^2$).
