Why
$$
\Pi\_+ \left(\frac{\overline{z}}{1-\overline{qz}}f\right)= \frac{f(z)-f(\bar{q})}{z-\overline{q}}, \quad f\in H^2(\mathbb D),$$
where


* $q\in \mathbb C$,
* $\Pi\_{+}$ is the Szegö projector:
$$\Pi\_{+}\left(\sum\_{k \in \mathbb{Z}} \widehat{f}(k) z^{k}\right)=\sum\_{k \in \mathbb{Z}\_{+}} \hat{f}(k) z^{k}
$$ and
* $H^2(\mathbb D)$ is the Hardy space of the unit disk?


I've started by introducing the Shift operator $S$ such that if $S e\_k =e\_{k+1}$: then  

$$
\Pi\_+ \left(\frac{\overline{z}}{1-\overline{qz}}f\right)=(\mathscr F^{-1} S(1-\overline{q}S)^{-1} \mathscr F )^\* f
$$ where $A^\*$ is the adjoint of $A$. But it remains to show that $$\big(\mathscr F^{-1} S(1-\overline{q}S)^{-1} \mathscr F \big)^\* f=\frac{f(z)-f(\overline{q})}{z-\overline{q}}$$ in $H^2(\mathbb D)$. How to to proceed?



Computation on the Hardy space
**Use nontangential boundary values.** Also, you want $|q| < 1$, since otherwise $f(\bar{q})$ need not be defined.


$f \in H^2(\mathbb{D})$ extends almost everywhere to $f \in L^2(\mathbb{T})$, using boundary values, where $\mathbb{T} = \{ z : |z| = 1 \}$.


On $\mathbb{T}$, we have $|z|=1$ and so $\bar{z} = 1/z$. So, as an element of $L^2(\mathbb{T})$,
$$
g(z) = \frac{\bar{z}}{1-\bar{q}\bar{z}}f(z) = \frac{1}{z-\bar{q}}f(z),
$$
for a.e. $z \in \mathbb{T}$. **Important:** we must be careful here, since $f(z) = f(e^{i \theta})$ is not necessarily defined for all $|z| = 1$, only for almost all $z$.


We must project $g$ onto $H^2$. By standard Hilbert space geometry, we must find an element $h \in L^2$ with $h \in (H^2)^\perp$, such that $g-h \in H^2$. Then $g-h$ is the desired projection of $g$ into $H^2$.


Now if we regard the formula for $g \in L^2(\mathbb{T})$ as defining a *meromorphic* function on $\mathbb{D}$, we see that there is a singularity at $z = \bar{q}$. However, if we let
$$
h(z) = \frac{f(\bar{q})}{z - \bar{q}}, \qquad z \in \mathbb{D}
$$
then
$$
g(z)-h(z) = \frac{f(z) - f(\bar{q})}{z - \bar{q}}, \qquad z \in \mathbb{D}
$$
defines an analytic function on $\mathbb{D}$, because the pole of $g$ at $\bar{q}$ has been cancelled, and it is differentiable everywhere else on $\mathbb{D}$. It is obvious that $h \in L^2(\mathbb{T})$, regardless of the value of $f(\bar{q})$, because $|q|<1$. In fact, just $|q| \neq 1$ is enough.


Finally, for $z \in \mathbb{T}$, the function
$$
\frac{1}{z-\bar{q}} = \frac{\bar{z}}{1-\bar{q}\bar{z}} = \sum\_{n=0}^\infty \bar{q}^n \bar{z}^{n+1}
$$
consists only of negative powers of $z = e^{i \theta}$, so all its Fourier series terms $a\_n e^{i n \theta}$ with $n \geq 0$ are zero. (And the sum converges absolutely and uniformly for each fixed $q$ with $|q|<1$).


Thus that function, and so also $h$, is indeed orthogonal to $H^2$ within $L^2$ as required (since, $H^2$ is exactly the closed subspace of $L^2$ for which all Fourier coefficients with $n<0$ vanish).


What makes everything work about these boundary values is that you can map from $H^2$ to $L^2$, via nontangential boundary values, and then back to $H^2$ via harmonic extensions (using the Poisson kernel) and the functions and Fourier coefficients (and Taylor series coefficients) are preserved.


Remember that any $f \in L^2(\mathbb{T})$ can be extended to a harmonic function on $\mathbb{D}$ using the Poisson kernel: $e^{i m \theta}$ for $\theta \in [0, 2\pi]$, as a function on $\mathbb{T}$, corresponds to $r^{|m|} e^{im\theta}$ for $z = r e^{i \theta}$ as a function on $\mathbb{D}$, which is harmonic, and also analytic if $m \geq 0$. The Hardy space is just the subspace with Fourier coefficients for negative $n$ being zero, and the Taylor series coefficients are exactly the Fourier series coefficients.


