Riesz projection as a Cauchy-type integral

Question

Let

\begin{equation*} f(\zeta)=\sum_{k\in\mathbb{Z}}\widehat{f}(k)\zeta^k \end{equation*}

be a complex-valued function on unit circle $\mathbb{T}=\{ \zeta\in\mathbb{C}:|\zeta|=1\},$ where $\{\widehat{f}(k)\}_{k\in\mathbb{Z}}$ denote the Fourier coefficients of $f\in L^p(\mathbb{T}).$

Define the Riesz projection as

\begin{equation*} P_+f(\zeta)=\sum_{k\geq 0}\widehat{f}(k)\zeta^k,~\zeta\in\mathbb{T}. \end{equation*}

Q: Why can the Riesz projection be written as a Cauchy type integral

\begin{equation*} P_+f(z)=\frac{1}{2\pi i}\int_{\mathbb{T}}\frac{f(\lambda)}{\lambda -z}d\lambda,~z\in\mathbb{D}, \end{equation*}

with density $f\in L^p(\mathbb{T})?$

Answer 1

I think the basic reason is that you can write $$ \frac{1}{\lambda-z} = \frac{1}{\lambda} \sum_{n=0}^{\infty} \left( \frac{z}{\lambda} \right)^{n}, $$ and then (via a careful argument probably involving the partial sums) interchange the sum and integral to find $$ P_+ f(z) = \sum_{n=0}^{\infty} z^n \frac{1}{2\pi i} \int_{|\lambda|=1} \frac{f(\lambda)}{\lambda^{n+1}} \, d\lambda. $$ You can then change variables and show that the coefficient of $z^n$ is just the $n$th Fourier coefficient of $f$.

The keys, of course, are the interchange of sum and integral, as well as the problem of convergence of the integral in the first place.