I am working through Theory of $H^p$ Spaces by Peter L. Duren, and I am having some difficulty with proof of the following theorem:
Theorem. Every function $f \in H^1$ can be expressed as the Cauchy integral of its boundary function. In fact, $$f^{(k)}(z) = \frac{k!}{2\pi i}\int_{|\zeta|=1} \frac{f(\zeta)d\zeta}{(\zeta - z)^{k+1}},\, |z| < 1;\; k=0,1,2,\ldots$$ Each of these integrals vanishes identically in $|z|>1$
Proof. It suffices to establish the result for $k=0$; the rest will then follow by the familiar procedure of " differentiation under the integral sign." Let $F(z)$ be the Cauchy integral of $f(e^{it})$ and let $\{P_n(z)\}_{n=1}^{\infty}$ be a sequence of polynomials such that $\lVert f - P_n\rVert_1 \to 0$. Writing $$F(z) = \frac{1}{2\pi i}\int_{|\zeta|=1}\frac{f(\zeta) - P_n(\zeta)}{\zeta - z}d\zeta, \quad |z| >1,$$ we find that $F(z) \equiv 0$ in $|z|>1$. Hence it follows that $$F(z) = \frac{1}{2\pi}\int_{0}^{2\pi} P(r,\theta - t)f(e^{it})dt = f(z), \quad |z| < 1 $$ This completes the proof.
The following line is where is I am struggling a bit
Let $F(z)$ be the Cauchy integral of $f(e^{it})$ and let $\{P_n(z)\}_{n=1}^{\infty}$ be a sequence of polynomials such that $\lVert f - P_n\rVert_1 \to 0$. Writing $$F(z) = \frac{1}{2\pi i}\int_{|\zeta|=1}\frac{f(\zeta) - P_n(\zeta)}{\zeta - z}d\zeta, \quad |z| >1,$$
I don't understand why we can write $F(z)$ as that above integral. Please help.
Fix $z, |z|>1.$ If $g$ is any entire function, then
$$\int_{|\zeta|=1}\frac{g(\zeta)}{\zeta - z}d\zeta = 0.$$
That follows from Cauchy's theorem, since the integrand as a function of $\zeta$ is holomorphic on $D(0,|z|).$ Apply this with $g=P_n$ to see that for this $z,$ the Cauchy integral of $f$ is the same as the Cauchy integral of $f-P_n.$