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Consider the following situation: a simple, closed, piecewise smooth curve $\gamma$ in the complex plane and $\Omega$ the bounded connected component of the complement of $\gamma$ in $\mathbb{C}$; a continuous function $\psi:\gamma \rightarrow \mathbb{C}$; a function $f:\Omega \rightarrow \mathbb{C}$ defined by $f(z) = \frac{1}{2\pi i}\int_{\gamma} \frac{\psi(w)}{w-z} dw$ for all $z\in\Omega$.

It's a simple result (than can be found in Rudin's Complex and Real Analysis in a slightly more general version) that $f$ is holomorphic in $\Omega$. I have been thinking in the following related questions:

1 . Is it the case that for every region $\Omega$, obtained as before, and for every holomorphic function $g$ defined there one can find a function $\psi$ defined on $\gamma$ such that the function $f$ constructed as before is $g$? The answer is yes for the case in which the function $g$ admits a holomorphic extension $h$ to an open set $U$ that contains $\overline{\Omega}$, taking $\psi=h$. I made no progress beyond that trivial case.

2 . Is it true that if $\Omega$ is as before and $g:\overline{\Omega}\rightarrow \mathbb{C}$ is holomorphic in $\Omega$ and continuous in $\overline{\Omega}$ then, when performing the previous construction with $\psi=g$, it turns out that $f=g$ in $̣\Omega$? The same trivial case of 1 applies here too.

3 . What conditions can\must be imposed to $\psi$ (or maybe $\gamma$) to obtain in the previous process a function $f$ admitting a continuous extension to $\overline{\Omega}$? No progress.

Thanks.

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1 . Is it the case that for every region $\Omega$, obtained as before, and for every holomorphic function $g$ defined there one can find a function $\psi$ defined on $\gamma$ such that the function $f$ constructed as before is $g$?

No. For

$$f(z) = \frac{1}{2\pi i}\int_\gamma \frac{\psi(w)}{w-z}\,dw,$$

we have the growth condition

$$\lvert f(z)\rvert \leqslant \frac{K(\psi,\gamma)}{\operatorname{dist}(z,\gamma)},$$

and a function that grows faster - like $g(z) = (z-w_0)^{-2}$ for a $w_0$ on the trace of $\gamma$ - cannot be an integral of the specified form.

2 . Is it true that if $\Omega$ is as before and $g\colon \overline{\Omega}\to\mathbb{C}$ is holomorphic in $\Omega$ and continuous in $\overline{\Omega}$ then, when performing the previous construction with $\psi = g$, it turns out that $f = g$ in ̣$\Omega$?

Yes, since the boundary is sufficiently smooth, the continuity on the closure of $\Omega$ is sufficient for $g$ to be the Cauchy integral of its boundary values. You can approximate the boundary curve $\gamma$ by curves in $\Omega$ with the same parameter interval, such that the integrand $\dfrac{g(\gamma_n(t))}{\gamma_n(t)-z}\gamma_n'(t)$ converges dominatedly to $\dfrac{g(\gamma(t))}{\gamma(t)-z}\gamma'(t)$, and that means the Cauchy formula holds.

For the special case of a disk, much is known about which functions are the Cauchy integral of $L^p$-functions on the boundary circle (Rudin treats some results on $H^p$ spaces in his book), and they have a simple characterisation. For domains with a piecewise smooth boundary, less is known, but experts in the field probably still know much, I don't unfortunately.

3 . What conditions can/must be imposed to $\psi$ (or maybe $\gamma$) to obtain in the previous process a function $f$ admitting a continuous extension to $\overline{\Omega}$?

A necessary condition is that

$$\int_\gamma \psi(w)\cdot w^n\,dw = 0$$

for all $n \geqslant 0$, since if $f$ admits a continuous extension to $\overline{\Omega}$, so does $g_n(z) = f(z)\cdot z^n$, and the integral over its boundary values must vanish by Cauchy's integral theorem and the approximation argument. For nice enough $\gamma$ at least (circles), that condition is also sufficient, if memory serves, but it may not be sufficient for badly behaved curves.

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