We know from Cauchy's Integral formula that if $f:D \rightarrow \mathbb C$ is holomorphic and $\gamma$ is some closed simple curve in the disc $D$, that $$ f(z) = \frac{1}{2\pi i}\int_\gamma \! \frac{f(\zeta)}{\zeta-z} \, d\zeta $$ for all $z \in D$ with $z$ inside of the image of $\gamma$. If instead of being holomorphic $f$ is merely continuous on the image of $\gamma$, we still get a holomorphic function $$ F(z) = \int_\gamma \! \frac{f(\zeta)}{\zeta-z} \, d\zeta. $$ My question is: what do we know about how $f$ and $F$ will relate?

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    $\begingroup$ The first sentence, "We know $\dots$", isn't quite right. TIf $\gamma$ doesn't cross itself, then the formula is correct only when $z$ is inside the curve $\gamma$. More generally, the formula becomes correct (whether or not $\gamma$ crosses itself) if you multiply the left side by the winding number of $\gamma$ around $z$. $\endgroup$ Jan 8, 2013 at 20:05

1 Answer 1


The transform $$F(z)=\frac{1}{2\pi i}\int_{\mathbb{T}}\frac{f(\zeta)}{\zeta-z}d\zeta,$$ where $\mathbb{T}$ denotes the unit circle, is strongly related to the so called Cauchy transform.

You will find a lot of information in the expository paper The Cauchy transform by Joseph A. Cima, Alec Matheson, and William T. Ross.

In order to say something nice about the transform on general curves $\gamma$ you would certainly need condition on the parametrization of the curve.

Also note that for $|z|<1$ we may consider the geometric series expansion $$\frac{1}{\zeta-z}=\frac{1}{\zeta}\frac{1}{1-\bar{\zeta}z}=\frac{1}{\zeta}\sum_{k=0}^\infty \bar{\zeta}^kz^k$$ (here we assume $\zeta\in\mathbb{T}$ so that $\bar{\zeta}\zeta=|\zeta|^2 =1$ and hence $\bar{\zeta}=1/\zeta$) which converges uniformly in $\zeta$. Using that in the above integral formula leads to $$\frac{1}{2\pi i}\int_\mathbb{T} \frac{f(\zeta)}{\zeta-z}d\zeta = \frac{1}{2\pi i}\int_\mathbb{T} \frac{1}{\zeta}\sum_{k=0}^\infty f(\zeta)\bar{\zeta}^kz^k d\zeta$$ Then, if we set $\zeta=e^{it}$ so that $d\zeta=ie^{it}dt =i\zeta dt$, we arrive at $$F(z)=\frac{1}{2\pi}\int_{-\pi}^\pi \sum_{k=0}^\infty f(e^{it})e^{-ikt}z^k dt$$ switching the order of summation and integration leads to $$F(z)=\sum_{k=0}^\infty\frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})e^{-ikt}z^k dt =\sum_{k=0}^\infty\hat{f}(k)z^k$$ That is $F$ is the analytic projection of the Fourier expansion of $f$.


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