$ D=D_{r}(c), r > 0 .$ Show that if $f$ is continuous in $\overline{D}$ and holomorphic in $D$, then for all $z\in D$: $$f(z)=\frac{1}{2\pi i} \int_{\partial D}\frac{f(\zeta)}{\zeta - z} d\zeta$$

I don't understand this question because I don't see how it is different to the special case of the Cauchy Integral formula. I would be very glad if somebody could tell me what the difference is and how to show that it is true.

  • 1
    $\begingroup$ What's the "special case" of the Cauchy Integral formula? $\endgroup$ – rfauffar Nov 22 '11 at 14:10
  • $\begingroup$ with $D= D_{r}(c)$, $f(c) = \frac{1}{2\pi i } \int_{\partial D} \frac{f(\zeta)}{\zeta-c}$ $\endgroup$ – Tashi Nov 22 '11 at 14:12

If I understand correctly, $D_r(c)$ is the open ball centered in $c$ with radius $r$? If this is the case, the difference between the two is that above your $c$ is fixed, and in the special case your $c$ "moves" with the ball. Fix $c$ then; we want to show that for every $z\in D_r(c)$ we have that $$f(z)=\frac{1}{2\pi i}\int_{\partial D_r(c)}\frac{f(\zeta)}{\zeta-z}d\zeta$$ (we assume that $f$ is holomorphic in the ball). Well, if we choose $s>0$ such that $D_s(z)\subseteq D_r(c)$, then by the special case we have that $$f(z)=\frac{1}{2\pi i}\int_{\partial D_s(z)}\frac{f(\zeta)}{\zeta-z}d\zeta,$$ and we see that $\partial D_s(z)$ is homologous to $D_r(c)$. We then have that $$f(z)=\frac{1}{2\pi i}\int_{\partial D_s(z)}\frac{f(\zeta)}{\zeta-z}d\zeta=\frac{1}{2\pi i}\int_{\partial D_r(c)}\frac{f(\zeta)}{\zeta-z}d\zeta.$$ Does that answer your question? Is that what your question was?

The nice thing about the "general" Cauchy formula is that the curve that we're integrating over no longer depends on the point you want to evaluate.

  • $\begingroup$ shifted center, ok. Thanks. $\endgroup$ – Tashi Nov 22 '11 at 14:42

I am not sure what the "usual Cauchy integral formula" is for you but I am assuming it is the following result:

Theorem If $f$ is holomorphic in an open set containing the closure of $D_{r}(c)=\{z\in\mathbb{C}:\left|z-c\right|<r\}$ for some $r>0$ and $c\in\mathbb{C}$, then $f(z)=\frac{1}{2\pi i}\int_{\left|\zeta-c\right|=r} \frac{f(\zeta)}{\zeta - z}d\zeta$ for all $z\in D_{r}(c)$.

Note that the hypothesis of the theorem is that $f$ is holomorphic in an open set containing the closure of $D_{r}(c)$ (in particular, there is a smoothness condition on the restriction of $f$ to the boundary of $D_{r}(c)$), whereas in the result stated in your question, we are only given continuity on the closure of $D_{r}(c)$.

In any case, the general result stated in your question is easy to prove. I will give a hint: fix $z\in D_{r}(c)$ and apply the usual Cauchy integral formula (i.e., the theorem stated above) to conclude that:

$f(z)=\int_{\left|\zeta-c\right|=\epsilon} \frac{f(\zeta)}{\zeta-z}d\zeta$

for all $\epsilon < r$ sufficiently close to $r$. You can now take the limit $\epsilon\to r$ in the above equality and apply the Lebesgue dominated convergence theorem to conclude the proof of the general result. (Please do not forget to verify that the hypothesis of the dominated convergence theorem is satisfied!)

I hope this helps!

  • $\begingroup$ Hi, I don't understand what you mean with applying the lebesgue dominated convergence theorem and hypothesis of the dominated convergence. Thanks. $\endgroup$ – Tashi Nov 22 '11 at 14:43
  • $\begingroup$ @Tashi Could you please see the relevant Wikipedia page? $\endgroup$ – Amitesh Datta Nov 24 '11 at 5:25

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