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I want to show the following:

Suppose $f\in C(|z|\leq 1)\cap O(|z|< 1)$, where $O(|z|< 1)$ means that $f$ is holomorphic in the open unit disk $D$. Then

$\int_{|z|=1} fdz=0$

(Note: I already have a version of Cauchy's integral theorem which tells me that for any piece-wise smooth, closed curve $\gamma$ in $D$, it's true that $\int_{\gamma}f(z)dz=0$.)

Here's my attempted proof, using Lebesgue's dominated convergence theorem:

Let $r_n=1-\frac{1}{n+1}$ for $n=1,2,..$.

Then, of course, $\lim_{n \to \infty} r_n =1$.

Consider the path integral

$\int_{|z|=r_n} f(z)dz = \int_{0}^{2\pi}ir_ne^{it}f(r_ne^{it})dt$

Let $f_n(t)=ir_ne^{it}f(r_n e^{it})$, the integrand of the RHS. Then $\lim_{n \to \infty}f_n(t)=\lim_{n \to \infty}ir_ne^{it}f(r_n e^{it})=ie^{it}f(e^{it})$ by continuity, and

$\int_{0}^{2\pi}ie^{it}f(e^{it})dt=\int_{|z|=1}f(z)dz$

I just need to bound each $|f_n(t)|$ above for the hypotheses of Lebesgue's dominated convergence theorem to hold. I think it should work to simply take $g=\max\{|f(z)|: z \in \bar{D}\}$, where $\bar{D}$ is the closed unit disk, since $\bar{D}$ is compact.

So, all the hypotheses hold, and we get that

$\lim_{n \to \infty} \int_{|z|=r_n}f_n(z)dz=\int_{|z|=1} f(z)dz. $

By the version of Cauchy's integral theorem which we already assume (for curves in the open disk), we have that the integral on the LHS is zero for ever $n$. Thus, the limit is 0, and we have

$0=\int_{|z|=1} f(z)dz.$

  • 0
    What you did seems correct to me.2012-02-09

0 Answers 0