# A proof (?) of a Cauchy Integral Theorem type claim

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

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

-
I forgot to ask a question. The question is, have I erred? I would guess this isn't at all the proof which is expected, and I haven't had much experience using LDCT "in the wild" before, so I am, as usual, a bit uncertain. – user18297 Feb 7 '12 at 23:21
What you did seems correct to me. – Davide Giraudo Feb 9 '12 at 22:01