Question

Hi, I have been trying to prove the following:

Suppose $\gamma$ is a simple rectifiable curve and let $z_0 \in \gamma$ be a fixed point. If the function $f(z)$ is defined and continuous on $\gamma$ except at $z_0$ and $\lim_{z \to z_0} (z-z_0)^{\alpha}f(z)=L$ exists, where $\alpha <1$, then $\int_{\gamma} f(z)dz$ exists.

I decided to prove this using Cauchy's criterion. My proof goes like this:

Assume that $z_0$ is the end-point of the curve $\gamma$. Suppose $z_1 \in \gamma$ is choosen so that $|(z-z_0)^{\alpha}f(z)-L|<\epsilon$, provided $|z-z_0| \leq |z_1-z_0|$ and let $z_2 \in \gamma$ be such that $|z_2-z_0| \leq |z_1-z_0|$. Then if $\gamma_{12}$ denotes the part of the curve $\gamma$ from $z_1$ to $z_2$: $$ \left | \int_{\gamma_{12}} f(z)dz \right | \leq \int_{\gamma_{12}} |f(z)||dz| \leq \int_{\gamma_{12}} \frac{|L|+\epsilon}{|z-z_0|^{\alpha}} ds$$ where $ds=|dz|$ is the arc-length parameter.

And this where I'm stuck. I don't know how to evaluate or estimate the last integral. In the book, where I found this theorem, the author just writes: $$\int_{\gamma_{12}} \frac{|L|+\epsilon}{|z-z_0|^{\alpha}} ds \leq \int_{0}^{k} \frac{|L|+\epsilon}{{\rho}^{\alpha}} d\rho = \frac{|L|+\epsilon}{1-\alpha} k^{1-\alpha} $$ where $k=|z_1-z_0|$ and $\rho=|z-z_0|$. But I don't understand how he got this estimate.

Answer 1

I think something wrong with the theorem you found,and the right expression maybe \int_{\gamma_{12}} \frac{|L|+\epsilon}{|z-z_0|^{\alpha}} ds \leq \int_{0}^{k} \frac{|L|+\epsilon}{{\rho}^{\alpha}} d\rho = \frac{|L|+\epsilon}{1-\alpha} k^{\1-alpha}, and for |z_2-z_0| \leq |z_1-z_0|,|(z-z_0)^{\alpha}f(z)-L|<\epsilon, I guess you omit an assumption that \forall \epsilon , \exists \delta s.t. |z_1-z_0|\leq \delta, right?