I can’t seem to wrap my head around this old qual problem, despite it seeming rather straightforward. We are given a function $f$ that is continuous in a neighborhood around a point a. We are asked to show:
\begin{align*} f(a) = \left(\frac{1}{2\pi i}\right) \; \text{lim}_{\epsilon \rightarrow 0} \int_{C_{\epsilon}(a)} \frac{f(z)}{z-a}dz \end{align*}
Basically, show the Cauchy Integral Theorem holds. Now I know it holds when $f$ is analytic inside the disk and continuous up to the boundary, and in this case we are essentially removing the interior by shrinking the radius to 0, so it’s like the analytic requirement is mute and we just need the “continuous up to the boundary” portion, at least that is my intuition.
If someone could (a) tell me if my intuition is correct, and (b) help me start how I might argue this formally, that would be a great help. Thanks!
Alternatively, parametrize $C_{\epsilon}(a)$ by $z = a + \epsilon e^{2\pi i\theta}$ to write
$$ \frac{1}{2\pi i} \int_{C_{\epsilon}(a)} \frac{f(z)}{z - a} \, \mathrm{d}z = \int_{0}^{1} f(a + \epsilon e^{2\pi i \theta}) \, \mathrm{d}\theta. $$
So it follows that
$$ \left| \frac{1}{2\pi i} \int_{C_{\epsilon}(a)} \frac{f(z)}{z - a} \, \mathrm{d}z - f(a) \right| \leq \int_{0}^{1} \left| f(a + \epsilon e^{2\pi i \theta}) - f(a) \right| \, \mathrm{d}\theta. $$
I guess now you know how to proceed.