If $f: \mathbb{C} \to \mathbb{C}$ is continuous and analytic off $[-1,1]$ then is entire.

Solution 1:

Like Daniel Fischer, I used a Cauchy integral formula approach. Let $\gamma (t) = 2e^{it}, 0\le t \le 2\pi.$ For small $\epsilon>0$ let $R_\epsilon=[-1-\epsilon,1+\epsilon]\times [-\epsilon,\epsilon].$ Consider $\partial R_\epsilon$ as a contour in the natural way, oriented clockwise.

Let $z \in D(0,2)\setminus [-1,1].$ Then $z\notin R_\epsilon$ for small $\epsilon >0.$ For such $\epsilon$ we have by Cauchy

$$f(z) = \frac{1}{2\pi i}\left ( \int_{\gamma + R_\epsilon} \frac{f(w)}{w-z}\,dw\right ).$$

Verify that the integral over $R_\epsilon \to 0$ as $\epsilon\to 0$ (here using the continuity of $f$). Thus we have

$$f(z) = \frac{1}{2\pi i}\left ( \int_\gamma \frac{f(w)}{w-z}\,dw\right )$$

for all $z \in D(0,2)\setminus [-1,1].$ But the last integral defines a holomorphic function in all of $D(0,2).$ Simply by continuity, $f$ equals this integral in all of $D(0,2)$ and we're done.