Fall 2010 - Problem 7

Solution.

By Morera's theorem, it's enough to show that the circulation of $f$ around any rectangle $R$ is $0$. Let $R = \br{a, b} \times \br{ic, id}$. If $R \cap \R = \emptyset$, then by connectedness, the rectangle lies entirely in $\C \setminus \R$, so by Cauchy's theorem, $\int_{\partial R} f\p{z} \,\diff{z} = 0$.

Now suppose $R \cap \R \neq \emptyset$ and let $\epsilon > 0$. Consider the rectangles $R_\epsilon^+ = \br{a, b} \times \br{ic, -i\epsilon}$ and $R_\epsilon^- = \br{a, b} \times \br{i\epsilon, id}$ with boundaries oriented counter-clockwise. Observe that both of these rectangles lie entirely in $\C \setminus \R$, so

by Cauchy's theorem. Let $U_\epsilon = \br{a, b} \times \set{i\epsilon}$ oriented from left to right, i.e., the bottom edge of $R_\epsilon^+$. Similarly, let $L_\epsilon = \br{a, b} \times \set{-i\epsilon}$ be oriented from right to left. We have

Since $f$ is continuous on the compact strip $\br{a, b} \times \br{-i\epsilon, i\epsilon}$, it is uniformly continuous on this strip. Thus, for any $\eta > 0$, if $\epsilon$ is small enough, then $\abs{f\p{t + i\epsilon} - f\p{t - i\epsilon}} < \eta$ uniformly in $t \in \br{a, b}$. Hence, for small enough $\epsilon$,

Hence, the contributions from $U_\epsilon$ and $L_\epsilon$ cancel out as $\epsilon \to 0$. It follows that

so $f$ is entire.