Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Analysis / 4

For the preliminary definition, put $S_N(z)={\sum }_{k=1}^N{f}_k(z)$. The series converges uniformly to $f$ on $U$ when, for every $ϵ>0$, some $N_0$ satisfies $|S_N(z)-f(z)|<ϵ$ for all $z\in U$ and all $N\ge N_0$. Equivalently, ${\sup }_{z\in U}|S_N(z)-f(z)|\to 0$.

Let $L(\gamma )$ be the finite length of the piecewise-smooth path. Finite linearity of the contour integral and the integral bound giveThe uniform limit is continuous by the stated allowed result, so its integral is defined. Consequently

Cauchy's integral theorem states that if a function is holomorphic on a simply connected open set, its integral around every closed piecewise-smooth path in that set is zero. A local form sufficient here is: if a function is holomorphic on an open set containing a triangle together with its interior, thenFor a general domain with holes, an arbitrary closed path need not have zero integral; the triangle or simple-connectedness hypothesis supplies the required interior.

Morera's theorem states that if $f$ is continuous on an open set $U$ and its integral around every triangle whose interior and boundary lie in $U$ is zero, then $f$ is holomorphic on $U$.

For a proof, choose a disk $B\subseteq U$ and a base point $z_0\in B$. Define $F(z)={\int }_{[z_0,z]}f(\zeta )d\zeta$ along the straight segment. Convexity keeps every triangle with vertices $z_0,z,z+h$ inside $B$. Its zero boundary integral impliesBy continuity, the last integral tends to $f(z)$ as $h\to 0$, so $F^{\prime }(z)=f(z)$. Thus $F$ is holomorphic on $B$. The allowed result that the derivative of a holomorphic function is holomorphic now gives that $f=F^{\prime }$ is holomorphic there. Such disks cover $U$, proving

For each triangle compactly contained in $U$, part (i) permits the interchange of the series and its boundary integral. Cauchy's integral theorem makes every term zero. The uniform limit is continuous, so Morera's theorem yields that $f$ is holomorphic on $U$.

For the final application, each function $(z-k{)}^{-2}$ is holomorphic away from its pole. Fix a positive integer $n$. For $z\in U_n$ and $k>2n$,The numerical series ${\sum }_{k>2n}4/k^2$ converges, so the Weierstrass M-test gives uniform convergence of the tail on $U_n$. Adding the finitely many initial terms preserves uniform convergence. Those initial terms need not be bounded near their excluded poles; only the tail needs the uniform bound. By the first conclusion, the full sum is holomorphic on each $U_n$. Their union is $\mathbb{C}\setminus \mathbb{N}$, hence