Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 8 / 3

For infinite products, a nonzero limiting product requires that its factors tend to one. A useful sufficient condition isAfter finitely many factors, $|u_j|<1/2$ and the principal holomorphic logarithm satisfies $|\log (1+u_j)|\le 2|u_j|$. Therefore the sum of logarithms converges, and exponentiating it gives a finite nonzero product. The same argument on compact sets proves infinite product convergence from logarithmic tails: a locally uniformly absolutely convergent tail of holomorphic logarithms gives a holomorphic nonvanishing tail product. Finite factors then determine all zeros and their orders.

The Weierstrass elementary factors areFor $|w|<1$,Their only zero is a simple zero at $w=1$.

Work with entire functions on $\mathbb{C}$. As usual for prescribed exact zero orders, the distinct zero locations must have consistent multiplicities. The literal statement allows repeated locations with conflicting orders; that cannot be true, for example if the same point is prescribed order one and order two. Remove consistent repetitions rather than adding their orders, and separate a possible zero at the origin.

List the distinct nonzero locations as $a_j$, with prescribed orders $n_j$, and write $m$ for the prescribed order at zero, or zero if the origin is not prescribed. Choose $p_j$ large enough thatThen the Weierstrass factorization theorem construction isOn every compact set, $|z/a_j|\le 1/2$ for all sufficiently large $j$ because $|a_j|\to \infty$. Its logarithmic tail is bounded byThus the product converges locally uniformly, is entire, and has exactly the specified zeros with exactly their orders. At a prescribed zero, only its own finite factor vanishes; all other finite factors and the tail product are nonzero.

Without escape to infinity the conclusion fails in general. Distinct proposed zeros $1/j$ accumulate at zero. The identity theorem would force an entire function with those zeros to vanish identically, which does not have precisely the prescribed isolated zeros. A finite zero set can of course be realized by a polynomial.

For entire functions with the same zero divisor, the quotient $q=f/g$ extends through every common zero by cancelling equal local powers. It is entire and nowhere zero. Hence $q^{\prime }/q$ is entire and has a primitive on the simply connected plane. Choose $h(0)$ with $e^{h(0)}=q(0)$ and putThe derivative of $q{e}^{-h}$ vanishes, and its value at zero is one. ThereforeIf also $f=e^{k}g$, then $e^{h-k}=1$ everywhere. The continuous difference $h-k$ takes values in the discrete set $2\pi i\mathbb{Z}$ and so is constant on the connected plane:The whole-plane hypothesis matters: a zero-free holomorphic quotient on a multiply connected domain need not possess a global holomorphic logarithm.

Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.