Let $(a_k)_{k\geq 0}$ be a sequence of complex numbers such that for
all $C>0$, $\lim_{k\to\infty}\mid a_k\mid C^k=0$. Show that
$$f(z)=\sum\limits^{\infty}_{k=0}a_k \exp(2i\pi kz)$$ is defined for
all $z$; that $f$ is entire, and that $f(z+1)=f(z)$ for all
$z\in\mathbb{C}$
I have proved the part $f(z+1)=f(z)$. To show that $f(z)$ converges and entire, I want to use the theorem
If $\{f_n\}^{\infty}_{n=1}$ is a sequence of holomorphic functions
that converges uniformly to a function $f$ in every compact subset of
$\Omega$, then $f$ is holomorphic in $\Omega$.
So I need to see what function it converges to, and $\sum\limits^{\infty}_{k=0}a_k e^{2i\pi kz}$ converges uniformly to that function in every compact subset of complex plane. But I am not sure how to proceed. Can you explain if this is the right idea, and if not, could you give me other suggestions? Thanks!
Best Answer
It suffices to show that $$ \tag{*} F(w) = \sum_{k=0}^\infty a_k w^k $$ defines an entire function, because then $f(z) = F(e^{2\pi i z})$ is entire as the composition of entire functions.
Given $R > 0$ choose any $C > R$, then $$ |a_k w^k| \le |a_k| R^k = |a_k| C^k \left( \frac{R}{C} \right)^k \le \left( \frac{R}{C} \right)^k $$ for all $|w| \le R$ and all sufficiently large $k$, which (by the Weierstrass M-test) implies that the series $(*)$ converges uniformly on the disk $|w| \le R$.
In other words, the condition $\lim_{k\to \infty} |a_k | C^k = 0$ for all $C > 0$ implies (and actually is equivalent to) that the radius of convergence of the power series $(*)$ is infinity.
So your idea of proving uniform convergence on compact sets is correct, but (in the general case of arbitrary coefficients $a_k$) there is no closed form for the limit function. The Weierstrass M-test helps because it can be used to prove uniform convergence without knowing the limit function. This is closely related to the Cauchy criterion which can be used to prove convergence of a sequence without knowing the limit.