Given a function which Fourier coefficient decay fast as $k^{-k}$, for example \begin{align} f(x):= \sum_{k=1}^\infty \frac{1}{k^k} \exp(2\pi \, i\,k\,x) . \end{align} How can we prove this function is entire?
Denoting by $R$ the convergence radius of the above power series, recently, I can prove that for arbitrary $r>0$ the convergence radius of the above power series is greater than $r$, i.e. $R >r$. Could I conclude that the function is entire?
Indeed, we have \begin{align*} f^{(n)}(x) = \sum_{k=1}^\infty \frac{(2 \pi \, i \, k)^n}{k^k} \exp(2\pi \, i\,k\,x) . \end{align*} Hence, the Taylor expansion at $x_0$ is given by \begin{align*} f(x) = \sum_{n=0}^\infty \left[ \sum_{k=1}^\infty \frac{(2 \pi \, i \, k)^n}{k^k} \exp(2\pi \, i\,k\,x)\right] \frac{1}{n!} (x-x_0)^n . \end{align*} It is easy to check that for arbitrary $r>1$ ( by considering the maximum of the function $f(x) = x^n r^{-x}$ and Stirling's inequality) \begin{align} \frac{k^n}{r^k} \leq \frac{2^n n!}{(\log r)^n}, \end{align} then \begin{align} |f(x)| &\leq \sum_{n=0}^\infty \left[ \sum_{k=1}^\infty \frac{(2 \pi k)^n}{k^k}\right] \frac{1}{n!} (x-x_0)^n\\ & \leq \sum_{n=0}^\infty \left[ \sum_{k=1}^\infty \frac{(4 \pi)^n r^k}{k^k (\log r)^n}\right] (x-x_0)^n . \end{align} It is sufficient to consider the last series. By Cauchy theorem, the radius of convergence is given by \begin{align} 1/R = \lim_{n \to \infty} |a_n|^{1/n} &=\frac{4\,\pi}{\log r} \lim_{n \to \infty} \left( \sum_{k=1}^\infty \frac{r^k}{k^k}\right)^{1/n} = \frac{4\,\pi}{\log r} , \end{align} since the summation converges for any $r$.
We conclude that $R= \frac{\log r}{4\,\pi}$, for arbitrary $r>0$. Thus, the function is entire?
Please give any comment on the proof if you find any bug, or suggest any other solution. I will really appreciate. Thank you.
Let $$f_k(z)=\frac{exp(2\pi i z)}{k^k}.$$ Let $R>0$ and assume $|z|\leq R$. We have $$|f_k(z)| = \frac{1}{k^k}exp(-2\pi y) \leq \frac{1}{k^k} exp(2\pi R).$$ The numerical series $$\sum_{k=1}^{\infty}\frac{1}{k^k}$$ has the convergence radius $+\infty$ (from the radical test). This implies that $$f(z)=\sum_{k=1}^{\infty} \frac{exp(2\pi i z)}{k^k}$$ is uniformly (absolute) convergent on the closed disc $\{|z|\leq R\}$. I guess this implies that $f(z)$ is an entire function, since any compact subset of $\mathbb C$ is a subset in a large enough disc.