Is the series $f(a)=\sum_{k} c_k a^{ib_k}$ an analytic function of $a\in \mathbb{R},a>0$ when $b_k\in \mathbb{R}, c_k\in \mathbb{C}$ is given as the parameter.
Since $a^{ib_k}=\exp(ib_k\log(a))$, it is easy to see that $a^{ib_k}$ is analytic. But not all the infinite sum of analytic functions is analytic. So what I want to ask is that $f(z)$ analytic? Thank you advance for your idea!
It is analytic function of $z = a^i$ when $0 < b_k < b_{k+1}$, $b_k \in \mathbb{Z}$ and and $\limsup |c_k|^{1/b_k} < 1$. You can write your function as $f(z) = \sum_k c_k z^{b_k}$ and under above confition $f$ is an analytic function of $z$. Now put $z = a^i$.
I dont think it is an analytic function of $a$ under above conditions, since $z = g(a) = a^i = \exp(i \log(a))$ is analytic function of $\log(a)$ when $a>0$ but $\log(a)$ is not analytic. $\log(a)$ is analytic only in a certain neighbourhood. Check Taylor series for $\log(a)$, it does not uniformly converge for all $a>0$.
A necessary condition for the function to be analytic is if $f(a) = \sum_k c_k a^{ib_k}$ is such that $\limsup_n |f^n(a^*)|^{1/n} < 1/|a-a^*|$ for all $a>0$ for some fixed $a^*$ where $f^n(a^*)$ is the $n^{th}$ derivative of $f(a)$ at $a^*$. Check this condition for the function $f(a) = a^i$. Non uniform convergence of $\log(a)$ should cause an issue in satisfying this necessary condition. If you prove $f(a) = a^i$ is not analytic then you are done. Actually if you prove $f(a) = a^i$ as analytic then also you are done under the condition specified in the first point of this answer.