Let $\{c_k\} \subset \mathbb C$ sch that $|c_k|\leq e^{-|k|}$ $(k\in \mathbb Z).$ We also assume that $f(\mathbb R) \subset \mathbb R.$
Put $f(x):= \sum_{k\in \mathbb Z} c_k e^{ikx}$ $x\in \mathbb R.$
(We note that $f$ is periodic function on $\mathbb R,$ and its Fourier series is absolutely convergent. Thus, $f$ is a continuous periodic on $\mathbb R.$
Question: Is $f$ is real analytic at some point on $\mathbb R$? Is $g(x)= f(\arcsin (x/r))$ ($|x/r|<1$) is real analytic in the neighborhood of origin?
My thought: (1) We note that series $\sum_{k\in \mathbb Z} c_{k}e^{i k(s+it)}$... (A) converges absolutely for $|t|<1, s \in \mathbb R$? Can we say the sum of series (A) is analytic extension of $f$? (2) We also observe that Fourier coefficient of $f:$ $|\hat{f}(k)|= |c_k| \leq e^{-|k|}$, have very nice decay.
If $|c_k| < e^{-|k|}$ then $\sum_{k=0}^\infty c_k z^k$ and $\sum_{k=0}^\infty c_{-k} z^k$ are complex analytic on $\{z \in \mathbb{C}, |z| < e\}$ so that $\displaystyle F(z) =\sum_{k=-\infty}^\infty c_k z^k$ is complex analytic on $\{z \in \mathbb{C}, 1/e <|z| < e\}$.
Thus $f(x) = \sum_{k=-\infty}^\infty c_k e^{i kx} = F(e^{ix})$ is complex analytic on $\{ x \in \mathbb{C}, Im(x) \in (-1,1)\}$
(complex-analytic is much stronger than real-analytic)
If you start from a continuous $2\pi$ periodic function $g : \mathbb{R} \to \mathbb{C}$ such that $\hat{g}(k) = c_k$, it means that $\forall x \in \mathbb{R}, \ g(x) = f(x)$ and $g$ is real-analytic and extends to a complex analytic function $\{x \in \mathbb{C}, Im(x) \in (-1,1)\} \to \mathbb{C}$
I'm not sure why you want to look at $f(\arcsin (x/r))$, you should look at the composition of analytic functions.