In Exercise I.4.4 of Katznelson's book An Introduction to Harmonic Analysis we learn that the Fourier series $$f(z) = \sum_{k=-\infty}^{+\infty} f_k z^k$$ defines an analytic function in the neighborhood of the unit circle if and only if there are constants $A_f , \alpha_f>0$ so that $$|f_k| \leq A_f e^{ - \alpha_f |k|}.$$ In Chapter 7, Katznelson mentions that the compactness of the circle makes this characterization much simpler than the non-compact case of real line and the Paley-Wiener theorem.
To the question at hand: the exponential decay above implies that the $L^2$ Sobolev norms
$$||f||_s = \sum_{k=0}^{\infty} k^{2s} |f_k|^2 \leq (2s)! A_f B_f^{2s}$$
of regularity $s=0,\tfrac{1}{2}, 1, \tfrac{3}{2}, 2, \ldots$ can all be bounded as above using some constants $A_f, B_f$. For the implication, one plugs the first inequality into the expression for the Sobolev norm and uses geometric series to get $B_f = (1 - e^{-2\alpha_f})^{-1}$. Notice that the radius $B_f$ goes to infinity as $\alpha_f \rightarrow 0$ as expected.
Question: is the converse true? That is, can we characterize the analytic $f$ among all smooth $f$ as being exactly those for which the sequence of Sobolev norms admits a bound $||f||_{s} \leq (2s)! A_f B_f^{2s}$ for all $s$?