The following exercise appears in Katznelson's Harmonic Analysis.
Show that $f$ is analytic on $\mathbb T$ if and only if there exist constants $K>0$ and $a>0$ such that $|\widehat f(j)|\le K e^{-a|j|}$.
Of course, the analyticity of $f$ is defined in the usual sense - a function $f$ is analytic on $\Bbb T$ if in a neighborhood of every $t_0 \in \Bbb T$, $f(t)$ can be represented by a power series of the form $\sum_{n=0}^\infty a_n(t-t_0)^n$.
For the first direction, suppose $f$ is analytic on $\Bbb T = \Bbb R/\Bbb Z$, i.e., for every $t_0 \in \Bbb T$, $f$ can be represented by a power series $\sum_{n=0}^\infty a_n(t-t_0)^n$. How should I produce the exponential bound on the Fourier coefficients? I have considered using the compactness of $\Bbb T$, to come up with finitely many points $t_0,t_1,\ldots,t_k$ and subsets $U_0,U_1,\ldots,U_k$ such that $t_i \in U_i$ for each $1\le i\le k$, the $U_i$'s are disjoint, and $\Bbb T = \bigcup_{i=0}^k U_i$. Then, $$|\widehat f(j)| = \left|\int_\Bbb T f(t) e^{-2\pi i jt} \ dt\right| \le \sum_{i=0}^k \left|\int_{U_i} f(t) e^{-2\pi i jt} \ dt \right|$$ Also, there are constants $\{a_{n,i}\}_{0\le i\le k, n\ge 0}$ such that $f(t) = \sum_{n=0}^\infty a_{n,i}(t-t_i)^n$ on $U_i$, so $$\int_{U_i} f(t) e^{-2\pi i jt} \ dt = \int_{U_i}e^{-2\pi i jt} \sum_{n=0}^\infty a_{n,i}(t-t_i)^n \, dt$$ I'm stuck here.
Another approach for the first direction - as $f$ can be locally represented by a power series, $f$ is infinitely differentiable, and so $\widehat f(j) = o(j^{-k})$ for all $k\in \Bbb N$, i.e., $\lim_{|j|\to\infty} j^k \widehat f(j) = 0$ for all $k \in \Bbb N$. This shows that the Fourier coefficients have faster-than-polynomial decay, but it doesn't immediately imply the required exponential decay.
I'm clueless about the other direction.
I'd appreciate any hints or solutions - thank you!
Related: Post 1.
Sketching an answer as requested, we note that $f$ originally defined on the unit circle is complex analytic there (in the sense that for any point on the unit circle, there is a converging power series representing $f$ on this neighborhood) if and only if $f$ extends to a holomorphic function in an annulus $1/R <|z| <R$ for some $R>0$. This follows from the compacity of the unit circle, so we can cover it with finite many disks where such power series exists etc
But a function $f$ is analytic on $1/R < |z| <R$ if and only if it has a Laurent series $f(z)=\sum_{n \in \mathbb Z}a_n z^n$ converging (locally uniformly and absolutely on any compact within) and considering $z=e^{i\theta}$, it follows that $a_n$ are precisely the Fourier coefficients of the original $f$ on the unit circle.
But now for $n \ge 0$ we have that $\sum_{n \ge 0} a_n z^n$ is a converging regular power series in $|z| < R$ so for any $1<\rho<R$ we have $|a_n|\rho^n \to 0$ so the estimate required holds with $a=\log \rho >0$
Similarly $\sum_{n \ge 0}a_{-n}z^{-n}$ converges on $|z| >1/R$ so with $w=1/z$ we get that $\sum_{n \ge 0}a_{-n}w^{n}$ converges on $|w| <R$ so we have the same estimate $|a_{-n}|\rho^n \to 0$ for the coefficients indexed by negative integers and we are done with one implication.
Conversely, if the required estimate holds and we take $R =\log a$ it follows that the Laurent series $g(z)=\sum_{n \in \mathbb Z}a_n z^n$ converges on $1/R <|z| <R$, hence is analytic there and since now $a_n$ are the Fourier series of $f$ (which also converges to $f$ by the given estimates on the unit circle), it follows that $g=f$ on the unit circle, so $g$ is the required analytic extension of $f$ and we are done!