- Suppose \(\hat { f }\) has compact support contained in \([ - M , M ]\) and let \(\begin{array} { r } { f ( z ) = \sum _ { n = 0 } ^ { \infty } a _ { n } z ^ { n } } \end{array}\) Show that
\[ a _ {n} = \frac {(2 \pi i) ^ {n}}{n !} \int_ {- M} ^ {M} \hat {f} (\xi) \xi^ {n} d \xi , \]
and as a result
\[ \limsup _ {n \to \infty} (n! | a _ {n} |) ^ {1 / n} \leq 2 \pi M. \]
In the converse direction, let f be any power series \(\textstyle f ( z ) = \sum _ { n = 0 } ^ { \infty } a _ { n } z ^ { n }\) with lim \(\begin{array} { r } { \operatorname* { s u p } _ { n \to \infty } ( n ! | a _ { n } | ) ^ { 1 / n } \leq 2 \pi M } \end{array}\) . Then, f is holomorphic in the complex plane, and for every \(\epsilon > 0\) there exists \(A _ { \epsilon } > 0\) such that
\[ | f (z) | \leq A _ {\epsilon} e ^ {2 \pi (M + \epsilon) | z |}. \]
Solution
Assume \(\operatorname{supp}\widehat f\subset[-M,M]\). Fourier inversion gives, for real \(x\) and hence by analytic continuation for complex \(z\), \[ f(z)=\int_{-M}^{M}\widehat f(\xi)e^{2\pi i\xi z}\,d\xi. \] Because the interval is compact, differentiation under the integral sign is justified to every order. Thus \[ f^{(n)}(0)=(2\pi i)^n\int_{-M}^{M}\widehat f(\xi)\xi^n\,d\xi. \] Since \(a_n=f^{(n)}(0)/n!\), \[ a_n=\frac{(2\pi i)^n}{n!} \int_{-M}^{M}\widehat f(\xi)\xi^n\,d\xi. \] Hence \[ n!|a_n| \le (2\pi M)^n\|\widehat f\|_{L^1([-M,M])}, \] and therefore \[ \limsup_{n\to\infty}(n!|a_n|)^{1/n}\le2\pi M. \]
Conversely, suppose \[ L:=\limsup_{n\to\infty}(n!|a_n|)^{1/n}\le2\pi M. \] Fix \(\varepsilon>0\) and put \[ R=2\pi(M+\varepsilon). \] Since \(R>L\), there is \(N\) such that for \(n\ge N\), \[ n!|a_n|\le R^n. \] After enlarging a constant \(C_\varepsilon\) to absorb the finitely many indices \(n<N\), we have for every \(n\ge0\) \[ |a_n|\le C_\varepsilon\frac{R^n}{n!}. \] Thus the power series converges absolutely for every \(z\in\mathbb C\) and \[ |f(z)| \le C_\varepsilon\sum_{n=0}^{\infty}\frac{(R|z|)^n}{n!} =C_\varepsilon e^{R|z|} =C_\varepsilon e^{2\pi(M+\varepsilon)|z|}. \] So \(f\) is entire and has the stated exponential-type bound.