Extract of an article:
"the Laplace transform of $T(t)f$ is an entire function and since the resolvent is meromorphic of finite exponential type it must be of finite exponential type, say $\nu$, too. An application of the Paley-Wiener theorem [2, 6.9.1] now yields $$\int_0^{\infty} e^{-zt} T(t)f \, dt=\int_0^{\nu} e^{-zt} T(t)f \, dt.$$"
The notations are as follows:
- $T(t)$ is the semigroup of an operator $A$ on a Hilbert space $H \ni f$, so that the Laplace transform of $T(t)f$, i.e. $\int_0^{\infty} e^{-zt} T(t)f \, dt$, is equal to the resolvent $(z-A)^{-1}f$ for $z$ large. In this article, the resolvent is assumed beforehand to be meromorphic of finite exponential type and $f$ is assumed to be such that $z \mapsto (z-A)^{-1}f$ is entire.
- The reference [2] is Boas "Entire functions" and the theorem 6.9.1 is the following: " Let $f(z)=\int_a^b e^{izt} g(t) \, dt$, where $g(t)$ is integrable, $|a| \leq b$, and $g(t)$ does not vanish almost everywhere in any neighborhood of $b$; then $f(z)$ is an entire function of order $1$ and type $b$ (and not of smaller type)."
I have no idea how the author uses the theorem of this reference. Any help would be appreciated.