Showing that a Fourier transform is holomorphic

Question

Let $f\in \mathcal{S}(\mathbb{R})$ be a Schwartz function. I would like to show that the Fourier function $$F(z)=\int_{\mathbb{R}}f(t)e^{itz}\, dt$$ is an entire function.

Here is my approach:

1. Write $z=x+iy$ and note that $itz=it(x+iy)=-ty+itx$. Thus, $$F(z)=\int_{\mathbb{R}}f(t)e^{-ty}e^{itx}\, dt$$ which shows that $F(z)$ exists for each $z$, because $f(t)e^{-ty}$ is also a Schwartz function and $F(z)$ is written as the (real) Fourier transform of a Schwartz function which is convergent.
2. Use Morera's theorem (together with Cauchy integral formula and Fubini's theorem) to conclude.

Of course, to complete the second step I need to check the continuity of $F$ which probably requires the dominated convergence theorem.

My questions is whether there is an alternative/neater way of proving that $F$ is entire. (For instance, can Morera's theorem be avoided by an application of the dominated convergence theorem or one of its cousins to differentiate under the integral sign?)

Answer 1

For this answer, I will suppose that $x \mapsto f(x)e^{a|x|}$ is $L^1$ for every $a \geq 0$.

The theorem of holomorphy under the integral sign is :

Theorem :

Let $(E,\mu)$ be a measured space, $U$ an open set of $\mathbb{C}$ and $f$ a function from $E \times U$ to $\mathbb{C}$. We suppose that :

• $x \mapsto f(x,z)$ is measurable on $E$ for every $z\in U$
• $z \mapsto f(x,z)$ is holomorphic on $U$ for $\mu$-almost every $x \in E$
• there exists $g \in L^1(E)$ such that for every $z\in U$ and $\mu$-almost every $x \in E$, $|f(x,z)|\leq|g(x)|$

Then, $F : z \mapsto \int_E f(x,z) \text{d}\mu(x)$ is holomorphic on $U$ and for every $k \in \mathbb{N}$ : $$F^{(k)}(z)=\int_E \dfrac{\partial^k f}{\partial z^k}(x,z) \text{d}\mu(x)$$ and these integrals are well defined and finite.

To come back to the initial problem, let's consider $E=\mathbb{R}$ with the Lebesgue measure and $f(x,z)=f(x)e^{izx}$. Fix $a>0$ and define $U_a=\lbrace z \in \mathbb{C} \; | \; -a<\text{Im}(z)<a \rbrace$.

For every $x \in \mathbb{R}$ and $z \in U_a$, we have : $$|f(x,z)|=|f(x)e^{izx}|=|f(x)e^{-\text{Im}(z)x}|\leq |f(x)|e^{a|x|}$$

Thus, we can apply the theorem with the control function $g_a(x)=f(x)e^{a|x|}$, which is $L^1$ by hypothesis. Hence, $F$ is holomorphic on any open band $U_a$ and thus, is holomorphic on $\mathbb{C}$.

Edit : I'll add a little remark on the theorem. The great difference between the theorem of holomorphy under the integral and the one of differentiation under the integral is that here, we only have to find a control function $g$ on the initial function $f$, not on its derivative, to have the existence and the expression of $F$ and its derivatives (this is due to the Cauchy inequalities, which allows to control the derivatives of $f$ just by controling $f$).