Use of Fubini's Theorem in Papa Rudin's Holomorphic Fourier Transforms

Question

I am starting to read on chapter 19, Holomorphic Fourier Transforms from Real and Complex Analysis by Walter Rudin. In the first page of that chapter I came across the function $$f(z) = \int_0^\infty F(t)e^{itz} dt$$ where $z\in \mathbb{C}$ is in the upper half plane, $F\in L^2(\mathbb{R})$ and $F$ vanishes on $(-\infty, 0)$. I want to prove that $f$ is analytic. I proved that $f$ is continuous in $\mathbb{C}$. To prove that it is holomorphic, the author hints to use Morera's theorem, for which I have to show that for any closed path $\gamma$ in $\mathbb{C}$, the integral $$\int_\gamma f(z) dz =0$$ My course of proof is to show \begin{align*} \int_\gamma f(z)\ dz &= \int_\gamma \int_0^\infty F(t) e^{2\pi i t z} \ dt \ dz \\ & = \int_0^\infty F(t) \int_\gamma e^{2\pi itz} \ dz \ dt \\ & = \int_0^\infty F(t) \cdot 0 \ dt \\ & = 0 \end{align*} But I can't see how the use of Fubini's theorem is justified in the change of order of integration.

Answer 1

Since $z$ lies in the upper half plane, the integral converges exponentially, so everything works out very nicely. More explicitly, let us parameterize, for $s\in [a,b]$, $\gamma(s)=x(s)+iy(s)$ with $y(s)>0$ for all $s$, and suppose $\gamma$ is $C^1$. Let $M=\sup\limits_{s\in [a,b]}|\gamma'(s)|$. Since $y$ is continuous, it has a strictly positive minimum, say $m>0$. Now, we have \begin{align} \int_{\gamma}\int_{0}^{\infty}|F(t)e^{itz}|\,dt\,d|z|&= \int_a^b\int_{0}^{\infty}|F(t)|e^{-ty(s)}\,dt\,|\gamma'(s)|\,ds\\ &\leq M\int_a^b\int_{0}^{\infty}|F(t)|e^{-mt}\,dt\,ds\\ &=M(b-a)\int_{0}^{\infty}|F(t)|e^{-mt}\,dt\\ &\leq M(b-a)\|F\|_{L^2((0,\infty))}\cdot\left(\int_{0}^{\infty}(e^{-mt})^2\,dt\right)^{1/2}, \end{align} where I used Cauchy-Schwarz in the end. Clearly the exponential integral is finite, and also $F\in L^2(\Bbb{R})$ by hypothesis. Hence, our original integral on the left is finite. This is enough for you to apply Fubini to interchange the order of integration.

So, the point is that because we're in the upper half plane, the $e^{itz}$ term decays exponentially, so it belongs to every $L^p((0,\infty))$ space for $1\leq p\leq \infty$ so we can afford (using Holder's inequality) to put $F$ in any $L^{p'}((0,\infty))$ space (where $p'$ is the Holder-conjugate of $p$). The above argument simply used Cauchy-Schwarz ($p=p'=2$).