# Use of Plancherel theorem in classical Paley Wiener theorem

I have been reading on holomorphic Fourier transforms from chapter 19, Real and Complex Analysis, Walter Rudin. In the beginning of the chapter he discusses functions of the form $$f(z) = \int_{-A}^A F(t)e^{itz} \ dz$$ where $$0 and $$F \in L^2(-A, A)$$. Rudin hints that the restrictions of the function $$f$$ onto real axis lies in $$L^2$$. He also hints at using Plancherel's theorem. I worked through some of it. I'll detail it here.

By definition of $$f$$, writing $$z=x+iy$$, we get $$f(x+iy) = \int_{-A}^A F(t)e^{-ty}e^{itx} \ dt$$ Restricting $$f$$ to real line, we get $$f(x) = \int_{-A}^A F(t) e^{itx} \ dt$$ But by definition of Fourier transform, we get that $$f(x)$$ is the Fourier transform of $$F(t)$$ in the domain $$(-A, A)$$. Now since $$F\in L^2(-A, A)$$, we can use Plancherel's theorem which will give $$\int_\mathbb{R} |f(x)|^2 \ dx = \int_{-A}^A |F(t)|^2 \ dt$$ and hence will prove that $$f \in L^2(\mathbb{R})$$.

I have two questions.

1. How is Plancherel's theorem justified above? I understand the use of Plancherel's theorem when $$F \in L^2(\mathbb{R})$$ or if the Fourier transform give discrete frequencies.
2. Isn't the same true for any fixed horizontal line, not just real axis?

Sure, think of $$y$$ as fixed and define the function $$\Phi_y:\Bbb{R}\to\Bbb{C}$$, \begin{align} \Phi_y(t)&:= \begin{cases} F(t)e^{-ty}&\text{if t\in (-A,A)}\\ 0&\text{else.} \end{cases} \end{align} Then, $$\Phi_y$$ is certainly measurable, and we have \begin{align} \int_{\Bbb{R}}|\Phi_y(t)|^p\,dt=\int_{-A}^A|F(t)e^{-ty}|^p\,dt\leq e^{Ap|y|}\int_{-A}^A|F(t)|^p\,dt, \end{align} so it is finite for all $$1\leq p\leq 2$$ (I’m using the fact that on a finite-measure space, if you belong to a higher Lebesgue space, then you belong to all lower ones by using Holder’s inequality with the constant function $$1$$). Next, we have obviously from the definitions, \begin{align} f(x+iy)=\int_{-A}^AF(t)e^{-ty}e^{itx}\,dt=\int_{\Bbb{R}}\Phi_y(t)e^{itx}. \end{align}
So, putting these facts together, we have that $$\Phi_y\in L^1(\Bbb{R})\cap L^2(\Bbb{R})$$, and we have that the function $$\phi_y:\Bbb{R}\to\Bbb{C}$$, $$\phi_y(x):=f(x+iy)$$ is indeed the Fourier-transform of $$\Phi_y$$ in both the $$L^1(\Bbb{R})$$ and $$L^2(\Bbb{R})$$ sense. Thus, $$\phi_y\in C_0(\Bbb{R})\cap L^2(\Bbb{R})$$, by the Riemann-Lebesgue lemma and Plancharel. In other words, yes the restriction of $$f$$ to each horizontal line lies in $$C_0\cap L^2$$.