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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 $0<A<\infty$ 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?
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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 $C_0\cap L^2$.

In particular see Rudin’s exercise 1 of that same chapter.

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