Formulation of Fourier transform

Question

I would like to know about Fourier transform more. I attended a standard lecture of mathematics, but we did not talk about Fourier transform on $L^2$ much, nor the theory of $L^2$. We only defined it as $$(\mathcal{F}f)(y)=\lim_{k\to\infty}\int_{C_k}e^{-2\pi i (x,y)}f(x)\mathrm{d}x$$ where $$C_k=\{x\in\mathbb{R}^n|\,|x|\leq k\}$$ is some circle with radius $k$. I would like to ask why does there have to be the limit through the circle? I guess it is some implication of 2-norm, but I don't really understand what and how.

Answer 1

A square-integrable function on $\mathbb{R}^{n}$ is not necessarily integrable, but it is integrable on any bounded set $S$ because of the Cauchy-Schwarz inequality: $$ \begin{align} \int_{S} |e^{-2\pi i(x,y)}f(x)|\,dx & = \int_{S}|f(x)|\,dx \\ & \le \left(\int_{S}1\,dx\right)^{1/2}\left(\int_{S}|f(x)|^{2}\,dx\right)^{1/2} \\ & \le |S|^{1/2}\|f\|_{L^{2}}. \end{align} $$ However, the Fourier transform is a unitary map from $L^{2}$ to $L^{2}$, which is implemented by the Fourier transform integral when the integral exists. Truncating the Fourier transform to a bounded set $S$ guarantees that the Fourier transform exists as a classical $L^{2}$ function, and the $L^{2}$ norm of that classical transform equals the $L^{2}$ norm of the truncated function being transformed.

Therefore, truncating to a finite region, and allowing that region to fill space will result in a Fourier transform function that converges in $L^{2}$ to the $L^{2}$ transform. You can use spheres or cubes, or whatever other convenient space-filling type of sequence that you like. Spheres and cubes are the easiest to succintly describe. But you could use just as easily use a sequence of bounded sets $\{ S_{n}\}$ over which to integrate, so long as $\chi_{S_{n}}$ converges pointwise everywhere to $1$.