\chapter{Fourier Transforms} \label{chapter-fouriertransforms}



\section{Fourier Series}



The Fourier transform serves as the basis of much of the work in this thesis, due in no small part to its many nice properties. The Fourier transform is, in some sense, a continuum analogue of the Fourier series \cite[p. 92]{boggess} and so we choose to briefly investigate the nature of Fourier series. The Fourier series of a sufficiently nice compactly supported (or periodic) function extracts the amplitude and frequency information by expanding the function as a (possibly infinite) linear combination of sines and cosines. Historically, Fourier series were developed to approximate general solutions to the heat equation on a bounded domain \cite{fourier1,fourier2}.

Fourier series play an integral role in communications, largely due to Shannon's sampling theorem \cite{shannon} which states that a band-limited signal can be completely reconstructed from a sufficient number of discrete samples. Much real-world data are continuous, however only discrete data can be stored or analyzed on computers, making the sampling theorem a very crucial theorem in signal reconstruction. The sampling theorem guarantees that the reconstructed data agrees with the original signal within a given tolerance. To this end, Fourier series are quite important in real-world applications. We now formally define the Fourier series.

Given a function $f\in L^1([0,2\pi])$, we may define its (formal) Fourier series (provided that it exists) as the function
%
\begin{equation}
 \frac{1}{\sqrt{2\pi}}\sum_{n=-\infty}^{\infty} e^{inx} \mathcal{F}f(n),
\end{equation}

\noindent where $x\in[0,2\pi]$ and
%
\begin{equation}
 \mathcal{F}f(n) = \frac{1}{\sqrt{2\pi}} \int_0^{2\pi} e^{-iny} f(y)\,dy.
\end{equation}

\noindent Indeed this is the largest domain for which Fourier series can be defined since
%
\begin{equation*}
 \int_0^{2\pi} |e^{-iny}f(y)|\,dy = \int_0^{2\pi} |f(y)|\,dy. 
\end{equation*}

\noindent Both integrals converge if and only if $f\in L^1([0,2\pi])$. Note that the $L^p$ spaces are nested on $[a,b]$ where $-\infty < a < b < \infty$. (In general, if $\X$ is a finite measure space and $p_2 > p_1$, then $L^{p_2}(\X) \subset L^{p_1}(\X)$ \cite[p. 186]{follandreal}.)

\begin{remark}
 The integrals that appear herein are to be interpreted as Lebesgue integrals. However, in many situations of interest in this thesis, the Lebesgue integral may be treated as a Riemann integral.
\end{remark}

Carleson's theorem \cite{carleson,hunt} shows that for $f\in L^p([0,2\pi])$ with $1 < p < \infty$, then the Fourier series of $f$ converges almost everywhere to $f$ when the Fourier series is summed as a principal value sum, that is to say that for $f\in L^p([0,2\pi])$,
%
\begin{equation}
 f(x) = \lim_{N\to\infty} \frac{1}{\sqrt{2\pi}} \sum_{n=-N}^N \mathcal{F}f(n) e^{inx}
\end{equation}

\noindent for almost every $x\in[0,2\pi]$.



\section{The Fourier Transform}



An inherent fault in the theory of Fourier series is that it is built on functions defined on bounded intervals, or more generally on periodic functions. Many functions that are of general interest are not defined on bounded intervals and are not periodic, e.g., $\exp(-x^2/2)$ for $x\in\R$. This leads to the theory of the Fourier transform. In some sense, the Fourier transform may be viewed as a continuum analogue of Fourier series.

The Fourier transform can be abstractly defined via group theoretic techniques \cite{follandharmonic,rudingroups,simon}. Given a locally compact abelian group $G$, one may define the set of \emph{continuous} characters, denoted $\widehat{G}$, to be the group of functions $\chi:G\to\mathbb{T}$, where $\mathbb{T}$ is the unit circle, satisfying $\chi(g_1g_2) = \chi(g_1)\chi(g_2)$. The set of continuous characters forms a group under multiplication called the \emph{dual group}. One can define the Fourier transform from $G$ to $\widehat{G}$ by $\mathcal{F}f(\chi) = \int_G \overline{\chi(g)} f(g)\,dg$ and a Fourier transform from $\widehat{G}$ back to $G$ similarly. These Fourier transforms are in some sense inverses of each other. Particularly, for suitably nice functions, the Fourier transforms are indeed inverses of each other. Notably, Fourier transforms come in pairs in the context of locally compact groups. For more information see \cite{follandharmonic,rudingroups,simon}. We will see in the next chapter that our generalizations of the Fourier transform also come in pairs.

It is a simple exercise to show that the dual group of the group $[0,2\pi)$ under addition modulo $2\pi$ is the group of integers \cite[p. 90]{follandharmonic}. In the defintion of the Fourier series appeared an integral over $[0,2\pi)$ and a sum over $\Z$. Hence, the Fourier series is a specific realization of the abstract Fourier transform on $[0,2\pi)$ and $\Z$. Considering instead the group of real numbers $\R$ with addition, a straightforward argument shows that if $f:\R\to\mathbb{T}$ is a continuous character, then $f(x) = \exp(ixy)$ for some fixed $y\in\R$ \cite[p. 90]{follandharmonic}. Thus, the dual group of $\R$ is again $\R$, and so we expect that the Fourier and inverse Fourier transforms on $\R$ are closely related. This leads into our first definition.

\begin{definition}
 Let $f\in L^1(\R)$ and $y\in\R$, then its \emph{Fourier transform}, denoted $\mathcal{F}f$, is defined to be
 %
 \begin{equation}
  \mathcal{F}f(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ixy}f(x)\,dx.
 \end{equation}
\end{definition}

\noindent This is well-defined for $f\in L^1(\R)$ since
%
\begin{equation*}
 \int_{-\infty}^{\infty} |e^{-ixy}f(x)|\,dx = \int_{-\infty}^{\infty} |f(x)|\,dx < \infty.
\end{equation*}

\noindent Both integrals $\int_{-\infty}^{\infty} e^{-ixy}f(x)\,dx$ converge for $y\in\R$ as it is absolutely integrable. Furthermore, this is the largest domain of definition for the Fourier transform as an integral operator.

The Fourier transform has numerous properties which are direct consequences of the Fourier kernel $\exp(-ixy)$ being a continuous character on $\R$, the Lebesgue measure, and the $L^1(\R)$ space. Some of these properties are summarized below. Proofs may be found in \cite[p. 179]{rudinreal}.
%
\begin{theorem}
 Let $f,g\in L^1(\R)$, $x',y,y'\in\R$, $\alpha\in\R^*$, $f_{x'}(x) = f(x-x')$, $(f\ast g)(x) = \int_{-\infty}^{\infty} f(x-\widetilde{x})g(\widetilde{x})\,d\widetilde{x}$, and $\D_{\alpha}f(x) = \sqrt{|\alpha|} f(\alpha x)$, then
 %
 \begin{enumerate}
  \item $\mathcal{F}f\in C_0(\R)$, i.e., $\mathcal{F}f$ is continuous on $\R$ and goes to zero at infinity,
  \item $\mathcal{F}(f_{x'})(y) = e^{-ix'y}\mathcal{F}f(y)$,
  \item $\mathcal{F}(e^{ixy'}f)(y) = \mathcal{F}f(y-y')$,
  \item $\mathcal{F}(f\ast g)(y) = \sqrt{2\pi}\mathcal{F}f(y)\cdot\mathcal{F}g(y)$,
  \item $\mathcal{F}(\D_{\alpha}f)(y) = \D_{\alpha^{-1}}\mathcal{F}f(y)$,
  \item If $f\in AC(\R)$, then $\mathcal{F}(f')(y) = iy\mathcal{F}f(y)$,
  \item If $xf\in L^1(\R)$, then $\mathcal{F}(xf)(y) = -i\frac{d}{dy}\mathcal{F}f(y)$.
 \end{enumerate}
\end{theorem}

The first property is known as the Riemann-Lebesgue lemma. The second and third properties are restatements of the fact that the function $x,y\mapsto \exp(-ixy)$ is a group homomorphism of $\R$ to $\mathbb{T}$. The fourth property relies on this fact, in addition to the Fubini-Tonelli theorem. The fifth is a consequence of the fact that the Fourier kernel is a product kernel, i.e., $f(x,y) = g(xy)$ for some function $g$. The last two are consequences of the fact that the function $x\mapsto \exp(\lambda x)$ is an eigenfunction of the derivative operator.

\begin{definition}
 Let $f\in L^1(\R)$ and $y\in\R$. The inverse Fourier transform of $f$, denoted $\mathcal{F}^{-1}f$, is defined to be
 %
 \begin{equation}
  \mathcal{F}^{-1}f(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ixy} f(x)\,dx.
 \end{equation}
\end{definition}

Indeed, as the name suggests, the operator $\mathcal{F}^{-1}$ does act as the inverse of the Fourier transform for such functions $f$. Comparing this to the definition of the Fourier series, we see that $\mathcal{F}f$ plays an analogous role to the integral $(2\pi)^{-1/2}\int_0^{2\pi} e^{-iny}f(x)\,dx$, whereas $\mathcal{F}^{-1}$ plays the role of the summation and the following identity holds for almost every $x\in\R$ provided that $f,\mathcal{F}f\in L^1(\R)$ \cite[p. 185]{rudinreal}:
%
\begin{equation}
 f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{ixy}\mathcal{F}f(x)\,dx.
\end{equation}

If $f\in L^1(\R)$ it is not guaranteed that $\mathcal{F}f\in L^1(\R)$ and so the above integral may not exist, i.e., $f$ may not be invertible. A typical example of such an $f$ is $f = \frac{1}{2}\chi_{[-1,1]}\in L^1(\R)$. For this choice of $f$, $\mathcal{F}f(y) = \sqrt{\frac{2}{\pi}} \frac{\sin y}{y}$. $\mathcal{F}f$ is not absolutely integrable on $\R$ and is therefore not in $L^1(\R)$. Instead one may argue that if $\mathcal{F}f\in L^1(\R)$, then we could realize $\mathcal{F}^{-1}\mathcal{F}f(y)$ as $\mathcal{F}^2f(-y)$. $\mathcal{F}^2f(-y)$ will be continuous by the Riemann-Lebesgue lemma, however the original definition of $f$ is not continuous. An adaptation of the latter argument works quite well for more general locally compact abelian groups.

The $L^1$ theory of the Fourier transform is quite nice, but it is perhaps the $L^2$ theory that is most appealing as its functional analytic properties on $L^2(\R)$ are even richer. Many of the $L^1$ properties can be adapted to the $L^2$ theory. To develop the Fourier transform on $L^2(\R)$, one first considers the dense subspace $L^1(\R)\cap L^2(\R)$ in $L^2(\R)$. That this is a dense subspace of $L^2(\R)$ is fairly straightforward. Given $f\in L^2(\R)$, define the sequence $f_n = \chi_{[-n,n]}f$. These functions are compactly supported and in $L^2(\R)$. By the nesting of $L^p$ spaces on finite measure spaces \cite[p. 186]{follandreal} (of which $[-n,n]$ under the Lebesgue measure is an example), each of the $f_n$ are in $L^1(\R)$. A straightforward calculation shows that $\|f-f_n\|_2 \to 0$ as $n$ tends to infinity. The heuristic argument is that in order for $f\in L^2(\R)$, the integral of its tail, represented by $f-f_n$, must go to zero.

The Fourier transform on $L^1(\R)\cap L^2(\R)$ is an $L^2$ isometry. That is to say that if $f\in L^1(\R)\cap L^2(\R)$, then $\mathcal{F}f\in L^2(\R)$ and $\|\mathcal{F}f\|_2 = \|f\|_2$. The standard proof for this involves regularizing the integrals with exponentially decaying functions \cite{rudinreal}, be they of the form $\exp(-|t|)$ or $\exp(-x^2)$. Since $L^1(\R)\cap L^2(\R)$ is dense in $L^2(\R)$ and the Fourier transform is an $L^2$ isometry from $L^1(\R)\cap L^2(\R)$ to $L^2(\R)$, it extends to an isometry on $L^2(\R)$. This extension cannot be realized as an integral operator directly as the integral of $\exp(-ixy)$ against an $L^2$ function need not exist.

The extension of the Fourier transform from $L^1(\R)\cap L^2(\R)$ to all of $L^2(\R)$ is traditionally also called the Fourier transform, though the understanding is not lost as the Fourier transforms on $L^1(\R)\cap L^2(\R)$ and on $L^2(\R)$ are quite different in nature. The Fourier transform on $L^2(\R)$ has numerous properties, sharing many with the Fourier transform on $L^1(\R)$. Let $\langle\cdot,\cdot\rangle$ denote the usual inner product on $L^2(\R)$. Some unique properties of the $L^2(\R)$ Fourier transform are
%
\begin{enumerate}
 \item $\langle \mathcal{F}f,\mathcal{F}g\rangle = \langle f,g\rangle$,
 \item $\mathcal{F}$ is onto $L^2(\R)$,
 \item $\mathcal{F}$ is a unitary,
 \item $\mathcal{F}^4 = I$,
 \item $\sigma(\mathcal{F}) = \{\pm 1,\pm i\}$.
\end{enumerate}

That $\mathcal{F}^4 = I$ follows from the fact that $\mathcal{F}^2$ acts as a reflection operator, i.e., $\mathcal{F}^2 f(x) = f(-x)$ on $L^2(\R)$. This is ultimately a consequence of the symmetry between the Fourier and inverse Fourier transforms, particularly, the two differ only insofar that $\exp(-ixy)$ and $\exp(ixy)$ differ by an overall sign change. Perhaps the most surprising and enticing property of the $L^2(\R)$ Fourier transform is the uncertainty principle associated to it.

\begin{theorem}
 Let $f\in L^2(\R)\cap AC(\R)$ be such that $x\mapsto xf(x)\in L^2(\R)$ and $y\mapsto y\mathcal{F}f(y)\in L^2(\R)$, then
 %
 \begin{equation}
  \left(\int_{-\infty}^{\infty} x^2|f(x)|^2\,dx\right)\left(\int_{-\infty}^{\infty} y^2|\mathcal{F}f(y)|^2\,dy\right) \ge \frac{1}{4}\left(\int_{-\infty}^{\infty}|f(x)|^2\,dx\right)^2.
 \end{equation}
\end{theorem}

There are many proofs for this theorem \cite{galindo,grochenig,stein}, most of which make use of some form of the Cauchy-Schwarz inequality. We provide one proof below.

\begin{proof}
 Let $f\in L^2(\R)\cap AC(\R)$ be such that $x\mapsto xf(x)\in L^2(\R)$ and $y\mapsto y\mathcal{F}f(y)\in L^2(\R)$, then $f'$ exists and $f'\in L^2(\R)$. By integration by parts,
 %
 \begin{equation*}
  \left(\int_{-\infty}^{\infty} |f(x)|^2\,dx\right)^2 = 4\left(\operatorname{Re}\int_{-\infty}^{\infty} xf(x)f'(x)\,dx\right)^2.
 \end{equation*}

 Note that $f$ goes to zero at infinity since $f\in AC(\R)$ and $|f|^2\in L^1(\R)$. Using Cauchy-Schwarz with the auxiliary functions $\eta(x) = xf(x)$ and $\xi(x) = f'(x)$ as well as the established equivalence $\mathcal{F}(f')(y) = iy\mathcal{F}f(y)$,
 %
 \begin{align*}
  4\left(\operatorname{Re}\int_{-\infty}^{\infty} \eta(x)\xi(x)\,dx\right)^2 & \le 4 \left(\int_{-\infty}^{\infty} |\eta(x)|^2\,dx\right)\left(\int_{-\infty}^{\infty} |\xi(x)|^2\,dx\right)\\
  &= 4\left(\int_{-\infty}^{\infty} x^2|f(x)|^2\,dx\right)\left(\int_{-\infty}^{\infty} y^2 |\mathcal{F}f(y)|^2\,dy\right).
 \end{align*}

 \noindent Thus
 %
 \begin{equation*}
  \left(\int_{-\infty}^{\infty} x^2|f(x)|^2\,dx\right)\left(\int_{-\infty}^{\infty} y^2 |\mathcal{F}f(y)|^2\,dy\right) \ge \frac{1}{4} \left(\int_{-\infty}^{\infty} |f(x)|^2\,dx\right)^2.
 \end{equation*}
 
 Equality holds above if and only if $\xi = -\lambda\eta$ for some $\lambda \in \C$, i.e., $f'(x) = -\lambda xf(x)$ and so $f(x) = \exp(-\lambda x^2/2)$. Provided that $\operatorname{Re}(\lambda) > 0$, $f\in L^2(\R)$ and results in a modulated Gaussian. A simple computation shows that for such a choice of $f$, equality indeed holds.
\end{proof}

The function $\exp(-x^2/2)$ is not just a minimizer for the uncertainty product, it is also an eigenfunction of the Fourier transform. We prove this below. Define $f$ by
%
\begin{equation}
 f(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ixy} e^{-\frac{x^2}{2}}\,dx.
\end{equation}

\noindent Taking a derivative of both sides with respect to $y$ and noting that we may differentiate under the integral sign, we get
%
\begin{equation*}
 f'(y) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} -ix e^{-ixy} e^{-\frac{x^2}{2}}\,dx.
\end{equation*}

\noindent $-x\exp(-x^2/2)$ can readily be viewed as the derivative of $\exp(-x^2/2)$, and an integration by parts yields
%
\begin{equation*}
 f'(y) = -\frac{y}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-ixy} e^{-\frac{x^2}{2}}\,dx = -y f(y).
\end{equation*}

\noindent The solution to this is clearly of the form $C\exp(-x^2/2)$.

Another standard technique for showing that the Gaussian is an eigenfunction of the Fourier transform makes use of rewriting $\exp(-ixy)\exp(-x^2/2)$ as a single exponential with complex argument, completing the square, and making use of the entirety of the exponential function to do a complex change of variables.

As noted above, the Fourier transform interchanges the roles of the multiplication operator $\mathcal{M}_{ix}$ and the derivative operator $\frac{d}{dy}$. The Fourier transform also interchanges the roles of $\mathcal{M}_y$ and $i\frac{d}{dx}$; as such the operator $\frac{d}{dx}+x$ is form-invariant under the Fourier transform, that is to say that
%
\begin{equation}
 \mathcal{F}\left(\left(\frac{d}{dx}+x\right)f\right)(y) = i\left(\frac{d}{dy}+y\right)\mathcal{F}f(y).
\end{equation}

\noindent Likewise,
%
\begin{equation}
 \mathcal{F}\left(\left(-\frac{d}{dx}+x\right)f\right)(y) = -i\left(-\frac{d}{dy}+y\right)\mathcal{F}f(y).
\end{equation}

\noindent The operators $\frac{d}{dx} + x$ and $-\frac{d}{dx}+x$ will play an important role in Chapter 3.

The Fourier transform is, unsurprisingly, critical to the spectral theory of the operator $-\frac{d^2}{dx^2}$. This should be expected since the Fourier transform is unitary and also enjoys the property that its kernel $\exp(-ixy)$ is a (non-$L^2$) eigenfunction of the operator $-\frac{d^2}{dx^2}$. The image of the operator $-\frac{d^2}{dx^2}$ under the Fourier transform is $y^2$, and likewise the image of the multiplication operator $\mathcal{M}_{x^2}$ under the Fourier transform is the operator $-\frac{d^2}{dy^2}$. As such, it is natural to construct the operator
%
\begin{equation*}
 -\frac{1}{2} \frac{d^2}{dx^2} + \frac{1}{2} x^2.
\end{equation*}

\noindent This operator will be invariant under the Fourier transform, much like the operators $\frac{d}{dx}+x$ and $-\frac{d}{dx}+x$. Such operators are called eigenoperators and some theory has been worked out for them \cite{mccallum}.









