
\chapter{Fourier transform}




Fourier series are a representation for periodic functions, however what about extending these techniques to non-periodic functions? This can be done by upgrading from a Fourier series to a 'Fourier integral'.

We will require different tools; hence the Fourier transform will become our analogue for the Fourier coefficients, acting as a kernel for the Fourier integral.


If $f$ equals some Fourier series, we want to find some sequence $\widehat{f}(n)$ such that

\[ f(x) = \sum^{\infty}_{n=-\infty} \widehat{f}(n) e^{inx} \]

if $f$ equals some 'Fourier integral', we want to find some function $\widehat{f}(\xi)$ such that

\[ f(x) = \int_{\mathbb{R}} \widehat{f}(\xi) e^{ix\xi} d\xi \]

The basic idea is that instead of considering a countably infinite, discrete set of frequency functions (in our case, $e^{inx}$ with $n \in \mathbb{Z}$), we now consider an uncountably infinite, continuous set of frequency functions ($e^{ix\xi}$ with $\xi \in \mathbb{R}$). Due to this continuous nature we end up studying an integral representation instead of a sum representation like in the Fourier series.


In our study of Fourier series, we started with the question of finding a trigonometric series representation if a function and finished with our solution; a method to compute the appropriate Fourier series coefficients.

For 'Fourier integrals', we will start the other way around; we will define the Fourier transform and then show that it is the appropriate weight for the 'Fourier integral' as a result called the Fourier inversion theorem.

\section{Fourier transform}

As alluded to, the Fourier transform will be the kernel that weighs how much each $e^{i\xi x}$ wave should be included in the Fourier integral to formulate the function.








\begin{definition}[Fourier transform]
For a real function $f$ defined on $\mathbb{R}$, the \emph{Fourier transform} (continuous spectrum) is the integral transform $\mathcal{F}$ operating on real functions that describes how trigonometric functions of frequency $\xi $ are present in a function (if the Fourier transform exists).
%\[  \mathcal{F} \{ f \} (\xi) =  \widehat{f}(\xi ) = \int^{\infty}_{-\infty} f(x) e^{-i 2 \pi \xi x} dx \]
\[  \mathcal{F} \{ f \} (\xi) =  \widehat{f}(\xi ) = \int^{\infty}_{-\infty} f(x) e^{-i \xi  x} dx \]
\begin{itemize}
	\item $f $ is a real function
	\item $\mathcal{F}\{f\}$ is the Fourier transform
\end{itemize}
\end{definition}






\begin{proposition}
%\[ f,g \in L^1 (\mathbb{R}) \]
\[ f,g  \text{ are real functions}\]
\begin{itemize}
	\item $ \mathcal{F} \{ af+bg \} = a \mathcal{F}\{f\} + b \mathcal{F}\{g\} $
	\item $ \mathcal{F} \{ f' \}   =  i \xi \mathcal{F} \{ f \} $
	\item $ \mathcal{F} \{ f^{*} \}(\xi)   =  [\mathcal{F} \{ f \}(-\xi)]^{*} $
	\item $ \mathcal{F}\{f\} \text{ is uniformly continuous}$
	\item $ \mathcal{F} \{ f(x-a) \} = e^{ - i a \xi} \mathcal{F} \{ f \} $
	\item $ \int_{-\infty}^{\infty} \widehat{f}(x)g(x) dx = \int_{-\infty}^{\infty} f(x)\widehat{g}(x) dx  $
	\item $ \mathcal{F}\{f\}( \xi - a) = \mathcal{F}\{ e^{i a x} f\}( \xi ) $
	\item $ \mathcal{F}\{f(ax)\}( \xi ) = \frac{1}{a} \mathcal{F}\{f\}( \frac{\xi}{a} ) $
\end{itemize}
\end{proposition}



These elementary properties may seem similar to those for Fourier coefficients. Indeed (no book on mathematics is complete without copious proliferation of the word 'indeed'), we will discuss many properties of the Fourier transform that are also applicable for Fourier series coefficients!


Both being integral transforms with similar kernels, the Fourier coefficients are related to the Fourier transform in the following way!

$\hat{f}(n)= \frac{1}{T} \mathcal{F}\{f\}( n\omega ) $


The selfduality of the Foruier transform
$\int^{\infty}_{\infty} \mathcal{F}\{f\}(\xi) g(\xi) d\xi = \int^{\infty}_{\infty} f(\xi) \mathcal{F}\{g\}(\xi) d\xi$

If $f$ is a real function that is absolutely integrable on $\mathbb{R}$ then its Fourier transform exists. Futhermore, the Fourier transform of such a function would be continuous.


If $f$ is a continuously differentiable real function whose continuous spectrum (Fourier transform) exists then the Fourier transform of $f'$ exists and follows $ \mathcal{F} \{ f' \}   =  i \xi \mathcal{F} \{ f \} $


\section{}


The Schwartz space is discussed in the advanced section of Real Analysis; it is a space that happens to find very nice interactions with the Fourier transform.










\section{Convolution theorem (Fourier transform)}


Now we will lead to the big result relating to the convolution operator in Fourier analysis. We know that $\mathcal{F}\{f\} +\mathcal{F}\{g\}=\mathcal{F}\{f+g\}$ due to the linearity of the Fourier transform, but is there a similar result for compactifying $\mathcal{F}\{f\} \mathcal{F}\{g\}$ into 1 Fourier transform?

The answer is 'yes', and our friend convolution comes in clutch.

\begin{theorem}[Convolution theorem (Fourier transform)]
Let $f,g$ be absolutely integrable on $\mathbb{R}$, then the following holds
\[ \mathcal{F}\{f * g\}(\xi) = \mathcal{F}\{f\}(\xi)\mathcal{F}\{g\}(\xi) \]
\end{theorem}

This is good news for signal processing engineers, since they now have an easier way to calculate the Fourier transform for their superimposed signals.It also facilitates problems in PDEs.


\section{Riemann-Lebesgue lemma}

Similar to our study on fFourier series, we will require an analogue of the Riemann-Lebesgue lemma.

\begin{lemma}[Riemann-Lebesgue lemma (Fourier transform)]
Let $f$ be an absolutely integrable, piecewise continuous function on $\mathbb{R}$, then the following holds.
\[ \lim_{\xi \to \pm\infty} \mathcal{F}\{f\}(\xi) = 0 \]
\end{lemma}



\section{Fundamental theorem of Fourier integrals}


When studying Fourier series of periodic functions, the ultimate question was when the Fourier series would converge to its own function. We will ask ourselves the same thing with the Fourier integral, when (and how, recalling that there exist different notions of convergence) does the Fourier integral converge to its own function?

The Fourier transform must exist in the first place, but even if it does, it does notguarantee that the Fourier integral will converge, yet alone back to the original function!




\begin{definition}[Inverse Fourier transform (Fourier integral)]
	\[ \mathcal{F}^{-1} \{f \} (x) = \frac{1}{2\pi}\int^{\infty}_{-\infty} f( \xi ) e^{i\xi x} d\xi \]
\end{definition}

\begin{theorem}[Fundamental theorem of Fourier integrals]
If $f$ is absolutely integrable and piecewise smooth function on $\mathbb{R}$, then the following CPV holds.
\[ \mathcal{F}^{-1} \{ \mathcal{F}\{f\} \} (x) =\frac{f(x+)+f(x-)}{2}\]
\end{theorem}

\begin{proposition}
Let$f,g$ be absolutely integrable piecewise smooth functions such that $\mathcal{F}\{f\}=\mathcal{F}\{g\}$, then $f=g$ whenever they are both continuous.
\end{proposition}

This proposition justifies us as talking about the Fourier integral as the inverse Fourier transform.

\begin{proposition}
	\[\mathcal{F}\{f(-x)\}(\xi) =2\pi f(\xi)\]
\end{proposition}



\section{Plancherel's theorem}


\[\int^{\infty}_{-\infty} |f(x)|^2 dx = \frac{1}{2\pi} \int^{\infty}_{-\infty} \mathcal{F}\{f\}(\xi)|^2 d\xi\]



\section{Fourier transform for distributions}

Like with the derivative of distributions, we would like the Fourier transform for distributions and real functions to coincide for real functions that are also distributions.

\begin{definition}[Fourier transform for (Schwartz) distributions]
\[ \langle \mathcal{F}\{T\},\phi \rangle = \langle T , \mathcal{F}\{\phi\} \rangle \]
\end{definition}

\subsection{Fourier transform on Schwartz functions}
Since we are dealing with Schwartz distributions, it is useful to understand how the Fourier transform works on Schwartz functions.
We first note that the fundamental theorem of Fourier integrals applies to Schwartz functions, and that the following fact holds.

\begin{proposition}
The Fourier transform as an operator on Schwartz space is injective.
\[\mathcal{F}(\mathcal{S}(\mathbb{R}))=\mathcal{S}(\mathbb{R})\]
\end{proposition}



\[ \mathcal{F}\{1\}(\xi) = 2\pi \delta (\xi) \]
\[ \mathcal{F}\{e^{iax}\}(\xi) = 2\pi \delta (\xi - a) \]


\begin{definition}[Dirac comb]
\[ \mathrm{comb}_T (x) = \sum^{\infty}_{n = -\infty} \delta (x-nT)\]
\end{definition}
%\newcommand\sh[1]{\ensuremath{\mathop{\text{\normalfont\fontencoding{T2A}\selectfont ш}}#1}}


We now define what it means to shift a distribution.

it comes from the idea of again thinking of distributions as idealized integral transform kernels; if a kernel was translated, the reverse chain rule would relate it such that the following would hold

\[ \langle T(x-c) , \phi \rangle = \langle T , \phi(x+c) \rangle \]
\[ \langle \mathcal{F}\{T'\} , \phi \rangle = \langle \mathcal{F}\{T\} , ix\phi(x) \rangle \]
\[ \langle \mathcal{F}\{T'\} , \phi \rangle = \langle \mathcal{F}\{T\} , ix\phi(x) \rangle \]
\[ \langle \mathcal{F}\{T\}(-\xi) , \phi \rangle = \langle 2\pi T , \phi x) \rangle \]
\[ \langle \mathcal{F}\{\mathcal{F}\{T(-x)\}\}, \phi \rangle = \langle 2\pi T , \phi x) \rangle \]















%\chapter{More Fourier series}

%Although we're familiar with the motive, definition, and derivation of Fourier series, we haven't talked about the interesting results relating to it like we've done with the Fourier transform.

%We'll start by porting all the Fourier transform results that apply to Fourier series coefficients. Recall that we calculate Fourier series for functions on compact intervals, so the general idea to port such results over is to consider the function on $\mathbb{R}$, however mapping all domain elements outside this compact interval to 0 (this is called giving the function 'support'). This essentially makes the Fourier transform the Fourier series coefficient function!

%Periodic L^1 space
%\begin{theorem}[Plancherel theorem (Fourier coefficients)]
%\end{theorem}

%\begin{theorem}[Parseval's theorem (Fourier coefficients)]
%\[ f \in L^2 \implies  \frac{1}{2\pi} \int^{\pi}_{-\pi} |f (x)|^2 dx =  \sum^{\infty}_{n=-\infty} | \widehat{f}(n)|^2 \]
%\end{theorem}

%One can actually solve the Basel problem by applying Parseval's theorem to $|x|$ on $[-\pi,\pi]$.

%\begin{lemma}[Riemann-Lebesgue lemma (Fourier coefficients)]
%\[f \text{ is continuous on } I \land f \text{ is } 2\pi \text{periodic on } \mathbb{R} \implies \lim_{n\to\infty} |\widehat{f}(n)|=0 \]
%\end{lemma}









\part{Advanced}

\chapter{$L^p$ analysis on Fourier series}

The fundamentals part of this book develops the theory of Fourier analysis with reference only to fundamental real analysis and basic complex analysis, however te introduction of more advanced tools such as Lebesgue integration, $L^p$ spaces and ideas from functional analysis allows for a much more advanced study of Fourier series and integrals.

On a less important note, the domain of periodic functions on $[-\frac{T}{2}, \frac{T}{2}]$ can be considered as the circle group when adding domain elements; Fourier series essentially exist on the circle group.

In this chapter, we will employ the Lebesgue integral rather than the Riemann integral so that we can study an alternatives to the fundamental theorem of Fourier series, notably the fact that the Fourier series of $L^2$ functions converges pointwise almost everywhere.

This is a slightly weaker type of convergence being considered, however it applies to a larger class of functions as well as being a Hilbert space, meaning that we can apply functional analysis to study Fourier series!




\begin{theorem}[Carleson's theorem]
Let $f$ be a function in $L^2$, then the Fourier series of $f$ converges to $f$ pointwise almost everywhere.
\end{theorem}


\chapter{$L^p$ analysis on Fourier integral}

Functional analysis have since become a major tool in the study of Fourier analysis, since (Lebesgue) integrable functions form their own space (called a Banach space) on which the Fourier transform is a functor.

In the same vein as the previous chapter, we apply functional analysis and measure theory to commence a deeper study on the Fourier integral, and use them to generalize the Fourier integral for other spaces.

$\mathcal{F}\{f\} \in L^\infty$
$\| \widehat{f} \|_{\infty} \leq \| f \|_{1} $
\begin{theorem}[Plancherel theorem]
\[ f \in L^1(\mathbb{R}) \cap L^2(\mathbb{R}) \implies \| \mathcal{F}\{f\} \|_2 = \| f \|_2 \]
\end{theorem}

\section{Fourier integral on distributions}

We define distributions in Functional Analysis, and now we generalize the Fourier integral to them.


\subsection{Poisson's summation formula}


\section{Fourier integral on $\mathbb{R}^n$}

\begin{lemma}[Riemann-Lebesgue lemma for $\mathbb{R}^n$]
Let $f$ be a $L^1(\mathbb{R}^n)$ function, then the following holds (i.e its Fourier transform's tails tend to 0).
\[ \lim_{\xi \to \pm\infty} \mathcal{F}\{f\}(\xi) = 0 \]
\[ f \in L^1 (\mathbb{R}^n) \implies \lim_{\xi \to \pm\infty} \mathcal{F}\{f\}(\xi) = 0 \]
\end{lemma}
