\headline={\tenrm\lefthead\hfill
\righthead\hfill Page \folio} 
\def\lefthead{{\bf Math 5467,  Spring 2003}}
\def\righthead{{\bf Fourier Transform Facts: $L^1$}}
{\bf A list of Fourier Transform formulas and properties}
\gap
The list that follows is derived or obtained from references (numbers 20 more) 
following the list.
\gap
$$
\hat f(\xi):=\int f(x)e^{-i\xi x}dx,\put{where}\int |f(x)|\,dx<\infty.
\eqn{1}$$
\gap
\dft{1'}{\hskip 1in}\sla{the Fourier transform of an integrable \fn is a \bdd\
function.}
$$
(f(t-h))_{dt}^{\widehat{}}\,(\xi)
=e^{-i\xi h}\hat f(\xi)
\eqn{2}$$
$$
(\tau_h f)^{\widehat{}}\,(\xi)=e^{-i\xi h}\hat f(\xi),
\put{where}\tau_h f(t):=f(t-h)\put{and}h\put{is a constant.}
\leqno{\ (2')}$$
$$
(f(\lam t))_{dt}^{\widehat{}}\,(\xi)
={1\over \lam}\hat f\left({\xi\over \lam}\right)
\eqn{3}$$
$$
(S_\lam f)^{\widehat{}}\,(\xi)={1\over \lam}\hat f\left({\xi\over \lam}\right),
\put{where}S_\lam f(t)=f(\lam t)\put{and}\lam\put{ia a positive constant.}
\eqn{3'}$$
$$
(S_\lam\tau_h f)^{\widehat{}}\,(\xi)
={1\over \lam}e^{-i(\xi h/\lam)}\hat f\left({\xi\over \lam}\right)
=e^{-i(\xi h/\lam)}{1\over \lam}S_{1\over \lam}\hat f(\xi).
\eqn{4}$$
$$
(f(2^j t-n))_{dt}^{\widehat{}}\,(\xi)
=e^{-i\xi n/2^j}2^{-j}\hat f(2^{-j}\xi).
\leqno{\ (4')}$$
\gap
\dft{5}{\hskip .75 in}\sla{If $f(t)$ is integrable, then $\hat f(\xi)$ is
continuous in
$\xi$ and tends to zero at infinity.}
\gap
The \its{convolution} of $f$ and $g$ is denoted $f*g$ and defined by
$$
f*g(t):=\int f(t-s)g(s)\,ds.
\eqn{6}$$
\dft{7} \sla{The convolution of two integrable functions is given by an absolutely
convergent integral for a.e. $t,$ and the convolution of two integrable functions
is itself an integrable function.}
$$
(f*g)^{\widehat{}}(\xi)=\hat f(\xi)\hat g(\xi).
\eqn{8}$$
$$
\hat g(\xi)=\widehat {f'}(\xi)=i\xi \hat f(\xi)
\put{if}f\in L^1\put{and}g=f'\in L^1.
\eqn{9}$$
$$
(tf(t))_{dt}^{\widehat{}}=\int e^{-i\xi t}tf(t)\,dt={d\over d\xi}\hat f(\xi)
\put{if}f\in L^1\put{and}tf(t)\in L^1.
\eqn{10}$$
$$
\int \hat f(\xi)g(\xi)\,d\xi=\int f(t)\hat g(t)\,dt
\put{if}f\in L^1\put{and}g\in L^1.
\eqn{11}$$
$$
f(t)={1\over 2\pi}\int e^{it\xi}\hat f(\xi)\,d\xi
\put{if}f\in L^1\put{and}\hat f\in L^1.
\eqn{12}$$
$$
\widehat {\overline {f}}(\xi)
=\overline {\widehat f(-\xi)}
=\widetilde {\widehat f}(\xi)\put{and}\widehat {\widetilde f}(\xi)
=\overline {\widehat f(\xi)}\put{and}\widehat {f(-t)}(\xi)=\hat f(-\xi).
\eqn{13}$$
$$
\int \hat f(\xi)g(\xi)e^{-i\xi {\bf x}}\,d\xi=\int f(t-{\bf x})\hat g(t)\,dt
\put{if}f\in L^1\put{and}g\in L^1.
\eqn{14}$$
\gap
{\bf Derivations and sources}
\gap
{\bf Foreword: a note on the integrals used here}
\gap
A function $f(t)$ is \its{integrable} if $f(t)$ is measurable and if 
$\int |f(x)|\,dx<\infty.$ The integral meant here is a Lebesgue integral. However,
if the integral $\int |f(x)|\,dx$ is an improper Riemann integral, the usual one
from Calculus, the value of that integral is the same as that of the Lebesgue
integral of the same function. This works because the integrand is
non-negative. There are examples where the improper Riemann integral exists, in
which the Lebesgue integral does not exist. The example is
$$
\int_0^\infty {\sin t\over t} \,dt
:=\lim_{R\to\infty}\int_0^R {\sin t\over t} \,dt={\pi \over 2},
$$
and $\int_0^\infty {\sin t\over t} \,dt$ does not exist as a Lebesgue integral
because the integral of the positive part and the integral of the negative part
are both infinite. The ``improper Lebesgue integral'' exists of course because
the Riemann and Lebesgue integrals over $[0,\,R]$ coincide. But nobody ever seems
to talk about improper Lebesgue integrals! The details of the example are based on
Zygmund, Ch. II Lemma $(8.2).$
\gap
We often say $``f\in L^1\quot$ when $f$ is integrable, and we often write
$\Vert f\Vert_1$ for $\int|f(x)|\,dx.$ More information, that I hope will be
useful to you, about Lebesgue integral theory is in the note 
\gap
{\bf A definition of the Fourier transform}
\gap
The \its{Fourier transform} of an integrable function
$f(x)$ is
$$
\hat f(\xi):=\int f(x)e^{-i\xi x}dx,\put{where}\int |f(x)|\,dx<\infty.
\eqn{21}$$
By putting absolute-value bars under the integral sign,
$$
\big|\hat f(\xi)\big|\le\int |f(x)|\,dx=\Vert f\Vert_1,
$$
\dft{21'}{\hskip 1in}\sla{the Fourier transform of an integrable \fn is a \bdd\
function.}
\gap
{\bf The Fourier transform of a translate}
\gap
If $f(t)$ is integrable so is $f(t-h)$ for any fixed $h.$ Moreover, 
$\int f(t-h)\,dt=\int f(t)\,dt.$ There is a
formula for the Fourier transform of $f(t-h)$ in terms of the Fourier transform
of $f(t).$ We calculate, using the literal change of variables $t\to t+h$ in the
following integral.
$$
(f(t-h))_{dt}^{\widehat{}}\,(\xi)
=\int f(t-h)e^{-i\xi t}dt
=\int f(t)e^{-i\xi (t+h)}dt
=e^{-i\xi h}\int f(t)e^{-i\xi t}dt
=e^{-i\xi h}\hat f(\xi)
$$
This gives us the {\bf translation formula}
$$
(f(t-h))_{dt}^{\widehat{}}\,(\xi)
=e^{-i\xi h}\hat f(\xi)
\eqn{22}$$
Sometimes it is convenient to treat translation as a linear operator. Then we
write $\tau_h f(t)$ for $f(t-h),$ and the translation formula becomes 
$$
(\tau_h f)^{\widehat{}}\,(\xi)=e^{-i\xi h}\hat f(\xi),
\put{where}\tau_h f(t):=f(t-h)\put{and}h\put{is a constant.}
\eqn{22'}$$
We also have $\|\tau_h f\|_1=\|f\|_1.$
\gap
{\bf The Fourier transform of a dilate}
\gap
If $f(t)$ is integrable so is $f(\lam t)$ for any fixed positive $\lam,$ 
and $\int f(\lam t)\,dt=(1/\lam)\int f(t)\,dt.$ There is
a formula for the Fourier transform of $f(\lam t)$ in terms of the Fourier
transform of $f(t).$ We calculate, using the literal change of variables $t\to
t/\lam$ in the following integral.
$$
(f(\lam t))_{dt}^{\widehat{}}\,(\xi)
=\int f(\lam t)e^{-i\xi t}dt
=\int f(t)e^{-i(\xi/\lam) t}dt/\lam
={1\over \lam}\hat f\left({\xi\over \lam}\right)
$$
This gives us the {\bf dilation formula} (for the Fourier transform)
$$
(f(\lam t))_{dt}^{\widehat{}}\,(\xi)
={1\over \lam}\hat f\left({\xi\over \lam}\right)
\eqn{23}$$
Somtimes it is convenient to treat dilation as a linear operator. Then we
write $S_\lam f(t)$ for $f(\lam t),$ and the dilation formula becomes 
$$
(S_\lam f)^{\widehat{}}\,(\xi)={1\over \lam}\hat f\left({\xi\over \lam}\right),
\put{where}S_\lam f(t)=f(\lam t)\put{and}\lam\put{ia a positive constant.}
\eqn{23'}$$
We have $\|S_\lam f\|_1=(1/\lam)\|f\|_1.$
\gap
Just for the record, let us compare $S_\lam\tau_h f$ and $\tau_h S_\lam f.$
\gap
To be sure we know what's going on, let's let $g(t):=f(t-h)=\tau_h f(t).$
\gap
We then have $S_\lam\tau_h f(t)=S_\lam g(t)=g(\lam t)=f(\lam t -h).$ 
\gap
On the other
hand $\tau_h S_\lam f(t)=S_\lam f(t-h)
=f(\lam (t-h))=f(\lam t-\lam h)=\tau_{\lam h} f(\lam t)
=S_\lam\tau_{\lam h} f(t).$
\gap
We will tend to use the first order of the operator products: $S_\lam\tau_h.$
\gap
{\bf The Fourier transform of $S_\lam\tau_h f$}
$$
(S_\lam\tau_h f)^{\widehat{}}\,(\xi)
={1\over \lam}e^{-i(\xi h/\lam)}\hat f\left({\xi\over \lam}\right)
=e^{-i(\xi h/\lam)}{1\over \lam}S_{1\over \lam}\hat f(\xi).
\eqn{24}$$
Here are the calculations:
$$
(S_\lam\tau_h f)^{\widehat{}}\,(\xi)
=\int f(\lam t-h)e^{-i\xi t}dt
=\int f(t-h)e^{-i(\xi/\lam) t}dt/\lam
=\int f(t)e^{-i(\xi/\lam) (t+h)}dt/\lam
={1\over \lam}e^{-i(\xi h/\lam)}\hat f\left({\xi\over \lam}\right).
$$
We will use this case most often:
$$
(f(2^j t-n))_{dt}^{\widehat{}}\,(\xi)
=e^{-i\xi n/2^j}2^{-j}\hat f(2^{-j}\xi).
\eqn{24'}$$
{\bf The Fourier transform of an integrable function is a continuous function
that is zero at infinity}
\gap
\dft{25}\sla{If $f(t)$ is integrable, then $\hat f(\xi)$ is continuous in $\xi$
and tends to zero at infinity.}
\gap
For the continuity we use the Lebesgue's Dominated Convergence Theorem, $(1)$
in the classnotes ``Lebesgue theory - an overview.''
\gap
Let $\xi_n\to \xi_o.$ Then
$$
\hat f(\xi_n)-\hat f(\xi_o)=\int (e^{i\xi_nt}-e^{i\xi_ot})f(t)\,dt.
$$
The integrand is dominated pointwise by the integrable function $g(t):=2|f(t)|$
and converges pointwise (in $t)$ to zero as $n\to \infty.$ The hypotheses of
Lebesgue's Dominated Convergence Theorem are thus satisfied, so the integral of
the limit, which is zero, is equal to the limit of the integrals, which is also
the limit of $\hat f(\xi_n)-\hat f(\xi_o)$ as $n\to\infty.$ Since $\{\xi_n\}$ was
an \sla{arbitrary} \seq tending to $\xi_o,$ 
$\dsp\lim_{\xi\to\xi_o}\hat f(\xi)=\hat f(\xi_o),$ so that continuity holds.
Again, $\xi_o$ was arbitrary, so continuity is true at every $\xi.$
\gap
That the Fourier transform of an integrable function has limit zero at infinity
is called the \sla{Riemann-Lebesgue Lemma}. We will use one of the Lebesgue facts
here,
$(9)$ in the notes on Lebesgue theory: $\lim_{h\to 0}\int |f(x+h)-f(x)|\,dx=0.$
A device of interest is used, namely that of finding two formulas for some
quantity and then averaging them to create a third formula. The device depends
on the identity $e^{\pm i\pi}=-1.$ We suppose that $\xi\ne 0.$ Then
$$\ea{
\hat f(\xi)
&=\int e^{-i\xi t}f(t)\,dt
=-\int e^{-i\pi}e^{-i\xi t}f(t)\,dt\cr
&=-\int e^{-i\xi\pi/\xi}e^{-i\xi t}f(t)\,dt\cr
&=-\int e^{-i\xi(t+\pi/\xi)}f(t)\,dt\cr
&=-\int e^{-i\xi t}f(t-\pi/\xi)\,dt\cr
&={ 1\over 2}\int e^{-i\xi t}f(t)\,dt
-{ 1\over 2}\int e^{-i\xi t}f(t-\pi/\xi)\,dt\cr 
&={ 1\over 2}\int e^{-i\xi t}(f(t)-f(t-\pi/\xi))\,dt.
}$$ 
Hence $|\hat f(\xi)|
\le { 1\over 2}\Big|\int e^{-i\xi t}(f(t)-f(t-\pi/\xi))\,dt\Big|
\le { 1\over 2}\int |f(t)-f(t-\pi/\xi)|\,dt\to 0$ as $|\xi|\to\infty.$ The second
inequality here is $(16)$ and the limit is $(9)$ in the notes on Lebesgue theory.
\gap
{\bf Convolution and the Fourier Transform}
\gap
The \its{convolution} of $f$ and $g$ is denoted $f*g$ and is given by
$$
f*g(t):=\int f(t-s)g(s)\,ds.
\eqn{26}$$
As always, we have to ask whether the integral makes sense. It is not at all
obvious but it is true that the integral is a finite Lebesgue integral for almost
all $t.$ The proof of this depends on Fubini's Theorem, $(4)$ in the notes on
Lebesgue theory.
\gap
First, we have by $(16)$ in the notes on Lebesgue theory that.
$$
|f*g(t)|\le\int |f(t-s)g(s)|\,ds.
$$
Since we now have non-negative functions, the integrals on both sides of the
inequality will exist, whether finite or not. By Tonelli's Theorem, $(4')$ in the
Lebesgue notes,
$$
\int |f*g(t)|\,dt
\le\int \int |f(t-s)g(s)|\,ds\,dt
=\int \left(\int |f(t-s)|\,dt\right)|g(s)|\,ds
=\int |f(t)|\,dt \int |g(s)|\,ds<\infty.
$$
By $(31)$ in the Lebesgue notes,  $\int |f(t-s)g(s)|\,ds<\infty$ a.e., so by
Fubini's Theorem, $(4)$ in the notes on Lebesgue theory, the integral defining
the convolution is an absolutely convergent integral for a.e. $t.$ What this
amounts to is
\gap
\dft{27} \sla{The convolution of two integrable functions is given by an
absolutely convergent integral for a.e. $t,$ and the convolution of two
integrable functions is itself an integrable function.}
\gap
All this gives a ``multiplication'' on $L^1,$ namely the convolution. Convolution
is associative and commutative, so $L^1$ becomes an \its{algebra} -- a vector
space with a multiplication that distributes over addition.
\gap
The Fourier transform transforms the convolution of $f$ and $g$ into the product
of their Fourier transforms:
$$
(f*g)^{\widehat{}}(\xi)=\hat f(\xi)\hat g(\xi).
\eqn{28}$$
Checking this is done using Fubini's theorem again:
$$\ea{
(f*g)^{\widehat{}}(\xi)
&=\int e^{-i\xi t}f*g(t)\,dt
=\int e^{-i\xi t}\int f(t-s)g(s)\,ds\,dt\cr
&=\int \int e^{-i\xi (t-s)}f(t-s)e^{-i\xi s}g(s)\,ds\,dt\cr
&=\int \left(\int e^{-i\xi s}g(s)\,ds\right)e^{-i\xi (t-s)}f(t-s)\,dt\cr
&=\int \left(\int e^{-i\xi s}g(s)\,ds\right)e^{-i\xi (t)}f(t)\,dt,
\put{using the literal substitution}t\to t+s, \cr
&=\hat f(\xi)\hat g(\xi).\cr
}$$
\gap
{\bf Derivatives and the Fourier Transform}
\gap
The Fundamental Theorem of Calculus has a counterpart in Lebesgue theory, but
for the next item we will just assume that $f$ is integrable, and that there is an
integrable function $g$ \st $f(t)=\int_{-\infty}^t g(s)\,ds.$ Then $f$ is
continuous, differentiable almost everywhere and $f'(t)=g(t)$ a.e. {\bf
The Fourier Transform of the Derivative}
$$
\hat g(\xi)=\widehat {f'}(\xi)=i\xi \hat f(\xi)
\put{if}f\in L^1\put{and}g=f'\in L^1.
\eqn{29}$$
In other words, the Fourier transform transforms differentiation into
multiplication of $\hat f(\xi)$ by $i\xi.$
\gap
If $f(t)$ and $tf(t)$ are integrable then {\bf Derivative of the Fourier
Transform}
$$
(tf(t))_{dt}^{\widehat{}}=\int e^{-i\xi t}tf(t)\,dt=-i{d\over d\xi}\hat f(\xi)
\put{if}f\in L^1\put{and}tf(t)\in L^1.
\eqn{30}$$
\gap
\gap
{\bf ``Moving the hat:'' an identity}
\gap
If $f$ and $g$ are integrable then
$$
\int \hat f(\xi)g(\xi)\,d\xi=\int f(t)\hat g(t)\,dt
\put{if}f\in L^1\put{and}g\in L^1.
\eqn{31}$$
Both integrals make sense because the Fourier transforms are bounded, by \rend{1'}
The proof of the formula is an exercise in using the Fubini Theorem.
\gap
{\bf The Fourier Inversion Formula}
\gap
If $f(t)$ and $\hat f(\xi)$ are integrable then
$$
f(t)={1\over 2\pi}\int e^{it\xi}\hat f(\xi)\,d\xi
\put{if}f\in L^1\put{and}\hat f\in L^1.
\eqn{32}$$
The proof of this involves an ``approximation of the identity'' argument that
will be essential in the process of extending the Fourier transform to functions
in $L^2.$
\gap
{\bf Some important ``minor'' facts and formulas}
\gap
{\bf Complex conjugation, reflection  and the Fourier transform}
\gap
Sometimes we have to switch ``bar'' and ``hat.'' Reflection shows up then. We
have, if $f\in L^1,$
$$
\widehat {\overline {f}}(\xi)
=\int e^{-i\xi t}{\overline {f}}(t)\,dt
=\overline {\int e^{i\xi t}f(t)\,dt}
=\overline {\widehat f(-\xi)}\put{and}
\widehat {f(-t)}(\xi)
=\int e^{-i\xi t}f(-t)\,dt
=\int e^{i\xi t}f(t)\,dt
=\widehat f(-\xi).
$$
The ``conjugated reflection'' of $f(t)$ is $\overline{f(-t)},$ a combination that
occurs often enough to cause people to want a notation for it. None has been
widely adopted. If we need it here, we'll use 
$\widetilde {f}(t):=\overline{f(-t)}$ for it. This operation is an
\its{involution,} an operation that is its own inverse. The same can be said for
``reflection.'' But we notice that reflection commutes with the Fourier transform
and conjugation does not. Now that we have introduced tha ``tilde'' operation, we
have to write down how it does in conjunction with the Fourier transform. We list
them all in one formula:
$$
\widehat {\overline {f}}(\xi)
=\overline {\widehat f(-\xi)}
=\widetilde {\widehat f}(\xi)\put{and}\widehat {\widetilde f}(\xi)
=\overline {\widehat f(\xi)}\put{and}\widehat {f(-t)}(\xi)=\hat f(-\xi).
\eqn{34}$$
\gap
{\bf A formula for the Fourier transform of one kind of product}
\gap
Since $\hat f(\xi)e^{-i\xi {\bf x}}=\int f(t-{\bf x})e^{-i\xi t}\,dt,$ we can, in
the first integral below, ``move the hat'' and get
$$
\int \hat f(\xi)g(\xi)e^{-i\xi {\bf x}}\,d\xi=\int f(t-{\bf x})\hat g(t)\,dt
\put{if}f\in L^1\put{and}g\in L^1.
\eqn{35}$$
\bye

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