DMat0101, Notes 3: The Fourier transform on L^1

1. Definition and main properties.

For {f\in L^1({\mathbb R}^n)}, the Fourier transform of {f} is the function

\displaystyle \mathcal{F}(f)(\xi)=\hat{f}(\xi)=\int_{{\mathbb R}^n}f(x)e^{-2\pi i x\cdot \xi}dx,\quad \xi\in{\mathbb R}^n.

Here {x\cdot y} denotes the inner product of {x=(x_1,\ldots,x_n)} and {y=(y_1,\ldots, y_n)}:

\displaystyle x\cdot y=\langle x,y\rangle=x_1y_1+\cdots x_n y_n.

Observe that this inner product in {{\mathbb R}^n} is compatible with the Euclidean norm since {x\cdot x=|x|^2}. It is easy to see that the integral above converges for every {\xi\in{\mathbb R}^n} and that the Fourier transform of an {L^1} function is a uniformly continuous function.

Theorem 1 Let {f,g\in L^1({\mathbb R}^n)}. We have the following properties.

(i) The Fourier transform is linear {\widehat{f+g}=\hat f + \hat g} and {\widehat{cf}=c \hat f} for any {c\in{\mathbb C}}.

(ii) The function {\hat f(\xi)} is uniformly continuous.

(iii) The operator {\mathcal F} is bounded operator from {L^1({\mathbb R}^n)} to {L^\infty({\mathbb R}^n)} and

\displaystyle \|\hat f \|_{L^{\infty}({\mathbb R}^n)}\leq \|f\|_{L^1({\mathbb R}^n)}.

(iv) (Riemann-Lebesgue) We have that

\displaystyle \lim _{|\xi|\rightarrow +\infty} \hat f(\xi)=0.

Proof: The properties (i), (ii) and (iii) are easy to establish and are left as an exercise. There are several ways to see (iv) based on the idea that it is enough to establish this property for a dense subspace of {L^1({\mathbb R}^n)}. For example, observe that if {f} is the indicator function of an interval of the real line, {f=\chi_{[a,b]}}, then we can calculate explicitly to show that

\displaystyle |\hat{f}(\xi)|=\bigg|\int_a ^b e^{-2\pi i x\xi}dx\bigg| =\bigg| \frac{e^{-2\pi i \xi a}-e^{-2\pi i \xi b}}{2\pi i \xi }\bigg|\lesssim \frac{1}{|\xi|}\rightarrow 0\quad\mbox{as}\quad |\xi|\rightarrow +\infty.

Tensoring this one dimensional result one easily shows that {\lim_{|\xi|\rightarrow +\infty}f(\xi)=0} whenever {f} is the indicator function of an {n}-dimensional interval of the form {[a_1,b_1]\times\cdots\times [a_n,b_n]}. Obviously the same is true for finite linear combinations of {n}-dimensional intervals since the Fourier transform is linear.

Now let {f} be any function in {L^1({\mathbb R}^n)} and {\epsilon >0} and consider a simple function which is a finite linear combination of {n}-dimensional intervals, such that {\|f-g\|_1<\epsilon/2}. Let also {M>0} be large enough so that {|\hat g (\xi)|<\epsilon/2} whenever {|\xi|>M}. Using (iii) and the linearity of the Fourier transform we have that

\displaystyle \begin{array}{rcl} |\hat f(\xi)|\leq |\widehat{(f-g)}\hat (\xi)|+|\hat g (\xi)|\leq \|f-g\|_{L^1}+|\hat g (\xi)|<\epsilon, \end{array}

whenever {|\xi|>M}, which finishes the proof. \Box

In view of (ii) and (iv) we immediately get the following.

Corollary 2 If {f\in L^1({\mathbb R}^n)} then {\hat f\in C_o({\mathbb R}^n)}.

Exercise 1 Show the properties (ii) and (iii) in the previous Theorem.

The discussion above and especially Corollary 2 shows that a necessary condition for a function {g} to be a Fourier transform of some function in {L^1({\mathbb R}^n)} is {g\in C_o({\mathbb R}^n)}. However, this condition is far from being sufficient as there are functions {g\in C_o({\mathbb R}^n)} which are not Fourier transforms of {L^1} functions. See Exercise 8.

Let us now see two important examples of Fourier transforms that will be very useful in what follows.

Example 1 For {a>0} let {f(x)=e^{-\pi a|x|^2}}. Then

\displaystyle \hat f(\xi)=a^{-\frac{n}{2}}e^{-\frac{\pi|\xi|^2}{a}}.

Proof: Observe that in one dimension we have

\displaystyle \begin{array}{rcl} \hat f(\xi)&=&\int_{\mathbb R} e^{-\pi ax^2}e^{-2\pi i x\xi}dx=\int_{\mathbb R} e^{-\pi a(x+i\frac{\xi}{a})^2}dx\ e^{-\frac{\pi\xi^2}{a}}\\ \\ &=&\int_{\mathbb R} e^{-\pi ax^2} dx \ e^{-\frac{\pi\xi^2}{a}}= a^{-\frac{1}{2}} e^{-\frac{\pi^2\xi^2}{a}}, \end{array}

where the third equality is a consequence of Cauchy’s theorem from complex analysis. The {n}-dimensional case is now immediate by tensoring the one dimensional result. \Box

Remark 1 Replacing {a=1} in the previous example we see that {e^{-\pi |x|^2}} is its own Fourier transform.

Example 2 For {a>0} let {g(x)=e^{-2\pi a |x|}}. Then

\displaystyle \hat g(\xi)=c_n\frac{a}{(a^2+|\xi|^2)^\frac{n+1}{2}},

where {c_n=\Gamma((n+1)/2)/\pi^\frac{n+1}{2}}.

Proof: The first step here is to show the subordination identity

\displaystyle e^{-\beta}=\frac{1}{\sqrt{\pi}}\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}}e^{-\beta^2/4u}du,\quad \beta>0, \ \ \ \ \ (1)

which is a simple consequence of the identities

\displaystyle \begin{array}{rcl} e^{-\beta}&=&\frac{2}{\pi}\int_0 ^\infty\frac{\cos \beta x}{1+x^2}dx,\\ \\ \frac{1}{1+x^2}&=&\int_0 ^\infty e^{-(1+x^2)u}du. \end{array}

Using (1) we can write

\displaystyle \begin{array}{rcl} \hat g(\xi)&=&\int_{{\mathbb R}^n} e^{-2\pi a|x|} e^{-2\pi i x\cdot \xi}dx \\ \\ &=&\frac{1}{\sqrt{\pi}}\int_{{\mathbb R}^n}\bigg(\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}}e^{-4\pi^2a^2|x|^2/4u}du\bigg) e^{-2\pi i x\cdot \xi} dx\\ \\ &=&\frac{1}{\sqrt{\pi}}\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}} \frac{1}{a^n}\bigg(\sqrt{\frac{u}{\pi}}\bigg)^\frac{n}{2} e^{-\frac{u|\xi|^2}{a^2}}du\\ \\ &=& \frac{1}{\pi^\frac{n+1}{2} a^n}\int_0 ^\infty u^\frac{n-1}{2}e^{-u\frac{|\xi|^2}{a^2}}e^{-u}du\\ \\ &=& \frac{1}{\pi^\frac{n+1}{2} a^n} \frac{1}{\big(1+\frac{|\xi|^2}{a^2}\big)^\frac{n+1}{2}}\int_0 ^\infty u^\frac{n-1}{2}e^{-u}du\\ \\ &=& \frac{\Gamma(\frac{n+1}{2})}{\pi^\frac{n+1}{2} }\frac{a}{\big(a^2+|\xi|^2 \big)^\frac{n+1}{2}}, \end{array}

by the definition of the {\Gamma}-function.\Box

Exercise 2 This exercise gives a first (qualitative) instance of the uncertainty principle. Prove that there does not exist a non-zero integrable function on {{\mathbb R}} such that both {f} and {\hat f} have compact support.

Hint: Observe that the function

\displaystyle \hat f(\xi)=\int_{\mathbb R} f(x) e^{-2\pi i x\xi}dx ,

extends to an entire function (why ?).

The definition of the Fourier transform extends without difficulty to finite Borel measures on {{\mathbb R}^n}. Let us denote by {\mathcal M({\mathbb R}^n)} this class of finite Borel measures and let {\mu\in \mathcal M({\mathbb R}^n)}. We define the Fourier transform of {\mu} to be the function

\displaystyle \mathcal{F}(\mu)(\xi)=\hat{\mu}(\xi)=\int_{{\mathbb R}^n}e^{-2\pi i x\cdot \xi}d\mu(x),\quad \xi\in{\mathbb R}^n.

We have the analogues of (i), (ii) and (iii) of Theorem 1 if we replace the {L^1} norm by the total variation of the measure. However property (iv) fails as can be seen by consider the Fourier transform of a Dirac mass at the point {0}. Indeed observe that

\displaystyle \hat{\delta_0}(\xi)=\int_{{\mathbb R}^n} e^{-2\pi i x\cdot \xi}d\delta_0(x)=1,

which is a constant function.

The Fourier transform interacts very nicely with convolutions of functions, turning them to products. This turns out to be quite important when considering translation invariant operators as we shall see later on in the course.

Proposition 3 Let {f,g\in L^1({\mathbb R}^n)}. Then {\widehat{f*g}=\hat f \hat g}.

Exercise 3 Prove Proposition 3.

Another important property of the Fourier transform is the multiplication formula.

Proposition 4 (Multiplication formula) Let {f,g\in L^1({\mathbb R}^n)}. Then

\displaystyle \int_{{\mathbb R}^n}\hat f(\xi)g(\xi)d\xi=\int_{{\mathbb R}^n}f(x)\hat g(x)dx.

We will now describe some easily verified symmetries of the Fourier transform. We introduce the following basic operations on functions:

Translation operator: { (\tau_{x_o} f)(x)=f(x-x_o),\quad x,x_o\in {\mathbb R}^n}

Modulation operator: {\textnormal{Mod}_{x_o}(f)(x)=e^{2\pi i x\cdot x_o} f(x),\quad x,x_o\in {\mathbb R}^n}

Dilation operator: {\textnormal{Dil}_\lambda ^p(f)(x)={\lambda^{-\frac{n}{p}}}f(x/\lambda),\quad x,\in {\mathbb R}^n,\lambda>0,1\leq p\leq\infty}.

Proposition 5 Let {f\in L^1({\mathbb R}^n)} We have the following symmetries:

(i) { \mathcal F \tau_{x_o}=\textnormal{Mod} _{-x_o}\mathcal F},

(ii) {\mathcal F \textnormal{Mod}_{\xi_o}=\tau_{\xi_o}\mathcal F},

(iii) {\mathcal F \textnormal {Dil} _\lambda ^p = \textnormal {Dil} _{ \lambda^{-1} } ^{p'} \mathcal F}, where {\frac{1}{p}+\frac{1}{p'}=1}.

Exercise 4 Prove the symmetries in Proposition 5 above. Also, let {U:{\mathbb R}^n\rightarrow {\mathbb R}^n} be an invertible linear transformation, that is, {U\in GL({\mathbb R}^n)}. Define the general dilation operator

\displaystyle ( \textnormal{Dil} _U ^pf)(x) =|\det{U}|^{-\frac{1}{p}}f(U^{-1}x), \quad x\in {\mathbb R}^n, 1\leq p\leq \infty.

Prove that

\displaystyle \mathcal F \textnormal{Dil}_U ^p = \textnormal{Dil}_{(U^*)^{-1}} ^{p'}\mathcal F,

where {U^*} is the (real) adjoint of {U}, that is the matrix for which we have {\langle Ux, y\rangle =\langle x , U^*y\rangle} for all {x,y\in {\mathbb R}^n}.

We now come to one of the most interesting properties of the Fourier transform, the way it commutes with derivatives.

Proposition 6 (a) Suppose that {f\in L^1({\mathbb R}^n)} and that {x_kf(x)\in L^1({\mathbb R}^n)} for some {1\leq k \leq n}. The {\hat f} is differentiable with respect to {\xi_k} and

\displaystyle \frac{\partial}{\partial \xi_k} \mathcal F(f)(\xi) = \mathcal F(- 2\pi i x_k f)(\xi).

(b) We will say that a function {f} has a partial derivative in the {L^p} norm with respect to {x_k} if there exists a function {g\in L^p({\mathbb R}^n)} such that

\displaystyle \lim_{h_k\rightarrow 0 }\bigg(\int_{{\mathbb R}^n}\bigg|\frac{f(x+h)-f(x)}{h_k}-g\bigg|^p dx\bigg)^\frac{1}{p}=0,

where {h=(0,\ldots,0,h_k,0,\ldots,0)} is a non-zero vector along the {k}-th coordinate axis. If {f} has a partial derivative {g} with respect to {x_k} in the {L^1}-norm, then

\displaystyle \hat g(\xi)=2\pi i \xi_j \mathcal {F}(\xi).

Exercise 5 Prove Proposition 6.

A similar result that involves the classical derivatives of a function is the following:

Proposition 7 For {k} a non-negative integer, suppose that {f\in C^k({\mathbb R}^n)} and that {\partial^\alpha f \in L^1({\mathbb R}^n)} for all {|\alpha|\leq k}, and {\partial^\alpha f\in C_o({\mathbb R}^n)} for {|\alpha|\leq k-1}. Show that

\displaystyle \widehat {\partial^\alpha f}(\xi)=(2\pi i \xi)^\alpha \hat f(\xi).

Exercise 6 Prove Proposition 7.

Several remarks are in order. First of all observe that Propositions 6,7 assert that the following commutation relations are true

(i) {\mathcal F (-2\pi i x_k) = \frac{\partial}{\partial \xi_k} \mathcal F ,}

(ii) { \mathcal F \frac{\partial}{\partial x_k} = 2\pi i \xi_k \mathcal F,}

where here we abuse notation and denote by {2\pi i x_k } the operator of multiplication by {2\pi i x_k}. Thus the Fourier transform turns derivatives to multiplication by the corresponding variable, and vice versa, it turns multiplication by the coordinate variable to a partial derivative, whenever this is technically justified. This is a manifestation of the heuristic principle that smoothness of a function translates to decay of the Fourier transform and on the other hand, decay of a function at infinity translates to smoothness of the Fourier transform.

A second remark is that these commutation relations generalize, in an obvious way, to higher derivatives. To make this more precise, let {P} be a polynomial on {{\mathbb R}^n}:

\displaystyle P(x)=\sum_{|\alpha|\leq d}c_\alpha x^\alpha .

Slightly abusing notation again we write {P(\partial ^\alpha)} for the differential operator

\displaystyle P(\partial^\alpha)=\sum_{|\alpha|\leq d}c_\alpha {\partial^\alpha} .

We then have that the following commutation relations are true

(i’) {\mathcal F P(-2\pi i x) = P(\partial^\alpha) \mathcal F ,}

(ii’) { \mathcal F P(\partial^\alpha) =P( 2\pi i \xi) \mathcal F.}

Observe that for `nice’ functions, for example {f\in C_c({\mathbb R}^n)} or {f\in \mathcal S(R^n)}, Propositions 6 and 7 are automatically satisfied.

2. Inverting the Fourier transform

On of the most important problems in the theory of Fourier transforms is that of the inversion of the Fourier transform. That is, given the Fourier transform {\hat f} of an {L^1} function, when can we recover the original function {f} from {\hat f}? We begin with a simple case where the recovery is quite easy.

Proposition 8 Let {f\in L^1({\mathbb R}^n)} be such that {\hat f \in L^1({\mathbb R}^n)}. Then the inversion formula holds true. In particular we have that

\displaystyle f(x)=\int_{{\mathbb R}^n} \hat f(\xi) e^{2\pi i x\cdot \xi} d\xi,

for almost every {x\in {\mathbb R}^n}.

Proof: The proof is based on the following calculation. For {a>0} we have that

\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n}\hat f(\xi)e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi&=&\int_{{\mathbb R}^n}\int_{{\mathbb R}^n} f(y) e^{-2\pi i y\cdot \xi}dy e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi\\ \\ &=&\int_{{\mathbb R}^n} f(x+y) \int_{{\mathbb R}^n} e^{-2\pi i y}e^{-a|\xi|^2} d\xi dy\\ \\ &=&(\frac{\pi}{a})^\frac{n}{2}\int_{{\mathbb R}^n}f(x+y) e^{-\frac{\pi^2|y|^2}{a}} dy \\ \\ &=& \int_{{\mathbb R}^n}f(x+\sqrt{a}y) e^{-\pi|y|^2}dy, \end{array}

where in the last equality we have used Example 1. We can thus write

\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n}\bigg|\int_{{\mathbb R}^n} \hat f(\xi)e^{-a|\xi|^2 e^{2\pi i x\cdot \xi} }d\xi -f(x) \bigg| dx&=& \int_{{\mathbb R}^n}\bigg|\int_{{\mathbb R}^n} f(x+\sqrt{a}y) e^{-\pi|y|^2}dy-f(x)\bigg| dx\\ \\ &=& \int_{{\mathbb R}^n}\bigg|\int_{{\mathbb R}^n} \{f(x+\sqrt{a}y) -f(x) \}e^{-\pi |y|^2} dy \bigg|dx \\ \\ &\leq &\int_{{\mathbb R}^n}\int_{{\mathbb R}^n} |f(x+\sqrt{a}y)-f(x)|dx e^{-\pi|y|^2}dy \\ \\ &=& \int_{{\mathbb R}^n}\|f-\tau_{-\sqrt{a}y}f\|_{L^1({\mathbb R}^n)}e^{-\pi|y|^2}dy. \end{array}

Since {\|f-\tau_{-\sqrt{a}y}f\|_{L^1({\mathbb R}^n)}\rightarrow 0} as {a\rightarrow 0} and {\|f-\tau_{-\sqrt{a}y}f\|_{L^1({\mathbb R}^n)}\leq 2\|f\|_{L^1({\mathbb R}^n)}}, Lebesgue’s dominated convergence theorem shows that {f} is almost everywhere equal to the {L^1}-limit of the sequence of functions

\displaystyle g_a(x)=\int_{{\mathbb R}^n}\hat f(\xi)e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi ,

as {a\rightarrow 0} (technically speaking we need to consider a sequence {a_k\rightarrow0}). On the other hand since {\hat f\in L^1({\mathbb R}^n)}, another application of Lebesgue’s dominated theorem shows that the {L^1}-limit of the functions {g_a} is also equal to {\int_{{\mathbb R}^n}\hat f(\xi)e^{2\pi i x\cdot \xi}d\xi}. This completes the proof of the proposition. \Box

An immediate corollary is that the Fourier transform is a one-to-one operator:

Corollary 9 Let {f_1,f_2\in L^1({\mathbb R}^n)} and suppose that {\hat f_1(\xi)=\hat f_2(\xi)} for all {\xi\in{\mathbb R}^n}. The we have that {f_1(x)=f_2(x)} for almost every {x\in{\mathbb R}^n}.

The proof is an obvious application of Proposition 8.

Exercise 7 (i) Suppose that {f\in C_c ^{n+1}({\mathbb R}^n)}. Show that

\displaystyle |\hat f(\xi)| \lesssim (1+|\xi|^2)^{-(n+1)/2}.

Conclude that whenever {f\in C^{n+1} _c({\mathbb R}^n)}, we have that

\displaystyle f(x)=\int_{{\mathbb R}^n} \hat f(\xi) e^{2\pi i x\cdot \xi} d\xi.

(ii) Show that {\mathcal F} maps the Schwartz space {\mathcal S({\mathbb R}^n)} onto {\mathcal S({\mathbb R}^n)}.

Exercise 8 The purpose of this exercise is to show that {\mathcal F(L^1({\mathbb R}^n))} is a proper subset of {C_o({\mathbb R}^n)} but also that it is a dense subset of {C_o({\mathbb R}^n)}.

(i) Show that {\mathcal F(L^1({\mathbb R}))} is a proper subset of {C_o({\mathbb R})}.

Hint: While there are different ways to do that, a possible approach is the following. For simplicity we just consider the case {n=1}:

(a) Show that {\big| \int_a ^b \frac{\sin x}{x}dx\big| \leq B} for all {0\leq |a|<|b|<\infty} where {B>0} is a numerical constant that does not depend on {a,b}.

(b) Suppose that {f\in L^1({\mathbb R})} is such that {\hat f} is an odd function. Use (a) to show that for every {b>0} we have that

\displaystyle \bigg|\int_1 ^b \frac{\hat f(\xi)}{\xi} d\xi\bigg| <A,

for some numerical constant {A>0} which does not depend on {b}.

(c) Construct a function {g\in C_o({\mathbb R})} which is not the Fourier transform of an {L^1} function. To do this note that it is enough to find a function {g\in C_o({\mathbb R})} which does not satisfy the condition in (b).

(ii) Show that {\overline{\mathcal F(L^1({\mathbb R}^n))}=C_o({\mathbb R}^n)} where the closure is taken in the {C_o} topology.

Hint: Observe that {C_c ^\infty({\mathbb R}^n)} is dense in {C_o({\mathbb R}^n)}, in the topology of the supremum norm.

It is convenient to define the formal inverse of the Fourier transform in the following way. For {f\in L^1({\mathbb R}^n)} we set

\displaystyle \mathcal F^{-1}(f)(\xi)=\mathcal F^*(f)(\xi)=\check f(\xi)=\int_{{\mathbb R}^n}f(x) e^{2\pi i x\cdot \xi}d\xi=\hat f(-\xi)=\tilde {\hat f}(\xi)=\hat{\tilde f}(\xi).

Here we denote by {\tilde g} the reflection of a function {g}, that is, {\tilde g(x)=g(-x)}. Observe that {\mathcal F^*} is the conjugate of the Fourier transform. Thus the operator {\mathcal F^*} is very closely connected to the operator {\mathcal F} and enjoys essentially the same symmetries and properties.

As we shall see later on, it is also the adjoint of the Fourier transform with respect to the {L^2} inner product

\displaystyle \langle f,g\rangle =\int_{{\mathbb R}^n} f\bar g.

Although we haven’t yet defined the Fourier transform on {L^2} we can calculate for {f,g\in L^1\cap L^2({\mathbb R}^n)} that

\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n} (\mathcal F f)\bar g &=&\int_{{\mathbb R}^n}\int_{{\mathbb R}^n}f(x)e^{-2\pi i x\cdot \xi} dx \bar g(\xi) d\xi \\ \\ &=&\int_{{\mathbb R}^n} f(x)\overline{\int_{{\mathbb R}^n} g(\xi)e^{2\pi i x\cdot \xi}d\xi} \ dx\\ \\ &=&\int_{{\mathbb R}^n} f \overline{ (\mathcal F^*(g))} \end{array}

Proposition 8 claims that {\mathcal F^*} is also the inverse of the Fourier transform in the sense that

\displaystyle \mathcal F^* \mathcal F f=f,

whenever {f,\mathcal F f\in L^1({\mathbb R}^n)}.

The proof of Proposition 8 is quite interesting in the following ways. First of all observe that we have actually showed that whenever {f\in L^1({\mathbb R}^n)}, {f} is equal (a.e.) to the {L^1} limit of the functions

\displaystyle \int_{{\mathbb R}^n}\hat f(\xi)e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi ,

as {a\rightarrow 0}. This does not require any additional hypothesis and actually provides us with a method of inverting the Fourier transform of any {L^1} function, at least in the {L^1} sense. The second remark is that the proof of Proposition 8 can be generalized to different methods of summability. Indeed, let {\Phi\in L^1({\mathbb R}^n)} be such that {\phi=\hat \Phi\in L^1({\mathbb R}^n)} and {\Phi(0) }. For {\epsilon>0} we consider the integrals

\displaystyle \int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot \xi}d\xi, \ \ \ \ \ (2)

which we will call the {\Phi}-means of the integral {\int_{{\mathbb R}^n}\hat f (\xi) e^{2\pi i x\cdot \xi}}, or just the {\Phi}-means of {\check f}. Using the multiplication formula in Proposition 4 we can rewrite the means (2) as

\displaystyle \int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot \xi} d\xi =(f*\tilde \phi_\epsilon)(x), \quad x\in {\mathbb R}^n. \ \ \ \ \ (3)

The following more general version of Proposition 8 is true.

Proposition 10 Let {\Phi\in L^1({\mathbb R}^n)} be such that {\phi=\hat \Phi\in L^1({\mathbb R}^n)} with {\int \phi =1}. We then have that the {\Phi}-means of {\int \hat f(\xi)e^{2\pi i x\cdot \xi}d\xi},

\displaystyle \int_{{\mathbb R}^n}\hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot\xi}d\xi,

converge to {f} in {L^1}, as {\epsilon\rightarrow 0}.

Proof: The proof is just a consequence of formula (3). Indeed, {\tilde \phi_\epsilon} is an approximation to the identity since {\tilde \phi\in L^1} and {\int \tilde {\phi}(x)dx=1} and thus {f*\tilde \phi_\epsilon} converges to {f} in the {L^1} norm as {\epsilon\rightarrow 0}. \Box

Proposition 8 says that the inversion formula is true whenever {f,\hat f\in L^1({\mathbb R}^n)}. This however is not the most natural assumption since the Fourier transform of an {L^1} function need not be integrable. The idea behind Proposition 10 is to `force’ {\hat f} in {L^1} by multiplying it by the {L^1} function {\Phi(\epsilon\xi)}. Thus, we artificially impose some decay on {\hat f}. This is equivalent to smoothing out the function {f} itself by convolving it with a smooth function {\tilde \phi_\epsilon}. Although no smoothness is explicitly assumed in Proposition 10, there is a hidden smoothness hypothesis in the requirement {\Phi, \phi \in L^1}. Indeed, we could have replaced this assumption by directly assuming that {\phi} is (say) a smooth function with compact support and taking {\Phi=\hat \phi}; then the conclusion {\hat \phi\in L^1({\mathbb R}^n)} would follow automatically. The trick of multiplying the Fourier transform of a general {L^1} function with an appropriate function in {L^1} or, equivalently, smoothing out the function {f} itself allows us then to invert the Fourier transform, at least in the {L^1}-sense. This process is usually referred to as a summability method.

As we shall see now, the inversion of a Fourier transform by means of a summability method is also valid in a pointwise sense. Because of formula (3), in order to understand the pointwise convergence of the {\Phi}-means of {\check{f}} we have to examine the pointwise convergence of the convolution {f*\phi_\epsilon} to {f}, whenever {\phi } is an approximation to the identity.

Definition 11 Let {f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}. The Lebesgue set of {f} is the set of points {x\in{\mathbb R}^n} such that

\displaystyle \lim_{r\rightarrow 0}\frac{1}{r^n}\int_{|y|< r}|f(x-y)-f(x)|dy=0.

The Lebesgue set of a locally integrable function {f} is closely related to the set where the integral of {f} is differentiable:

Definition 12 Let {f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}. The set of points where the integral of {f} is differentiable is the set of points {x\in{\mathbb R}^n} such that

\displaystyle \lim_{r\rightarrow 0} \frac{1}{\Omega_n r^n}\int_{|y|< r}f(x-y)=f(y),

where {\Omega_n} is the volume of the unit ball {B(0,1)} in {{\mathbb R}^n}. In other words, we say that the integral of {f} is differentiable at some point {x\in {\mathbb R}^n} if the average of {f} with respect to Euclidean balls centered at {x} the value of {f} at the point {x}.

We shall come back to these notions a bit later in the course when we will introduce the maximal function of {f} which is just the maximal average of {f} around every point. For now we will use as a black box the following theorem:

Theorem 13 Let {f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}. Then the integral of {f} is differentiable at almost every point {x\in {\mathbb R}^n}.

While postponing the proof of this theorem for later on in the course, we can already see the following simple proposition connecting the Lebesgue set of {f} to to the set of points where the integral of {f} is differentiable. In particular we see that almost every point in {{\mathbb R}^n} is Lebesgue point of {f}.

Corollary 14 Let {f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}. Then almost every {x\in{\mathbb R}^n} is a Lebesgue point of {f}.

Proof: For any rational number {q} we have that the function {f(x)-q} is locally integrable. Theorem 13 then implies that

\displaystyle \lim_{r\rightarrow 0} \frac{1}{r^n}\int_{|y|\leq r}\big\{|f(x-y)-q|-|f(x)-q|\big\}dy=0,

for almost every {x\in {\mathbb R}^n}. Thus the set {F_q} where the previous statement is not true has measure zero and so does the set {F:=\cup_{q\in{\mathbb Q}} F_q}. Now let {x\in {\mathbb R}^n \setminus F}. Indeed, let {\epsilon>0} and {q\in {\mathbb Q}} be such that {|f(x)-q|<\epsilon/2}. We then have

\displaystyle \begin{array}{rcl} \frac{1}{\Omega_n r^n}\int_{|y|<r}|f(x-y)-f(x)|dy &\leq& \frac{1}{\Omega_n r^n}\int_{|y|<r}|f(x-y)-q|dy\\ \\ && +\frac{1}{\Omega_n r^n}\int_{|y|<r}|f(x)-q|dy. \end{array}

The first summand converges to {|f(x)-q|<\epsilon/2} as {r\rightarrow 0} since {x\notin F} while the second summand is smaller than {\epsilon/2}. This shows that the Lebesgue set of {f} is contained in {{\mathbb R}^n\setminus F} and thus that almost every point in {{\mathbb R}^n} is a Lebesgue point of {f}. \Box

We can now give the following pointwise convergence result for approximations to the identity.

Theorem 15 Let {\phi \in L^1({\mathbb R}^n)} with {\int \phi=1}. We define {\psi(x):={\mathrm{esssup}}_{|y|\geq |x|}|\phi(y)|}. If {\psi\in L^1({\mathbb R}^n)} and {f\in L^p({\mathbb R}^n)} for {1\leq p \leq \infty} then

\displaystyle \lim_{\epsilon\rightarrow 0} (f*\phi_\epsilon)(x)=f(x),

whenever {x} is a Lebesgue point for {f}. If in addition \hat \phi \in L^1(\mathbb R^n) then the \hat \phi-means of \check f,

\displaystyle \int_{\mathbb R^n} \hat f(\xi) \hat \phi(\epsilon \xi) e^{2\pi i x\cdot \xi} d\xi,

converge to f (x) as \epsilon \to 0 for almost every x \in \mathbb R ^n .

Proof: Let {x} be a Lebesgue point of {f} and fix {\delta>0}. By Corollary 14 there exists {\eta>0} such that

\displaystyle \frac{1}{r^n}\int_{|y|<r}|f(x-y)-f(x)|dy<\delta, \ \ \ \ \ (4)

whenever {|r|<\eta}.

We can estimate as usual

\displaystyle \begin{array}{rcl} |(f*\phi_\epsilon)(x)-f(x)|&=&\bigg|\int_{{\mathbb R}^n}[f(x-y)-f(x)]\phi_\epsilon(y)dy\bigg|\\ \\ &\leq& \bigg|\int_{|y|<\eta}[f(x-y)-f(x)]\phi_\epsilon(y)dy\bigg| \\ \\ &&+\bigg|\int_{|y|\geq \eta}[f(x-y)-f(x)]\phi_\epsilon(y)dy\bigg| \\ \\ &=:& I_1+I_2. \end{array}

We claim that

\displaystyle \psi(x)\lesssim_{n,\phi} |x|^{-n}, \quad x\in{\mathbb R}^n. \ \ \ \ \ (5)

First of all observe that {\psi} is radially decreasing. We will abuse notation and write {\psi(x)=\psi(|x|)}. For every {r>0} we have that

\displaystyle \int_{r/2\leq |x|<r}\psi(x)dx\geq \psi(r)(r^n-(r/2)^n)\Omega_n.

Now since {\psi\in L^1}, the left hand side in the previous estimate tends to {0} when {r\rightarrow 0} and when {r\rightarrow \infty} we get the claim.

We write (4) in polar coordinates to get

\displaystyle \frac{1}{r^n}\int_{S^{n-1}}\int_0 ^r |f(x-sy')-f(x)|s^{n-1}ds d\sigma_{n-1}(y')<\delta.

Setting {g(s)=\int_{S^{n-1}}|f(x-sy')-f(x)| d\sigma_{n-1}(y')} we can rewrite the previous estimate in the form

\displaystyle G(r):=\int_0 ^r g(s)s^{n-1}ds \leq \delta r^n,

whenever {|r|<\eta}. We now estimate {I_1} as follows

\displaystyle \begin{array}{rcl} I_1&\leq& \int_{S^{n-1}}\int_0 ^\eta |f(x-ry')-f(x)|\psi_\epsilon(r)|d\sigma_{n-1}(y')r^{n-1}dr \\ \\ &=&\int_0 ^\eta g(r)r^{n-1}\frac{1}{\epsilon^n}\psi(r/\epsilon)dr\\ \\ &=&\int_0 ^\eta G'(r)\frac{1}{\epsilon^n}\psi(r/\epsilon) dr. \end{array}

At this point the proof simplifies a bit if we assume that {\psi} is differentiable. In this case we have that {\psi'\leq 0} and we can estimate the last integral by

\displaystyle \begin{array}{rcl} \int_0 ^\eta G'(r)\frac{1}{\epsilon^n}\psi(r/\epsilon) dr&=&\frac{1}{\epsilon^n}G(\eta)\psi(\frac{\eta}{\epsilon})-\int_0 ^\eta G(r)\frac{1}{\epsilon^{n+1}}\psi'(\frac{r}{\epsilon})dr\\ \\ &\lesssim_{n,\phi}& \delta - \delta \frac{1}{\epsilon^{n+1}}\int_0 ^\eta r^n\psi'(\frac{r}{\epsilon})dr \\ \\ &=&\delta +\delta \frac{n}{\epsilon^n}\int_0 ^\eta r^{n-1}\psi(r/\epsilon)dr \\ \\ &\leq & \delta\bigg(1+\frac{n}{\omega_{n-1}}\int_{{\mathbb R}^n}\psi(x)dx\bigg). \end{array}

The argument actually goes through without the assumption that {\psi} is differentiable by a clever use of the Riemann-Stieljes integral. Note that the function {\psi} is decreasing thus almost everywhere differentiable. This shows that {I_1\lesssim_{n,\phi} \delta}.

For {I_2} we estimate as follows

\displaystyle \begin{array}{rcl} I_2\leq \|f\|_p\|\psi_\epsilon\chi_{\{|x|\geq \eta \}}\|_{p'}+|f(x)|\|\chi_{\{|x|\geq \eta \}}\psi_\epsilon\|_1. \end{array}

For the second summand we have that

\displaystyle \|\chi_{\{|x|\geq \eta \}}\psi_\epsilon\|_1=\frac{1}{\epsilon^n}\int_{|x|\geq \eta}\psi_\epsilon(x/\epsilon)dx=\int_{|x|\geq \eta/\epsilon}\psi(x)dx\rightarrow 0,

as {\epsilon\rightarrow 0}, since {\psi\in L^1}.

On the other hand, we have

\displaystyle \begin{array}{rcl} \| \psi_\epsilon\chi_{\{|x|\geq \eta \}}\|_{p'}&=&\bigg(\int_{|x|\geq \eta}[\psi_\epsilon(x)]^{p'}dx\bigg)^\frac{1}{p'}=\bigg(\int_{|x|\geq \eta}[\psi_\epsilon(x)]^\frac{p'}{p}\psi_\epsilon(x)dx\bigg)^\frac{1}{p'}\\ \\ &\leq & \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_\infty ^\frac{1}{p} \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_1\leq \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_\infty ^\frac{1}{p}\|\psi\|_1. \end{array}

Now since {\psi_\epsilon} is decreasing we have

\displaystyle \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_\infty \leq \psi_\epsilon(\eta)=\frac{1}{\epsilon^n}\psi(\eta/\epsilon)=\eta^{-n}\big(\frac{\eta}{\epsilon}\big)^n\psi(\eta/\epsilon)\rightarrow 0,

when {\epsilon\rightarrow 0}.

We have showed that

\displaystyle \limsup_{\epsilon\rightarrow 0} |(f*\phi_\epsilon)(x)-f(x)|\lesssim_{n,\phi} \delta,

whenever {x} is a Lebesgue point of {f}. Since {\delta>0} was arbitrary this completes the proof of the theorem.\Box

Remark 2 The previous theorem is true in the case that {\phi} is a radially decreasing function in {L^1} or, in general, a function that satisfies a bound of the form {|\phi(x)|\lesssim (1+|x|)^{-(n+\delta)}} for some {\delta>0}.

We conclude the discussion on the inversion of the Fourier transform with a useful corollary.

Corollary 16 Let {f\in L^1({\mathbb R}^n)} and assume that {f} is continuous at {0} and that {\hat f \geq 0}. Then {\hat f\in L^1({\mathbb R}^n)} and

\displaystyle f(x)=\int_{{\mathbb R}^n}\hat f(\xi) e^{2\pi i x\cdot \xi}d\xi,

for almost every {x\in{\mathbb R}^n}. In particular,

\displaystyle f(0)=\int_{{\mathbb R}^n} \hat f(\xi)d\xi.

Proof: By identity (3) we have that

\displaystyle \int_{{\mathbb R}^n}\hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot \xi} d\xi=(f*\tilde \phi_\epsilon)(x),

for all {x\in{\mathbb R}^n}. Observe that the functions on both sides of this identity are continuous functions of {x}. Now let {\phi,\Phi} satisfy the conditions of Theorem 15. Assume furthermore that {\Phi} is non-negative and continuous at {0}. For example we can consider the function {\Phi(\xi)=\phi(\xi)=e^{-\pi |\xi|^2}}. Now since the point {0} is a point of continuity of {f}, it certainly belongs to the Lebesgue set of {f}. Thus we have that {\lim_{\epsilon\rightarrow 0} (f*\tilde \phi_\epsilon)(0)=f(0)} which gives

\displaystyle \lim_{\epsilon\rightarrow 0}\int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) d\xi=f(0).

Since {\hat f \Phi} is positive, we can use Fatou’s lemma to write

\displaystyle \int_{{\mathbb R}^n}\hat f(\xi) d\xi= \int_{{\mathbb R}^n}\liminf_{\epsilon_k\rightarrow 0} \hat f(\xi) \Phi(\epsilon _k\xi) d\xi\leq f(0),

so {\hat f \in L^1({\mathbb R}^n)}. Thus the inversion formula holds true for {f} and we get

\displaystyle f(x)=\int_{{\mathbb R}^n} \hat f(\xi) e^{2\pi i x\cdot \xi } d\xi,

for almost every {x\in {\mathbb R}^n}. However

\displaystyle f(0)=\lim_{\epsilon\rightarrow 0}\int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) d\xi=\int_{{\mathbb R}^n}\lim_{\epsilon \rightarrow 0}\hat f(\xi) \Phi(\epsilon \xi) d\xi=\int_{{\mathbb R}^n}\hat f(\xi) d\xi,

since {\hat f\in L^1}. \Box

2.1. Two special summability methods

We describe in detail two summability methods that are of special interest. These are based on the Examples 1 and 2 in the beginning of this set of notes.

The Gauss-Weierstrass summability method. By dilating the function {W(x)=e^{-\pi|x|^2}} we get

\displaystyle W(x,t):= W_{\sqrt{4\pi t}}(x)=(4\pi t)^{-\frac{n}{2}}e^{- \frac{|x|^2}{4t} }.

The function {W(x,t),\ t>0,} is called the Gauss kernel and it gives rise to the Gauss-Weierstrass method of summability. The Fourier transform of {W} is

\displaystyle \widehat{W_{\sqrt{4\pi t}} }(\xi)=\widehat W(\sqrt{2\pi t}\xi)=e^{-4\pi^2t|\xi|^2}.

It is also clear that

\displaystyle \int_{{\mathbb R}^n} W(x,t)dx=1,

for all {t>0}. The discussion in the previous sections applies to the Gauss-Weierstrass summability method and we have that the means

\displaystyle w(x,t):=\int_{{\mathbb R}^n}f(y)W(y-x,t)dy=\int_{{\mathbb R}^n} \hat f(\xi)e^{-4\pi^2 t |\xi|^2}e^{2\pi i x\cdot \xi} d\xi

convergence to {f} in {L^1({\mathbb R}^n)} and also in the pointwise sense, for every {x} in the Lebesgue set of {f}. One of the aspects of Gauss-Weierstrass summability is that the function {w(x,t)} defined above satisfies the heat equation:

\displaystyle \begin{array}{rcl} \frac{\partial w}{\partial t}-\Delta w &=& 0,\quad \mbox{ on }{\mathbb R}^{n+1} _+,\\ w(x,0) &=&f(x),\quad x\in{\mathbb R}^n. \end{array}

To see that the Gauss-Weierstrass means of {\check f} satisfy the Heat equation with initial data {f}, one can use the formula for {w(x,t)} and calculate everything explicitly. However it is easier to consider the Fourier transform of the solution {u(x,t)} of the Heat equation in the {x} variable and show that it must agree with the Fourier transform of {w(x,t)}, again in the {x} variable. Observe that under suitable assumptions on the initial data {f} we get that the solution {w(x,t)} converges to the initial data {f} as `time’ {t\rightarrow 0}.

Exercise 9 Let {f(x)=e^{-\pi x^2}}, {x\in{\mathbb R}}. Using the properties of the Fourier transform show that the function {\hat f} satisfies the initial value problem

\displaystyle \begin{array}{rcl} u'+2\pi x u&=&0,\\ \\ u(0)&=&1. \end{array}

Solve the initial value problem to give an alternative proof of the fact that {\hat f(\xi)=e^{-\pi \xi^2}}. Observe that the differential equation above is invariant under the Fourier transform.

The Abel summability method. We consider the function {P(x)=c_n\frac{1}{(1+|x|^2)^\frac{n+1}{2}}} where {c_n=\frac{\Gamma((n+1)/2}{\pi^\frac{n+1}{2}}}. By dilating the function {P} we have

\displaystyle P(x,t):= P_t(x)=c_n\frac{t}{(t^2+|x|^2)^\frac{n+1}{2}}.

The function {P(x,t),\ t>0,} is called the Poisson kernel (for the upper half plane) and it gives rise to the Abel method of summability. The Fourier transform of {P} is

\displaystyle \widehat{P_t}(\xi)=\hat P(t\xi)=e^{-2\pi t|\xi|}.

This is just a consequence of the calculation in Example 2, the inversion formula and the easily verified fact that {P\in L^1({\mathbb R}^n)}. It is also clear by a direct calculation or through the previous Fourier transform relation that

\displaystyle \int_{{\mathbb R}^n} P(x,t)dx=1,

for all {t>0}. Everything we have discussed in these notes applies to the Abel summability method. In particular we have that whenever {f\in L^1({\mathbb R}^n)}, the means

\displaystyle u(x,t):=\int_{{\mathbb R}^n}f(y)P(y-x,t)dy=\int_{{\mathbb R}^n} \hat{f}(\xi) e^{-2\pi t|\xi|}e^{-2\pi i x\cdot \xi} d\xi,

converge to {f} in {L^1} as {t\rightarrow 0} and also in the pointwise sense for all {x} in the Lebesgue set of {f}. The function {u(x,t)} is also called the Poisson integral or extension of {f}. It is not difficult to see that it satisfies the Dirichlet problem

\displaystyle \begin{array}{rcl} \Delta u &=&0, \quad \mbox{ on }{\mathbb R}^{n+1} _+,\\ u(x,0) &=&f(x),\quad x\in{\mathbb R}^n. \end{array}

Here we denote by {{\mathbb R}^n _+} the upper half plane {{\mathbb R}^n _+=\{(x,y):x\in {\mathbb R}^n, y>0\}}. Thus, if we are given an {L^1} function on the `boundary’ {{\mathbb R}^n}, the Poisson integral of {f} provides us with a harmonic function {u(x,t)} in the upper half plane which has boundary value {f} in the sense that {u(x,t)} converges to {f} as {t\rightarrow 0} both in the {L^1} sense as well as almost everywhere.

Remark 3 The Poisson extension of {f\in L^p({\mathbb R}^n)}, {1\leq p\leq \infty}

\displaystyle u(x,t)=\int_{{\mathbb R}^n} f(y)P(x-y,t)dy,

is harmonic in {{\mathbb R}_+ ^{n+1}}, that is, that it satisfies the Laplace equation:

\displaystyle \Delta_{x,t} u(x,t) =\sum_{j=1} ^n \frac{\partial}{\partial x_k}u(x,t)+\frac{\partial}{\partial t} u(x,t)=0.

This is essentially a consequence of the fact that {\Delta_{x,t}P(x,t)=0} for {(x,t)\in{\mathbb R}^n _+}.

In general, we can ask for a harmonic function {u(x,t)} in {{\mathbb R}^{n+1} _+} which has boundary value {\lim_{t\rightarrow 0}u(\cdot,t)=f\in L^p({\mathbb R}^n)} where the limit is taken {L^p}-sense. In the case {p<\infty} this extension is uniquely given by the Poisson integral of {f}. Also, the same is true if {p=\infty} and {f\in C_o({\mathbb R}^n)\subset L^\infty({\mathbb R}^n)}. On the other hand, if we ask for a function which is harmonic in {{\mathbb R}^{n+1} _+}, continuous in {\overline {{\mathbb R}^{n+1} _+}} and has boundary function {f}, then no assumption on {f} can guarantee that this extension is unique. Take for example {f=0} and {u_1(x,t)=0}, {u(x,t)=t}. The solution of the Dirichlet problem becomes unique though if we require in addition that the harmonic extension is a bounded function in {{\mathbb R}^{n+1} _+}. See [SW] for more information.

Exercise 10 Prove the subordination identity:

\displaystyle e^{-\beta}=\frac{1}{\sqrt{\pi}}\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}}e^{-\beta^2/4u}du,\quad \beta>0. \ \ \ \ \

For this, first prove the identities

\displaystyle \begin{array}{rcl} e^{-\beta}&=&\frac{2}{\pi}\int_0 ^\infty\frac{\cos \beta x}{1+x^2}dx,\\ \\ \frac{1}{1+x^2}&=&\int_0 ^\infty e^{-(1+x^2)u}du. \end{array}

The second identity above is obvious. In order to prove the first, use the theory of residues for the function

\displaystyle f(z)=\frac{e^{i\beta z}}{1+z^2}.


[Update 11 Mar 2011: Exercise 10 added.]

[Update 11 Mar 2011: Statement of Theorem 15 completed with the corollary about the convergence of the means of \check f.]

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About ioannis parissis

I'm a postdoc researcher at the Center for mathematical analysis, geometry and dynamical systems at IST, Lisbon, Portugal.
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