Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following is from Stein and Shakarchi's Complex Analysis:

For each $a>0$ we denote by ${\mathcal F}_a$ the class of all functions $f$ that satisfy the following two conditions:
1. The function $f$ is holomorphic in the horizontal strip $$S_a=\{z\in{\Bbb C}:|Im(z)|<a\}$$ 2. There exists a constant $A>0$ such that $$ |f(x+iy)|\leq\frac{A}{1+x^2}\quad\text{for all}\quad x\in{\Bbb R}, |y|<a. $$

Denote by ${\mathcal F}$ the class of all functions that belong to ${\mathcal F}_a$ for some $a$. Then the Fourier inversion holds in this class.

My questions are: is there a name for this class? Does it have anything to do with the Schwartz space on which the Fourier transform is a linear isomorphism?

share|improve this question
1  
The Fourier transform is an isomorphism also on $\mathscr S'(\mathbb R^d)$, the space of tempered distributions. –  Jochen Wengenroth Feb 24 '13 at 17:14

3 Answers 3

up vote 3 down vote accepted

This space has no name (no that I know), and it is probably invented by Stein and Shakarchi for some pedagogical purpose, or for convenience of exposition.

As Serge says it is not contained in the Schwartz space $S$ (of test functions). Neither it contains the Schwartz space $S$.

It is contained in the space of Schwartz distributions $S^\prime$ but probably the authors did not want to introduce distributions in the elementary Complex Analysis textbook. As it contains all examples they want to consider in this book, they introduced it.

There are several other spaces where Fourier transform can be defined, and even the image can be explicitly described, for example, $L^2$, or more general distributions than Schwartz, but I agree with Bazin that Schwartz distributions is perhaps the most useful class.

In an elementary textbook, one has to begin with something, so they invented this artificial class. In the later volumes of the course they address $L^2$ and Schwartz distributions.

share|improve this answer

The optimal space for the Fourier transform is the space of tempered distributions $\mathscr S'(\mathbb R^n)$, i.e. the dual space of the Schwartz functions $\mathscr S(\mathbb R^n)$. The latter is the Fréchet space of $C^\infty$ functions on $\mathbb R^n$ decreasing faster than any polynomial as well as all their derivatives; the semi-norms are $$ p_{\alpha\beta}(\phi)=\sup_{\mathbb R^n}\vert x^\beta\partial^\alpha\phi\vert. $$ Defining $\hat \phi(\xi)=\int e^{-2i\pi x\cdot\xi} \phi(x) dx$ for $\phi\in\mathscr S(\mathbb R^n)$, it is easy to prove that the Fourier transform is an isomorphism of $\mathscr S(\mathbb R^n)$ and $$\phi(x)=\int e^{2i\pi x\cdot\xi} \hat\phi(\xi) d\xi.\tag{1}$$

The dual space of $\mathscr S(\mathbb R^n)$ is huge, contains for instance all the spaces $L^p(\mathbb R^n)$ and all the distribution derivatives of any function in these spaces. The space $\mathscr S'(\mathbb R^n)$ contains also the homogeneous distributions and the distributions with compact support. It is easy to define by duality the Fourier transform on $\mathscr S'(\mathbb R^n)$. With brackets of duality, we define $$ \langle\hat T,\phi\rangle=\langle T,\hat \phi\rangle\quad\text{and we have $\hat{\hat T}= \tilde T$,} $$ where $\tilde T(x)=T(-x)$, that is the very same formula as in (1).

share|improve this answer

The function $f(x)=1/(1+x^2)$ belongs to this class but not to the Schwartz space of test functions.

Any $C^\infty$ function with bounded support belongs to the Schwartz space but not to this class.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.