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Apologies if my question is poorly phrased. I'm a computer scientist trying to teach myself about generalized functions. (Simple explanations are preferred. -- Thanks.)

One of the references I'm studying states that the space of Schwartz test functions of rapid decrease is the set of infinitely differentiable functions: $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ such that for all natural numbers $n$ and $r$,

$\lim_{x\rightarrow\pm\infty} |x^n \varphi^{(r)}(x)|$

What I would like to know is why is necessary or important for test functions to decay rapidly in this manner? i.e. faster than powers of polynomials. I'd appreciate an explanation of the intuition behind this statement and if possible a simple example.


EDIT: the OP is actually interested in a particular 1994 paper on "Spatial Statistics" by Kent and Mardia, 1994 Link between kriging and thin plate splines (with J. T. Kent). In Probability, Statistics and Optimization (F. P. Kelly ed.). Wiley, New York, pp 325-339.

Both are in Statistics at Leeds,




Scanned article: http://www.gigasize.com/get.php?d=90wl2lgf49c

FROM THE OP: Here is motivation for my question: I'm trying to understand a paper that replaces an integral $$\int f(\omega) d\omega$$ with $$\int \frac{|\omega|^{2p + 2}}{ (1 + |\omega|^2)^{p+1}} \; f(\omega) \; d\omega$$ where $p \ge 0$ ($p = -1$ yields to the unintegrable expression) because $f(\omega)$ contains a singularity at the origin i.e. is of the form $\frac{1}{\omega^2}.$

LATER, ALSO FROM THE OP: I understand some parts of the paper but not all of it. For example, I am unable to justify the equations (2.5) and (2.7). Why do they take these forms and not some other form?

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The faster a function decays, the more functions you can integrate it against. –  Qiaochu Yuan Jul 10 '10 at 17:19
I would say that «because with that condition things work as one wants them to» in some contexts. Notice that there are other spaces of test functions which are useful (and in fact, generalized functions are most generally introduced using not the ones you mention but $C^\infty$ functions of compact support) –  Mariano Suárez-Alvarez Jul 10 '10 at 17:23
Somewhere below in the comments ( mathoverflow.net/questions/31308/… ) the OP gave a link to the paper he is looking at. Somehow I feel like this rather open-ended question here is getting nowhere. I think it may be more productive if the OP points out (in perhaps a new question) the precise statement that is giving him/her trouble. That way he/she is likely to get a more focused and to the point response. Just my 2 pence. –  Willie Wong Jul 10 '10 at 22:57
My primary objective is to understand the paper by Kent and Mardia. But in order to begin to do so, I think need to become familiar with the mathematics they use which I "assumed" was from the area of generalized functions and specifically Schwartz spaces, which is why I've been reading up on both subjects. But it would be helpful if someone would kindly confirm my "suspicions" and point me in the right direction. I apologize for any confusion I may have caused. –  Olumide Jul 11 '10 at 1:59
Let me put it this way: if you intend to make new theories and construct new theorems, then you do need to understand the intricacies of the details of generalized functions, especially what you are allowed to do and what you are not. If you intend to be a user of results, or are just surveying the literature, all that suffices you know is that for tempered distributions (a subset of generalized functions which grows at most polynomially near infinity; sort of opposite of Schwartz functions), many things that are defined for functions can be done for them... –  Willie Wong Jul 11 '10 at 11:48

7 Answers 7

From the Fourier analysis point of view, the reason is the property of the Fourier transform to interchange derivatives and multiplications, which you can read more about on Wikipedia. The crucial point is that the smoothness of a function is directly related to the decay rate of its (inverse) Fourier transform. So if you want a family of infinitely differentiable functions whose Fourier transform is also infinitely differentiable, you are necessarily led to consider the Schwarz class.

As a by product of the definition, you also have that the Schwarz class is closed under pointwise multiplication and under convolutions.

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And of course, also the Schwartz class is closed under the Fourier transform. –  Robin Chapman Jul 10 '10 at 18:13
These are good ideas, but I think your concluding sentence is a bit too strong. One could easily define the space of test functions to be the space of compactly supported smooth functions. The Fourier transform of such a function, being a very special kind of Schwarz function, is also smooth with extreme decay at infinity. The problem is that while any locally integrable function defines a linear functional over $C_c^{\infty}$, the formula $\hat{T}(\phi)=T(\hat{\phi})$ does not make sense since $\hat{\phi}$ will never be compactly supported when $\phi$ is compactly supported. –  Peter Luthy Jul 10 '10 at 21:13
@Peter: see also Robin's comment. $C^\infty_c$ is not closed under Fourier transform, which is implicitly what I meant. I also thought about giving the fact that $\mathcal{S}$ is invariant under $\mathcal{F}$ implies that the Fourier transform is well defined for $\mathcal{S}'$, but I got lazy (especially since the OP is a computer scientist and I don't know how much background he has). So no, I don't think the concluding sentence is too strong. I just left a gap. Which you nicely filled. Thanks :) –  Willie Wong Jul 10 '10 at 21:54

If you want to extend differentiation to all continuous functions, then (provided you have some convenient mathematical properties of the extension) you are FORCED to use distributions or roughly equivalent things; you have no choice! Similarly, to extend the Fourier transform you are forced to consider tempered distributions.

Speaking as a pure mathematician: the main purpose of general distributions is to extend differentiation, not integration (since integration makes things nicer; it is differentiation which is the nastier operation). They are fine as long as you aren't using the Fourier transform.

Thus, every locally integrable function can be regarded as a distribution, and therefore differentiated; so, when you're considering differential equations, this might be all you need (you don't have to worry whether the functions are differentiable or not, because distributions always are). You find distribution solutions, then try to prove that they're actually functions.

It's similar to solving polynomial equations by using complex numbers; even if all the roots are real, it's still sometimes easier to solve them with complex numbers, then try to prove they're real (e.g. by showing they're self-conjugate).

However, if you want to do Fourier Transforms then you have to consider tempered distributions (or Schwartz distributions), since general distributions are sometimes too nasty to have Fourier transforms.

Note that even genuine locally integrable functions need not represent tempered distributions, so general distributions are not appropriate for Fourier transforms even when you only want to consider functions.

But Fourier inversion works perfectly for tempered distributions, no further restrictions are needed, unlike, say, $L^1$. If $f \in L^1$ then $\widehat{f}$ is usually not in $L^1$, so you can't do Fourier inversion theory nicely on $L^1$ (you would have to assume that also $\widehat{f} \in L^1$, which is often not true!)

Extension in mathematics is very powerful; when you don't have to worry about restrictions and annoying details, it is easier! For example, complex numbers are easier than real numbers, complex analysis is easier than real analysis, and Lebesgue integration is easier than Riemann integration!! Students never believe this, but it's true if you actually want to use it (rather than do toy problems in books)...

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Just a quick clue. The example you want is essentially the Gaussian normal distribution from probability, $$ \frac{1}{\sqrt {2 \pi}} \; \; e^{- x^2 / 2} $$ and probably the simplest motivation is that the Fourier transform of this function is just itself (well, up to a constant multiple, depends on whose definition you have).

These are a stand-in for functions of compact support. A function and its Fourier transform cannot both have compact support, that is a fact of life.



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Your statement about functions of compact support is the correct problem with defining the Fourier transform of distributions with smooth compactly supported functions as the test functions. –  Peter Luthy Jul 10 '10 at 21:14

The Schwartz space $\mathcal{F}$ is just one space, one could use to define distributions. Two other common examples are smooth functions $C^{\infty}$ and smooth functions with compact support $C^{\infty} _c$. Then one has the inclusions $$ C^{\infty} _{c} \subseteq \mathcal{S} \subseteq C^{\infty} $$ Now distributions are just taking the topological dual of these spaces, so one has then $$ (C^{\infty})' \subseteq \mathcal{S}' \subseteq (C^{\infty} _{c})' $$ So the inclusions get reversed. So imposing a less restrictive decay condition would lead you to a small space of distributions. In fact, $(C^{\infty})'$ consists of distributions of compact support.

The other issue mentioned in the other posts, is that the Fourier transform takes the Schwartz space into itself. It is much less obvious what the Fouriertransform does on $C_c^{\infty}$, and the Fourier transform is not even defined on $\mathcal{C}^{\infty}$.

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While I'm not saying anything new, I feel the responses thus far either miss the point or are not very complete. Generalized functions (aka distributions) are defined as linear functionals on some class of functions, typically referred to as test functions. To begin with, one usually wants any locally integrable function to be a generalized function. If $f$ is any locally integrable function then the generalized function corresponding to $f$ is just the linear functional $\int f\phi$ when $\phi$ is a test function. So the obvious first choice for the space of test functions is the space of compactly supported functions since integrating a locally integrable function against a smooth compactly supported always makes sense. Then one can define the derivative of a generalized function, say T, to be the functional T' which satisfies $T'(\phi)=-T(d\phi/dx)$ whenever $\phi$ is a smooth, compactly supported function. If T can be represented by a smooth function, then this is just the integration by parts formula, which makes sense since $\phi$ is compactly supported. So the function $e^{e^{e^x}}$ is a perfectly reasonable generalized function in this case.

As said a number of times above, one would also like to define the Fourier transform of a generalized function via the formula $\hat{T}(\phi)=T(\hat{\phi})$. The problem with the space of compactly supported functions is that the Fourier transform of a nonzero compactly supported function is never compactly supported. So $T(\hat{\phi})$ might not make sense if T is allowed to be any locally integrable function. In particular, suppose that $\phi$ was some smooth function of compact support whose Fourier transform goes to zero slower than something like $e^{-x^{10}}$. The function $e^{x^{11}}$ is locally integrable and hence a linear functional on the space of compactly supported smooth functions, but it is easy to see that $\int e^{x^{11}}\hat{\phi}$ isn't going to be a finite number.

The Schwarz space is nice because the Fourier transform of a Schwarz function is a Schwarz function. So given any linear functional T on the Schwarz space (such a T is called a tempered distribution), one can define the Fourier transform $\hat{T}$ of $T$ via the formula $\hat{T}(\phi)=T(\hat{\phi})$ when $\phi$ is a Schwarz function. This formula will always make sense when T is a tempered distribution.

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Thanks everyone for answers given so far. Now for some really ignorant questions from me. I'm really trying to make sense of generalized functions, so here goes:

Its often said that the concept of generalized functions helps to assign integrals to otherwise integrable functions (pardon my phrasing). What confuses me is why multiplying an otherwise unintegrable function with an "arbitrary test function" and then integrating the product is a valid. This seems to me to be the reason for the Schwartz class of test functions; namely functions that can "cool down" faster than any polynomial can blow up. Or in other words, given an ill-behaved, ready-to-blow-up function, a test function that can "tame it" can always be chosen ...

Is this right?

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Here is motivation for my question: I'm trying to understand a paper that replaces an integral $\int f(\omega) d\omega$ with $\int \frac{|\omega|^{2p + 2}}{ (1 + |\omega|^2)^{p+1}} f(\omega) d\omega$ where $p \ge 0$ ($p = -1$ yields to the unintegrable expression) because $f(\omega)$ contains a singularity at the origin i.e. is of the form $frac{1}{\omega^2}$. The paper in question is: "The link between kriging and thin-plate splines" by J.T. Kent and K.V. Madria. The paper however makes no mention of Schwartz spaces. –  Olumide Jul 10 '10 at 19:07
Google scholar doesn't show a paper by that name. Do you have a link or a journal reference? –  Willie Wong Jul 10 '10 at 19:23
Its a paper in the collection: "Probability, statistics, and optimisation: a tribute to Peter Whittle" : Wiley, 1994. I'll send you a link to the scanned copy. –  Olumide Jul 10 '10 at 20:14
Here is a link to the scan of the paper: gigasize.com/get.php?d=90wl2lgf49c –  Olumide Jul 10 '10 at 20:34
@Will Jagy: I disagree. Splines sometimes have something to do with harmonic analysis. The linked paper has some rudimentary singular integral type stuff, I wouldn't say that it has nothing to do with Fourier transform. @Olumide: if you are trying to understand Schwartz functions, you are taking the roundabout route to understand all the background material before reading the paper. While I laud such work ethic in general, this is not the way to go if you need to understand the paper in a hurry. –  Willie Wong Jul 10 '10 at 22:54
up vote 0 down vote accepted

I believe I now have the answer to the question. The power of $\omega$ appear from the taylor expansion of $e^{i\omega.t_j}$ (in section 2.3 of Kent and Mardia's paper)


(Apologies for the seeming bit of self promotion, but I've tagged this as the correct answer.)

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