Splines, harmonic analysis, singular integrals. - MathOverflow

Splines, harmonic analysis, singular integrals.

2010-07-10T17:14:04Z

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.

Thanks.

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,

http://www.amsta.leeds.ac.uk/~sta6kvm/

http://www.maths.leeds.ac.uk/~john/

http://www.amsta.leeds.ac.uk/~sta6kvm/SpatialStatistics.html

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?

Answer by Willie Wong for Splines, harmonic analysis, singular integrals.

2010-07-10T17:24:09Z

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.

Answer by Will Jagy for Splines, harmonic analysis, singular integrals.

2010-07-10T17:27:16Z

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.

See:

http://en.wikipedia.org/wiki/Schwartz_space

Answer by Olumide for Splines, harmonic analysis, singular integrals.

2010-07-10T18:07:40Z

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?

Answer by Helge for Splines, harmonic analysis, singular integrals.

2010-07-10T18:18:11Z

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 $\mathcal{C}^{\infty}$.

Answer by Zen Harper for Splines, harmonic analysis, singular integrals.

2010-07-10T18:41:49Z

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)...

Answer by Peter Luthy for Splines, harmonic analysis, singular integrals.

2010-07-10T21:45:15Z

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.

Answer by Olumide for Splines, harmonic analysis, singular integrals.

2010-07-23T12:44:15Z

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)

Thanks.

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