Singular integral

From Infogalactic: the planetary knowledge core
(Redirected from Singular integrals)
Jump to: navigation, search

In mathematics, singular integrals are central to harmonic analysis and are intimately connected with the study of partial differential equations. Broadly speaking a singular integral is an integral operator

T(f)(x) = \int K(x,y)f(y) \, dy,

whose kernel function K : Rn×Rn → R is singular along the diagonal x = y. Specifically, the singularity is such that |K(xy)| is of size |x − y|n asymptotically as |x − y| → 0. Since such integrals may not in general be absolutely integrable, a rigorous definition must define them as the limit of the integral over |y − x| > ε as ε → 0, but in practice this is a technicality. Usually further assumptions are required to obtain results such as their boundedness on Lp(Rn).

The Hilbert transform

<templatestyles src="Module:Hatnote/styles.css"></templatestyles>

The archetypal singular integral operator is the Hilbert transform H. It is given by convolution against the kernel K(x) = 1/(πx) for x in R. More precisely,

H(f)(x) = \frac{1}{\pi}\lim_{\varepsilon \to 0} \int_{|x-y|>\varepsilon} \frac{1}{x-y}f(y) \, dy.

The most straightforward higher dimension analogues of these are the Riesz transforms, which replace K(x) = 1/x with

K_i(x) = \frac{x_i}{|x|^{n+1}}

where i = 1, …, n and x_i is the i-th component of x in Rn. All of these operators are bounded on Lp and satisfy weak-type (1, 1) estimates.[1]

Singular integrals of convolution type

<templatestyles src="Module:Hatnote/styles.css"></templatestyles>

A singular integral of convolution type is an operator T defined by convolution with a kernel K that is locally integrable on Rn\{0}, in the sense that

T(f)(x) = \lim_{\varepsilon \to 0} \int_{|y-x|>\varepsilon} K(x-y)f(y) \, dy.

 

 

 

 

(1)

Suppose that the kernel satisfies:

1. The size condition on the Fourier transform of K

\hat{K}\in L^\infty(\mathbf{R}^n)

2. The smoothness condition: for some C > 0,

\sup_{y \neq 0} \int_{|x|>2|y|} |K(x-y) - K(x)| \, dx \leq C.

Then it can be shown that T is bounded on Lp(Rn) and satisfies a weak-type (1, 1) estimate.

Property 1. is needed to ensure that convolution (1) with the tempered distribution p.v. K given by the principal value integral

\operatorname{p.v.}\,\, K[\phi] = \lim_{\epsilon\to 0^+} \int_{|x|>\epsilon}\phi(x)K(x)\,dx

is a well-defined Fourier multiplier on L2. Neither of the properties 1. or 2. is necessarily easy to verify, and a variety of sufficient conditions exist. Typically in applications, one also has a cancellation condition

\int_{R_1<|x|<R_2} K(x) \, dx = 0 ,\ \forall R_1,R_2 > 0

which is quite easy to check. It is automatic, for instance, if K is an odd function. If, in addition, one assumes 2. and the following size condition

\sup_{R>0} \int_{R<|x|<2R} |K(x)| \, dx \leq C,

then it can be shown that 1. follows.

The smoothness condition 2. is also often difficult to check in principle, the following sufficient condition of a kernel K can be used:

  • K\in C^1(\mathbf{R}^n\setminus\{0\})
  • |\nabla K(x)|\le\frac{C}{|x|^{n+1}}

Observe that these conditions are satisfied for the Hilbert and Riesz transforms, so this result is an extension of those result.[2]

Singular integrals of non-convolution type

These are even more general operators. However, since our assumptions are so weak, it is not necessarily the case that these operators are bounded on L'p.

Calderón–Zygmund kernels

A function K : Rn×Rn → R is said to be a CalderónZygmund kernel if it satisfies the following conditions for some constants C > 0 and δ > 0.[2]

(a) \qquad |K(x,y)| \leq \frac{C}{|x-y|^n}
(b) \qquad |K(x,y) - K(x',y)| \leq \frac{C|x-x'|^\delta}{\bigl(|x-y|+|x'-y|\bigr)^{n+\delta}}\text{ whenever }|x-x'| \leq \frac{1}{2}\max\bigl(|x-y|,|x'-y|\bigr)
(c) \qquad |K(x,y) - K(x,y')| \leq \frac{C|y-y'|^\delta}{\bigl(|x-y|+|x-y'|\bigr)^{n+\delta}}\text{ whenever }|y-y'| \leq \frac{1}{2}\max\bigl(|x-y'|,|x-y|\bigr)

Singular integrals of non-convolution type

T is said to be a singular integral operator of non-convolution type associated to the Calderón–Zygmund kernel K if

\int g(x) T(f)(x) \, dx = \iint g(x) K(x,y) f(y) \, dy \, dx,

whenever f and g are smooth and have disjoint support.[2] Such operators need not be bounded on Lp

Calderón–Zygmund operators

A singular integral of non-convolution type T associated to a Calderón–Zygmund kernel K is called a Calderón–Zygmund operator when it is bounded on L2, that is, there is a C > 0 such that

\|T(f)\|_{L^2} \leq C\|f\|_{L^2},

for all smooth compactly supported ƒ.

It can be proved that such operators are, in fact, also bounded on all Lp with 1 < p < ∞.

The T(b) theorem

The T(b) theorem provides sufficient conditions for a singular integral operator to be a Calderón–Zygmund operator, that is for a singular integral operator associated to a Calderón–Zygmund kernel to be bounded on L2. In order to state the result we must first define some terms.

A normalised bump is a smooth function φ on Rn supported in a ball of radius 10 and centred at the origin such that |∂α φ(x)| ≤ 1, for all multi-indices |α| ≤ n + 2. Denote by τx(φ)(y) = φ(y − x) and φr(x) = r−nφ(x/r) for all x in Rn and r > 0. An operator is said to be weakly bounded if there is a constant C such that

 \left|\int T\bigl(\tau^x(\varphi_r)\bigr)(y) \tau^x(\psi_r)(y) \, dy\right| \leq Cr^{-n}

for all normalised bumps φ and ψ. A function is said to be accretive if there is a constant c > 0 such that Re(b)(x) ≥ c for all x in R. Denote by Mb the operator given by multiplication by a function b.

The T(b) theorem states that a singular integral operator T associated to a Calderón–Zygmund kernel is bounded on L2 if it satisfies all of the following three conditions for some bounded accretive functions b1 and b2:[3]

(a) M_{b_2}TM_{b_1} is weakly bounded;

(b) T(b_1) is in BMO;

(c) T^t(b_2), is in BMO, where Tt is the transpose operator of T.

Notes

  1. Lua error in package.lua at line 80: module 'strict' not found.
  2. 2.0 2.1 2.2 Lua error in package.lua at line 80: module 'strict' not found.
  3. Lua error in package.lua at line 80: module 'strict' not found.

References

  • Lua error in package.lua at line 80: module 'strict' not found..
  • Lua error in package.lua at line 80: module 'strict' not found..
  • Lua error in package.lua at line 80: module 'strict' not found..
  • Lua error in package.lua at line 80: module 'strict' not found. (in Russian).
  • Lua error in package.lua at line 80: module 'strict' not found..
  • Lua error in package.lua at line 80: module 'strict' not found., (European edition: ISBN 3-540-15967-3).
  • Lua error in package.lua at line 80: module 'strict' not found.