Singular integral

Singular integrals of convolution type

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

Suppose that the kernel satisfies:

1. The size condition on the Fourier transform of ${\displaystyle K}$

   ${\displaystyle \hat{K}\in L^{\mathrm{\infty }}({\mathbb{R}}^n)}$

2. The smoothness condition: for some ${\displaystyle C>0}$,

   ${\displaystyle \underset{y\ne 0}{sup}{\int }_{|x|>2|y|}|K(x-y)-K(x)|dx\le C.}$

Then it can be shown that ${\displaystyle T}$ is bounded on ${\displaystyle L^p({\mathbb{R}}^n)}$ and satisfies a weak-type ${\displaystyle (1,1)}$ estimate.

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

${\displaystyle \mathrm{p}.\mathrm{v}.K[\varphi ]=\underset{ϵ\to 0^{+}}{lim}{\int }_{|x|>ϵ}\varphi (x)K(x)dx}$

is a well-defined Fourier multiplier on ${\displaystyle L^2}$. 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

${\displaystyle {\int }_{R_1<|x|<R_2}K(x)dx=0,\mathrm{\forall }{R}_1,R_2>0}$

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

${\displaystyle \underset{R>0}{sup}{\int }_{R<|x|<2R}|K(x)|dx\le C,}$

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 ${\displaystyle K}$ can be used:

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

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

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 ${\displaystyle L^p}$.

Calderón–Zygmund kernels

A function ${\displaystyle K:{\mathbb{R}}^n\times {\mathbb{R}}^n\to \mathbb{R}}$ is said to be a Calderón–Zygmund kernel if it satisfies the following conditions for some constants ${\displaystyle C>0}$ and ${\displaystyle \delta >0}$.

1. ${\displaystyle |K(x,y)|\le \frac{C}{|x-y{|}^n}}$

2. ${\displaystyle |K(x,y)-K(x^{\prime },y)|\le \frac{C|x-x^{\prime }{|}^{\delta }}{(|x-y|+|x^{\prime }-y|{)}^{n+\delta }}\text{ whenever }|x-x^{\prime }|\le \frac{1}{2}max(|x-y|,|x^{\prime }-y|)}$

3. ${\displaystyle |K(x,y)-K(x,y^{\prime })|\le \frac{C|y-y^{\prime }{|}^{\delta }}{(|x-y|+|x-y^{\prime }|{)}^{n+\delta }}\text{ whenever }|y-y^{\prime }|\le \frac{1}{2}max(|x-y^{\prime }|,|x-y|)}$

Singular integrals of non-convolution type

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

${\displaystyle \int g(x)T(f)(x)dx=\iint g(x)K(x,y)f(y)dydx,}$

whenever ${\displaystyle f}$ and ${\displaystyle g}$ are smooth and have disjoint support. Such operators need not be bounded on ${\displaystyle L^p}$

Calderón–Zygmund operators

A singular integral of non-convolution type ${\displaystyle T}$ associated to a Calderón–Zygmund kernel ${\displaystyle K}$ is called a Calderón–Zygmund operator when it is bounded on ${\displaystyle L^2}$, that is, there is a ${\displaystyle C>0}$ such that

${\displaystyle \Vert T(f){\Vert }_{L^2}\le C\Vert f{\Vert }_{L^2},}$

for all smooth compactly supported ƒ.

It can be proved that such operators are, in fact, also bounded on all ${\displaystyle L^p}$ with ${\displaystyle 1<p<\mathrm{\infty }}$.

The T ( b ) {\displaystyle T(b)} theorem

The ${\displaystyle T(b)}$ 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 ${\displaystyle L^2}$. In order to state the result we must first define some terms.

A normalised bump is a smooth function ${\displaystyle \phi }$ on ${\displaystyle {\mathbb{R}}^n}$ supported in a ball of radius 1 and centred at the origin such that ${\displaystyle |{\mathrm{\partial }}^{\alpha }\phi (x)|<1}$, for all multi-indices ${\displaystyle |\alpha |\le n+2}$. Denote by ${\displaystyle {\tau }^x(\phi )(y)=\phi (y-x)}$ and ${\displaystyle {\phi }_r(x)=r^{-n}\phi (x/r)}$ for all ${\displaystyle x}$ in ${\displaystyle {\mathbb{R}}^n}$ and ${\displaystyle r>0}$. An operator is said to be weakly bounded if there is a constant ${\displaystyle C}$ such that

${\displaystyle \left|\int T({\tau }^x({\phi }_r))(y){\tau }^x({\psi }_r)(y)dy\right|\le C{r}^{-n}}$

for all normalised bumps ${\displaystyle \phi }$ and ${\displaystyle \psi }$. A function is said to be accretive if there is a constant ${\displaystyle c>0}$ such that ${\displaystyle Re(b)(x)\ge c}$ for all ${\displaystyle x}$ in ${\displaystyle \mathbb{R}}$. Denote by ${\displaystyle M_b}$ the operator given by multiplication by a function ${\displaystyle b}$.

The ${\displaystyle T(b)}$ theorem states that a singular integral operator ${\displaystyle T}$ associated to a Calderón–Zygmund kernel is bounded on ${\displaystyle L^2}$ if it satisfies all of the following three conditions for some bounded accretive functions ${\displaystyle b_1}$ and ${\displaystyle b_2}$:

1. ${\displaystyle M_{b_2}T{M}_{b_1}}$ is weakly bounded;
2. ${\displaystyle T(b_1)}$ is in BMO;
3. ${\displaystyle T^t(b_2),}$ is in BMO, where ${\displaystyle T^t}$ is the transpose operator of ${\displaystyle T}$.