Singular integral operators of convolution type explained

In mathematics, singular integral operators of convolution type are the singular integral operators that arise on Rⁿ and Tⁿ through convolution by distributions; equivalently they are the singular integral operators that commute with translations. The classical examples in harmonic analysis are the harmonic conjugation operator on the circle, the Hilbert transform on the circle and the real line, the Beurling transform in the complex plane and the Riesz transforms in Euclidean space. The continuity of these operators on L² is evident because the Fourier transform converts them into multiplication operators. Continuity on L^p spaces was first established by Marcel Riesz. The classical techniques include the use of Poisson integrals, interpolation theory and the Hardy–Littlewood maximal function. For more general operators, fundamental new techniques, introduced by Alberto Calderón and Antoni Zygmund in 1952, were developed by a number of authors to give general criteria for continuity on L^p spaces. This article explains the theory for the classical operators and sketches the subsequent general theory.

L² theory

Hilbert transform on the circle

See also: Harmonic conjugate. The theory for L² functions is particularly simple on the circle. If f ∈ L²(T), then it has a Fourier series expansion$$f(\backslash theta)=\backslash sum\_\; a\_n\; e^.$$

Hardy space H²(T) consists of the functions for which the negative coefficients vanish, a_n = 0 for n < 0. These are precisely the square-integrable functions that arise as boundary values of holomorphic functions in the open unit disk. Indeed, f is the boundary value of the function

$$F(z)=\backslash sum\_\; a\_n\; z^n,$$

in the sense that the functions

$$f\_r(\backslash theta)=F(re^),$$

defined by the restriction of F to the concentric circles |z| = r, satisfy

$$\backslash |f\_r-f\backslash |\_2\; \backslash rightarrow\; 0.$$

The orthogonal projection P of L²(T) onto H²(T) is called the Szegő projection. It is a bounded operator on L²(T) with operator norm 1. By Cauchy's theorem

$$F(z)=\; \backslash int\_$$

=1

\frac \,d\zeta= \int_^ \, d\theta.

Thus

$$F(re^)=\; \backslash int\_^\; \backslash ,\; d\backslash theta.$$

When r = 1, the integrand on the right-hand side has a singularity at θ = 0. The truncated Hilbert transform is defined by

$$H\_\backslash varepsilon\; f(\backslash varphi)\; =\; \backslash int\_\; \backslash ,\; d\backslash theta=\; \backslash int\_$$

\ge \delta

\, d\zeta,

where δ = |1 – e^iε|. Since it is defined as convolution with a bounded function, it is a bounded operator on L²(T). Now

$$H\_\backslash varepsilon=\backslash int\_\backslash varepsilon^\backslash pi\; 2\; \backslash Re\; (1-e^)^\; \backslash ,\; d\backslash theta\; =\backslash int\_\backslash varepsilon^\backslash pi\; 1\; \backslash ,\; d\backslash theta\; =\; i\; -\; .$$

If f is a polynomial in z then

$$H\_\backslash varepsilon\; f(z)\; -\; f(z)=\; \backslash int\_$$

\ge \delta

\, d\zeta.

By Cauchy's theorem the right-hand side tends to 0 uniformly as ε, and hence δ, tends to 0. So

$$H\_\backslash varepsilon\; f\; \backslash rightarrow\; if$$

uniformly for polynomials. On the other hand, if u(z) = z it is immediate that

$$\backslash overline\; =\; -\; u^\; H\_\backslash varepsilon(u\; \backslash overline).$$

Thus if f is a polynomial in z⁻¹ without constant term

H _{\varepsilon} f → - i f uniformly.

Define the Hilbert transform on the circle by$$H\; =\; i(2P-I).$$

Thus if f is a trigonometric polynomial

H _{\varepsilon} f → Hf uniformly.

It follows that if f is any L² function

H _{\varepsilon} f → Hf in the L² norm.

This is an immediate consequence of the result for trigonometric polynomials once it is established that the operators H_ε are uniformly bounded in operator norm. But on [–''π'',''π'']

$$(1-e^)^=\; [(1-e^\{i\backslash theta\})^\{-1\}\; -i\backslash theta^\{-1\}]\; +i\backslash theta^.$$

The first term is bounded on the whole of [–π,π], so it suffices to show that the convolution operators S_ε defined by

$$S\_\backslash varepsilon\; f(\backslash varphi)\; =\; \backslash int\_\; f(\backslash varphi-\backslash theta)\backslash theta^\backslash ,d\backslash theta$$

are uniformly bounded. With respect to the orthonormal basis e^inθ convolution operators are diagonal and their operator norms are given by taking the supremum of the moduli of the Fourier coefficients. Direct computation shows that these all have the form

$$\backslash frac\backslash left\; |\backslash int\_a^b\; \backslash ,\; dt\backslash right|$$

with 0 < a < b. These integrals are well known to be uniformly bounded.

It also follows that, for a continuous function f on the circle, H_εf converges uniformly to Hf, so in particular pointwise. The pointwise limit is a Cauchy principal value, written

$$Hf=\; \backslash mathrm\backslash ,\; \backslash int\; \backslash ,\; d\backslash zeta.$$

If f is just in L² then H_εf converges to Hf pointwise almost everywhere. In fact define the Poisson operators on L² functions by

$$T\_r\; \backslash left\; (\backslash sum\; a\_n\; e^\; \backslash right)=\backslash sum\; r^$$

a_n e^,

for r < 1. Since these operators are diagonal, it is easy to see that T_rf tends to f in L² as r increases to 1. Moreover, as Lebesgue proved, T_rf also tends pointwise to f at each Lebesgue point of f. On the other hand, it is also known that T_rHf − H_{1 − r} f tends to zero at each Lebesgue point of f. Hence H_{1 – r} f tends pointwise to f on the common Lebesgue points of f and Hf and therefore almost everywhere.

Results of this kind on pointwise convergence are proved more generally below for L^p functions using the Poisson operators and the Hardy–Littlewood maximal function of f.

The Hilbert transform has a natural compatibility with orientation-preserving diffeomorphisms of the circle.^[1] Thus if H is a diffeomorphism of the circle with

$$H(e^)=e^,\backslash ,\backslash ,\backslash ,\; h(\backslash theta+2\backslash pi)=h(\backslash theta)+2\backslash pi,$$

then the operators

$$H\_\backslash varepsilon^h\; f(e^)=\backslash frac\backslash int\_$$

\ge \varepsilon

\frace^\, d\theta,

are uniformly bounded and tend in the strong operator topology to H. Moreover, if Vf(z) = f(H(z)), then VHV⁻¹ − H is an operator with smooth kernel, so a Hilbert–Schmidt operator.

In fact if G is the inverse of H with corresponding function g(θ), then

$$(VH^h\_\backslash varepsilon\; V^-\; H\_\backslash varepsilon)\; f(e^)\; =\; \backslash int\_$$

\ge \varepsilon

\left[{g^\prime(\theta) e^{ig(\theta)} \over e^{ig(\theta)} - e^{ig(\varphi)}} - {e^{i\theta} \over e^{i\theta} - e^{i\varphi} }\right]\,f(e^)\, d\theta.

Since the kernel on the right hand side is smooth on T × T, it follows that the operators on the right hand side are uniformly bounded and hence so too are the operators H_ε^h. To see that they tend strongly to H, it suffices to check this on trigonometric polynomials. In that case

$$H^h\_\backslash varepsilon\; f(\backslash zeta)=\; \backslash int\_$$

\ge \varepsilon

\frac dz= \int_

\ge \varepsilon

\, dz + \frac \int_

\ge \varepsilon

.

In the first integral the integrand is a trigonometric polynomial in z and ζ and so the integral is a trigonometric polynomial in ζ. It tends in L² to the trigonometric polynomial$$\backslash int\; \backslash ,\; dz.$$

The integral in the second term can be calculated by the principle of the argument. It tends in L² to the constant function 1, so that

$$\backslash lim\_\; H\_\backslash varepsilon^h\; f(\backslash zeta)\; =\; f(\backslash zeta)\; +\; \backslash int\; \backslash ,\; dz,$$

where the limit is in L². On the other hand, the right hand side is independent of the diffeomorphism. Since for the identity diffeomorphism, the left hand side equals Hf, it too equals Hf (this can also be checked directly if f is a trigonometric polynomial). Finally, letting ε → 0,

$$(VH\; V^-\; H)\; f(e^)\; =\; \backslash frac\; \backslash int\; \backslash left[\{g^\backslash prime(\backslash theta)\; e^\{ig(\backslash theta)\}\; \backslash over\; e^\{ig(\backslash theta)\}\; -\; e^\{ig(\backslash varphi)\}\}\; -\; \{e^\{i\backslash theta\}\; \backslash over\; e^\{i\backslash theta\}\; -\; e^\{i\backslash varphi\}\}\backslash right]\backslash ,f(e^)\backslash ,\; d\backslash theta.$$

The direct method of evaluating Fourier coefficients to prove the uniform boundedness of the operator H^ε does not generalize directly to L^p spaces with 1 < p < ∞. Instead a direct comparison of H^εf with the Poisson integral of the Hilbert transform is used classically to prove this. If f has Fourier series

$$f(e^)=\backslash sum\_\; a\_n\; e^,$$

its Poisson integral is defined by

$$P\_rf(e^)=\backslash sum\_\; a\_n\; r^$$

e^=\int_0^ \,d\theta =K_r\star f(e^),

where the Poisson kernel K_r is given by$$K\_r(e^)=\backslash sum\_\; r^$$

e^ =.

In f is in L^p(T) then the operators P_r satisfy$$\backslash |P\_rf\; -\; f\backslash |\_p\backslash rightarrow\; 0.$$

In fact the K_r are positive so$$\backslash |K\_r\backslash |\_1\; =\; \backslash int\_0^\; K\_r(e^)\backslash ,\; d\backslash theta\; =\; 1.$$

Thus the operators P_r have operator norm bounded by 1 on L^p. The convergence statement above follows by continuity from the result for trigonometric polynomials, where it is an immediate consequence of the formula for the Fourier coefficients of K_r.

The uniform boundedness of the operator norm of H_ε follows because HP_r − H_1−r is given as convolution by the function ψ_r, wherefor 1 − r ≤ |θ| ≤ π, and, for |θ| < 1 − r,$$\backslash psi\_r(e^)=1+\; .$$

These estimates show that the L¹ norms ∫ |ψ_r| are uniformly bounded. Since H is a bounded operator, it follows that the operators H_ε are uniformly bounded in operator norm on L²(T). The same argument can be used on L^p(T) once it is known that the Hilbert transform H is bounded in operator norm on L^p(T).

Hilbert transform on the real line

See also: Hilbert transform. As in the case of the circle, the theory for L² functions is particularly easy to develop. In fact, as observed by Rosenblum and Devinatz, the two Hilbert transforms can be related using the Cayley transform.^[2]

The Hilbert transform H_R on L²(R) is defined by$$\backslash widehat\; =\; \backslash left\; (i\backslash chi\_\; \backslash right)\; \backslash widehat,$$where the Fourier transform is given by$$\backslash widehat(t)=\backslash int\_^\backslash infty\; f(x)\; e^\; \backslash ,\; dx.$$

Define the Hardy space H²(R) to be the closed subspace of L²(R) consisting of functions for which the Fourier transform vanishes on the negative part of the real axis. Its orthogonal complement is given by functions for which the Fourier transform vanishes on the positive part of the real axis. It is the complex conjugate of H²(R). If P_R is the orthogonal projection onto H²(R), then

$$H\_=i(2P\_-I).$$

The Cayley transform$$C(x)=$$carries the extended real line onto the circle, sending the point at ∞ to 1, and the upper halfplane onto the unit disk.

Define the unitary operator from L²(T) onto L²(R) by$$Uf(x)=\backslash pi^\; (x+i)^\; f(C(x)).$$

This operator carries the Hardy space of the circle H²(T) onto H²(R). In fact for |w| < 1, the linear span of the functions$$f\_w(z)=\; \backslash frac$$is dense in H²(T). Moreover,$$Uf\_w(x)\; =\; \backslash frac\; \backslash frac$$where$$z=C^(\backslash overline).$$

On the other hand, for z ∈ H, the linear span of the functions$$g\_z(t)=e^\backslash chi\_$$

Notes and References

1. See:
2. See: