The Integral Operator
Introduction
An integral operator is the map that averages a function against a weight in two variables. For a measurable kernel $K$ on the product of two measure spaces it is $$ (T_Kf)(x)=\int_YK(x,y)\,f(y)\,d\nu(y), $$ and the equation $T_Kf=g$ is the integral equation of the second kind when the unknown appears also outside the integral, $f-T_Kf=g$. The operator is the classical inverse of a differential operator: the solution of a linear differential equation is written as an integral against a fundamental solution, and the resolvent of a differential operator is, where it exists as an integral operator, an operator of this form. This article develops the operator itself, with no hypothesis beyond the integral and the $L^2$ inner product: the algebra its kernels carry, its boundedness on $L^2$ under the three hypotheses that supply it, the Hilbert–Schmidt case in which the kernel is square-integrable, and the compactness that the square-integrable case yields.
The prerequisites are Measure Theory and Integration for the Lebesgue integral, the $L^p$ spaces, the
Cauchy–Schwarz and Hölder inequalities and the Fubini–Tonelli theorem, and Modes of Convergence for
convergence in $L^p$. The $L^2$ inner product is the one of that article, $\langle f,g\rangle=\int f\bar g$,
linear in the first argument and conjugate-linear in the second. Nothing here uses the general theory of
operators on a Hilbert space beyond the inner product and Cauchy–Schwarz: the boundedness, the
Hilbert–Schmidt norm and the compactness are all proved concretely from the integral. The operator norm,
the adjoint of a bounded operator, the spectral theorem, the Fredholm alternative and the general theory
of compact operators are the subject of Banach and Hilbert Spaces and of Compact Operators, later in
this Part; the theorem that every continuous operator on test functions is an integral operator with a
distribution kernel is The Schwartz Kernel Theorem, later in this Part, and no such representation is
claimed here for a kernel that fails the integrability hypotheses below. The kernel of the adjoint and the
Hermitian and positive definite kernels are the subject of the * group of this category; the
translation-invariant kernel, which defines a convolution operator, is Convolution Operators, the next
article of this group, and is only an example here.
The Operator and Its Kernel
Definition
Definition. Let $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ be $\sigma$-finite measure spaces, $\mathbb K=\mathbb R$ or $\mathbb C$, and let $K:X\times Y\to\mathbb K$ be measurable for $\mathcal A\otimes\mathcal B$. The integral operator with kernel $K$ is $$ (T_Kf)(x)=\int_YK(x,y)f(y)\,d\nu(y), $$ defined for those $f\in L^2(Y,\nu)$ for which the integral converges for $\mu$-almost every $x$, and $T_Kf$ is then defined as a class in $L^2(X,\mu)$ wherever it lies in $L^2$. The function $K$ is the kernel, and when $X=Y$ and $\mu=\nu$ the kernel is a kernel on $X$. When the two variables of the kernel must be told apart, the first is the variable of the image and the second the variable of the source, so that $K(x,y)$ multiplies the value of $f$ at $y$ and the result is read at $x$.
Two kernels agreeing $\mu\otimes\nu$-almost everywhere define the same operator on the functions for which both integrals converge, and this is the only ambiguity; the article works with kernels as functions and records the equality almost everywhere where it matters. The definitions are arranged so that the Hilbert–Schmidt hypothesis $K\in L^2(X\times Y,\mu\otimes\nu)$ below makes $T_K$ a bounded operator defined on all of $L^2(Y,\nu)$.
The Algebra of Kernels
Theorem (composition of integral operators). Let $K\in L^2(X\times Z,\mu\otimes\lambda)$ and $L\in L^2(Z\times Y,\lambda\otimes\nu)$. Then the function $$ (K\star L)(x,y)=\int_ZK(x,z)L(z,y)\,d\lambda(z) $$ is defined for almost every $(x,y)$, lies in $L^2(X\times Y,\mu\otimes\nu)$, and satisfies $T_KT_L=T_{K\star L}$.
Proof. For almost every $(x,y)$ the integrand is the product of the two functions $z\mapsto K(x,z)$ and $z\mapsto L(z,y)$, which lie in $L^2(Z,\lambda)$ for almost every $x$ and almost every $y$ respectively; Cauchy–Schwarz gives $$ \int_Z\lvert K(x,z)L(z,y)\rvert\,d\lambda(z) \le\Bigl(\int_Z\lvert K(x,z)\rvert^2d\lambda(z)\Bigr)^{1/2} \Bigl(\int_Z\lvert L(z,y)\rvert^2d\lambda(z)\Bigr)^{1/2}, $$ so $(K\star L)(x,y)$ is defined and finite almost everywhere and measurable. Integrating its square and applying Cauchy–Schwarz once more, then Tonelli, $$ \iint_{X\times Y}\lvert(K\star L)(x,y)\rvert^2dx\,dy \le\iint_{X\times Y}\Bigl(\int_Z\lvert K(x,z)\rvert^2dz\Bigr) \Bigl(\int_Z\lvert L(z,y)\rvert^2dz\Bigr)dx\,dy =\lVert K\rVert_2^2\,\lVert L\rVert_2^2 , $$ which is finite; hence $K\star L\in L^2(X\times Y)$. For the equality of operators, let $f\in L^2(Y,\nu)$ and compute formally and then justify by Fubini–Tonelli applied to the absolutely convergent triple integral, which the estimate above bounds: $$ (T_KT_Lf)(x)=\int_ZK(x,z)\Bigl(\int_YL(z,y)f(y)\,d\nu(y)\Bigr)d\lambda(z) =\int_Y\Bigl(\int_ZK(x,z)L(z,y)\,d\lambda(z)\Bigr)f(y)\,d\nu(y)=(T_{K\star L}f)(x). \qquad\blacksquare $$
Proposition (the kernel product is associative). For kernels $K\in L^2(X\times W)$, $L\in L^2(W\times Z)$ and $M\in L^2(Z\times Y)$ one has $(K\star L)\star M=K\star(L\star M)$ almost everywhere. The product is bilinear, and the square-integrable kernels on $X$ form an associative algebra under $\star$ with the kernel $K_0(x,y)$ of the identity of the algebra acting as unit wherever such a kernel exists.
Proof. Both sides are the triple integral $\iiint K(x,w)L(w,z)M(z,y)\,dw\,dz$ over the product, which is finite almost everywhere by two applications of Cauchy–Schwarz and is symmetric in the bracketing by Fubini–Tonelli. Bilinearity is linearity of the integral. $\blacksquare$
Examples
Example (rank-one kernels). For $u\in L^2(X,\mu)$ and $v\in L^2(Y,\nu)$ put $K(x,y)=u(x)\overline{v(y)}$. Then $$ (T_Kf)(x)=u(x)\int_Y\overline{v(y)}f(y)\,d\nu(y)=u(x)\,\langle f,v\rangle , $$ so $T_K$ has rank at most one, its image being the span of $u$; it has rank exactly one when $u$ and $v$ are both nonzero.
Example (the finite-dimensional case). For $X=\{1,\dots,m\}$ and $Y=\{1,\dots,n\}$ with counting measure, every measurable kernel is a matrix $K=(K_{ij})$ and $T_K$ is multiplication by that matrix; the composition $K\star L$ is the matrix product, and the Hilbert–Schmidt norm below is the Euclidean norm of the matrix entries. The general theory is thus the infinite-dimensional extension of the algebra of matrices acting on column vectors.
Example (the Volterra kernel). On $X=Y=\mathbb R$ with Lebesgue measure, $K(x,y)=\mathbf 1_{\{x>y\}}$ gives $(T_Kf)(x)=\int_{-\infty}^xf(y)\,dy$, the indefinite integral; the operator is not translation invariant and is a standard example of a quasinilpotent operator, the spectral theory beyond the present Article.
Example (the translation-invariant kernel). On $X=Y=\mathbb R^n$ with Lebesgue measure, a kernel of the form $K(x,y)=k(x-y)$ with $k\in L^1(\mathbb R^n)$ gives $(T_Kf)(x)=\int k(x-y)f(y)\,dy=(k*f)(x)$, a convolution operator. These are the kernels whose operators commute with all translations, and they are the subject of Convolution Operators, the next article of this group.
Boundedness on $L^2$
The Square-Integrable Kernel
Theorem (the Hilbert–Schmidt bound). If $K\in L^2(X\times Y,\mu\otimes\nu)$, then $T_K$ maps $L^2(Y,\nu)$ boundedly into $L^2(X,\mu)$, with $$ \lVert T_K\rVert\leq\lVert K\rVert_{L^2(X\times Y)}, \qquad \lVert K\rVert_{L^2(X\times Y)}^2=\iint_{X\times Y}\lvert K(x,y)\rvert^2d\mu(x)\,d\nu(y). $$
Proof. For $f\in L^2(Y,\nu)$ and almost every $x$ the function $y\mapsto K(x,y)$ lies in $L^2(Y,\nu)$ by Tonelli, so Cauchy–Schwarz gives $\lvert(T_Kf)(x)\rvert^2\le\lVert K(x,\cdot)\rVert_2^2\lVert f\rVert_2^2$ and hence, after integrating in $x$ and applying Tonelli, $$ \lVert T_Kf\rVert_2^2 \le\int_X\lVert K(x,\cdot)\rVert_2^2\,d\mu(x)\,\lVert f\rVert_2^2 =\lVert K\rVert_{L^2(X\times Y)}^2\lVert f\rVert_2^2 . $$ The operator is linear and defined on all of $L^2(Y,\nu)$ by this estimate, so it is bounded with the stated bound. $\blacksquare$
The inequality is the reason the square-integrable kernels are the class the corpus works with: the operator norm is controlled by a double integral, and every kernel that satisfies $\lVert K\rVert_2<\infty$ defines a bounded operator with no further hypothesis.
The Schur Test
Theorem (Schur test). Let $X=Y$ and let $K\ge0$ be a measurable kernel for which there is $C\ge0$ with $$ \int_XK(x,y)\,d\mu(y)\leq C\ \text{for almost every }x, \qquad \int_XK(x,y)\,d\mu(x)\leq C\ \text{for almost every }y . $$ Then $T_K$ maps $L^p(\mu)$ boundedly into itself for every $1\leq p\leq\infty$, with $\lVert T_K\rVert_{L^p\to L^p}\leq C$.
Proof. The case $p=1$ is Fubini and the case $p=\infty$ is the first hypothesis. For $1 Remark. The test also holds with two weights: if $p_0,q_0>0$ are measurable and
$\int_XK(x,y)q_0(y)\,dy\leq Cp_0(x)$, $\int_XK(x,y)p_0(x)\,dx\leq Cq_0(y)$ almost everywhere, then $T_K$ is
bounded on the weighted space $L^p(p_0^p\,d\mu)$ with the same constant. The unweighted statement is the
one used in this corpus, and the weighted one is recorded because the kernel of a Riemann–Liouville
fractional integral satisfies it. Theorem (Young). Let $K(x,y)=k(x-y)$ on $\mathbb R^n$ with $k\in L^1(\mathbb R^n)$. Then for
$1\leq p\leq\infty$,
$$
\lVert T_Kf\rVert_p=\lVert k*f\rVert_p\leq\lVert k\rVert_1\lVert f\rVert_p .
$$ Proof. For $p=1$ the inequality is Fubini, $p=\infty$ the trivial bound, and for $1 The Young bound is the reason a kernel that decays like an integrable function gives a bounded operator
even when it is not square-integrable: $k$ may be the Poisson kernel, which lies in $L^1$ and not in $L^2$.
The two hypotheses of this section overlap and neither contains the other; the square-integrable case is
the one that gives compactness below, and the $L^1$ case need not. Definition. An integral operator $T_K$ is Hilbert–Schmidt if
$K\in L^2(X\times Y,\mu\otimes\nu)$, and its Hilbert–Schmidt norm is
$$
\lVert T_K\rVert_{\mathrm{HS}}=\lVert K\rVert_{L^2(X\times Y)}
=\Bigl(\iint_{X\times Y}\lvert K(x,y)\rvert^2\,d\mu(x)\,d\nu(y)\Bigr)^{1/2}.
$$
The operator is of finite rank if its image is finite-dimensional, and of rank one if the image is
one-dimensional. By the Hilbert–Schmidt bound, $\lVert T_K\rVert_{\mathrm{op}}\leq\lVert T_K\rVert_{\mathrm{HS}}$, so the
Hilbert–Schmidt norm dominates the operator norm; the two are not equal in general, and the inequality is
strict for a rank-one kernel whose factor has norm one, where $\lVert T_K\rVert_{\mathrm{op}}=1$ while
$\lVert T_K\rVert_{\mathrm{HS}}$ is the product of the two factor norms and may be larger. Theorem (the Hilbert–Schmidt norm is basis independent). Let $(X,\mu)$ and $(Y,\nu)$ be separable
measure spaces, with orthonormal bases $\{e_i\}$ of $L^2(X,\mu)$ and $\{f_j\}$ of $L^2(Y,\nu)$. Then
$$
\lVert T_K\rVert_{\mathrm{HS}}^2=\sum_{i,j}\lvert\langle T_Kf_j,e_i\rangle\rvert^2 ,
$$
the series converging and the value not depending on the two bases. Proof. The functions $e_i(x)\overline{f_j(y)}$ form an orthonormal basis of
$L^2(X\times Y,\mu\otimes\nu)$, and the identity
$$
\langle K,e_i\otimes\overline{f_j}\rangle
=\iint K(x,y)\overline{e_i(x)}\,f_j(y)\,d\mu(x)\,d\nu(y)=\langle T_Kf_j,e_i\rangle
$$
holds by Fubini–Tonelli, the integrand being summable. Parseval's identity in $L^2(X\times Y)$ gives the
statement, and the basis-independence is the basis-independence of the Parseval sum. $\blacksquare$ Theorem. Let $K\in L^2(X\times Z)$ and $L\in L^2(Z\times Y)$. Then $K\star L\in L^2(X\times Y)$ and
$$
\lVert K\star L\rVert_{\mathrm{HS}}\leq\lVert K\rVert_{\mathrm{HS}}\,\lVert L\rVert_{\mathrm{HS}} .
$$
Consequently the Hilbert–Schmidt operators are closed under composition and under multiplication on either
side by an integral operator, and the composition of two Hilbert–Schmidt operators is Hilbert–Schmidt. Proof. The first two statements are the composition theorem of the first section together with the
estimate $\lVert K\star L\rVert_2\leq\lVert K\rVert_2\lVert L\rVert_2$ proved there. $\blacksquare$ Remark. The Hilbert–Schmidt operators form a two-sided ideal of the algebra of bounded operators on
$L^2$, and this is where the general theory begins; the concrete statement above is all that the corpus
uses before Banach and Hilbert Spaces, later in this Part, where the ideal is placed in the algebra
$B(L^2)$ and the trace-class operators are introduced. Definition. A bounded operator $T:L^2(X,\mu)\to L^2(X,\mu)$ is compact if the closure of the image
of the closed unit ball is compact, equivalently if every bounded sequence $(f_n)$ has a subsequence whose
images $Tf_n$ converge; if the image has finite dimension the operator is of finite rank and compact. The
definition and its consequences — the spectral theory of a compact operator, the Fredholm alternative —
are the subject of Compact Operators and Banach and Hilbert Spaces, later in this Part; the property
is used here only in the concrete case proved below. Theorem (square-integrable kernels are compact). Let $(X,\mu)$ and $(Y,\nu)$ be separable measure
spaces with orthonormal bases $\{e_i\}$ and $\{f_j\}$ and let $K\in L^2(X\times Y)$. For $N\geq1$ put
$$
K_N(x,y)=\sum_{i,j\leq N}\langle T_Kf_j,e_i\rangle\,e_i(x)\overline{f_j(y)} .
$$
Then each $T_{K_N}$ has finite rank at most $N^2$, the kernels converge, $\lVert K-K_N\rVert_2\to0$, and
consequently $\lVert T_K-T_{K_N}\rVert_{\mathrm{op}}\to0$. Hence $T_K$ is compact. Proof. Each $K_N$ is a finite sum of rank-one kernels $e_i\otimes\overline{f_j}$, so $T_{K_N}$ has
finite rank; the convergence $\lVert K-K_N\rVert_2\to0$ is Parseval's identity in the orthonormal basis
$\{e_i\otimes\overline{f_j}\}$, and the operator-norm convergence follows from the Hilbert–Schmidt bound
applied to $K-K_N$. It remains to see that a norm limit of finite-rank operators is compact. Let $S_n$ be
finite rank with $\lVert S-S_n\rVert\to0$. Given a bounded sequence $f_m$ with $\lVert f_m\rVert\leq1$,
extract by the compactness of $S_1$ a subsequence on which $S_1f_m$ converges, then a further subsequence
on which $S_2f_m$ converges, and so on diagonally, obtaining a subsequence on which $S_nf_m$ converges for
every $n$. On that subsequence
$$
\lVert Sf_m-Sf_{m'}\rVert\leq\lVert S-S_n\rVert\lVert f_m-f_{m'}\rVert+\lVert S_nf_m-S_nf_{m'}\rVert
\leq2\lVert S-S_n\rVert+\lVert S_nf_m-S_nf_{m'}\rVert ,
$$
and choosing $n$ first and then $m,m'$ large makes both terms small. Hence $Sf_m$ is Cauchy, and the
diagonal subsequence converges; $S$ is compact. $\blacksquare$ Corollary. Every Hilbert–Schmidt integral operator is compact, and so is every finite-rank operator.
The Hilbert–Schmidt norm is not the operator norm: the family $K_N$ converges to $K$ in the
Hilbert–Schmidt norm and therefore in the operator norm, and the limit is compact although a general
bounded operator need not be. Remark (what compactness does not follow from). The Young bound of the previous section gives a
bounded operator for every $k\in L^1(\mathbb R^n)$, and such an operator is never compact unless it is
zero. A translation-invariant operator on $L^2(\mathbb R^n)$ is unitarily equivalent to multiplication by
its multiplier, and a multiplication operator by a function $m$ on $L^2$ of a non-atomic space of infinite
measure is compact exactly when $m=0$ almost everywhere: if $\lvert m\rvert\geq c>0$ on a set of positive
finite measure then the restriction to the $L^2$ of that set is bounded below by $c$, so its unit ball
would be relatively compact only if that space had finite dimension. Hence a nonzero convolution operator
on $L^2(\mathbb R^n)$ is not compact, and compactness is a property of the kernels that are not
translation invariant. The criterion is developed with the convolution operators themselves in
Convolution Operators, and in general in Compact Operators, later in this Part. The interest of compactness is that it makes the integral equation solvable. If $K\in L^2(X\times X)$,
then $T_K$ is compact, and the equation $f-T_Kf=g$, with $g\in L^2$, falls under the Fredholm
alternative: either $f-T_Kf=g$ has a unique solution for every $g$, or the homogeneous equation
$f-T_Kf=0$ has a nonzero finite-dimensional solution space, and in the second case the equation is
solvable exactly for those $g$ orthogonal to the solutions of the homogeneous adjoint equation. The
adjoint kernel that enters the second alternative is the subject of The Adjoint of an Integral Operator,
later in this category; the alternative itself is stated and proved in Compact Operators, later in this
Part, and the present Article supplies only the compactness on which it rests. An integral operator $T_K$ with kernel $K$ on the product of two $\sigma$-finite measure spaces is the map
$f\mapsto\int K(x,\cdot)f$, linear and defined on all of $L^2$ once $K$ is square-integrable. The kernels
compose by $(K\star L)(x,y)=\int K(x,z)L(z,y)\,dz$, the product is associative and bilinear, and
$T_KT_L=T_{K\star L}$; the finite-dimensional instance is matrix multiplication, and a kernel of the form
$k(x-y)$ gives a convolution operator. The operator is bounded on $L^2$ with $\lVert T_K\rVert\leq\lVert K\rVert_{L^2(X\times Y)}$ whenever the
kernel is square-integrable, by Cauchy–Schwarz and Tonelli; it is bounded on every $L^p$ with
$\lVert T_K\rVert\leq C$ whenever a nonnegative kernel has row and column integrals bounded by $C$, by the
Schur test; and it is bounded on every $L^p$ with $\lVert T_K\rVert\leq\lVert k\rVert_1$ when the kernel is
translation invariant with an integrable profile, by Young's inequality. The Hilbert–Schmidt operators are
the square-integrable kernels, with norm $\lVert T_K\rVert_{\mathrm{HS}}=\lVert K\rVert_2$ dominating the
operator norm, basis-independent and multiplicative under composition; they are the ideal of the bounded
operators that this article reaches without the general theory. A square-integrable kernel is the
$L^2$-limit of its finite-rank truncations and the operator the operator-norm limit of finite-rank
operators, hence compact; and compactness is exactly what makes the integral equation $f-T_Kf=g$ amenable
to the Fredholm alternative.Kernels in $L^1$ and the Young Bound
The Hilbert–Schmidt Case
The Hilbert–Schmidt Norm
The Ideal Property
Compactness
Compactness
Finite-Rank Approximation
The Integral Equation
Summary
Summary of Notation
Symbol
Meaning
$(X,\mathcal A,\mu)$, $(Y,\mathcal B,\nu)$
$\sigma$-finite measure spaces, source and image
$K(x,y)$
The kernel; first variable of the image, second of the source
$T_K$
The integral operator, $(T_Kf)(x)=\int_YK(x,y)f(y)\,d\nu(y)$
$K\star L$
Kernel product $(K\star L)(x,y)=\int_ZK(x,z)L(z,y)\,dz$, with $T_{K\star L}=T_KT_L$
$\lVert K\rVert_2$
$L^2(X\times Y)$ norm of the kernel
$\lVert T_K\rVert$
Operator norm on $L^2$
$\lVert T_K\rVert_{\mathrm{HS}}=\lVert K\rVert_2$
Hilbert–Schmidt norm
$C$
Schur constant, bounding the row and column integrals
$k$
Profile of a translation-invariant kernel $K(x,y)=k(x-y)$
compact
Image of the closed unit ball relatively compact
Further Reading