Integral operators

Jordan Bell
April 26, 2016

1 Product measures

Let (X,𝒜,μ) be a σ-finite measure space. Then with 𝒜𝒜 the product σ-algebra and μμ the product measure on 𝒜𝒜, (X×X,𝒜𝒜,μμ) is itself a σ-finite measure space.

Write Fx(y)=F(x,y) and Fy(x)=F(x,y). For any measurable space (X,𝒜), it is a fact that if F:X×XX is measurable then Fx is measurable for each xX and Fy is measurable for each yX.11 1 Heinz Bauer, Measure and Integration Theory, p. 138, Lemma 23.5.

Suppose that F1(X×X), F:X. Fubini’s theorem tells us the following.22 2 Heinz Bauer, Measure and Integration Theory, p. 139, Corollary 23.7. 33 3 Suppose that F:X×X[0,] is measurable. Tonelli’s theorem, Heinz Bauer, Measure and Integration Theory, p. 138, Theorem 23.6, tells us that the functions xXFx𝑑μ,yXFy𝑑μ are measurable X[0,], and that X×XFd(μμ)=X(XFy𝑑μ)𝑑μ(y)=X(XFx𝑑μ)𝑑μ(x). There are sets N1,N2𝒜 with μ(N1)=0 and μ(N2)=0 such that if xN1c then Fx1(X) and if yN2c then Fy1(X). Define

I1(x)={XFx(y)𝑑μ(y)xN1c0xN1

and

I2(y)={XFy(x)𝑑μ(x)yN2c0yN2.

I11(X) and I21(X), and

X×XFd(μμ)=XI2(y)𝑑μ(y)=XI1(x)𝑑μ(x).

2 Integral operators in L2

Let k2(X×X) and let g2(X). By Fubini’s theorem, there is a set Z𝒜 with μ(Z)=0 such that if xZc then kx2(X). For xZnc, by the Cauchy-Schwarz inequality,

X|kxg|𝑑μ(X|kx|2𝑑μ)1/2(X|g|2𝑑μ)1/2=kxL2gL2,

so kxg1(X).

Since μ is σ-finite, there are An𝒜, μ(An)<, with AnX. For each n, the function (x,y)1An(x)g(y) belongs to 2(X×X) and hence, by the Cauchy-Schwarz inequality, (x,y)k(x,y)1An(x)g(y) belongs to 1(X×X). Applying Fubini’s theorem, there is a set Nn𝒜 with μ(Nn)=0 such that if xNnc then yk(x,y)1An(x)g(y) belongs to 1(X), and the function In:X defined by

In(x)={Xkx(y)1An(x)g(y)𝑑μ(y)xNnc0xNn

belongs to 1(X).

Let M=n(ZNn), for which

μ(M)nμ(ZNn)n(μ(Z)+μ(Nn))=0.

We note

Mc=n(ZcNnc).

For g2(X), define KMg:X by

KMg(x)={Xkx(y)g(y)𝑑μ(y)xMc0xM. (1)

For xMc,

In(x)=Xkx(y)1An(x)g(y)𝑑μ(y)=1An(x)Xkx(y)g(y)𝑑μ(y)=1An(x)KMg(x).

Then

1AnKMg=1Mc1AnKg=1McIn,

which shows that fn=1AnKMg is measurable X. For any xX, for sufficiently large n we have fn(x)=KMg(x), thus fnKMg pointwise, which implies that KMg:X is measurable.44 4 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 142, Lemma 4.29.

Using the Cauchy-Schwarz inequality and then Fubini’s theorem,

X|KMg(x)|2𝑑μ(x) =Mc|Xkx(y)g(y)𝑑μ(y)|2𝑑μ(x)
gL22Mc(X|kx(y)|2𝑑μ(y))𝑑μ(x)
=gL22kL22.

This shows that KMg2(X), with

KMgL2kL2gL2.

Recapitulating, for g2(X) there is some M𝒜 with μ(M)=0 such that for xMc, kx2(X), and such that KMg:X defined by (1) belongs to 2(X). If N is any set satisfying these conditions, then for xMcNc,

KMg(x)=Xkx(y)g(y)𝑑μ(y)=KNg(x),

and μ((McNc)c)=μ(MN)=0. Therefore, for g2(X) it makes sense to define KgL2(X) by Kg=KMg.

If f,g2(X) and f=g in L2(X), check that Kf=Kg in L2(X). We thus define K:L2(X)L2(X) for gL2(X) as

Kg(x)=Xkx(y)g(y)𝑑μ(y)=g,kx¯,

where

f,g=Xfg¯𝑑μ.
Theorem 1.

Let (X,A,μ) be a σ-finite measure space. For kL2(X×X), it makes sense to define KgL2(X) by

Kg(x)=Xkx(y)g(y)𝑑μ(y)=g,kx¯.

K:L2(X)L2(X) is a bounded linear operator with KkL2.

3 Integrals of functions

Suppose that f:X is a function, which we do not ask to be measurable, and that Z1,Z2𝒜, μ(Z1)=0, μ(Z2)=0, satisfy 1Z1cf,1Z2cf1(X). We have

X1Z1cf𝑑μ =X1Z1c(1Z2+1Z2c)f𝑑μ
=X1Z1cZ2f𝑑μ+X1Z1cZ2cf𝑑μ
=X1Z1cZ2cf𝑑μ
=X1Z2cZ1cf𝑑μ
=X1Z2cf𝑑μ.

Therefore if there is some Z𝒜 with μ(Z)=0 and 1Zf1(X), it makes sense to define

Xf𝑑μ=X1Zf𝑑μ.

However, only if f is itself measurable do we write f1(X).

4 Self-adjoint operators

Theorem 2.

Let (X,A,μ) be a σ-finite measure space. For kL2(X×X) satisfying kx=kx¯, K:L2(X)L2(X) is self-adjoint.

Proof.

For f,gL2(X),

Kf,g =XKf(x)g(x)¯𝑑μ(x)
=X(Xkx(y)f(y)𝑑μ(y))g(x)¯𝑑μ(x)
=X(Xky(x)g(x)¯𝑑μ(x))f(y)𝑑μ(y)
=X(Xky(x)¯g(x)¯𝑑μ(x))f(y)𝑑μ(y)
=XKg(y)¯f(y)𝑑μ(y)
=f,Kg.

It follows that K:L2(X)L2(X) is self-adjoint. ∎

5 Hilbert-Schmidt operators

Let (X,𝒜,μ) be a measure space and let 1p<. It is a fact that if μ is σ-finite and 𝒜 is countably generated, then the Banach space Lp(X) is separable.55 5 Donald L. Cohn, Measure Theory, second ed., p. 102, Proposition 3.4.5.

Theorem 3.

Let (X,A,μ) be a σ-finite countably generated measure space. For kL2(X×X), K:L2(X)L2(X) is a Hilbert-Schmidt operator with

KHS=kL2.
Proof.

L2(X) is separable, so there is an orthonormal basis {en} for L2(X). Using Parseval’s formula and then Fubini’s theorem,

nKen,Ken =nX|Ken(x)|2𝑑μ(x)
=nX|en,kx¯|2𝑑μ(x)
=X(n|en,kx¯|2)𝑑μ(x)
=Xkx¯,kx¯𝑑μ(x)
=X(X|kx|2𝑑μ(y))𝑑μ(x)
=X×X|k|2d(μμ)
=kL22.

This shows that

KHS=(nKen,Ken)1/2=kL2.

If T is a compact linear operator on L2(X), then T*T is a positive compact operator on L2(X). Then |T|=T*T is a positive compact operator.66 6 See Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 109, Theorem 5.1.3 Let sj be the nonzero eigenvalues of |T| repeated according to geometric multiplicity, with sj+1sj, j1, called the singular values of T. By the spectral theorem, there is an orthonormal basis for {ej:j1} for L2(X) such that |T|ej=sjej for each j1. Then

THS2 =j1Tej,Tej
=j1T*Tej,ej
=j1|T|2ej,ej
=j1|T|ej,|T|ej
=j1sj,sj
=j1|sj|2.

Summarizing,

kL22=KHS2=j1|sj(T)|2.

6 Trace class operators

A compact operator T on L2(X) is called trace class if Ttr<, where

Ttr=j1sj(T).

For a trace class operator it makes sense to define

tr(T)=nTen,en,

which does not depend on the orthonormal basis {en} of L2(X).

Let X be a locally compact Hausdorff space and let be the Borel σ-algebra of X. A Borel measure on X is a measure on . We say that a Borel measure μ on X is locally finite if for each xX there is an open set Ux with xUx and μ(Ux)<. A Radon measure on X is a locally finite Borel measure μ on X such that for each A and for any ϵ>0 there is an open set Uϵ with AUϵ and

μ(A)>μ(Uϵ)-ϵ

and for each open set U and for any ϵ>0 there is a compact set Kϵ with KϵU and

μ(U)<μ(Kϵ)+ϵ.

By definition, if μ is a Radon measure then μ(U) can be approximated by μ(K) for compact sets K contained in U. We prove that this holds for μ(A) if μ(A)<.77 7 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 291, Lemma B.2.1.

Lemma 4.

Let X be a locally compact Hausdorff space and let μ be a Radon measure on X. If AB with μ(A)<, there for any ϵ>0 there is a compact set Kϵ, KϵA, such that

μ(A)<μ(Kϵ)+ϵ.
Proof.

If L is a compact set, B, and BL, let T=LB. For δ>0 there is an open set Wδ, TWδ, such that μ(Wδ)<μ(T)+δ. Let Kδ=LWδ, and because X is Hausdorff, L is closed and hence Kδ is closed and therefore compact. Now, as BL,

LWδLT=L(LB)=B

and

μ(BKδ)=μ(B(LWδ))μ(Wδ(LB))=μ(WδT)<δ.

We have proved that if L is a compact set and B is a Borel set contained in L, then for any δ>0 then there is a compact set Kδ with KδB and

μ(BKδ)<δ.

Now let U be an open set with AU and μ(U)<, say μ(U)<μ(A)+1. Let L be a compact set with LU and

μ(U)<μ(L)+ϵ.

A=(AL)(AL), so

μ(A)=μ(AL)+μ(AL),

and

μ(AL)μ(UL)<ϵ.

Let B=AL. Because B is a Borel set contained in a compact set L, there is a compact set K contained in B such that

μ(BK)<ϵ.

As A=B(AL) and KB,

μ(AK)=μ((BK)(AL))=μ(BK)+μ(AL)<2ϵ.

Let X be a locally compact Hausdorff space and let μ be a Radon measure on X. An admissible kernel is a function kC(X×X)2(X×X) for which there is some gC(X)2(X) such that |k(x,y)|g(x)g(y) for all (x,y)X×X. We call S:L2(X)L2(X) an admissible integral operator if there is an admissible kernel k such that

Sg(x)=Xkx(y)g(y)𝑑μ(y).

The following gives conditions under which we can calculate the trace of an integral operator.88 8 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 172, Proposition 9.3.1.

Theorem 5.

Let X be a first-countable locally compact Hausdorff space and let μ be a Radon measure on X. Let kC(X×X)L2(X) and let

Kg(x)=Xkx(y)g(y)𝑑μ(y).

If there are admissible integral operators S1 and S2 such that K=S1S2, then K is of trace class and

tr(K)=Xk(x,x)𝑑μ(x).

The following is Mercer’s theorem.99 9 E. Brian Davies, Linear Operators and their Spectra, p. 156, Proposition 5.6.9.

Theorem 6 (Mercer’s theorem).

If kC(X×X)L2(X×X) and K:L2(X)L2(X) is a positive operator, then

tr(K)=Xk(x,x)𝑑μ(x).