Integral operators. Jordan Bell Department of Mathematics, University of Toronto. April 22, X F x(y)dµ(y) x N c 1 0 x N 1

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1 Integral operators Jordan Bell Department of Mathematics, University of Toronto April 22, Product measures Let (, A, µ be a σ-finite measure space. Then with A A the product σ- algebra and µ µ the product measure on A A, (, A A, µ µ is itself a σ-finite measure space. Write F x (y F (x, y and F y (x F (x, y. For any measurable space (, A, it is a fact that if F : is measurable then F x is measurable for each x and F y is measurable for each y. 1 Suppose that F L 1 (, F : C. Fubini s theorem tells us the following. 2 3 There are sets N 1, N 2 A with µ(n 1 0 and µ(n 2 0 such that if x N1 c then F x L 1 ( and if y N2 c then F y L 1 (. Define { I 1 (x F x(ydµ(y x N c 1 0 x N 1 and I 2 (y { F y (xdµ(x y N c 2 0 y N 2. I 1 L 1 ( and I 2 L 1 (, and F d(µ µ I 2 (ydµ(y I 1 (xdµ(x. 1 Heinz Bauer, Measure and Integration Theory, p. 138, Lemma Heinz Bauer, Measure and Integration Theory, p. 139, Corollary Suppose that F : [0, ] is measurable. Tonelli s theorem, Heinz Bauer, Measure and Integration Theory, p. 138, Theorem 23.6, tells us that the functions x F xdµ, y F y dµ are measurable [0, ], and that ( F d(µ µ F y dµ ( dµ(y F xdµ dµ(x. 1

2 2 Integral operators in L 2 Let k L 2 ( and let g L 2 (. By Fubini s theorem, there is a set Z A with µ(z 0 such that if x Z c then k x L 2 (. For x Z c n, by the Cauchy-Schwarz inequality, ( k x g dµ 1/2 ( 1/2 k x 2 dµ g dµ 2 k x L 2 g L 2, so k x g L 1 (. Since µ is σ-finite, there are A n A, µ(a n <, with A n. For each n, the function (x, y 1 An (xg(y belongs to L 2 ( and hence, by the Cauchy-Schwarz inequality, (x, y k(x, y1 An (xg(y belongs to L 1 (. Applying Fubini s theorem, there is a set N n A with µ(n n 0 such that if x Nn c then y k(x, y1 An (xg(y belongs to L 1 (, and the function I n : C defined by { I n (x k x(y1 An (xg(ydµ(y x Nn c 0 x N n belongs to L 1 (. Let M n (Z N n, for which µ(m n µ(z N n n (µ(z + µ(n n 0. We note M c n (Z c N c n. For g L 2 (, define K M g : C by { K M g(x k x(yg(ydµ(y x M c 0 x M. (1 For x M c, I n (x k x (y1 An (xg(ydµ(y 1 An (x Then 1 An K M g 1 M c 1 An Kg 1 M c I n, k x (yg(ydµ(y 1 An (x K M g(x. which shows that f n 1 An K M g is measurable C. For any x, for sufficiently large n we have f n (x K M g(x, thus f n K M g pointwise, which implies that K M g : C is measurable. 4 4 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhikers Guide, third ed., p. 142, Lemma

3 Using the Cauchy-Schwarz inequality and then Fubini s theorem, 2 K M g(x 2 dµ(x k x (yg(ydµ(y dµ(x M c ( g 2 L k 2 x (y 2 dµ(y dµ(x M c g 2 L 2 k 2 L 2. This shows that K M g L 2 (, with K M g L 2 k L 2 g L 2. Recapitulating, for g L 2 ( there is some M A with µ(m 0 such that for x M c, k x L 2 (, and such that K M g : C defined by (1 belongs to L 2 (. If N is any set satisfying these conditions, then for x M c N c, K M g(x k x (yg(ydµ(y K N g(x, and µ((m c N c c µ(m N 0. Therefore, for g L 2 ( it makes sense to define Kg L 2 ( by Kg K M g. If f, g L 2 ( and f g in L 2 (, check that Kf Kg in L 2 (. We thus define K : L 2 ( L 2 ( for g L 2 ( as Kg(x k x (yg(ydµ(y g, k x, where f, g f gdµ. Theorem 1. Let (, A, µ be a σ-finite measure space. For k L 2 (, it makes sense to define Kg L 2 ( by Kg(x k x (yg(ydµ(y g, k x. K : L 2 ( L 2 ( is a bounded linear operator with K k L 2. 3 Integrals of functions Suppose that f : C is a function, which we do not ask to be measurable, and that Z 1, Z 2 A, µ(z 1 0, µ(z 2 0, satisfy 1 f, 2 f L 1 (. 3

4 We have 1 fdµ 1 (1 Z2 + 2 fdµ 1 Z 2 fdµ + 1 Z2 c fdµ 1 Z c 2 fdµ 2 Z c 1 fdµ 2 fdµ. Therefore if there is some Z A with µ(z 0 and 1 Z f L 1 (, it makes sense to define fdµ 1 Z fdµ. However, only if f is itself measurable do we write f L 1 (. 4 Self-adjoint operators Theorem 2. Let (, A, µ be a σ-finite measure space. For k L 2 ( satisfying k x k x, K : L 2 ( L 2 ( is self-adjoint. Proof. For f, g L 2 (, Kf, g Kf(x g(xdµ(x ( k x (yf(ydµ(y g(xdµ(x ( k y (x g(xdµ(x f(ydµ(y ( k y (xg(xdµ(x f(ydµ(y f, Kg. Kg(y f(ydµ(y It follows that K : L 2 ( L 2 ( is self-adjoint. 5 Hilbert-Schmidt operators Let (, A, µ be a measure space and let 1 p <. It is a fact that if µ is σ-finite and A is countably generated, then the Banach space L p ( is 4

5 separable. 5 Theorem 3. Let (, A, µ be a σ-finite countably generated measure space. For k L 2 (, K : L 2 ( L 2 ( is a Hilbert-Schmidt operator with K HS k L 2. Proof. L 2 ( is separable, so there is an orthonormal basis {e n } for L 2 (. Using Parseval s formula and then Fubini s theorem, Ke n, Ke n Ke n (x 2 dµ(x n n e n, k x 2 dµ(x n ( e n, k x 2 dµ(x This shows that n kx, k x dµ(x ( k x 2 dµ(y dµ(x k 2 L 2. k 2 d(µ µ ( 1/2 K HS Ke n, Ke n k L 2. n If T is a compact linear operator on L 2 (, then T T is a positive compact operator on L 2 (. Then T T T is a positive compact operator. 6 Let s j be the nonzero eigenvalues of T repeated according to geometric multiplicity, with s j+1 s j, j 1, called the singular values of T. By the spectral theorem, there is an orthonormal basis for {e j : j 1} for L 2 ( such that 5 Donald L. Cohn, Measure Theory, second ed., p. 102, Proposition See Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 109, Theorem

6 T e j s j e j for each j 1. Then T 2 HS j 1 T e j, T e j j 1 T T e j, e j j 1 T 2 e j, e j j 1 T e j, T e j j 1 s j, s j s j 2. j 1 Summarizing, k 2 L 2 K 2 HS s j (T 2. j 1 6 Trace class operators A compact operator T on L 2 ( is called trace class if T tr <, where T tr j 1 s j (T. For a trace class operator it makes sense to define tr (T n T e n, e n, which does not depend on the orthonormal basis {e n } of L 2 (. Let be a locally compact Hausdorff space and let B be the Borel σ-algebra of. A Borel measure on is a measure on B. We say that a Borel measure µ on is locally finite if for each x there is an open set U x with x U x and µ(u x <. A Radon measure on is a locally finite Borel measure µ on such that for each A B and for any ɛ > 0 there is an open set U ɛ with A U ɛ and µ(a > µ(u ɛ ɛ and for each open set U and for any ɛ > 0 there is a compact set K ɛ with K ɛ U and µ(u < µ(k ɛ + ɛ. 6

7 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 <. 7 Lemma 4. Let be a locally compact Hausdorff space and let µ be a Radon measure on. If A B with µ(a <, there for any ɛ > 0 there is a compact set K ɛ, K ɛ A, such that µ(a < µ(k ɛ + ɛ. Proof. If L is a compact set, B B, and B L, let T L \ B. For δ > 0 there is an open set W δ, T W δ, such that µ(w δ < µ(t + δ. Let K δ L \ W δ, and because is Hausdorff, L is closed and hence K δ is closed and therefore compact. Now, as B L, and L \ W δ L \ T L \ (L \ B B µ(b \ K δ µ(b \ (L \ W δ µ(w δ \ (L \ B µ(w δ \ T < 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 K δ with K δ B and µ(b \ K δ < δ. Now let U be an open set with A U and µ(u <, say µ(u < µ(a + 1. Let L be a compact set with L U and A (A L (A \ L, so and µ(u < µ(l + ɛ. µ(a µ(a L + µ(a \ L, µ(a \ L µ(u \ L < ɛ. Let B A L. Because B is a Borel set contained in a compact set L, there is a compact set K contained in B such that As A B (A \ L and K B, µ(b \ K < ɛ. µ(a \ K µ((b \ K (A \ L µ(b \ K + µ(a \ L < 2ɛ. 7 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 291, Lemma B

8 Let be a locally compact Hausdorff space and let µ be a Radon measure on. An admissible kernel is a function k C( L 2 ( for which there is some g C( L 2 ( such that k(x, y g(xg(y for all (x, y. We call S : L 2 ( L 2 ( an admissible integral operator if there is an admissible kernel k such that Sg(x k x (yg(ydµ(y. The following gives conditions under which we can calculate the trace of an integral operator. 8 Theorem 5. Let be a first-countable locally compact Hausdorff space and let µ be a Radon measure on. Let k C( L 2 ( and let Kg(x k x (yg(ydµ(y. If there are admissible integral operators S 1 and S 2 such that K S 1 S 2, then K is of trace class and tr (K k(x, xdµ(x. The following is Mercer s theorem. 9 Theorem 6 (Mercer s theorem. If k C( L 2 ( and K : L 2 ( L 2 ( is a positive operator, then tr (K k(x, xdµ(x. 8 Anton Deitmar and Siegfried Echterhoff, Principles of Harmonic Analysis, second ed., p. 172, Proposition E. Brian Davies, Linear Operators and their Spectra, p. 156, Proposition