Compact Operators in Hilbert Space

Hart Smith

Department of Mathematics University of Washington, Seattle

Math 526/556, Spring 2015 Hilbert-Schmidt integral kernels on L2(A)

2 2 If K (x, y) ∈ L (A × A), and H = L (A), define TK ∈ B(H) by Z TK g(x) = K (x, y) g(y) dy .

Then hf , TK gi = hTK f , gi iff Z Z  Z Z  f (x) K (x, y) g(y) dy dx = K (y, x)f (x) dx g(y) dy

TK is self-adjoint iff K (x, y) = K (y, x) for a.a. x, y ∈ A × A.

TK is compact since it’s a Hilbert-Schmidt operator. Self-adjoint Hilbert-Schmidt integral kernels on L2(A)

Corollary: assume K (x, y) = K (y, x) 2 ∃ orthonormal set {φj } ⊂ L (A)of eigenvectors for TK : Z K (x, y) φj (y) dy = λj φj (x) , λj 6= 0 ,

such that if {ψk } is an orthonormal basis for ker(TK ), then 2 {φj } ∪ {ψk } is an orthonormal basis for L (A).

It follows that

{φj (x)φk (y)} ∪ {φj (x)ψk (x)} ∪ {ψj (x)φk (y)} ∪ {ψj (x)ψk (y)}

is an orthonormal basis for L2(A × A). Expand K (x, y) in basis:

∞ X K (x, y) = λj φj (x) φj (y) j=1 Square root of T ∗T

Suppose T is compact on H (not necessarily self-adjoint).

T ∗T is compact, self-adjoint on H, and hT ∗Tx, xi = kTxk2 ≥ 0

Lemma There exists a unique compact, non-negative, self-adjoint operator S such that S2 = T ∗T .

∗ ⊥ Proof. Let N0 = ker(T T ) = ker(T ) , E = N0 , so H = E ⊕ N0.

T ∗T : E → E is compact, 1-1, so in some orthonormal basis

∗ T T = diag(µ1, . . . , µ1, µ2, . . . , µ2, µ3,...)

where µ1 > µ2 > µ3 > ··· > 0 and µj → 0 unless finitely many. Square root of T ∗T

Define S : E → E by the matrix √ √ √ √ √  S = diag µ1 ,..., µ1 , µ2 ,..., µ2 , µ3 ,...

Extend S to H = E ⊕ N0 by Sy = 0 if y ∈ N0 = ker(T ). 2 ∗ 2 ∗ S x = T Tx if x ∈ E, and S y = T Ty = 0 if y ∈ N0. 2 ∗ Any z ∈ H of form x + y ∈ E ⊕ N0, so S = T T on H. If S2 = T ∗T is as in the statement, then S commutes with 2 ∗ S = T T , so each Nµj , as well as N0, is invariant for S. p • S self-adjoint, non-negative, ⇒ S = µj on Nµj .

• Similarly, S = 0 on N0, so S unique.

Note: hSx, Sxi = hS2x, xi = hT ∗Tx, xi = hTx, Txi

therefore: kSxk = kTxk for all x ∈ H. Polar Decomposition for compact T on H.

Theorem Suppose T is compact on a Hilbert space H. Let E = (ker T )⊥. One can write T = US, where U : E → H is a norm preserving map, S is self-adjoint non-negative, and kSxk = kTxk ∀ x ∈ H.

Proof. Let S2 = T ∗T , with S ≥ 0 self-adjoint, so kSxk = kTxk. S : E → E is self-adjoint, 1-1, so dense range: S(E) = E.

Define U : S(E) → H by U(Sx) = Tx for x ∈ E.

Since kUyk=kyk and S(E) = E, U extends uniquely to a map E → H with kUyk = kyk for all y ∈ E. Indeed, U is an isometry of E onto T (E).

Since U(Sx) = Tx for all x ∈ E, and U(Sx) = Tx = 0 for x ∈ E⊥ = ker T , then U(Sx) = Tx for all x ∈ H. Remark: U and S are unique. The condition kSxk = kTxk forces S : ker T → {0}, and S = S∗ then implies S : E → E. Then T ∗T = SU∗US = S2, so S is the unique non-negative self-adjoint square root of T ∗T . Example:

1 1 1 1 T (x1, x2, x3,...) = (0, x1, 2 x2, 3 x3,...) , T = SR ◦ diag(1, 2 , 3 ,...)

Corollary

If T ∈ B(H) is compact, there exists a sequence {TN } ⊂ B(H) of finite-rank operators such that limN→∞ kT − TN kB(H) = 0.

Proof. Let S be the truncation of S to N√ ⊕ · · · ⊕ N√ . N √ µ1 µN Then kS − SN k ≤ µN+1 → 0. Let TN = USN , then

kT − TN k ≤ kUk kS − SN k → 0 . Singular value decomposition

Definition If µ > µ > µ ... are the eigenvalues of T ∗T , the numbers 1 √ 2 3 ρk = µk are called the singular values of T .

Remarks.

kT k = maxk ρk . ∞ X 2 T is Hilbert-Schmidt iff ρk mult(ρk ) < ∞ , and k=1 ∞ 2 X 2 kT kHS = ρk mult(ρk ) . k=1 ∗ Proof. Take {xn} = eigenvector basis for T T . Then ∞ ∞ ∞ X 2 X ∗ X 2 kTxnk = hT Txn, xni = ρk mult(ρk ) . n=1 n=1 k=1 Singular value decomposition

Singular Value Decomposition Theorem

If T ∈ B(H) is compact with singular values {ρk }, there exists ⊥ an orthonormal basis{xn}for (ker T ) , and an orthonormal basis {yn} for T (H) , such that ∞ X Tx = ρk(n) hxn, xi yn . j=0

⊥ Proof. Let {xn} be O.N. basis of E = (ker T ) in which

S = diag(ρ1, . . . , ρ1, ρ2, . . . , ρ2, ρ3,... )

Since U : E → T (H) is isometry, then {yn} = {Uxn} is an P O.N. basis for T (H). Then Txn = ρk(n) yn, and x = nhxn, xixn with convergence in H, so the result holds by continuity.