Chapter 5

Compact operators

In this chapter, we study a special class of linear operators, called compact operators.

Why should we study compact operators? One important reason is that they can be approximated by finite rank operators. So they play an important role in the numerical approximation of solutions to operator equations. We had seen that if H is an infinite-dimensional Hilbert space and A ∈ CL(H) with ||A|| < 1, then given y ∈ H, there is a unique x ∈ H such that

which is given by the Neumann series

But all of the operators A, A², A³, ··· act on the infinite-dimensional H, so that computing these powers may not at all be feasible, and the convergence of the series may be “slow” (see Exercise 2.23, page 87). But now imagine that we can approximate A by finite matrices A_n and consider instead

where the y_n → y as n → ∞, and the unknown x_n are obtained by solving the finite linear algebraic equation (I – A_n)x_n = y_n. Then we can easily compute x_n = (I – A_n)⁻¹y_n, and if x_n → x, then we are able to determine x approximately. This wishful thinking can be made a reality if A is “compact”, as we shall see later on in this chapter when we learn Theorem 5.6 (page 218) on Galerkin approximations.

We begin by giving the definition of a compact operator.

5.1Compact operators

Definition 5.1. (Compact operators). Let X, Y be normed spaces. A linear transformation T : X → Y is said to be compact if

for every bounded sequence (x_n)_n∈N contained in X, (Tx_n)_n∈N has a convergent subsequence.

We will denote the set of all compact operators from X to Y by K(X, Y).

Why do we call such operators compact? The following result answers this question. Recall firstly that a closed and bounded set in an infinite dimensional normed space may fail to be compact (Example 1.26, page 48, and Example 1.28, page 50). So if T ∈ CL(X, Y), and B is the closed unit ball in X with centre 0, then although we know that T(B) is closed and bounded, it needn’t be compact. However, compact operators T are special in the sense that T(B) is guaranteed to be compact!

Theorem 5.1.

Let X, Y be normed spaces, and T : X → Y be a linear transformation. Then the following are equivalent:

(1)T is compact.

(2)The closure T(B) of the image under T of the closed unit ball, B := {x ∈ X : ||x|| 1}, is compact.

Proof.

(1) (2): Let (z_n)_n∈N be a sequence in T(B). Then there is a sequence (y_n)_n∈N in T(B) such that

Let y_n = Tx_n, x_n ∈ B, n ∈ N. Since for all n we have ||x_n|| 1, and because T is compact, (y_n)_n∈N has a convergent subsequence, say (y_nk)_k∈N, converging to, say y. As each y_n ∈ T(B), we have y ∈ T(B). From (5.1), (z_nk)_k∈N converges to y too. Hence T(B) is compact.

(2) (1): Let (x_n)_n∈N be a bounded sequence in X, and let M > 0 be such that for all n ∈ N, ||x_n|| M. Then (x_n/M)_n∈N is in B, and (T(x_n)/M)_n∈N is in T(B) ⊂ T(B). As T(B) is compact, (T(x_n)/M)_n∈N has a convergent subsequence. Thus (Tx_n)_n∈N has a convergent subsequence. Consequently, T is compact.

5.2The set K(X, Y) of all compact operators

Corollary 5.1.

Every compact operator is continuous, that is K(X, Y) ⊂ CL(X, Y).

Proof. Let B := {x ∈ X : ||x|| 1}. If T is compact, then T(B) is compact, and in particular, bounded. So T(B) ⊂ T(B) is bounded too. So there is some M > 0 such that ||Tz|| M for all z ∈ B. But this gives ||Tx|| M ||x|| for all x ∈ X. (This is trivially true when x = 0, and if x ≠ 0, then by taking z = x/||x|| ∈ B, we have ||Tz|| M, yielding the desired inequality.) So T ∈ CL(X, Y).

Is it true that K(X, Y) = CL(X, Y)? No:

Example 5.1. (Not all continuous linear transformations are compact). Let X be any infinite dimensional inner product space, for example ℓ². We will show that the identity operator I ∈ CL(X) is not compact.

Let {u₁, u₂, u₃, ···} be any orthonormal set in X. (Start with any countably infinite independent set, and use Gram-Schmidt.) Then ||u_n|| = 1 for all n ∈ N, and so the sequence (u_n)_n∈N is bounded. However, the sequence (Iu_n)_n∈N has no convergent subsequence, since for all n, m ∈ N with n ≠ m, we have ||Iu_n – Iu_m||² = ||u_n – u_m||² = 1 + 0 + 0 + 1 = 2, and this can’t be made as small as we please. Hence I is not compact, but is continuous.

In contrast to the above, it turns out that all finite rank operators are compact. Recall that an operator T is called a finite rank operator if its range, ran(T), is a finite-dimensional vector space. For ease of exposition, we will just prove this when Y is an inner product space.

Theorem 5.2. Let X be a normed space and Y be an inner product space. Suppose that T ∈ CL(X, Y) is such that ran(T) is finite dimensional.
Then T is compact.

Proof. Let {u₁, ···, u_m} be an orthonormal basis for ran(T). Let (x_n)_n∈N be a bounded sequence in X. Suppose that M > 0 is such that ||x_n|| M for all n ∈ N. We want to show that (Tx_n)_n∈N has a convergent subsequence. (We will show that

for some subsequence (x_nk)_k∈N.) For all n ∈ N and each ℓ ∈ {1, . . . , m},

(x_n)_n∈N has some subsequence (x_n⁽¹⁾)_n∈N such that .

(x_n⁽¹⁾)_n∈N has some subsequence (x_n⁽²⁾)_n∈N such that .

(x_n^(m–1))_n∈N has some subsequence (x_n^(m))_n∈N such that .

Claim: (Tx_n^(m))_n∈N converges to α₁u₁ + ··· α_mu_m.

it follows that (Tx_n^(m))_n∈N is a convergent subsequence of the sequence (Tx_n)_n∈N. Consequently T is compact.

In elementary linear algebra, not only were all linear transformations from Cⁿ to C^m continuous, they were even compact!

Example 5.2. (L(Cⁿ, C^m) = CL(Cⁿ, C^m) = K(Cⁿ, C^m)).

If A ∈ C^n×m, then T_A ∈ CL(Cⁿ, C^m) given by T_Ax = Ax, x ∈ Cⁿ, is finite-rank because ran T_A ⊂ C^m, and so T_A is compact. In particular, the identity map I : C^d → C^d is compact.

We had seen that K(X, Y) ⊂ CL(X, Y). But CL(X, Y) is a vector space, with the usual pointwise operations. So it is natural to ask if K(X, Y) is a subspace of CL(X, Y). The answer is “yes”, and this is what we show next.

Theorem 5.3. .

If X, Y are normed spaces, then K(X, Y) is a subspace of CL(X, Y).

Proof.

(S1)0 is compact since (0x_n)_n∈N = (0)_n∈N is convergent for all bounded sequences (x_n)_n∈N in X.

(S2)If T, S are compact, and (x_n)_n∈N is bounded, then (T_n)_n∈N has some subsequence (Tx_nk)_k∈N that is convergent, and (Sx_nk)_k∈N has some subsequence (Sx_nkℓ)_ℓ∈N that is convergent. So ((T + S)x_nkℓ)_ℓ∈N is convergent. Thus T + S is compact.

(S3)If T, S are compact, α ∈ K and (x_n)_n∈N is bounded, then (T_n)_n∈N has some subsequence (Tx_nk)_k∈N that is convergent, and so it follows that is convergent. Thus αT is compact.

Since CL(X, Y) is a normed space, we can even ask if K(X, Y) is a closed subspace of CL(X, Y). We now show that if Y is a Banach space, then K(X, Y) is a closed subspace of CL(X, Y), or briefly:

“Limits of compact operators are compact.”

Theorem 5.4. Let X be a normed space, Y a Banach space, and (T_n)_n∈N be a sequence in K(X, Y) that converges in CL(X, Y) to T ∈ CL(X, Y). Then T is compact.

Proof. Suppose that (x_n)_n∈N is a bounded sequence in X, and let M > 0 be such that for all n ∈ N, ||x_n|| M. Since T₁ is compact, (T₁x_n)_n∈N has a convergent subsequence (T₁x_n⁽¹⁾)_n∈N, say. Again, since (x_n⁽¹⁾)_n∈N is a bounded sequence, and T₂ is compact, (T₂x_n⁽¹⁾)_n∈N has a convergent subse-quence, say (T₂x_n⁽²⁾)_n∈N. We continue in this manner to obtain the following:

Consider the diagonal sequence x₁, x₂⁽¹⁾, x₃⁽²⁾, ··· .

By meditating on the above picture, one can convince oneself that

, ··· is a subsequence of .

As , ··· converges, its subsequence,

converges too, and so ( converges.

For n, m ∈ N, we have

Hence is a Cauchy sequence in Y and since Y is complete, it converges in Y. So, starting from the bounded sequence (x_n)_n∈N in X, we have found a subsequence of the sequence (Tx_n)_n∈N, that converges in Y. Consequently, T is compact.

Corollary 5.2. Let X be a normed space, Y a Hilbert space, and (T_n)_n∈N be a sequence of finite rank operators in CL(X, Y) that converges in CL(X, Y) to T ∈ CL(X, Y). Then T is compact.

Example 5.3. (When is a diagonal operator on ℓ² compact?)

Let X = Y = ℓ², (λ_n)_n∈N be a bounded in K, and Λ ∈ CL(ℓ²) be “given by”

Then we had seen in Exercise 2.17 (page 76) that ||Λ|| = |λ_n|.

Claim: Λ is compact if and only if λ_n = 0.

(If part): Consider for n ∈ N, the operators Λ_n ∈ CL(ℓ²) given by

Each Λ_n is a finite rank operator because ran Λ_n ⊂ span{e₁, ··· , e_n}, where e_k is the sequence with kth term equal to 1, and all others equal to 0. Hence Λ_n is compact. Then

Consequently, Λ, being the uniform limit of a sequence of compact operators, is compact.

(Only if part): Suppose that Λ is compact, but it is not the case that

(λ_n)_n∈N is convergent with limit 0, that is,

that is,

Taking N = 1, there exists n₁ > 1 such that |λ_n1| .

Taking N = n₁, there exists n₂ > n₁ such that |λ_n2| .

· · ·

Proceeding in this manner, we can construct inductively a subsequence (λ_nk)_k∈N of (λ_n)_n∈N such that for all k ∈ N, |λ_nk| . Now consider the bounded sequence (e_nk)_k∈N in ℓ². We have (Λe_nk)_n∈N = (λ_nk e_nk)_k∈N. But for all . This shows that (λ_nk e_nk)_k∈N has no convergent subsequence, contradicting the compactness of Λ.

Exercise 5.1. (Hilbert Schmidt operators are compact.)

Let H be a Hilbert space with an orthonormal basis (u₁, u₂, u₃, ···}.

Let T ∈ CL(H) be Hilbert-Schmidt, that is, .

(1)If m ∈ N, then define T_m : H → H by T_mx = .

Prove that T^m ∈ CL(H) and that .

Hint:. and use the Cauchy-Schwarz inequality in ℓ².

(2)Show that every Hilbert-Schmidt operator T is compact.

Hint:Using (1), conclude that T is the limit in CL(H) of the sequence of finite rank operators T_m, m ∈ N.

Exercise 5.2. Let H be a Hilbert space, and x₀, y₀ ∈ H be fixed.

Define x₀ y₀ : H → H by (x₀ y₀)(x) = 〈x, y₀〉x₀, x ∈ H.

(1)Show that x₀ y₀ ∈ CL(H) and that ||x₀ y₀|| ||x₀||||y₀||.

(2)Is x₀ y₀ compact?

(3)Let A, B ∈ CL(H). Show that A(x₀ y₀)B = (Ax₀) (B*y₀).

Recall that CL(H) has the structure of a complex algebra with multiplication of T, S ∈ CL(H) taken as composition T S ∈ CL(H). What is the relation of K(H q as a subset of CL(H) with respect to this operation of multiplication? The answer is that K(H) forms an “ideal” in CL(H).

Definition 5.2. (Ideal in an algebra).

An ideal I of an algebra R is a subset I of R having the properties:

(I1) 0 ∈ I.

(I2) If a, b ∈ I, then a + b ∈ I.

(I3) If a ∈ I and r ∈ R, then ar ∈ I and ra ∈ I.

For example, if R = Z, the set of all integers, then I = 2Z, the set of all even integers, is an ideal in R. In algebra, ideals are important, since they serve as kernels of algebra homomorphisms.

Theorem 5.5. Let H be a Hilbert space. Then we have:

(1)If T ∈ K(H) is compact and S ∈ CL(H), then TS is compact.

(2)If T ∈ CL(H) is compact, then T* is compact.

(3)If T ∈ CL(H) is compact and S ∈ K(H), then ST is compact.

Proof.

(1)Let (x_n)_n∈N be a bounded sequence in H. Suppose M > 0 is such that ||x_n|| M for all n ∈ N. Since S ∈ CL(H), it follows that (Sx_n)_n∈N is also a bounded sequence (||Sx_n|| ||S||||x_n|| ||S||M). As T is compact, (T(Sx_n))_n∈N = (TSx_n)_n∈N has a convergent subsequence. Thus TS is compact.

(2)As T ∈ K(H) and T* ∈ CL(H), by part (1) above, TT* is compact.

Let (x_n)_n∈N be a bounded sequence in X and ||x_n|| M for all n.

Then (TT*x_n)_n∈N has some convergent subsequence, say (TT*x_nk)_k∈N. Hence, given an > 0,

So (T*x_nk)_k∈N is a Cauchy sequence, and as H is a Hilbert space, it is convergent. Consequently, T* is compact.

(3)Since T is compact, by part (2), it follows that T* is also compact.

Moreover, as S* ∈ CL(H), we have T*S* is compact, using part (1). From part (2) again, we get (T* S*)* = S**T** = ST is compact.

Summary: The set K(H) is a closed ideal of CL(H).

Example 5.4. (Compact operators on infinite dimensional Hilbert spaces are never invertible.) Let H be an infinite dimensional Hilbert space, and T ∈ K(H). If T were invertible in CL(H), then T⁻¹ ∈ CL(H), so that I = TT⁻¹ ∈ K(H), which is false, since we had seen that the identity operator on an infinite dimensional Hilbert space is not compact.

Exercise 5.3. Let T ∈ CL(H), where H is an infinite-dimensional Hilbert space.

(1)Give an example of H and T such that T² is compact, but T isn’t.

(2)Show that if T is self-adjoint and T² is compact, then T is compact.

Exercise 5.4. Determine if the following statements are true for all S, T ∈ CL(H), where H is an infinite dimensional Hilbert space.

(1)If S and T are compact, then S + T is compact.

(2)If S + T is compact, then S or T is compact.

(3)If S or T is compact, then ST is compact.

(4)If ST is compact, then S is compact or T is compact.

Exercise 5.5. Let H be a Hilbert space. Let A ∈ CL(H) be fixed.

We define Λ ∈ CL(CL(H)) by Λ(T) = A*T + TA, T ∈ CL(H).

Show that the subspace K(H) of CL(H) is Λ-invariant, that is, ΛK(H) ⊂ K(H).

5.3Approximation of compact operators

Compact operators play an important role in numerical analysis since they can be approximated by finite rank operators. This means that when we want to solve an operator equation involving a compact operator, then we can replace the compact operator by a sufficiently good finite-rank approximation, reducing the operator equation to an equation involving finite matrices. The solution can then be found using linear algebra. In this section we will prove Theorems 5.6 and 5.7, which form the basis of the Galerkin Method in numerical analysis.

Consider the equation (I – K)x = y, where K is a given operator on a Hilbert space H, y ∈ H is a given vector, and x ∈ H is the unknown. Suppose we consider instead the equation (I – K₀)x₀ = y₀, where K₀ is close to K, and y₀ is close to y. The following result describes how big ||x – x₀|| can get.

Theorem 5.6.

Let

(1)H be a Hilbert space,

(2)K ∈ CL(H) be such that I – K is invertible in CL(H),

(3)K₀ ∈ CL(H) be such that := ||(K – K₀)(I – K)⁻¹|| < 1.

Then for every y, y₀ ∈ H, there exist unique x, x₀ ∈ X such that

(a) (I – K)x = y,

(b) (I – K₀)x₀ = y₀, and

(c) .

Note that from part (c) we see that the upper estimate on ||x – x₀|| is small when y ≈ y₀ and K ≈ K₀. So the result is telling us that if we have a scheme of approximating the operator K and the vector y, then we can solve the equation

approximately by solving instead the equation

and moreover, we have a handle on how large the error ||x – x₀|| can get. Later on, in Theorem 5.7 we will see that for compact operators K, such an approximating scheme for producing K₀ does exist.

Proof. As ||(K – K₀) (I – K)⁻¹|| < 1, by the Neumann Series Theorem, we have I + (K – K₀)(I – K)⁻¹ is invertible, and so

is invertible as well. Moreover,

Furthermore, we have

and so

Let y, y₀ ∈ X. Since I – K and I – K₀ are invertible, there are unique x, x₀ ∈ X such that (I – K)x = y and (I – K₀)x₀ = y₀. Also,

and so , as desired.

Question: If K is compact, y ∈ H, then how do we find approximations K₀ to K and y₀ to y?

Answer: Via projections.

Theorem 5.7. (Galerkin approximation).

Let

(1)H be a Hilbert space,

(2)K be a compact operator on H,

(3)(P_n)_n∈N be a sequence of projections (P²_n = P*_n = ∈ CL(H)) of finite rank such that P_n converges strongly to I (for all x ∈ H, P_nx = x).

Then P_nKP_n K in CL(H).

We remark that a mere strong convergence assumption results in uniform convergence, and this miracle happens since we have a compact operator at hand. We also remark that a standard way of producing such a sequence of projections is via choosing an orthonormal basis {u₁, u₂, u₃, ···} for H, and then we can take P_n to be the projection onto the closed finite dimensional subspace Y = span{u₁, ···, u_n}:

Proof. We’ll prove the following claims:

(1)P_nK K in CL(H) (projection approximation),

(2)KP_n K in CL(H) (Sloan approximation),

(3)P_nKP_n K in CL(H) (Galerkin approximation).

(1): For all x ∈ H, we have

and so ||P_nx|| ||x|| for all x, that is, ||P_n|| 1. Suppose that it is not the case that P_nK – K converges to 0 in CL(H) as n → ∞. This means that

Thus there exists an > 0 such that for all N ∈ N, there exists an n > N, such that ||P_nK – K|| > .

Hence there exists an > 0 such that for all N ∈ N, there exists an n > N, such that sup{||(P_nK – K)x|| : x ∈ H, ||x|| 1} > .

So there exists an > 0 such that for all N ∈ N, there exists an n > N, such that there exists an x ∈ H with ||x|| 1, but ||(P_nK – K)x|| > .

The last statement allows us to construct a sequence (x_nk)_k∈N in X such that ||x_nk|| 1 and ||(P_nkK – K)x_nk|| > as follows.

Taking N = 1, there exists an n₁ > 1 and an x_n1 ∈ H with ||x_n1|| 1 but ||(P_n1K – K)x_n1|| > .

Taking N = n₁, there exists an n₂ > n₁ and an x_n2 ∈ H with ||x_n2|| 1 but ||(P_n2K – K)x_n2|| > .

Taking N = n₃, there exists an n₃ > n₂ and an x_n3 ∈ H with ||x_n3|| 1 but ||(P_n3K – K)x_n3|| > .

Thus (x_nk)_k∈N is bounded and ||(P_nkK – K)x_nk|| > for all ks.

As (x_nk)_k∈N is bounded and K is compact, there exists a subsequence, say (Kx_nkℓ)_ℓ∈N, of (K_nk)_k∈N, that is convergent to y, say. Then we have

a contradiction. This completes the proof of (1).

(2): As K is compact, so is K*. Thus by (1), in CL(H). But , and so KP_n K in CL(H).

(3): Finally,

This completes the proof

So, in Theorem 5.6, what is y₀, K₀? We can take K₀ = P_nKP_n, where P_n is the orthogonal projection onto span{u₁, ···, u_n}, and y₀ = P_ny. We note that ||y₀ – y|| = ||P_ny – y|| is small for large n, and ||K – P_nKP_n|| is small for large n. Thus = ||(K – K₀)(I – K)⁻¹|| is small. So if we look at the equation (I – P_nKP_n)x₀ = P_ny instead of (I – K)x = y, then ||x – x₀|| can be made as small as we please by taking n large enough. We give a simple toy example.

Example 5.5.

Consider the operator on ℓ². For all x = (x_n)_n∈N ∈ ℓ²,

So ||K|| , and I – K is invertible in CL(ℓ²).

K is Hilbert-Schmidt as . So K is compact.

Let . To find approximate solutions of the equation

we fix an n ∈ N, and solve instead x – P_nKP_nx = P_ny, that is, the system

The approximate solutions for n = 1, 2, 3, 4, 5 are given (correct up to four decimal places) by

while the exact unique solution to the equation (I – K)x = y is given by

To 4 decimal places, this is x = (0.5, 0.3333, 0.2500, 0.2000, 0.1667, ···).

5.4(∗) Spectral Theorem for Compact Operators

In elementary linear algebra, one learns about the Spectral Theorem, which says that every Hermitian matrix T ∈ C^d×d is diagonalisable, with a basis of orthonormal eigenvectors u₁, ··· , u_d ∈ C^d, and corresponding real eigenvalues λ₁ ··· λ_d, so that

Towards seeking a generalisation to the Hilbert space case, we’ll now show that while the spectrum of a general self-adjoint operator may be quite complicated, for a compact self-adjoint operator, things are quite similar to the finite-dimensional case.

Theorem 5.8. (Spectral Theorem for compact, self-adjoint operators).

Let H be a Hilbert space and T = T* ∈ K(H) have infinite rank.

Then there exist orthonormal eigenvectors u_n, n ∈ N, with corresponding eigenvalues λ_n, n ∈ N, such that λ_n = 0, and for all x ∈ H,

We had already seen that the eigenvalues of a self-adjoint operator must be real, and that the eigenvectors corresponding to distinct eigenvalues are orthogonal. It is also clear that for any eigenvalue λ of T, we have that |λ| ||T||, since if v ∈ H\{0} is a corresponding eigenvector, then

Let us make a few more observations which will be used in proving the spectral theorem.

Lemma 5.1. If T = T* ∈ CL(H), then ||T|| = .

Proof. Let M := .

If x ∈ H is such that ||x|| = 1, then we have by Cauchy-Schwarz that . Thus M ||T||.

It remains to show the reverse inequality. For any x, y ∈ H,

and so . We note that by the definition of M,

and so from the above, together with the Parallelogram Law, we obtain

Let θ ∈ R be such that 〈T x, y〉 = |〈T x, y〉|e^iθ. Replacing y by e^iθy yields

If Tx = 0 or x = 0, then ||Tx|| M ||x|| is trivially true.

If Tx ≠ 0 and x ≠ 0, then with in the above, we obtain

||Tx||||x|| M ||x||² and so ||Tx|| M ||x||. Thus ||T|| M.

Moreover if T is compact, then this bound M is achieved, thanks to the following result. Indeed, if x is the unit-norm eigenvector corresponding to the eigenvalue λ whose modulus |λ| = ||T||, then

Lemma 5.2. If H is a nontrivial Hilbert space and T = T* ∈ K(H), then either ||T|| or −||T|| is an eigenvalue of T.

Proof. If T = 0, then this is trivial. So let us suppose that T is nonzero. From the previous lemma, it follows that there is a sequence (x′_n)_n∈N of unit norm vectors in H such that . But as 〈Tx′_n, x′_n〉 is real, we have 〈Tx′_n, x′_n〉 is either |〈Tx′_n, x′_n〉| or −|〈Tx′_n, x′_n〉|. Thus either for infinitely many n, 〈Tx′_n, x′_n〉 is positive (and then the subsequence (〈Tx_n, x_n〉)_n∈N with these ns converges to ||T||), or for infinitely many n, 〈Tx′_n, x′_n〉 is negative (and then the subsequence (〈Tx_n, x_n〉)_n∈N with these ns converges to −||T||). So 〈Tx_n, x_n〉 λ, where λ is either ||T|| or −||T||. We have

So Tx_n – λx_n 0. As T is compact, there is a subsequence, say (Tx_nk)_k∈N of (Tx_n)_n∈N that converges, say, to y ∈ H. Then (x_nk)_k∈N converges to y/λ, because λx_nk = Tx_nk – (Tx_nk – λx_nk) y – 0 = y.

Thanks to the continuity of T, we obtain

Hence Ty = λy, and y ≠ 0 since .

Lemma 5.3. Let H be a Hilbert space, T = T* ∈ CL(H), and Y be a T-invariant closed subspace of H. Then:

(1)Y^⊥ is also T-invariant.

(2)The restriction T|_Y^⊥ : Y^⊥ → Y^⊥ of T to the Hilbert space Y^⊥ is also self-adjoint.

(3)If T is in addition compact, then T|_Y^⊥ is also compact.

Proof.

(1)Let z ∈ Y^⊥. For all y ∈ Y, we have that Ty ∈ Y (Y is T-invariant!), and so 〈Tz, y〉 = 〈z, T*y〉 = 〈z, Ty〉 = 0. Thus Tz ∈ Y^⊥.

(2)Y^⊥, being a closed subspace of a Hilbert space, is itself a Hilbert space. As Y^⊥ is T-invariant, the restriction T|_Y^⊥ : Y^⊥ → Y^⊥ is well-defined. Let us denote this restriction by . For z₁, z₂ ∈ Y^⊥,

Thus T|_Y^⊥ is self-adjoint.

(3)Finally, suppose that T is compact. Let (z_n)_n∈N be a bounded sequence in Y^⊥. Then ||z_n||_Y^⊥ = ||z_n||. So (z_n)_n∈N is a bounded sequence in H. As T is compact, (Tz_n)_n∈N = (z_n)_n∈N has a subsequence (x_nk)_k∈N that is convergent in H. In particular, (x_nk)_k∈N is Cauchy in H, and hence also Cauchy in Y^⊥ (because ||z_n – z_m|| = ||z_n – z_m||_Y^⊥). As Y^⊥ is complete, it follows that (x_nk)_k∈N is convergent in Y^⊥. So T|_Y^⊥ is compact.

Proof. (Of the spectral theorem). Let H₁ := H and T₁ := T.

By Lemma 5.2, there exists an eigenvalue λ₁ of T₁ and a corresponding eigenvector u₁ such that ||u₁|| = 1 and |λ₁| = ||T₁||.

Set H₂ = (span{u₁})^⊥. Then H₂ is a closed subspace of H₁, and it is also T-invariant: TH₂ ⊂ H₂. Let T₂ := T|_H2. Then T₂ is self-adjoint and compact. There exist an eigenvalue λ₂ of T₂ and a corresponding eigenvector u₂ such that ||u₂|| = 1 and |λ₂| = ||T₂||. So

Clearly, {u₁, u₂} are orthonormal, Tu₁ = λ₁u₁ and Tu₂ = λ₂u₂.

Now let H₃ := (span{u₁, u₂})^⊥. Then H₃ is a closed subspace of H, H₃ ⊂ H₂, and as span{u₁, u₂} is T-invariant, we obtain TH₃ ⊂ H₃. Let T₃ := T|_H3. Then T₃ is self-adjoint and compact. Thus there exist an eigenvalue λ₃ of T₃ and a corresponding eigenvector u₃ ∈ H₃, such that ||u₃|| = 1 and |λ₃| = ||T₃||. As u₃ ∈ H₃ ⊂ H₂, we have that

Continuing in this manner, we get a sequence λ₁, λ₂, λ₃, ··· of eigenvalues of T, and a corresponding set of eigenvectors u₁, u₂, u₃, ··· , such that

The process would stop at some n if H_n := (span{u₁, ···, u_n−1})^⊥ would become {0}, but we will now show that thanks to the infinite rank assumption on T, this case is impossible. Suppose, on the contrary, that H_n = {0}.

For any x ∈ H, we have .

As H_n = {0}, we obtain .

So ran T is spanned by Tu₁, ··· , Tu_n, a contradiction to the assumption that T has infinite rank. Thus one has an infinite sequence of eigenvectors u₁, u₂, u₃, ··· with eigenvalues λ₁, λ₂, λ₃, ··· .

Let us now show that (|λ_n|)_n∈N converges to 0. As it is decreasing, it converges to . If > 0, then for n ≠ m,

But this contradicts the fact that T is compact, since the sequence (Tu_n)_n∈N should have some convergent (and hence Cauchy) subsequence. So (|λ_n|)_n∈N converges to 0, and hence (λ_n)_n∈N also converges to 0.

For all x ∈ H, we have , and so

The last inequality above follows from Bessel’s Inequality (Exercise 4.15, page 172). Hence for all x ∈ H.

Exercise 5.6.

Let H be an infinite dimensional Hilbert space, and T = T* ∈ K(H) be one-to-one. Show that the eigenvectors of T form an orthonormal basis for H.

Exercise 5.7. Let H be a Hilbert space. Suppose that T = T* ∈ K(H) has infinite rank, and is positive, that is, 〈T x, x〉 0 for all x ∈ H.

Prove that T has a square root, that is, an operator ∈ CL(H) such that ()² = T.