Compactness

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8 Compactness

It is not easy to study linear operators “in general” and there are many questions about operators in Hilbert spaces raised many decades ago which are still unanswered. Therefore it is reasonable to single out classes of operators which have (relatively) simple properties. Such a class of operators more closed to finite dimensional ones will be studied here.

These operators are so compact that we even can fit them in our course

8.1 Compact operators

Let us recall some topological definition and results.

Definition 1 A compact set in a metric space is defined by the property that any its covering by a family of open sets contains a subcovering by a finite subfamily.

In the finite dimensional vector spaces ℝⁿ or ℂⁿ there is the following equivalent definition of compactness (equivalence of 1 and 2 is known as Heine–Borel theorem):

Theorem 2 If a set E in ℝⁿ or ℂⁿ has any of the following properties then it has other two as well:

E is bounded and closed;
E is compact;
Any infinite subset of E has a limiting point belonging to E.

Exercise^* 3 Which equivalences from above are not true any more in the infinite dimensional spaces?

Definition 4 Let X and Y be normed spaces, T∈B(X,Y) is a finite rank operator if Im T is a finite dimensional subspace of Y. T is a compact operator if whenever (x_i)₁^∞ is a bounded sequence in X then its image (T x_i)₁^∞ has a convergent subsequence in Y.

The set of finite rank operators is denote by F(X,Y) and the set of compact operators—by K(X,Y)

Exercise 5 Show that both F(X,Y) and K(X,Y) are linear subspaces of B(X,Y).

We intend to show that F(X,Y)⊂K(X,Y).

Lemma 6 Let Z be a finite-dimensional normed space. Then there is a number N and a mapping S: l₂^N → Z which is invertible and such that S and S⁻¹ are bounded.

Proof. The proof is given by an explicit construction. Let N=dimZ and z₁, z₂, …, z_N be a basis in Z. Let us define

S: l₂^N → Z by S(a₁,a₂,…,a_N)=

∑

k=1

a_k z_k,

then we have an estimation of norm:

⎪⎪
⎪⎪

⎪⎪
⎪⎪
⎪⎪
⎪⎪

∑

k=1

a_k z_k

⎪⎪
⎪⎪
⎪⎪
⎪⎪

≤

∑

k=1

⎪
⎪

a_k

⎪
⎪

⎪⎪
⎪⎪

z_k

⎪⎪
⎪⎪

≤

⎛
⎜
⎜
⎝

∑

k=1

⎪
⎪

a_k

⎪
⎪

⎞
⎟
⎟
⎠

1/2

⎛
⎜
⎜
⎝

∑

k=1

⎪⎪
⎪⎪

z_k

⎪⎪
⎪⎪

⎞
⎟
⎟
⎠

1/2

So ||S||≤ (∑₁^N ||z_k||²)^1/2 and S is continuous.

Clearly S has the trivial kernel, particularly ||Sa||>0 if ||a||=1. By the Heine–Borel theorem the unit sphere in l₂^N is compact, consequently the continuous function a↦ ||∑₁^N a_k z_k|| attains its lower bound, which has to be positive. This means there exists δ>0 such that ||a||=1 implies ||Sa||>δ , or, equivalently if ||z||<δ then ||S⁻¹ z||<1. The later means that ||S⁻¹||≤ δ⁻¹ and boundedness of S⁻¹. □

Corollary 7 For any two metric spaces X and Y we have F(X,Y)⊂ K(X,Y).

Proof. Let T∈F(X,Y), if (x_n)₁^∞ is a bounded sequence in X then ((Tx_n)₁^∞⊂ Z=Im T is also bounded. Let S: l₂^N→ Z be a map constructed in the above Lemma. The sequence (S⁻¹T x_n)₁^∞ is bounded in l₂^N and thus has a limiting point, say a₀. Then Sa₀ is a limiting point of (T x_n)₁^∞. □

There is a simple condition which allows to determine which diagonal operators are compact (particularly the identity operator I_X is not compact if dimX =∞):

Proposition 8 Let T is a diagonal operator and given by identities T e_n=λ_n e_n for all n in a basis e_n. T is compact if and only if λ_n→ 0.

Figure 16: Distance between scales of orthonormal vectors

Proof. If λ_n↛0 then there exists a subsequence λ_{n_k} and δ>0 such that | λ_{n_k} |>δ for all k. Now the sequence (e_{n_k}) is bounded but its image T e_{n_k}=λ _{n_k} e_{n_k} has no convergent subsequence because for any k≠ l:

⎪⎪
⎪⎪

λ _{n_k}e_{n_k}−λ _{n_l}e_{n_l}

⎪⎪
⎪⎪

= (

⎪
⎪

λ _{n_k}

⎪
⎪

² +

⎪
⎪

λ _{n_l}

⎪
⎪

²)^1/2≥

√

δ ,

i.e. T e_{n_k} is not a Cauchy sequence, see Figure 16. For the converse, note that if λ_n→ 0 then we can define a finite rank operator T_m, m≥ 1—m-“truncation” of T by:

T_m e_n =

⎧
⎨
⎩

Te_n=λ_n e_n,	1≤ n≤ m;
0 ,	n>m.

(45)

Then obviously

(T−T_m) e_n =

⎧
⎨
⎩

0,	1≤ n≤ m;
λ_n e_n ,	n>m,

and ||T−T_m||=sup_n>m| λ_n |→ 0 if m→ ∞. All T_m are finite rank operators (so are compact) and T is also compact as their limit—by the next Theorem. □

Theorem 9 Let T_m be a sequence of compact operators convergent to an operator T in the norm topology (i.e. ||T−T_m||→ 0) then T is compact itself. Equivalently K(X,Y) is a closed subspace of B(X,Y).

Figure 17: The є/3 argument to estimate | f(x)−f(y) |.

T₁x₁⁽¹⁾ T₁x₂⁽¹⁾ T₁x₃⁽¹⁾ … T₁x_n⁽¹⁾ … → a₁

T₂x₁⁽²⁾ T₂x₂⁽²⁾ T₂x₃⁽²⁾ … T₂x_n⁽²⁾ … → a₂

T₃x₁⁽³⁾ T₃x₂⁽³⁾ T₃x₃⁽³⁾ … T₃x_n⁽³⁾ … → a₃

… … … … … …

T_nx₁⁽ⁿ⁾ T_nx₂⁽ⁿ⁾ T_nx₃⁽ⁿ⁾ … T_nx_n⁽ⁿ⁾ … → a_n

… … … … … … ↓

↘

a

Table 2: The “diagonal argument”.

Proof. Take a bounded sequence (x_n)₁^∞. From compactness

of T₁	⇒ ∃	subsequence (x_n⁽¹⁾)₁^∞ of (x_n)₁^∞	s.t.	(T₁x_n⁽¹⁾)₁^∞ is convergent.
of T₂	⇒ ∃	subsequence (x_n⁽²⁾)₁^∞ of (x_n⁽¹⁾)₁^∞	s.t.	(T₂x_n⁽²⁾)₁^∞ is convergent.
of T₃	⇒ ∃	subsequence (x_n⁽³⁾)₁^∞ of (x_n⁽²⁾)₁^∞	s.t.	(T₃x_n⁽³⁾)₁^∞ is convergent.
…	…	…	…	…

Could we find a subsequence which converges for all T_m simultaneously? The first guess “take the intersection of all above sequences (x_n^(k))₁^∞” does not work because the intersection could be empty. The way out is provided by the diagonal argument (see Table 2): a subsequence (T_m x_k^(k))₁^∞ is convergent for all m, because at latest after the term x_m^(m) it is a subsequence of (x_k^(m))₁^∞.

We are claiming that a subsequence (T x_k^(k))₁^∞ of (T x_n)₁^∞ is convergent as well. We use here є/3 argument (see Figure 17): for a given є>0 choose p∈ℕ such that ||T−T_p||<є/3. Because (T_p x_k^(k))→ 0 it is a Cauchy sequence, thus there exists n₀>p such that ||T_p x_k^(k)−T_p x_l^(l)||< є/3 for all k, l>n₀. Then:

⎪⎪
⎪⎪

T x_k^(k)−T x_l^(l)

⎪⎪
⎪⎪

(T x_k^(k)−T_p x_k^(k))+(T_p x_k^(k)−T_p x_l^(l))+(T_p x_l^(l)−T x_l^(l))

⎪⎪
⎪⎪

≤

⎪⎪
⎪⎪

T x_k^(k)−T_p x_k^(k)

⎪⎪
⎪⎪

T_p x_k^(k)−T_p x_l^(l)

⎪⎪
⎪⎪

T_p x_l^(l)−T x_l^(l)

⎪⎪
⎪⎪

≤

Thus T is compact. □

8.2 Hilbert–Schmidt operators

Definition 10 Let T: H→ K be a bounded linear map between two Hilbert spaces. Then T is said to be Hilbert–Schmidt operator if there exists an orthonormal basis in H such that the series ∑_k=1^∞||T e_k||² is convergent.

Example 11

Let T: l₂→ l₂ be a diagonal operator defined by Te_n=e_n/n, for all n≥ 1. Then ∑ ||Te_n||²=∑n⁻²=π²/6 (see Example 16) is finite.
The identity operator I_H is not a Hilbert–Schmidt operator, unless H is finite dimensional.

A relation to compact operator is as follows.

Theorem 12 All Hilbert–Schmidt operators are compact. (The opposite inclusion is false, give a counterexample!)

Proof. Let T∈ B(H,K) have a convergent series ∑ ||T e_n||² in an orthonormal basis (e_n)₁^∞ of H. We again (see (45)) define the m-truncation of T by the formula

T_m e_n =

⎧
⎨
⎩

Te_n,	1≤ n≤ m;
0 ,	n>m.

(46)

Then T_m(∑₁^∞a_k e_k)=∑₁^m a_k e_k and each T_m is a finite rank operator because its image is spanned by the finite set of vectors Te₁, …, Te_n. We claim that ||T−T_m||→ 0. Indeed by linearity and definition of T_m:

(T−T_m)

⎛
⎜
⎜
⎝

∞

∑

n=1

a_n e_n

⎞
⎟
⎟
⎠

∞

∑

n=m+1

a_n (Te_n).

Thus:

⎪⎪
⎪⎪
⎪⎪
⎪⎪

(T−T_m)

⎛
⎜
⎜
⎝

∞

∑

n=1

a_n e_n

⎞
⎟
⎟
⎠

⎪⎪
⎪⎪
⎪⎪
⎪⎪

∞

∑

n=m+1

a_n (Te_n)

⎪⎪
⎪⎪
⎪⎪
⎪⎪

(47)

≤

∞

∑

n=m+1

⎪
⎪

a_n

⎪
⎪

⎪⎪
⎪⎪

(Te_n)

⎪⎪
⎪⎪

≤

⎛
⎜
⎜
⎝

∞

∑

n=m+1

⎪
⎪

a_n

⎪
⎪

⎞
⎟
⎟
⎠

1/2

⎛
⎜
⎜
⎝

∞

∑

n=m+1

⎪⎪
⎪⎪

(Te_n)

⎪⎪
⎪⎪

⎞
⎟
⎟
⎠

1/2

≤

⎪⎪
⎪⎪
⎪⎪
⎪⎪

∞

∑

n=1

a_n e_n

⎪⎪
⎪⎪
⎪⎪
⎪⎪

⎛
⎜
⎜
⎝

∞

∑

n=m+1

⎪⎪
⎪⎪

(Te_n)

⎪⎪
⎪⎪

⎞
⎟
⎟
⎠

1/2

(48)

so ||T−T_m||→ 0 and by the previous Theorem T is compact as a limit of compact operators. □

Corollary 13 (from the above proof) For a Hilbert–Schmidt operator

⎪⎪
⎪⎪

≤

⎛
⎜
⎜
⎝

∞

∑

n=m+1

⎪⎪
⎪⎪

(Te_n)

⎪⎪
⎪⎪

⎞
⎟
⎟
⎠

1/2

Proof. Just consider difference of T and T₀=0 in (47)–(48). □

Example 14 An integral operator T on L₂[0,1] is defined by the formula:

(T f)(x)=

∫

K(x,y)f(y) d y, f(y)∈L₂[0,1], (49)

where the continuous on [0,1]×[0,1] function K is called the kernel of integral operator.

Theorem 15 Integral operator (49) is Hilbert–Schmidt.

Proof. Let (e_n)_−∞^∞ be an orthonormal basis of L₂[0,1], e.g. (e^{2π i
nt})_n∈ℤ. Let us consider the kernel K_x(y)=K(x,y) as a function of the argument y depending from the parameter x. Then:

(T e_n)(x)=

∫

K(x,y)e_n(y) d y=

∫

K_x(y)e_n(y) d y= ⟨ K_x,ē_n ⟩.

So ||T e_n||²= ∫₀¹| ⟨ K_x,ē_n ⟩ |² d x. Consequently:

∞

∑

−∞

⎪⎪
⎪⎪

T e_n

⎪⎪
⎪⎪

∞

∑

−∞

∫

⎪
⎪

⟨ K_x,ē_n ⟩

⎪
⎪

² d x

∫

∞

∑

⎪
⎪

⟨ K_x,ē_n ⟩

⎪
⎪

² d x

(50)

∫

⎪⎪
⎪⎪

K_x

⎪⎪
⎪⎪

² d x

∫

⎪
⎪

K(x,y)

⎪
⎪

² d x d y < ∞

Exercise 16 Justify the exchange of summation and integration in (50).

□

Remark 17 The definition 14 and Theorem 15 work also for any T: L₂[a,b] → L₂[c,d] with a continuous kernel K(x,y) on [c,d]×[a,b].

Definition 18 Define Hilbert–Schmidt norm of a Hilbert–Schmidt operator A by ||A||_HS²=∑_n=1^∞||Ae_n||² (it is independent of the choice of orthonormal basis (e_n)₁^∞, see Question 27).

Exercise^* 19 Show that set of Hilbert–Schmidt operators with the above norm is a Hilbert space and find the an expression for the inner product.

Example 20 Let K(x,y)=x−y, then

(Tf)(x)=

∫

(x−y)f(y) d y =x

∫

f(y) d y −

∫

yf(y) d y

is a rank 2 operator. Furthermore:

⎪⎪
⎪⎪

_HS²

∫

(x−y)² d x d y =

∫

⎡
⎢
⎢
⎣

(x−y)³

⎤
⎥
⎥
⎦

x=0

 d y

∫

(1−y)³

y³

 d y=

⎡
⎢
⎢
⎣

−

(1−y)⁴

y⁴

⎤
⎥
⎥
⎦

On the other hand there is an orthonormal basis such that

Tf=

√

⟨ f,e₁ ⟩e₁−

√

⟨ f,e₂ ⟩e₂,

and ||T||=1/√12 and ∑₁² ||Te_k||²=1/6 and we get ||T||≤ ||T||_HS in agreement with Corollary 13.


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Last modified: November 6, 2024.

T₁x₁⁽¹⁾	T₁x₂⁽¹⁾	T₁x₃⁽¹⁾	…	T₁x_n⁽¹⁾	…	→	a₁
T₂x₁⁽²⁾	T₂x₂⁽²⁾	T₂x₃⁽²⁾	…	T₂x_n⁽²⁾	…	→	a₂
T₃x₁⁽³⁾	T₃x₂⁽³⁾	T₃x₃⁽³⁾	…	T₃x_n⁽³⁾	…	→	a₃
…	…	…	…	…	…
T_nx₁⁽ⁿ⁾	T_nx₂⁽ⁿ⁾	T_nx₃⁽ⁿ⁾	…	T_nx_n⁽ⁿ⁾	…	→	a_n
…	…	…	…	…	…		↓
						↘
							a