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- Compact sets and compact operators
3.1. Review of compactness in metric spaces.
Definition 3.1. The diameter of a subset A of a metric space X, d is defined to be supa,b∈A d(a, b). We say that A is precompact if, for every > 0, there is a finite covering of A by sets of diameter at most . It is equivalent to ask that, for each > 0 there exist a finite subset F of A such that
a∈F B(a;^ )^ ⊇^ A; such a set^ F^ is called an^ -net^ for^ A.
Definition 3.2. A topological space X is said to be sequentially compact if every sequence in X has a convergent subsequence.
Definition 3.3. We say that a subset A of a topological space X is relatively compact (in X) if the closure A, taken in X, is compact.
Theorem 3.4. For a metric space X, d the following are equivalent:
(1) X is compact; (2) X is complete and precompact; (3) X is sequentially compact.
Corollary 3.5. If X is a complete metric space and A ⊆ X then the following are equivalent:
(1) A is precompact; (2) A is relatively compact; (3) every sequence in A has a subsequence converging to a limit in X.
3.2. Compactness in Banach spaces. We recall from an earlier course that closed bounded subsets of a finite- dimensional normed space are compact, but that the unit ball of an infinite-dimensional normed space is not precompact.
Lemma 3.6. Let A be a subset of a normed space X. Then A is precompact if and only if A is bounded and, for all > 0 , there is a finite-dimensional subspace Y of X with dY (a) < for all a ∈ A.
It is useful to have criteria for compactness of subsets of familiar spaces. These criteria usually have a strong whiff of uniform convergence about them. In the case of sequence spaces, the most convenient formulation is in terms of the projection Pn introduced in the last chapter. Recall that
Pn(x(1), x(2), x(3),... ) = (x(1), x(2), x(3),... , x(n), 0 , 0 ,... ).
Theorem 3.7. Let A be a subset of `p ( 1 ≤ p < ∞) or of c 0. Then A is relatively compact if and only if A is bounded and ‖(I − Pn)(x)‖ → 0 uniformly over x ∈ A.
Proof. Suppose first that ‖(I − Pn)(x)‖ → 0 uniformly over x ∈ A, and let > 0 be given. There exists N such that ‖x − Pn(x)‖ < for all x ∈ X. This implies that dY (x) < , where Y is the finite-dimensional subspace sp〈e 1 ,... , en〉. We can now apply Lemma 1.6. For the converse, assume that A is precompact and let > 0 be given; there is a finite /2 net F for A. For each a ∈ F we choose na such that ‖a − Pna (a)‖ < /2. Take n = maxa∈F na and note that if x ∈ A, we have ‖x − a‖ < /2 for some a ∈ A, whence ‖(I − Pn)(x)‖ ≤ ‖(I − Pna )(x)‖ ≤ ‖I − Pna ‖‖x − a‖ + ‖a − Pna (a)‖ < .
Definition 3.8. Let L be a topological space and let A be a set of scalar-valued functions on L. We say that A is equicontinuous if for each t ∈ L and each > 0 there is an open set U 3 t such that |f (s) − f (t)| ≤ whenever s ∈ U and f ∈ A.
Theorem 3.9 (Ascoli). Let K be a compact space and let A be a subset of C (K). Then A is norm-precompact if and only if A is norm-bounded and equicontinuous.
Proof. First assume that A is norm-precompact; this certainly implies that A is norm-bounded so we just have to establish equicontinuity. Let > 0 be given and let {f 1 ,... , fn} be a finite /3-net for A. For each t ∈ L and each j there is an open set Uj 3 t such that |fj (s) − fj (t)| < whenever s ∈ Uj. Now take U =
⋂n j=1 Uj^ and let^ s^ ∈^ U^ ,^ f^ ∈^ A. For some^ j we have ‖f − fj ‖∞ < /3, from which we deduce
|f (t) − f (s)| ≤ |f (t) − fj (t)| + |fj (t) − fj (s)| + |f (s) − fj (s)| ≤ 2 ‖f − fj ‖∞ + |fj (t) − fj (s)| < ,
because s ∈ U ⊆ Uj. Notice that this implication did not involve compactness of K. For the converse implication, we assume that K is compact and that A is bounded and equicontinuous. Given > 0 and t ∈ K there is an open set Ut 3 t such that |x(s) − x(t)| < /3 for all s ∈ U and all x ∈ A. By compactness, there is a finite sub-covering Ut 1 ,... , Utn. Consider the map R : A → Rn, ‖ · ‖∞ given by (Rx)j = x(tj ); the image R[A] is bounded in Rn, hence precompact. Let x 1 ,... , xN be chosen so that {R(xk) : 1 ≤ k ≤ N } is an /3-net for R[A]. I claim that {x 1 ,... , xN } is an -net for A. Indeed, for any x ∈ A there exists k such that ‖R(x) − R(xk)‖∞ < /3. Thus,
19
20
for any j we have |x(tj ) − xk(tj )| < /3. For an arbitrary s ∈ K we have s ∈ Utj for some j and we rapidly obtain |x(s) − xj (s)| < .
3.3. Compact operators.
Definition 3.10. We say that a linear operator T from a normed space X into a normed space Y is compact if the image T [BX ] is relatively compact in Y. [Remember that this means that the closure T [BX ], taken in Y , is compact.] When Y is a Banach space, it is equivalent to demand that T [BX ] be precompact. We write K(X; Y ) for the set of all compact linear operators from X to Y , and abbreviate it to K(X) when X = Y.
Proposition 3.11. (1) If a bounded linear operator T : X → Y has finite rank (i.e. T [X] is finite-dimensional), then T is compact. (2) If T 1 and T 2 are compact operators from X to Y then so is T 1 + T 2. (3) If T : X → Y is compact then so is U T S : W → Z whenever W and Z are normed spaces and S : W → X, U : Y → Z are bounded.
Proof. (1) If T [X] = Z then Z is a finite-dimensional normed space and so bounded subsets of Z are relatively compact in Z (and hence in Y ). If T is bounded the the set T [BX ] is a bounded subset of Z. (2) The sets Ki = Ti[BX ] are compact and hence so is K 1 + K 2 , a set which contains (T 1 + T 2 )[BX ]. (3) Let K be the compact subset T [BX ] of Y. By continuity of U the image U [K] is a compact subset of Z, and hence so is λU [K] for any scalar λ. We have U T S[BW ] ⊆ λU [K] for λ = ‖S‖.
Theorem 3.12. If Y is a Banach space and X is a normed space, then K(X; Y ) is a closed linear subspace of L(X; Y ). When X = Y K(X) is a closed ideal of the ring L(X).
Proof. It is enough to prove that K(X; Y ) is closed; so let T ∈ K(X; Y ), taken in L(X; Y ). Given > 0 there exists a compact operator S such that ‖S − T ‖ < /3. Since S is compact there is a finite /3-net for S[BX ], S(x 1 ),... , S(xn) say, with xj ∈ BX. Now for any x ∈ BX we have
‖T (x) − T (xj )‖ ≤ ‖T (x) − S(x)‖ + ‖S(x) − S(xj )‖ + ‖T (xj ) − S(xj )‖ ≤ ‖S(x) − S(xj )‖ + 2‖T − S‖,
and this is less than for a suitably chosen j.
Corollary 3.13. If (Tn)n∈N is a sequence of bounded finite-rank operators and ‖T − Tn‖ → 0 then T is compact.
For all familiar Banach spaces, the converse is also true: every compact operator is norm-approximable by finite rank operators. In any case approximation by finite-rank is a good thing to try when you want to prove that some specific operator is compact.
Example 3.14. Let A = (ai,j )∞ i,j=1 be an infinite matrix satisfying
i,j |a 2 i,j <^ ∞. Then by Section B methods, there is a bounded linear operator T : 2 → 2 defined by
T (x) = y, where yi =
j
ai,j xj ,
and satisfying
‖T ‖^2 ≤
i,j
|a^2 i,j.
A neat way to establish compactness is to note that ‖T − Tn‖ → 0 where Tn is the finite-rank operator associated with
the truncated matrix A(n)^ = (a( i,jn) )∞ i,j=1, where
a( i,jn) =
ai,j if i, j ≤ n 0 otherwise.
3.4. Duals of compact operators.
Theorem 3.15. Let X and Y be normed spaces and let T ∈ L(X; Y ). If T is compact, so is T ∗. When Y is complete, the converse also holds.
Proof. Assume that T is compact, let K = T [BX ], a compact subset of Y and define R : Y ∗^ → C (K) to be the restriction mapping. Notice that ‖R(g)‖∞ = sup{|g(T (x))| : x ∈ BX } = ‖T ∗(g)‖.
Define A to be the set of all restrictions R(g) = g K with g ∈ BY ∗. Then A is bounded and equicontinuous, hence precompact by Ascoli’s theorem. Since the metric space A is isometric to T ∗[BY ∗ ], this is precompact too and by completeness of X∗, T ∗^ is a compact operator. If we now assume that T ∗^ is compact, we may use the first part of the question to show that T ∗∗^ : X∗∗^ → Y ∗∗^ is compact. The image T [BX ] is isometric to JY T [BX ] = T ∗∗JX [BX ] ⊆ T ∗∗[BX∗∗^ ], and hence precompact. This yields compactness of T when Y is a Banach space.
