# 1 Compact and Precompact Subsets of H

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1 Compact Sets and Compact Operators by Francis J. Narcowich November, 2014 Throughout these notes, H denotes a separable Hilbert space. We will use the notation B(H) to denote the set of bounded linear operators on H. We also note that B(H) is a Banach space, under the usual operator norm. 1 Compact and Precompact Subsets of H Definition 1.1. A subset S of H is said to be compact if and only if it is closed and every sequence in S has a convergent subsequence. S is said to be precompact if its closure is compact. Proposition 1.2. Here are some important properties of compact sets. 1. Every compact set is bounded. 2. Let S be bounded set. Then S si precompact if and only if every sequence has a convergent subsequence. 3. Let H be finite dimensional. Every closed, bounded subset of H is compact. 4. In an infinite dimensional space, closed and bounded is not enough. Proof. Properties 2 and 3 are left to the reader. For property 1, assume that S is an unbounded compact set. Since S is unbounded, we may select a sequence {v n } n=1 from S such that v n as n. Since S is compact, this sequence will have a convergent subsequence, say {v n } =1, which still be unbounded.(why?) Let v = lim v n. Thus, for ε = 1 there is a positive integer K for which v v n < 1 for all K. By the triangle inequality, v n v + 1. Now, the right side is bounded, but the left side isn t, since v n as. This is a contradiction, so S must be bounded. For property 4, let S = {f H: f 1}. Every o.n. basis {φ n } n=1 is in S. However, for such a basis φ m φ n = 2, n m. Again, this means there are no Cauchy subsequences in {φ n } n=1, and consequently, no convergent subsequences. Thus, S is not compact. 1

2 2 Compact Operators Definition 2.1. Let K : H H be linear. K is said to be compact if and only if K maps bounded sets into precompact sets. Equivalently, K is compact if and only if for every bounded sequence {v n } n=1 in H the sequence {Kv n } n=1 has a convergent subsequence. We denote the set of compact operators on H by C(H). Proposition 2.2. If K C(H), then K is bounded i.e., C(H) B(H). In addition, C(H) is a subspace of B(H). Proof. We leave this as an exercise for the reader. We now turn to giving some examples of compact operators. We start with the finite-ran operators. If the range of a bounded operator K is finite dimensional, then we say that K is a finite-ran operator. Proposition 2.3. Every finite-ran operator K is compact. Proof. The range of K is finite dimensional, so every bounded subset of the range is precompact. Let S {f H: f C}, where C is fixed. Note that the range of K restricted to S is also bounded: Kf K op f C K op. Thus, K maps a bounded set S into a bounded subset of a finite dimensional subspace of H, which is itself precompact. Hence, K is thus compact. To describe K explicitly, let {φ } n =1 be a basis for R(K). Then, Kf = n =1 a φ. We want to see how the a s depend on f. Consider Kf, φ j = f, K φ j = n =1 a φ, φ j. Next let ψ j = K φ j, so that f, K φ j = f, ψ j. Because {φ } n =1 is a basis, it is linear independent. Hence, the Gram matrix G j, = φ, φ j is invertible, and so we can solve the system of equations f, ψ j = n =1 G j,a. Doing so yields a = n j=1 (G 1 ),j f, ψ j. The a s are obviously linear in f. Of course, a different basis will give a different representation. Let H = L 2 [0, 1]. A particularly important set of finite ran operators in C(H) are ones given by finite ran or degenerate ernels, (x, y) = n =1 φ (x)ψ (y), where the functions involved are in L 2. The operator is then Kf(x) = 1 0 (x, y)f(y)dy. In the example that we did for resolvents, the ernel was (x, y) = xy 2, and the operator was Ku(x) = 1 0 (x, y)u(y)dy. Later, we will show that the Hilbert-Schmidt ernels also yield compact operators. Before, we do so, we will discuss a few more properties of compact operators. 2

3 Lemma 2.4. Let {φ n } n=1 be an o.n. set in H and let K C(H). Then, lim n Kφ n = 0. Proof. Suppose not. Then we may select a subsequence {φ m } of {φ n } n=1 for which Kφ m α > 0 for all m. Because K is compact, we can also select a subsequence {φ } of {φ m } for which {Kφ } is convergent to ψ H. Now, {φ } being a subsequence of {φ m } implies that Kφ α > 0. Taing the limit in this inequality yields ψ α > 0. Next, note that lim Kφ, ψ = ψ 2. However, lim Kφ, ψ = lim φ, K ψ = 0, by Bessel s inequality. Thus, ψ 2 = 0, which is a contradiction. This lemma is a special case of a more general result. We say that a sequence {f n } is wealy convergent to a f H if and only if for all g H we have lim n f n, g = f, g. For example, the o.n. set in the lemma wealy converges to 0. Proposition 2.5. Let {f n } wealy converge to f H. If K C(H), then lim n Kf n = Kf. That is, K maps wealy convergent sequences into strongly convergent ones. Proof. The proof is similar to that of Lemma 2.4. See exercise 4 in assignment 9. We remar that the converse is true, too. This leads to an alternative characterization of compact operators: K is compact if and only if K maps wealy convergent sequences into strongly convergent ones. See the boo Functional Analysis, by F. Riesz and B. Sz.-Nagy. Our next result is one of the most important theorems in the theory of compact operators. Theorem 2.6. C(H) is a closed subspace of B(H). Proof. Suppose that {K n } n=1 is a sequence in C(H) that converges to K B(H), in the operator norm. We want to show that K is compact. Assume the {v } is a bounded sequence in H, with v C for all. Compactness will follow if we can prove that {Kv } has a convergent subsequence. The technique for doing this is often called a diagonalization argument. We start with the full sequence and form {K 1 v }. Since K 1 is compact, we can select a subsequence {v (1) } such that {K 1v (1) } is convergent. We may carry out the same procedure with {K 2 v (1) }, selecting a subsequence of {K 2v (1) } that is convergent. Call it {v (2) }. Since this is a subsequence of {v(1) }, {K 1 v (2) } is convergent. Continuing in this way, we construct subsequences 3

4 {v (j) } for which {K mv (j) } is convergent for all 1 m j. Next, we let }, the diagonal sequence. This is a subsequence of all of the {u j := v (j) j {v (j) } s. Consequently, for n fixed, {K nu j } j=1 will be convergent. To finish up, we will use an up, over, and around argument. Note that for all l, m, Ku l Ku m Ku l K n u l + K n u l K n u m + K n u m Ku m Since Ku l K n u l K K n op u l 2C K K n op and, similarly, Ku m K n u m 2C K K n op, so Ku l Ku m 4C K K n op + K n u l K n u m. Let ε > 0. First choose N such that for n N, K K n op < ε/(8c). Fix n. Because {K n u l } is convergent, it is Cauchy. Choose N so large that K n u l K n u m < ε/2 for all l, m N. Putting these two together yields Ku l K n u l ε, provided l, m N. Thus {Ku l } is Cauchy and therefore convergent. Corollary 2.7. Hilbert-Schmidt operators are compact. Proof. Let H = L 2 [0, 1] and suppose (x, y) L 2 (R), R = [0, 1] [0, 1]. The associated Hilbert-Schmidt operator is Ku = 1 0 (x, y)u(y)dy. Let {φ n } n=1 be an o.n. basis for L2 [0, 1]. With a little wor, one can show that {φ n (x)φ m (y)} n,m=1 is an o.n. basis1 for L 2 (R). Also, from example 2 in the notes on Bounded Operators & Closed Subspaces, we have that K op L 2 (R). Expand (x, y) in the o.n. basis {φ n (x)φ m (y)} n,m=1 : (x, y) = n,m=1 α m,n φ n (x)φ m (y), α m,n = (x, y), φ n (x)φ m (y) L 2 (R) Next, let N (x, y) = N n,m=1 α m,nφ n (x)φ m (y) and also K N be the finite ran operator K N u(x) = 1 0 N(x, y)u(y)dy. By Parseval s theorem, we have that N 2 L 2 (R) = n,m=n+1 α m,n 2 and by example 2 mentioned above, K K N 2 op N 2 L 2 (R), so K K N 2 op n,m=n+1 α m,n 2 Because the series on the right above converges to 0 as N, we have lim N K K N = 0. Thus K is the limit in B(L 2 [0, 1]) of finite ran operators, which are compact. By the theorem above, K is also compact. 1 See Keener, Theorem 3.5 4

5 We now turn to some of the algebraic properties of C(H). Proposition 2.8. Let K C(H) and let L B(H). Then both KL and LK are in C(H). Proof. Let {v } be a bounded sequence in H. Since L is bounded, the sequence {Lv } is also bounded. Because K is compact, we may find a subsequence of {KLv } that is convergent, so KL C(H). Next, again assuming {v } is a bounded sequence in H, we may extract a convergent subsequence from {Kv }, which, with a slight abuse of notation, we will denote by {Kv j }. Because L is bounded, it is also continuous. Thus {LKv j } is convergent. It follows that LK is compact. Proposition 2.9. K is compact if and only if K is compact. Proof. Because K is compact, it is bounded and so is its adjoint K, in fact K op = K op. By Proposition 2.8, we thus have that KK is compact. It follows that if {u n } be a bounded sequence in H, then we may extract a subsequence {u j } such that the sequence {KK v j } is convergent. This of course means that this sequence is also Cauchy. Note that KK (v j v ), v j v = K (v j v ), K (v j v ) = K (v j v ) 2. From and the fact that {v j } is bounded, we see that KK (v j v ), v j v v j v KK (v j v ) C KK (v j v ). Thus, K (v j v ) 2 C KK (v j v ) Since {KK v j } is Cauchy, for every ε > 0, we can find N such that whenever j, N, KK (v j v ) < ε 2 /C. It follows that K (v j v ) < ε, if j, N. This implies that {K v j } is Cauchy and therefore convergent. We want to put this in more algebraic language. Taing L to be compact in Proposition 2.8, we have that the product of two compact operators is compact. Since C(H) is already a subspace, this implies that it is an algebra. Moreover, by taing L to be just a bounded operator, we have that C(H) is a two-sided ideal in the algebra B(H). Since K being compact implies K is compact, C(H) is closed under the operation of taing adjoints; thus, C(H) is a -ideal. Finally, by Theorem 2.6, we have that C(H) is a closed subspace of B(H). We summarize these results as follows. Theorem C(H) is a closed, two-sided, -ideal in B(H). 5

6 We remar that a closed *-algebra in B(H) is called a C*-algebra. So, C(H) is a C*-algebra that is also a two-sided ideal in B(H). Previous: Example of the Fredholm alternatve and resolvent Next: the closed range theorem 6

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