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MATH 675 – HOMEWORK 8
SOLUTIONS
Exercise 1. Prove that a finite rank operator on a Hilbert space is compact. Recall that a A ∈ L(H, H) has finite rank if ran(A) is finite dimensional.
Let {ϕk}Nk=1 be an orthonormal basis for ran(A), and let {xn} be a sequence in H with ‖xn‖ ≤ 1. Then for each n, Axn =
∑N k=1〈Axn, ϕk〉^ ϕk.^ Letting 〈Axn, ϕk〉 = akn we see that for each k the sequence {akn}∞ n=1 is bounded since
|akn| = |〈Axn, ϕk〉| ≤ ‖Axn‖‖ϕk‖ ≤ ‖A‖‖xn‖‖ϕk‖ ≤ ‖A‖.
Since every bounded sequence of numbers has a convergent subsequence, we let {xn 1 (m)}∞ m=1 ⊆ {xm}∞ m=1 satisfy limm a^1 n 1 (m) = a^1. Then choose a subsequence {xn 2 (m)}∞ m=1 ⊆ {xn 1 (m)}∞ m=1 such that limm a^2 n 2 (m) = a^2. Continue in this fashion and arrive at a subsequence {xnN (m)}∞ m=1 with the property that limm aknN (m) = ak for all 1 ≤ k ≤ N. Finally note that defining y =
∑N k=1 a
k (^) ϕk we have that
‖AxnN (m) − y‖^2 =
∑^ N
k=
|aknN (m) − ak|^2 → 0
as m → ∞. Hence {Axn}∞ n=1 has a convergent subsequence and A is compact.
Exercise 2.
(a) Prove that if {ϕn}∞ n=1 is an orthonormal system in the Hilbert space H, and if n 6 = m, then ‖ϕn − ϕm‖ =
(b) Prove that the identity operator on an infinite-dimensional Hilbert space H is not compact.
To see part (a), note that using orthogonality and the fact that ‖ϕk‖ = 1 for all k,
‖ϕn −ϕm‖^2 = 〈ϕn −ϕm, ϕn −ϕm〉 = 〈ϕn, ϕn〉−〈ϕn, ϕm〉−〈ϕm, ϕn〉+〈ϕm, ϕm〉 = 2.
To see part (b), note that {ϕk} is a bounded set and that its image under the identity operator is itself. However by part (a) we see that no subsequence of {ϕk} can be Cauchy because if 0 < <
2 is chosen then no matter how large n and m are we cannot have ‖ϕn − ϕm‖ < as long as n 6 = m. Hence the sequence Iϕk has no convergent subsequence and I is not compact.
Exercise 3. Kolmogorov, p. 251, Problem 2. (The first part only, i.e., prove that A is a compact operator, or in Kolmogorov’s words, a completely continuous operator.)
The idea will be to show that A can be written as the limit in the operator norm of a sequence of finite-rank operators. For each N , define the operator AN on l^2 by AN (x 1 , x 2 , x 3 ,.. .) = (x 1 ,
x 2 ,
x 3 ,... , 2 −N^ +1xN , 0 , 0 ,.. .).
Clearly AN has finite rank and note that for each x = (xn)n = 1∞^ ∈ l^2 ,
‖(A − AN )x‖^2 = ‖(0,... , 0 , 2 −N^ xN +1, 2 −N^ −^1 xN +2,.. .)‖^2
=
∑^ ∞
n=N +
2 −^2 n+2|xn|^2 ≤ 2 −^2 N
∑^ ∞
n=N +
|xn|^2 ≤ 2 −^2 N^ ‖x‖^2.
Therefore, ‖A − AN ‖ ≤ 2 −N^ and so limN ‖A − AN ‖ = 0.
Exercise 4. Prove that if A ∈ L(H, H) is compact, and B ∈ L(H, H) is arbitrary, then AB ∈ L(H, H) is compact.
Suppose that {xn} is a sequence such that ‖xn‖ ≤ 1. Since B is bounded, {Bxn}∞ n=1 satisfies ‖Bxn‖ ≤ ‖B‖‖xn‖ ≤ ‖B‖. Taking yn = Bxn/‖B‖ we have that ‖yn‖ ≤ 1 and since A is compact that {Ayn} has a convergent subsequence. But then also {‖B‖Ayn} = {‖B‖ABxn/‖B‖} = {ABxn} has a convergent sub- sequence. Hence AB is compact.
Exercise 5. Prove the following facts about orthogonal projectors. Assume that P is the orthogonal projector onto the closed subspace M of a Hilbert space H.
(1) P 2 = P. An operator which satisfies this criterion is said to be idempotent.
(2) I − P is the orthogonal projector onto M ⊥.
(3) P is a bounded linear transformation on H with ‖P x‖ ≤ ‖x‖, and in fact, ‖P ‖ = 1.
Recall that P x is defined as follows. Given x ∈ H x can be written uniquely as a sum x = w + z where w ∈ M and z ∈ M ⊥. We define P x = w. To see (1), note that P 2 x = P w = w = P x since z = 0 in the unique decomposi- tion of w. To see (2), note that z = x − w = x − P x = (I − P )x so that I − P is the orthogonal projector onto M ⊥. To see (3) note that by the orthogonality of w and z, ‖x‖^2 = ‖w‖^2 + ‖z‖^2. Therefore, ‖P x‖^2 = ‖w‖^2 ≤ ‖x‖^2 and P is bounded with ‖P ‖ ≤ 1. Also since P x = x for all x ∈ M it follows that ‖P x‖ = ‖x‖ for some x 6 = 0. Hence ‖P ‖ = 1.
