Questions: Compact Operators

Question 1

In an infinite-dimensional Hilbert space, the identity operator I (where I(x) = x for all x) is:

• A. Compact, because it maps the closed unit ball to itself, and the unit ball is a compact set
• B. Not compact, because in infinite dimensions there exist bounded sequences in the unit ball with no convergent subsequence under I
• C. Compact, because every bounded operator on a Hilbert space is compact
• D. Not bounded, because it maps unbounded sequences to unbounded sequences

Answer: (B)

The identity operator is bounded (||I|| = 1) but not compact. In infinite dimensions, the closed unit ball is not compact — there exist sequences of unit vectors with no convergent subsequence (e.g., an orthonormal sequence in a Hilbert space, which has no convergent subsequence by Bessel's inequality). Since I(xₙ) = xₙ, such a sequence provides a bounded sequence whose image has no convergent subsequence, violating the definition of compactness. This is the canonical example showing that boundedness does not imply compactness in infinite dimensions — a fundamental contrast with finite-dimensional linear algebra.

Question 2

A compact operator T on an infinite-dimensional Banach space has spectral behavior resembling a finite matrix in that:

• A. Its spectrum consists of exactly finitely many isolated eigenvalues
• B. Its spectrum outside {0} consists of discrete eigenvalues — either finitely many, or forming a sequence converging to 0 — with no continuous spectrum possible outside {0}
• C. All its eigenvalues are real and non-negative, as in a symmetric matrix
• D. It can always be diagonalized in an orthonormal eigenbasis, just like a symmetric matrix

Answer: (B)

For a compact operator, the spectrum outside {0} consists entirely of eigenvalues — either finitely many, or a countably infinite sequence converging to 0. This is exactly the discrete structure of a finite matrix's eigenvalue decomposition. A general bounded operator (like a multiplication operator or the identity) can have continuous spectrum with no eigenvalues at all. The discreteness of the compact operator's spectrum outside {0} is what makes it tractable: you can analyze it through eigenvalues and eigenvectors, in close analogy with matrix diagonalization, rather than needing the full machinery of spectral theory for general operators.

Question 3

In finite-dimensional vector spaces, every bounded linear operator is automatically compact.

• T. True
• F. False

Answer: True

In ℝⁿ (or any finite-dimensional normed space), a set is compact if and only if it is closed and bounded (Heine-Borel). Every bounded linear operator maps bounded sets to bounded sets. In finite dimensions, bounded sets are relatively compact — their closures are compact. Therefore, every bounded linear operator maps bounded sets to relatively compact sets, satisfying the definition of a compact operator. This is why compactness of operators is trivial in finite dimensions but non-trivial in infinite dimensions, where bounded sets are not generally relatively compact.

Question 4

A compact operator T: X → Y maps most bounded sequence (xₙ) to a sequence (Txₙ) that converges in Y.

• T. True
• F. False

Answer: False

The definition of compactness guarantees only that every bounded sequence has a convergent SUBSEQUENCE in the image — not that the full sequence converges. The operator extracts a convergent subsequence from any bounded sequence, but the original sequence need not converge. This is the standard compactness condition: relatively compact means the closure is compact, which is equivalent to every sequence having a convergent subsequence. Confusing 'has a convergent subsequence' with 'converges' is a common error in functional analysis.

Question 5

Why does the identity operator on an infinite-dimensional Hilbert space fail to be compact, and what does this failure reveal about the difference between finite and infinite dimensions?

Model answer: The identity operator fails to be compact because the closed unit ball in an infinite-dimensional Hilbert space is not compact. Concretely, an orthonormal sequence (e₁, e₂, e₃, ...) lies in the unit ball, but I(eₙ) = eₙ has no convergent subsequence — by Bessel's inequality, ||eₙ − eₘ|| = √2 for all n ≠ m, so no subsequence is Cauchy. In finite dimensions, Heine-Borel guarantees that every closed bounded set is compact, so the same argument fails: any bounded sequence in ℝⁿ automatically has a convergent subsequence, making the identity automatically compact. The failure in infinite dimensions reveals that the interplay between boundedness and compactness breaks down fundamentally — being bounded no longer implies being 'small enough' to have compact image.

This example is the gateway to understanding functional analysis as a distinct discipline from linear algebra. Finite-dimensional intuitions about bounded operators, closed sets, and convergence all require re-examination in infinite dimensions. Compact operators are the class that preserve finite-dimensional behavior, and understanding why the identity fails to be compact — and what property it lacks — is the key to understanding what compactness is actually measuring.