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Compactness is not preserved by strong operator limits

Statement refuted

The false general statement is: the strong operator limit of a sequence of compact operators is compact. On $H : = ℓ^{2} (N, K)$ , with $K = R$ or $C$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ), let

$P_{N} x : = \sum_{n < N} x_{n} e_{n} (N \in N)$

be the $N$ -th coordinate projection. Each $P_{N}$ is compact (Compact linear operator), the sequence converges to the identity in the strong operator topology (Strong and weak operator topologies), $∥ I - P_{N} ∥ = 1$ for every $N$ , and the identity is not compact.

Facts & Assumptions

[A1]

In $ℓ^{2} (N, K)$ one has $∥ x ∥_{2}^{2} = \sum_{n \in N} ∣ x_{n} ∣^{2}$ with the finite-subset meaning of the sum, $⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ , and the coordinate bound $∣ x_{n} ∣ \leq ∥ x ∥_{2}$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[A2]

A bounded finite-rank operator is compact (Bounded finite rank operators are compact, A bounded linear operator between normed spaces); the identity of $ℓ^{2} (N, K)$ is not compact (Identity is compact iff the space is finite dimensional).

[A3]

Strong operator convergence means $∥ (T_{j} - T) x ∥ \to 0$ for every fixed $x$ (Strong and weak operator topologies, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ); the operator norm satisfies $∥ S ∥ \geq ∥ S x ∥ / ∥ x ∥$ for $x \neq 0$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Open ball, closed ball and sphere in a metric space).

Counterexample

technique · direct

Given: $K \in {R, C}$ , the Hilbert space $H = ℓ^{2} (N, K)$ , and the coordinate projections $P_{N} x = \sum_{n < N} x_{n} e_{n}$ .

1.1

Each $P_{N}$ is linear with $∥ P_{N} x ∥_{2}^{2} = \sum_{n < N} ∣ x_{n} ∣^{2} \leq ∥ x ∥_{2}^{2}$ , hence bounded with $∥ P_{N} ∥ \leq 1$ , and its range is contained in the linear span of $e_{0}, \dots, e_{N - 1}$ , a finite-dimensional subspace; so $P_{N}$ is compact by [A2].

A1A2

1.2

For every $x \in H$ one has $∥ x - P_{N} x ∥_{2}^{2} = \sum_{n \geq N} ∣ x_{n} ∣^{2} \to 0$ : given $ε > 0$ , the convergence of the nonnegative sum gives a finite $F \subseteq N$ with $\sum_{n \notin F} ∣ x_{n} ∣^{2} < ε$ , and for $N > \max F$ the tail ${n : n \geq N}$ is contained in $N ∖ F$ , so the tail sum is below $ε$ .

A1A3

1.3

$∥ I - P_{N} ∥ = 1$ for every $N$ : the upper bound $∥ (I - P_{N}) x ∥_{2}^{2} = \sum_{n \geq N} ∣ x_{n} ∣^{2} \leq ∥ x ∥_{2}^{2}$ follows from [A1], and the lower bound holds because $(I - P_{N}) e_{N} = e_{N}$ with $∥ e_{N} ∥_{2} = 1$ .

A1A3

2.1

By [step 1.2] the sequence $(P_{N})$ converges to $I$ in the strong operator topology; by [step 1.3] the convergence is not in operator norm; and the limit $I$ is not compact by [A2].

step 1.1step 1.2step 1.3A2A3

3.1

With [step 1.1] and [step 1.2] this shows that a strong operator limit of compact operators need not be compact, refuting the general statement.

step 1.1step 2.1∎

Depends on

Used by

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis — §4.2, strong limits of finite-rank projections (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.1, compactness is not preserved by strong limits (standard reference, not scraped)