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A compact operator can have nondense range

Statement refuted

The false general statement is: every compact operator has dense range. In fact on the Hilbert space $ℓ^{2} (N, K)$ , $K = R$ or $C$ (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ), the operator

$P (x) : = x_{1} e_{1} (x = (x_{n})_{n \in N})$

is a nonzero compact operator (Compact linear operator) whose range is the closed one-dimensional subspace $span {e_{1}}$ , a proper subspace; hence the range is not dense.

Facts & Assumptions

[A1]

In $ℓ^{2} (N, K)$ one has $∥ a ∥_{2}^{2} = \sum_{n \in N} ∣ a_{n} ∣^{2}$ and $⟨ a, b ⟩ = \sum_{n \in N} a_{n} \overline{b_{n}}$ ; the standard vectors satisfy $⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ ; and $∥ a + b ∥_{2}^{2} = ∥ a ∥_{2}^{2} + 2 Re ⟨ a, b ⟩ + ∥ b ∥_{2}^{2}$ for the inner-product norm (Square-summable families on an arbitrary index set and the space $ℓ^{2} (I)$ ).

[A2]

A bounded operator with finite-dimensional range is compact (Bounded finite rank operators are compact, Compact linear operator); a finite-dimensional linear subspace of a normed space is closed (A finite-dimensional normed subspace is closed, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear subspace of a vector space).

[A3]

For nonempty $A$ in a metric space, the closure is $\overline{A} = {y : d (y, A) = 0}$ (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, Open ball, closed ball and sphere in a metric space); $∥ P x ∥_{2} \leq ∥ P ∥ ∥ x ∥_{2}$ for bounded $P$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Counterexample

technique · direct

Given: $K \in {R, C}$ and the operator $P (x) = x_{1} e_{1}$ on $H : = ℓ^{2} (N, K)$ .

1.1A1A3

$P$ is well defined and linear with $∥ P x ∥_{2} = ∣ x_{1} ∣ \leq ∥ x ∥_{2}$ , because $∣ x_{1} ∣^{2} \leq \sum_{n} ∣ x_{n} ∣^{2}$ ; hence $P$ is a bounded operator with $∥ P ∥ \leq 1$ .

1.2A1

$P$ is nonzero: $P (e_{1}) = e_{1} \neq 0$ , and $∥ e_{1} ∥_{2} = 1$ .

2.1step 1.1A2

The range of $P$ is $span {e_{1}}$ : every $P x = x_{1} e_{1}$ lies in it, and conversely $c e_{1} = P (c e_{1})$ ; this span is one dimensional, hence finite dimensional, and closed by [A2].

3.1step 1.1step 2.1A2

$P$ is compact by [A2], since it is bounded with finite-dimensional range.

3.2step 2.1A1A3

The range is not dense: for every scalar $c \in K$ one has $∥ e_{2} - c e_{1} ∥_{2}^{2} = ∥ e_{2} ∥_{2}^{2} + ∣ c ∣^{2} ∥ e_{1} ∥_{2}^{2} = 1 + ∣ c ∣^{2} \geq 1$ by the orthogonality $⟨ e_{1}, e_{2} ⟩ = 0$ , so $dist (e_{2}, span {e_{1}}) \geq 1$ and $e_{2} \notin \overline{span {e_{1}}}$ by [A3]; since the range is closed by [step 2.1], it is therefore a proper closed subspace.

4.1step 3.1step 1.2step 3.2∎

So $P$ is a nonzero compact operator with nondense range, which refutes the general statement; the claims are [step 3.1], [step 1.2] and [step 3.2].

Depends on

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Sources

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.1, finite-rank examples of compact operators (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis — §4.2 p.184, Example 4.23 (standard reference, not scraped)