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Spectrum of a multiplication operator

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $(X, A, μ)$ be a nonzero $σ$ -finite measure space and let $m : X \to C$ be measurable and essentially bounded, with

$∥ m ∥_{\infty} : = \inf {M \geq 0 : ∣ m (x) ∣ \leq M for almost every x} .$

On the complex Hilbert space $L^{2} (μ)$ (The space $L^{p} (μ)$ as the quotient by null functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ) the multiplication operator $M_{m} [f] : = [m f]$ is bounded with $∥ M_{m} ∥ \leq ∥ m ∥_{\infty}$ , and

$σ (M_{m}) = essran (m) : = {λ \in C : μ ({∣ m - λ ∣ < ε}) > 0 for every ε > 0},$

the spectrum taken in $B (L^{2} (μ))$ (Bounded operators form a Banach algebra, noncommutative in dimension at least two, Spectrum and resolvent set in a Banach algebra). If $μ$ is the zero measure then $L^{2} (μ) = {0}$ is not a nonzero algebra and the spectral convention of this page does not apply.

Facts & Assumptions

Given: The Axiom of Choice, a nonzero $σ$ -finite measure space $(X, A, μ)$ , an essentially bounded measurable $m$ , and the operator $M_{m}$ on $L^{2} (μ)$ .

[L1]

$L^{2} (μ)$ is a complex Banach space of almost-everywhere equivalence classes, with $∥ [f] ∥_{2} = ∥ f ∥_{2}$ ; convergence in norm and equality of classes are as in The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ and Complex completeness, density, and inner product: the consumer interface (The space $L^{p} (μ)$ as the quotient by null functions).

[L2]

Elements of $L^{2}$ are equivalence classes modulo a.e. equality; quotient operations are induced by pointwise operations (The space $L^{p} (μ)$ as the quotient by null functions). The multiplier estimate and composition identities will be proved on representatives below.

[L3]

If $T$ is invertible then $T$ is bounded below: $∥ x ∥ = ∥ T^{- 1} T x ∥ \leq ∥ T^{- 1} ∥ ∥ T x ∥$ ; and an operator that is not bounded below is not invertible (Spectrum and resolvent set in a Banach algebra).

[L4]

Under AC the bounded operators on a nonzero complex Banach space form a unital complex Banach algebra with composition as multiplication (Bounded operators form a Banach algebra, noncommutative in dimension at least two).

[L5]

Assume AC (The Axiom of Choice), which includes choice for families indexed by $N$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix a sigma-finite cover and replace it by its finite partial unions to obtain increasing measurable sets $E_{k}$ with finite measure and union $X$ .

Verification

technique · direct

1.1L1L2L4L5

Put $r = ∥ m ∥_{\infty}$ . Each set $N_{j} = {∣ m ∣ > r + 1 / j}$ is null: by the definition of the infimum there is an a.e. bound strictly below $r + 1 / j$ . A countable union of measurable null sets is null, by disjointifying the union and applying countable additivity. Outside $⋃_{j \geq 1} N_{j}$ , $∣ m ∣ \leq r$ . For any measurable $f$ of finite squared integral, the nonnegative integral therefore gives $\int ∣ m f ∣^{2} \leq r^{2} \int ∣ f ∣^{2}$ . Thus multiplication defines a bounded linear map $M_{m} [f] = [m f]$ , independent of representatives by [L2]. The same reasoning applies to every essentially bounded measurable symbol $h$ . If $h, k$ are such symbols, then on every class $M_{h} M_{k} [f] = [h (k f)] = [(h k) f] = M_{h k} [f]$ , and $M_{1} = I$ ; a general multiplier need not be the identity. Since the measure is nonzero, some $E_{k}$ from [L5] has positive measure (otherwise their union is null). Its indicator has nonzero finite $L^{2}$ norm. Hence $L^{2}$ is a nonzero complex Banach space by [L1], and [L4] supplies its operator algebra. AC is inherited from [L4] and supplies the Countable Choice required by [L1].

2.1step 1.1L2L4L3

If $λ \notin essran (m)$ , then there is $ε > 0$ with $μ ({∣ m - λ ∣ < ε}) = 0$ ; hence $∣ m - λ ∣ \geq ε$ almost everywhere and the measurable function $ϕ : = 1 / (m - λ)$ on ${∣ m - λ ∣ \geq ε}$ (arbitrary, say $0$ , on the null complement) is essentially bounded by $1 / ε$ . By step 1.1 its multiplication operator satisfies $M_{ϕ} (M_{m} - λ) = (M_{m} - λ) M_{ϕ} = 1$ on classes. Thus $M_{m} - λ I$ is invertible, and so is its negative $λ I - M_{m}$ , the shifted operator in [L3]; therefore $λ \in ρ (M_{m})$ .

2.2step 1.1L5algebra

If $λ \in essran (m)$ , then for every $n \geq 1$ the set $A_{n} : = {∣ m - λ ∣ < 1 / n}$ has positive measure; since $X = ⋃_{k} E_{k}$ with $μ (E_{k}) < \infty$ , some $A_{n} \cap E_{k}$ has positive measure, since otherwise their countable union $A_{n}$ would be null. Take the least such $k$ ; this deterministic choice uses no choice principle.

3.1step 2.2L3L1algebra

For this least $k$ the normalized indicator $f_{n} : = μ (A_{n} \cap E_{k})^{- 1 / 2} 1_{A_{n} \cap E_{k}}$ is a unit vector in $L^{2} (μ)$ , and $∥ (M_{m} - λ) f_{n} ∥_{2} \leq (1 / n) ∥ f_{n} ∥_{2} = 1 / n \to 0$ ; hence $M_{m} - λ$ is not bounded below, so by [L3] neither it nor its negative is invertible and $λ \in σ (M_{m})$ .

4.1step 1.1step 2.1step 3.1∎

Steps [step 2.1] and [step 3.1] together give both inclusions, so $σ (M_{m}) = essran (m)$ ; the boundedness assertion is [step 1.1].

Depends on

Used by

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Dependency tree · two levels

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Sources

Theo Bühler and Dietmar A. Salamon, Functional Analysis — Example 5.17 and §5.2.1, printed pp. 220–222 (standard reference, not scraped)