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Functional calculus for a multiplication operator

Example

Assume AC. Let $H = L^{2} (0, 1)$ with Lebesgue measure and let $T = M_{t}$ be multiplication by the coordinate, $(M_{t} f) (t) = t f (t)$ . Then $σ (T) = [0, 1]$ and $f (T) = M_{f}$ for every continuous $f$ on $[0, 1]$ .

Facts & Assumptions

[A1]

Complex $L^{2} (0, 1)$ with $⟨ f, g ⟩ = \int_{0}^{1} f \overline{g}$ is a Hilbert space, and $∥ f ∥_{2}^{2} = \int_{0}^{1} ∣ f ∣^{2}$ ( $L^{2}$ with the integral pairing is a Hilbert space, The space $L^{p} (μ)$ as the quotient by null functions, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ).

[A2]

Endpoints have Lebesgue measure zero by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, so continuous-function integrals on $(0, 1)$ agree with those on $[0, 1]$ . A continuous real function on $[0, 1]$ is Lebesgue integrable with the Lebesgue integral equal to its Riemann integral, primitives of continuous functions evaluate definite integrals, and the substitution rule and power rule give the polynomial integrals used below (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term).

[A3]

$z \in ρ (T)$ exactly when $z I - T$ is bijective with bounded inverse; a self-adjoint operator $T$ is normal, the defining adjoint pairing characterizes self-adjointness (Spectrum and resolvent of a bounded operator, Self-adjoint, positive, unitary and normal operators, The Hilbert-space adjoint of a bounded operator).

[A4]

For the self-adjoint calculus of a bounded self-adjoint operator there is a unique isometric unital star-homomorphism $C (σ (T)) \to B (H)$ with $z \mapsto T$ and range $C^{*} (I, T)$ ; complex polynomials are uniformly dense in $C ([0, 1])$ , since they form a unital point-separating self-adjoint algebra on the compact Hausdorff interval (Heine-Borel by bisection: every closed bounded interval $[a, b]$ is compact, Distinct points of a metric space have disjoint balls around them) (Continuous functional calculus for bounded self adjoint operators, Complex Stone–Weierstrass dichotomy for separating self-adjoint algebras; the unital case is dense).

[A5]

AC is the hypothesis of the calculus and Hilbert-space suppliers (The Axiom of Choice).

Verification

technique · direct

Given: The Hilbert space $H = L^{2} (0, 1)$ and the operator $M_{t}$ of multiplication by $t$ .

1.1

$M_{t}$ is well defined on a.e. classes, linear, bounded with $∥ M_{t} f ∥_{2} \leq ∥ f ∥_{2}$ , and self-adjoint with $M_{t}^{2} = M_{t^{2}}$ ; self-adjointness follows from $⟨ M_{t} u, v ⟩ = \int t u \overline{v} = ⟨ u, M_{t} v ⟩$ . For $λ \in C$ one has $M_{t} - λ I = M_{t - λ}$ .

A1A3

2.1

If $z \notin [0, 1]$ then $g (t) = 1 / (z - t)$ is continuous on $[0, 1]$ with $∣ g ∣ \leq 1 / dist (z, [0, 1])$ , so $M_{g}$ is bounded and $M_{g} (z I - M_{t}) = (z I - M_{t}) M_{g} = I$ ; hence $z \in ρ (M_{t})$ .

step 1.1A1A3

2.2

If $z \in [0, 1]$ and $0 < δ \leq 1$ , let $f_{δ} (t) : = (1 - ∣ t - z ∣ / δ)_{+}$ on $[0, 1]$ , a continuous function vanishing wherever $∣ t - z ∣ \geq δ$ , and put $φ_{δ} : = c_{δ} f_{δ}$ with $c_{δ}$ chosen so that $∥ φ_{δ} ∥_{2} = 1$ . For $z = 0$ or $z = 1$ take $0 < δ \leq 1$ ; for $0 < z < 1$ take $0 < δ \leq \min (z, 1 - z)$ . Respectively, the polynomial integrals give $\int f_{δ}^{2} = δ / 3$ or $2 δ / 3$ and $\int (t - z)^{2} f_{δ}^{2} = δ^{3} / 30$ or $δ^{3} / 15$ respectively, so the ratio of the second integral to the first is $δ^{2} / 10$ in both cases. Thus $∥ (M_{t} - z I) φ_{δ} ∥_{2}^{2} = δ^{2} / 10$ . A bounded inverse $B$ would imply $1 \leq ∥ B ∥ δ / \sqrt{10}$ for all such $δ > 0$ , which is impossible; hence $M_{t} - z I$ is not bounded below and $z \in σ (M_{t})$ .

step 1.1A2A3

3.1

Steps 2.1 and 2.2 give $σ (M_{t}) = [0, 1]$ , and $M_{t}$ is self-adjoint, so the continuous functional calculus for $M_{t}$ is defined on $C ([0, 1])$ .

step 1.1step 2.1step 2.2A3

4.1

For a continuous $f$ the multiplier $M_{f}$ is well defined on a.e. classes, linear, and bounded with $∥ M_{f} ∥ \leq ∥ f ∥_{\infty}$ , by $\int ∣ f u ∣^{2} \leq ∥ f ∥_{\infty}^{2} \int ∣ u ∣^{2}$ . Multiplication of multipliers and the unital algebra property of the supplied calculus give $p (M_{t}) = M_{p}$ for every complex polynomial $p$ . For each $ε > 0$ , [A4] supplies a polynomial $p$ with $∥ f - p ∥_{\infty} < ε$ . The calculus isometry and the multiplier bound give $∥ f (M_{t}) - M_{f} ∥ \leq ∥ f (M_{t}) - p (M_{t}) ∥ + ∥ M_{p} - M_{f} ∥ \leq 2 ∥ f - p ∥_{\infty} < 2 ε$ . Since this holds for every $ε > 0$ , $f (M_{t}) = M_{f}$ . In particular the isometry and range conclusions already established for the calculus give $∥ M_{f} ∥ = ∥ f ∥_{\infty}$ and range $C^{*} (I, M_{t})$ .

step 3.1A1A4

5.1

Therefore $σ (M_{t}) = [0, 1]$ and the calculus of $M_{t}$ is multiplication by $f$ .

step 3.1step 4.1A5∎

Depends on

Used by

Continuous calculus does not contain discontinuous spectral projections Counterexample

Dependency tree · two levels

138 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3, printed pp.235–245 (standard reference, not scraped)
Dana P. Williams, Lecture Notes on the Spectral Theorem, §4, pp.10–15 (standard reference, not scraped)