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Rescaling finite-measure kernel operators to probability spaces

Statement

Assume AC. Let $(X, μ)$ and $(Y, ν)$ be finite measure spaces with masses $0 < M_{X} : = μ (X) < \infty$ and $0 < M_{Y} : = ν (Y) < \infty$ . Put $d \hat{μ} = d μ / M_{X}$ and $d \hat{ν} = d ν / M_{Y}$ . For $a \in L^{2} (X \times Y, μ \times ν)$ , let $T_{a} : L^{2} (Y, ν) \to L^{2} (X, μ)$ be its square-integrable-kernel operator and let $\hat{T}$ be the operator with kernel $\hat{a} : = \sqrt{M_{X} M_{Y}} a$ against $\hat{ν}$ . The maps $U_{X} g = \sqrt{M_{X}} g$ and $U_{Y} h = \sqrt{M_{Y}} h$ are unitary and $U_{X} T_{a} U_{Y}^{- 1} = \hat{T} .$ Moreover $∥ \hat{a} ∥_{L^{2} (\hat{μ} \times \hat{ν})} = ∥ a ∥_{L^{2} (μ \times ν)}$ , so the operator norms and kernel $L^{2}$ norms agree. If either mass is zero, the corresponding $L^{2}$ space and kernel operator are zero.

For $X = Y = D$ with Lebesgue mass $M > 0$ , the normalized kernel is $\hat{a} = M a$ and the measure is $d \hat{x} = d x / M$ .

Facts & Assumptions

Given: AC, finite measure spaces $(X, μ)$ and $(Y, ν)$ , their positive masses, and a square-integrable kernel class $a$ .

[F1]

The complex $L^{2}$ norm is the quotient norm induced by $\int ∣ f ∣^{2}$ ; Cauchy–Schwarz and representative independence hold for this norm (The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions, The complex $L^{2}$ pairing is well-defined and satisfies Cauchy–Schwarz).

[F2]

Product integration and section integrals are valid almost everywhere on the completed product under Countable Choice, which AC supplies (The completed product measure, Tonelli and Fubini for the completed product, with only almost-everywhere section measurability, The Axiom of Choice).

Proof

Given: AC, the spaces, $M_{X}, M_{Y} > 0$ , and $a \in L^{2} (μ \times ν)$ .

1.1givenF1algebra

For every $g \in L^{2} (X, μ)$ , scaling the integral first for nonnegative simple functions and then by the definition of the nonnegative integral gives $∥ U_{X} g ∥_{L^{2} (\hat{μ})}^{2} = \int M_{X} ∣ g ∣^{2} d \hat{μ} = \int ∣ g ∣^{2} d μ = ∥ g ∥_{L^{2} (μ)}^{2}$ . Its inverse is $h \mapsto h / \sqrt{M_{X}}$ , so $U_{X}$ is unitary; the same calculation proves the assertion for $U_{Y}$ .

2.1givenF1F2step 1.1algebra

For $g \in L^{2} (Y, \hat{ν})$ , $U_{Y}^{- 1} g = g / \sqrt{M_{Y}}$ . Using $d ν = M_{Y} d \hat{ν}$ in the integral defining $T_{a}$ and then multiplying the output by $\sqrt{M_{X}}$ gives $U_{X} T_{a} U_{Y}^{- 1} g = \int_{Y} \sqrt{M_{X} M_{Y}} a (x, y) g (y) d \hat{ν} (y)$ . The equality holds almost everywhere by [F2], and is independent of representatives by [F1].

3.1F1F2step 2.1algebra∎

Scaling both factors of the product measure gives $∥ \hat{a} ∥_{2}^{2} = M_{X} M_{Y} \int ∣ a ∣^{2} d (\hat{μ} \times \hat{ν}) = \int ∣ a ∣^{2} d (μ \times ν)$ . Unitary equivalence in step 2.1 therefore preserves the operator norm and the kernel $L^{2}$ norm. If either mass is zero, every $L^{2}$ class on that factor is zero, so the integral operator is zero.

Depends on

Used by

The volume integral scattering operator is compact Theorem

Dependency tree · two levels

41 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

Andreas Neubauer and Ronny Ramlau, Lecture Notes: Integral Equations and Boundary Value Problems (standard reference, not scraped)
Otmar Scherzer and Leonidas Mindrinos, Inverse Problems: Lecture Notes, Sommersemester 2015, §5.2.6 (standard reference, not scraped)