ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Adjoints of shifts, multiplication and integral operators

Example

Assume the Axiom of Countable Choice. Then:

On $ℓ^{2} (N; K)$ the right shift $S (x_{0}, x_{1}, x_{2}, \dots) = (0, x_{0}, x_{1}, \dots)$ has Hilbert adjoint the left shift $L (y_{0}, y_{1}, y_{2}, \dots) = (y_{1}, y_{2}, y_{3}, \dots)$ .
For $m \in L^{\infty} (X, μ; C)$ the multiplication operator $M_{m} [f] = [m f]$ on complex $L^{2} (μ)$ is bounded and $M_{m}^{*} = M_{\overline{m}}$ .
Let $(X, A, μ)$ and $(Y, B, ν)$ be $σ$ -finite measure spaces and let $k$ represent a class in $L^{2} (μ \times ν; C)$ . Then $(K f) (x) = \int_{Y} k (x, y) f (y) d ν (y)$ defines a bounded operator $K : L^{2} (ν) \to L^{2} (μ)$ independently of the representatives of $k$ and $f$ , and its adjoint is $K^{*} g (y) = \int_{X} \overline{k (x, y)} g (x) d μ (x)$ .

Facts & Assumptions

[A1]

The Hilbert adjoint of $T$ is the unique operator with $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ (The Hilbert-space adjoint of a bounded operator).

[A2]

$ℓ^{2}$ carries the first-variable-linear pairing $⟨ x, y ⟩ = \sum_{k} x_{k} \overline{y_{k}}$ and is a Hilbert space; the complex $L^{2}$ pairing is $⟨ f, g ⟩ = \int f \overline{g}$ with Cauchy–Schwarz $∣ ⟨ f, g ⟩ ∣ \leq ∥ f ∥_{2} ∥ g ∥_{2}$ (The standard inner products make K n, ell two and quotient L two Hilbert spaces, Complex completeness, density, and inner product: the consumer interface, Complex Lp classes and Euclidean test-function conventions).

[A3]

For $m \in L^{\infty}$ and $f \in L^{2}$ the product $m f$ lies in $L^{2}$ with $∥ m f ∥_{2} \leq ∥ m ∥_{\infty} ∥ f ∥_{2}$ (Complex Holder, Minkowski, and the quotient norm).

[A4]

On a $σ$ -finite product, Tonelli applies to nonnegative measurable functions and Fubini to $L^{1}$ functions, with the iterated integrals equal to the product integral (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product, Finite, sigma-finite, and semifinite measures).

[A5]

Countable Choice is the hypothesis under which the adjoint and the $L^{2}$ completions are available (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

technique · direct

Given: The spaces and operators of the statement.

1.1

For $x, y \in ℓ^{2}$ the series in $⟨ S x, y ⟩ = \sum_{n \geq 0} (S x)_{n} \overline{y_{n}} = \sum_{n \geq 1} x_{n - 1} \overline{y_{n}} = \sum_{k \geq 0} x_{k} \overline{y_{k + 1}} = ⟨ x, L y ⟩$ converge absolutely by Cauchy–Schwarz, so $S^{*} = L$ by uniqueness of the adjoint.

A1A2

1.2

The multiplication operator satisfies $∥ M_{m} f ∥_{2} \leq ∥ m ∥_{\infty} ∥ f ∥_{2}$ by [A3], and $⟨ M_{m} f, g ⟩ = \int m f \overline{g} = \int f \overline{\overline{m} g} = ⟨ f, M_{\overline{m}} g ⟩$ for all $f, g \in L^{2}$ , so $M_{m}$ is bounded with $M_{m}^{*} = M_{\overline{m}}$ .

A1A2A3

1.3

For $f \in L^{2} (ν)$ put $h (x) : = (\int_{Y} ∣ k (x, y) ∣^{2} d ν (y))^{1 / 2} \in [0, + \infty]$ . Wherever the section integral converges absolutely, the Cauchy–Schwarz inequality in the $y$ -variable gives $∣ \int_{Y} k (x, y) f (y) d ν (y) ∣ \leq \int_{Y} ∣ k (x, y) f (y) ∣ d ν (y) \leq h (x) ∥ f ∥_{2}$ ; by Tonelli [A4] applied to the nonnegative $(μ \times ν)$ -measurable function $∣ k ∣^{2}$ , the function $h^{2}$ is $μ$ -measurable with $\int_{X} h (x)^{2} d μ (x) = ∥ k ∥_{2}^{2} < + \infty$ , so $h \in L^{2} (μ)$ and $h (x) < + \infty$ for $μ$ -almost every $x$ . Hence $(K f) (x) = \int_{Y} k (x, y) f (y) d ν (y)$ is defined and finite for $μ$ -almost every $x$ , and it is $μ$ -measurable after zero extension: Tonelli [A4] makes the section integrals of the positive and negative parts of $Re (k f)$ and $Im (k f)$ $μ$ -measurable, and on the conull set where the integral of $k f$ is finite the real and imaginary parts of $(K f) (x)$ are differences of these measurable functions. Since $∣ K f ∣ \leq h ∥ f ∥_{2}$ almost everywhere and $h ∥ f ∥_{2} \in L^{2} (μ)$ , the class of $K f$ lies in $L^{2} (μ)$ with $∥ K f ∥_{2} \leq ∥ k ∥_{2} ∥ f ∥_{2}$ . Finally, if $k = k^{'}$ and $f = f^{'}$ almost everywhere, then the function $(x, y) \mapsto ∣ k (x, y) f (y) - k^{'} (x, y) f^{'} (y) ∣$ is nonnegative and measurable with vanishing product integral, so for $μ$ -almost every $x$ its section vanishes $ν$ -almost everywhere by Tonelli [A4], that is, $K f = K f^{'}$ $μ$ -almost everywhere.

A2A4

2.1

For $f \in L^{2} (ν)$ and $g \in L^{2} (μ)$ the function $(x, y) \mapsto k (x, y) f (y) \overline{g (x)}$ lies in $L^{1} (μ \times ν)$ : with $h$ as in step 1.3, Tonelli and Cauchy–Schwarz in $L^{2} (μ)$ give $\int_{X \times Y} ∣ k (x, y) f (y) \overline{g (x)} ∣ d (μ \times ν) = \int_{X} ∣ g (x) ∣ (\int_{Y} ∣ k (x, y) f (y) ∣ d ν (y)) d μ (x) \leq ∥ f ∥_{2} \int_{X} h (x) ∣ g (x) ∣ d μ (x) \leq ∥ f ∥_{2} ∥ h ∥_{2} ∥ g ∥_{2} = ∥ k ∥_{2} ∥ f ∥_{2} ∥ g ∥_{2} < + \infty$ . Hence Fubini [A4] applies and $⟨ K f, g ⟩ = \int_{X} \int_{Y} k (x, y) f (y) \overline{g (x)} d ν (y) d μ (x) = \int_{Y} f (y) \overline{\int_{X} \overline{k (x, y)} g (x) d μ (x)} d ν (y) = ⟨ f, K^{*} g ⟩$ with $K^{*} g (y) = \int_{X} \overline{k (x, y)} g (x) d μ (x)$ ; applying the same estimate to the conjugate kernel $\overline{k}$ , which also lies in $L^{2} (μ \times ν)$ with the same norm, gives $∥ K^{*} g ∥_{2} \leq ∥ k ∥_{2} ∥ g ∥_{2}$ , so $K^{*}$ is a bounded operator and is the Hilbert adjoint of $K$ .

step 1.3A1A2A4

3.1

Steps 1.1, 1.2 and 2.1 exhibit the three displayed adjoints, so the claimed formulas hold, under the countable-choice hypothesis recorded in [A5].

step 1.1step 1.2step 2.1A5∎

Depends on

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Sources

Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3.1 and Example 5.35 ff. (standard reference, not scraped)
Andrew Lin and Casey Rodriguez, MIT 18.102 Introduction to Functional Analysis, Example 186 (standard reference, not scraped)