TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Picard range and pseudoinverse-domain criterion

Statement

Assume the Axiom of Countable Choice. Let $K : X \to Y$ be a compact linear map between real or complex Hilbert spaces, with Hilbert adjoint $K^{*} : Y \to X$ . Write $N_{> 0} : = {1, 2, 3, \dots}$ . If $R (K)$ has finite dimension $r \in N$ , put $J = {1, \dots, r}$ , interpreted as empty when $r = 0$ ; if $R (K)$ is infinite-dimensional, put $J = N_{> 0}$ . Thus $J$ indexes exactly the positive singular values $s_{j} > 0$ in nonincreasing order. Let $(v_{j})_{j \in J}$ and $(u_{j})_{j \in J}$ be complete orthonormal families in $N (K)^{⊥}$ and $\overline{R (K)}$ , respectively, with $K v_{j} = s_{j} u_{j}$ and $K^{*} u_{j} = s_{j} v_{j}$ . The empty sum is zero, finite $J$ uses a finite sum, and infinite $J$ uses the norm-convergent series whenever the stated square-summability condition holds. Use inner products linear in the first argument. Define the Picard sum $P_{K} (y) : = \sum_{j \in J} \frac{∣ ⟨ y, u_{j} ⟩_{Y} ∣^{2}}{s_{j}^{2}}, y \in Y .$ Then:

The exact equation $K x = y$ is solvable if and only if $y ⊥ N (K^{*})$ and $P_{K} (y) < \infty$ .
The domain of the Hilbert-space Moore–Penrose inverse is $D (K^{†}) = R (K) \oplus R (K)^{⊥} = {y \in Y : P_{K} (y) < \infty},$ and for every $y \in D (K^{†})$ , $K^{†} y = \sum_{j \in J} \frac{⟨ y, u_{j} ⟩_{Y}}{s_{j}} v_{j} .$ Thus the orthogonal component in $R (K)^{⊥} = N (K^{*})$ is discarded by $K^{†}$ ; exact range membership additionally requires that component to be zero.

Statement references

Compact Hilbert-space inverse model

Hilbert-space Moore–Penrose inverse and its domain

Compact-operator singular system and expansion

Adjoint kernel and range orthogonality

Clason, Theorem 3.10 and complete proof, §3.2, printed pp. 23–24

Scherzer and Mindrinos, Theorem 6.7, printed p. 87

Kaarnioja, “Solvability of $y = A x$ ,” PDF pp. 10–12

Singular-system diagram

For a fixed positive singular index $j$ , the diagram records the domain/codomain action on its one-dimensional singular directions:

The two composites agree because $K (λ v_{j}) = λ K v_{j} = λ s_{j} u_{j}$ . This diagram explains the singular-pair convention; no proof step relies on it.

Facts & Assumptions

Given: Countable Choice, the compact Hilbert-space model $K : X \to Y$ , its adjoint $K^{*}$ , and the positive singular system described in the statement.

[F1]

The model supplies real or complex Hilbert parameter/data spaces and a compact linear forward map (Compact linear inverse model between Hilbert spaces).

[F2]

The compact SVD supplies the positive singular values, complete orthonormal right and left singular families, the singular relations, and the operator expansion (Singular value decomposition for compact operators).

[F3]

$(R (K))^{⊥} = N (K^{*})$ and $\overline{R (K)} = (N (K^{*}))^{⊥}$ (Kernel–range orthogonality for Hilbert adjoints).

[F4]

$⟨ K x, y ⟩_{Y} = ⟨ x, K^{*} y ⟩_{X}$ (The Hilbert-space adjoint of a bounded operator).

[F5]

Finite Bessel inequality: $\sum_{j \in F} ∣ ⟨ x, e_{j} ⟩ ∣^{2} \leq ∥ x ∥^{2}$ for every finite orthonormal family (The finite Bessel inequality and best approximation by a finite orthonormal family).

[F6]

Finite orthogonal sums satisfy Pythagoras (Pythagoras and finite orthogonal sums).

[F7]

On a complete orthonormal family, the finite Fourier sums converge in norm to each vector in its closed span (Fourier expansion in a Hilbert space).

[F8]

Every closed subspace of a Hilbert space has an orthogonal decomposition (Orthogonal decomposition by a closed subspace).

[F9]

Hilbert spaces are complete and Cauchy–Schwarz makes the inner product continuous in either argument (Hilbert space, Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[F10]

For $r + z$ with $r \in R (K)$ and $z \in R (K)^{⊥}$ , the inverse returns the minimum-norm preimage of $r$ and annihilates $z$ (Hilbert-space Moore–Penrose inverse).

[F11]

A compact linear operator is bounded and hence continuous in norm (Compact linear operator).

[A1]

Countable Choice is the hypothesis in the compact SVD and Hilbert Fourier expansion used here (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Proof

technique · direct

1.1givenF2F3F4F5A1algebra

Suppose $y = K x$ . Then $y ⊥ N (K^{*})$ by [F3]. For each $j \in J$ , the adjoint identity and singular relation give $⟨ y, u_{j} ⟩_{Y} = ⟨ K x, u_{j} ⟩_{Y} = ⟨ x, K^{*} u_{j} ⟩_{X} = s_{j} ⟨ x, v_{j} ⟩_{X}$ . Applying [F5] to finite subsets of the orthonormal family $(v_{j})$ shows $P_{K} (y) = \sum_{j \in J} ∣ ⟨ x, v_{j} ⟩_{X} ∣^{2} \leq ∥ x ∥_{X}^{2} < \infty$ . This proves necessity in claim 1.

1.2F2F3F6F9A1constructalgebra

Now let $y ⊥ N (K^{*})$ and $P_{K} (y) < \infty$ . By [F3], $y \in \overline{R (K)}$ . For finite $J$ take the finite sum; for infinite $J$ let $x_{N} = \sum_{j = 1}^{N} s_{j}^{- 1} ⟨ y, u_{j} ⟩_{Y} v_{j}$ . For $M > N$ , [F6] gives $∥ x_{M} - x_{N} ∥_{X}^{2} = \sum_{j = N + 1}^{M} s_{j}^{- 2} ∣ ⟨ y, u_{j} ⟩_{Y} ∣^{2}$ , whose tails tend to zero because $P_{K} (y) < \infty$ . Thus $(x_{N})$ is Cauchy and converges to $x \in X$ by [F9]. The complete right singular family places $x$ in $N (K)^{⊥}$ .

2.1step 1.2F1F2F7F10F11A1algebra

By [F11], applying $K$ to the finite sums and passing to the limit gives $K x = \lim_{N \to \infty} K x_{N} = \lim_{N \to \infty} \sum_{j = 1}^{N} ⟨ y, u_{j} ⟩_{Y} u_{j} = y$ , where the last equality is [F7] on $\overline{R (K)}$ . The finite- $J$ case is the same equality with a finite sum. Since $x \in N (K)^{⊥}$ and $K x = y$ , [F10] identifies $x = K^{†} y$ . This proves sufficiency in claim 1 and the formula there.

3.1step 1.2step 2.1F3F7F8F10A1construct

For arbitrary $y \in Y$ , apply [F8] to the closed subspace $M = \overline{R (K)}$ and write $y = p + q$ with $p \in M$ and $q \in M^{⊥}$ . Since $R (K) \subseteq M$ , $q \in R (K)^{⊥}$ . The Fourier coefficients of $p$ are $⟨ p, u_{j} ⟩_{Y} = ⟨ y, u_{j} ⟩_{Y}$ , and [F7] expands $p$ in the complete left singular family. If $P_{K} (y) < \infty$ , the construction in steps 1.2–2.1 applied to $p \in M = (N (K^{*}))^{⊥}$ gives $p \in R (K)$ and $K^{†} p = \sum_{j} s_{j}^{- 1} ⟨ y, u_{j} ⟩_{Y} v_{j}$ ; hence $y = p + q \in R (K) \oplus R (K)^{⊥} = D (K^{†})$ and [F10] gives the displayed formula.

4.1step 1.1F3F10∎

Conversely, let $y = r + q \in D (K^{†})$ with $r \in R (K)$ and $q \in R (K)^{⊥}$ . By [F3], $q \in N (K^{*})$ and each $u_{j} \in N (K^{*})^{⊥}$ , so $⟨ q, u_{j} ⟩_{Y} = 0$ . Step 1.1 applied to $r$ shows $P_{K} (y) = P_{K} (r) < \infty$ ; [F10] gives $K^{†} y = K^{†} r$ and the same series formula. Thus the domain is exactly the set where the Picard sum is finite, completing claim 2.

Depends on

Used by

One-pass filtered reconstruction from finite Born data Algorithm
A compact diagonal map with dense nonclosed range Example

Dependency tree · two levels

70 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

Christian Clason, Regularization of Inverse Problems: Lecture Notes, arXiv:2001.00617v2, Theorem 3.10 (standard reference, not scraped)
Otmar Scherzer and Leonidas Mindrinos, Inverse Problems: Lecture Notes, University of Vienna, 2015, Theorem 6.7 (standard reference, not scraped)
Vesa Kaarnioja, Inverse Problems: Fourth Lecture, FU Berlin, 8 May 2023 (standard reference, not scraped)