Function Spaces, Inverse Problems, and Risk Certificates · Chapter III-10

Function Spaces and Legal Operator Algebra

An inverse formula is meaningful only after the domain, codomain, norm, and range are fixed. An operator can be injective while its inverse is unbounded or defined only on a strict range. Compositions that look algebraically valid may therefore be undefined on candidate functions used in an identification proof.

Conceptual map

  1. III-10.01Spaces and norms
  2. III-10.02Operator domains
  3. III-10.03Meaning of an inverse
  4. III-10.04Boundedness and theorem tools
  5. III-10.05Composition and graph cores

1. An operator formula begins with spaces

A symbol such as \(T^{-1}y\) has content only after the domain of \(T\), its range, and the norms on both spaces have been fixed.

Let \(X\) and \(Y\) be normed spaces and \(T:D(T)\subseteq X\to Y\) a linear operator. A full-domain map \(T:X\to Y\) is bounded when a finite constant \(M\) satisfies \(\lVert Tx\rVert_{Y}\le M\lVert x\rVert_{X}\). If \(T\) is injective, the inverse is the map

\[T^{-1}:\operatorname{range}(T)\to D(T).\](1)
Definition 1 · Legal composition

The composition \(ST\) is defined on \(\{x\in D(T):Tx\in D(S)\}\). Algebraic cancellation involving \(T^{-1}\) is valid only on vectors that belong to \(\operatorname{range}(T)\) and to every subsequent operator domain.

2. Bounded inversion is a lower-bound property

Proposition 1 · Stability of an injective operator

Let \(T:X\to Y\) be bounded and injective. Its inverse on \(\operatorname{range}(T)\) is bounded exactly when a constant \(c>0\) satisfies

\[\lVert Tx\rVert _{Y}\ge c\lVert x\rVert _{X} \quad \text{for every} x\in X.\](2)

If \(X\) is Banach, condition (2) also implies that \(\operatorname{range}(T)\) is closed in \(Y\).

Proof. If \(\lVert T^{-1}\rVert \le M\), then \(\lVert x\rVert=\lVert T^{-1}Tx\rVert \le M\lVert Tx\rVert\), giving (2) with \(c=1/M\). Conversely, (2) gives \(\lVert T^{-1}y\rVert \le c^{-1}\lVert y\rVert\) on the range. If \(Tx_{n}\) converges, (2) makes \(x_{n}\) Cauchy; completeness supplies a limit whose image is the proposed range limit. ∎

3. An injective compact operator with unstable inverse

Take \(X=Y=\ell ^{2}\) and define \(Te_{k}=k^{-1}e_{k}\). Then

\[(Tx)_{k}=x_{k}/k, \quad \operatorname{range}(T)=\{y\in \ell ^{2}: \sum k^{2}|y_{k}|^{2}<\infty \}.\](3)

The map is bounded, compact, and injective. Yet \(\lVert T^{-1}e_{k}\rVert=k\), so the inverse norm on successively larger coordinate bands diverges. The sequence \(y_{k}=1/k\) belongs to \(\ell ^{2}\), while its formal inverse is the constant sequence and lies outside \(\ell ^{2}\). Thus \(y\) is outside the range in (3).

In an \(N\)-coordinate truncation, the inverse norm is exactly \(N\). Noise \(\delta e_{N}\) in the measured object becomes \(N\delta e_{N}\) after inversion. With \(\delta=10^{-4}\) and \(N=10\),000, a measurement perturbation of size \(10^{-4}\) creates a latent error of size one.

4. Unbounded operators and graph cores

An unbounded operator can still be closed: convergence \(x_{n}\to x\) and \(Tx_{n}\to y\) then implies \(x\in D(T)\) and \(Tx=y\). Its graph norm \(\lVert x\rVert_{T}=\lVert x\rVert_{X}+\lVert Tx\rVert_{Y}\) records both the object and its image. A graph core is a subset dense in \(D(T)\) under this norm.

Failure case · Ambient convergence used through an inverse

Convergence \(y_{n}\to y\) in \(Y\) gives no convergence of \(T^{-1}y_{n}\) when the inverse is unbounded. The proof must supply convergence in the range norm or a regularization argument.

5. Implementation, exercises, and sources

For each displayed operator, create a type line containing domain, codomain, norm, range, and boundedness. For a numerical inverse, report the singular-value cutoff and the norm of the regularized inverse. A finite matrix result should be accompanied by its behavior as the discretization dimension grows.

Download the volume verification script →

Exercises

  1. Prove compactness of the diagonal operator in (3) by finite-rank truncation.
  2. Characterize the range when the diagonal sequence is \(2^{-k}\) and compute the inverse norm on the first \(N\) coordinates.
  3. Give operators \(S\) and \(T\) for which both are densely defined while \(D(ST)\) is a strict subset of \(D(T)\).
Partial solutions

1. Let \(T_{N}\) retain the first \(N\) diagonal coordinates. It has finite rank and \(\lVert T-T_{N}\rVert=1/(N+1)\), which tends to zero; norm limits of finite-rank operators are compact. 2. The range is {\(y:\sum 4^{k}|y_{k}|^{2}<\infty\)}, and the first-\(N\) inverse norm is \(2^{N}\).

  1. John B. Conway, A Course in Functional Analysis, Chapters II–III.Bounded operators, closed range, and compactness.
  2. Walter Rudin, Functional Analysis, Chapters 2, 4.Banach-space operator theorems.
  3. Heinz W. Engl, Martin Hanke, and Andreas Neubauer, Regularization of Inverse Problems, Chapter 2.Ill-posed inversion and regularization.

6. Audit checkpoint

State every domain, codomain, norm, range, inverse domain, and composition domain. Report whether the inverse is bounded, closed, regularized, or available only in a finite truncation.

7. Scope boundary

The chapter establishes operator-domain discipline for Banach and Hilbert space inverse problems. Operator algebras and general spectral calculus are outside its boundary.

Prerequisites