I am getting started with inverse problems in statistics. However, I didn't something related to it.

I was reading this paper http://math.uni-heidelberg.de/studinfo/reiss/CavalierInvProb.pdf.

It says

The classical problem is the following : let A be an operator from the Hilbert space H in G :

Given g ∈ G find f ∈ H such that Af = g.

This is really an inverse problem in the sense that one has to invert the operator A. A case of major interest is the case of ill-posed problems where the operator is not invertible. The problem is then to handle this inversion in order to obtain a precise reconstruction.

A problem is well posed if

  1. there exists a solution to the problem (existence)

  2. there is at most one solution to the problem (uniqueness)

  3. the solution depends continuously on the data (stability)

A problem which is not well-posed is called ill-posed.

If the data space is defined as the set of solutions, existence is clear. However, this could be modified if the data are perturbed by noise. Uniqueness of the solution is not easy to show. In case where it is not garanted by the data, then the set of a priori solutions can be restricted, and the problem is then reformulated. Nevertheless, the main issue is usually stability. Indeed, suppose $A^{−1}$ exists but is not bounded. Given a noisy version of g called $g_ε$, the reconstruction $f_ε$ = $A^{−1}g_ε$ may be far from the true f.

I actually didn't get it. Can you give some examples to help me clarify. I didn't get what they mean by $A^{-1}$ exists but is not bounded. And what about the uniqueness of the solution. Can you give an example where the solution is not unique. I am not being able to grasp the concept.


1 Answer 1


Linear map from an infinite-dimensional space need not be bounded, which in this context means continuous, or bounded in the sense of having finite operator norm, or bounded in the sense it maps every bounded set to a bounded set.

Consider, for example, a map $L \colon l^2(\mathbb{N}) \to l^2(\mathbb{N})$ defined by $L(e_n) = e_n/n$ for basis vectors $e_n$. It is continuous (i.e. bounded) linear operator, and has an inverse defined by $L^{-1}(e_k) = ke_k$. The inverse is not bounded (continuous), and consequently a small perturbation in input can result in huge changes in output. Consider what happens if the input of the inverse is perturbed by $\varepsilon e_N$ with $N= 1,10,100,1000,\ldots$.

For uniqueness, consider any non-injective operator, for example a coordinate projection $P(x,y) = x$. Then you have $P(x,0) = P(x,7) = P(x,-4)$, so you obviously can't invert $P$ uniquely.

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    $\begingroup$ Remark: Note that the inverse $L^{-1}$ is only defined on the range of $L$ (which is not the whole $\ell^2$ but is dense in it). The range contains finite linear combinations of basis elements and is therefore dense, but it does not contain, for example, the sequence $(n^{-3/2})\in\ell^2$. $\endgroup$ Sep 1, 2014 at 17:09

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