**American Journal of Computational Mathematics**

Vol.04 No.01(2014), Article ID:42295,4 pages

10.4236/ajcm.2014.41001

Logarithm of a Function, a Well-posed Inverse Problem

Silvia Reyes Mora, Víctor A. Cruz Barriguete, Denisse Guzmán Aguilar

Instituto de Física y Matemáticas, Universidad Tecnológica de la Mixteca, Huajuapan de León, Oax, México

Email: sreyes@mixteco.utm.mx

Copyright © 2014 Silvia Reyes Mora et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. In accordance of the Creative Commons Attribution License all Copyrights © 2014 are reserved for SCIRP and the owner of the intellectual property Silvia Reyes Mora et al. All Copyright © 2014 are guarded by law and by SCIRP as a guardian.

ABSTRACT

Received November 13, 2013; revised December 13, 2013; accepted December 20, 2013

It poses the inverse problem that consists in finding the logarithm of a function. It shows that when the function is holomorphic in a simply connected domain, the solution at the inverse problem exists and is unique if a branch of the logarithm is fixed. In addition, it’s demonstrated that when the function is continuous in a domain, where is Hausdorff space and connected by paths. The solution of the problem exists and is unique if a branch of the logarithm is fixed and is stable; for what in this case, the inverse problem turns out to be well-posed.

**Keywords:**

Logarithm function; Inverse problem; Stability

1. Introduction

The inverse problems generally are ill-posed in Hadamard sense. These words lead us to think that there exist inverse problems that are well-posed and which are possible to be solved analytically [1].

The ill-posed problems do not fulfill with at least one of the conditions of existence, uniqueness or stability of the solution. Nevertheless, if a problem does not have a solution, or the solution is not unique, it is possible to correct in spite of doing considerations on the domain and the co-domain of the operator who represents the problem.

Unlike a complex number different from zero for which, it is always possible to find his logarithm. The functions need certain conditions on his domain to guarantee the existence of his logarithm [2]. It appears that one of these conditions is that the domain is simply connected. To assure the uniqueness of the logarithm of a function, it is es- sential to take a branch of the logarithm.

In this paper, there appears the problem of finding the logarithm of a given function, by different techniques. It is demonstrated that the above-mentioned problem is an inverse stable problem in Hadamard sense [3]. Therefore, the problem of existence and uniqueness is solved. In addition, it is demonstrated that it is stable, which transforms it into a well-posed inverse problem. It is realized the analysis of the solution of the inverse problem for when the do- main of the inverse operator, it corresponds to the functions that are not annulled and in addition they are holomor- phic in some region [4]. It is demonstrated that when is a simply connected domain, the solution of the in- verse problem exists and is unique. In a similar way, when the space, it corresponds to the space of the conti- nuous functions that are not annulled on a region simply connected and where Z is a Hausdorff’s space and connected for paths, the solution of the inverse problem exists, is unique and is stable [5].

2. Exposition of the problem

It’s known that not all the real numbers have logarithm, nevertheless, all complex numbers have it.

The natural question that arises is under what conditions a function has logarithm. To answer to the previous question, an operator is considered, if, and, are spaces of functions, it is possible to pose the direct problem: From a function on, find to a function such that. The corresponding inverse problem is given a function, to find to a function, such that. It’s known that the inverse of the operator is the operator, so the inverse problem can be seen as the search of the logarithm of a given function. Notice that as is considered the operator, it has felt to take as space to the functions that are not annulled and in addition be continuous or holomorphic on some region; nevertheless they must determine the conditions that the region must fulfill to demonstrate the existence, uni- queness and stability of the inverse problem.

3. Holomorphic logarithm

In this section, it is considered that corresponds to the holomorphic functions and to the holomorphic functions that are not annulled on any region. There’s demonstrated that when Ω is a simply connected domain, the solution of the inverse problem exists and is unique when a branch of the logarithm is fixed. The problem consists of knowing if given a function holomorphic f can be a function, such that.

Definition: Given a domain, and be and. It is said that is a loga- rithm of if. If the functions and are holomorphic in Ω, it is said that is a ho- lomorphic logarithm of [6].

Theorem 1: Let be, where Ω is a simply connected set of the plane, then the so- lution of the inverse problem exists and is unique.

Proof: If it thinks that Ω it is a simply connected domain, by theorem of the Riemann application, the simply connected domains of the flat sound of two types: the conformal equivalent to the plane and the conformal equivalent to the unitary disc D. For such a motive, the existence of the solution of the inverse problem, are obtained as a consequence of the propositions 1, 2 and 3. As for the uniqueness, it is essential to take a branch of the logarithm.

Proposition 1: Every function admits a holomorphic logarithm.

Proof: It is known that for every. Let be a point such that. It’s de- fined by:

, (1)

where is the curve that joins to with. it’s holomorphic and is not annulled on, then is holomorphic on. Then, it´s definite as well. This way,. Consider

then

. Then, it is constant on. So,

Then.

Let be, then and the proof it is complete.

Proposition 2: Let Ω be a simply connected domain own of, admits a holomorphic logarithm.

Proof: Consider, since Ω be a simply connected domain own of, there is an conformal function, such that. Since, follows that. By the propo- sition 1, the function admits a holomorphic logarithm in. That is, there exists such that

Hereby and therefore admits a holomorphic logarithm in Ω.

Proposition 3: Every function admits a holomorphic logarithm.

Proof: Since it is not annulled in and. Then, that is, there exist a function such that is the primitive of, that is to say, for everything. Notice that

Then, if it follows that.

4. Continuous logarithm

In this section, the inverse problem it’s studied, when the condition weakens of being a holomorphic function to being a continuous function. Nevertheless, on having asked him only continuity to the functions and hav- ing tried to solve the inverse problem, it is needed to do more restrictions on the domain of the above men- tioned functions.

It is considered to be that and the space that correspond to the spaces of conti- nuous and continuous that are not annulled function respectively. There is demonstrated that when it’s a simply connected domain and is Hausdorff space and connected by paths, the solution of the inverse problem exists, is unique if a branch of the logarithm is fixed adapted and is stable.

Theorem 2 Let be, where a simply connected set is (not necessarily it is a subset of the plane) content in Hausdorff space and connected by paths; then the solution of the inverse problem exists, is unique and is stable.

Proof: The existence of the solution of the inverse problem, it’s demonstrated in the proposition 3 and the ex- istence in the proposition 4.

Since, where and are Banach spaces, in addition, the operator is continuous; then, by the open mapping theorem, the inverse operator is continuous and therefore, the solution of the inverse problem is stable.

Definition: Let and be topological spaces. Consider to a continuous function. Is say that is a covering map if it exists a neighborhood for, with following properties:

1) with

2) is a neighborhood for such that is a homeomorphism on,.

Proposition 4: The function is a covering map.

Definition: Let be a continuous function and let be a covering map. The ap- plication is a lifting for (in respect of) if.

Note that if it is known that the exp function is a covering map, then to define the lifting, and under the condi- tions for the existence and uniqueness of the lifting, it will be had that; and therefore on having demonstrated the existence and uniqueness of the lifting, there will be demonstrated the existence and unique- ness of the logarithm of a continuous function .

4.1. Uniqueness of the lifting

For the uniqueness of the lifting it is necessary that the topological space be connected and Hausdorff.

Theorem 3 Let be a local homeomorphism, let be a connected and Hausdorff space and let be a continuous application. Suppose that and are lifting of. Then, if exists, such that, it follows that in.

Proof: The set is defined. Note that and therefore. If it is shown that is open and closed it follow that. To show that is a closed set, enough to prove that is an open set. Let be, clearly and since is a Hausdorff space, exist

neighborhoods for and, respectively such that. Consider that, note that isn’t an empty set, then so, is open set since is a

finite intersection of open sets. Let be, then and. What implies that. Therefore and so, the open set and then the set is closed. To show that is an open set. Let be and. By hypothesis, there is a neighborhood for such that is open set and is an homeomorphism. Since and are continuous func- tion, there is an neighborhood for such that and. Since is an injective function, it follows that so, it’s a closed set and then the set is open. Since is connected space and is a non-empty, open and closed set, it follows that.

4.2. Existence of the lifting

The study of the existence of the lifting needs of the study of the fundamental group.

Let be, where, curves that begin and end in, it is to say, they are closed curves. Is said that it is related with, , if it exists a continuous function such that

,

The function is called continuous homotopy.

It is easy to see that the relation is an equivalence relation. Since it is known well, everything equivalence relation induces a partition. In this case the classes are formed by the set of curves that are homotopic to. Intuitively it is possible to define the operation join curved and the above mentioned operation gives a structure of group.

Definition: The first group of homotopy of is defined, with basis:

,

where is a closed curve.

We noted that is the curve.

The following result gives necessary and sufficient conditions for the existence of the lifting.

Theorem 4: Let be a covering map and let be a continuous function, with a connected by paths set. Then they are equivalent:

1) A lifting there exists for;

2).

Where is the induced mapping of fundamental groups.

Proof: The demonstration obviously, because.

There will be demonstrated that. Let be and let be a curve in from to. The curve in that it begins in it has a unique lifting that it begins. is defined. It is demonstrated that it is definite as well, independently of the choice of, let be another curve from to. Then is a closed curve of to with:

.

This means that there is a homotopy from to a closed curve that gets up to a closed curve in based in. Property homotopy covering is applied to to obtain a lifting. Since is a closed curve to, it’s too. By the uniqueness of the lift curve, the first half of is and the second half is route the other way around, with the common midpoint. This shows that it´s definite as well. Need to show that is a continuous function. Let be a neighborhood of that it has a lifting containing to such that is an homeomorphism. A connected by path neighborhood is chosen for with. The curves from to points, it is possible to take a given fixed curve from to follow by curves in from y to the points. Then, the curves in it has lifting where and is the inverse of. So, and, therefore is continuous in.

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