Keywords:
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
.
References