Logarithm of a Function, a Well-Posed Inverse Problem

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1 American Journal o Computational Mathematics, 4, 4, -5 Published Online February 4 ( Logarithm o 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 sreyes@mixteco.utm.mx Received November 3, 3; revised December 3, 3; accepted December, 3 Copyright 4 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 o the Creative Commons Attribution License all Copyrights 4 are reserved or SCIRP and the owner o the intellectual property Silvia Reyes Mora et al. All Copyright 4 are guarded by law and by SCIRP as a guardian. ABSTRACT It poses the inverse problem that consists in inding the logarithm o a unction. It shows that when the unction is holomorphic in a simply connected domain Ω, the solution at the inverse problem exists and is unique i a branch o the logarithm is ixed. In addition, it s demonstrated that when the unction is continuous in a domain Ω X, where X is Hausdor space and connected by paths. The solution o the problem exists and is unique i a branch o the logarithm is ixed and is stable; or what in this case, the inverse problem turns out to be well-posed. KEYWORDS Logarithm Function; Inverse Problem; Stability. 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 []. The ill-posed problems do not ulill with at least one o the conditions o existence, uniqueness or stability o the solution. Nevertheless, i a problem does not have a solution, or the solution is not unique, it is possible to correct in spite o doing considerations on the domain and the co-domain o the operator who represents the problem. Unlike a complex number dierent rom zero or which, it is always possible to ind his logarithm. The unctions need certain conditions on his domain to guarantee the existence o his logarithm []. It appears that one o these conditions is that the domain is simply connected. To assure the uniqueness o the logarithm o a unction, it is essential to take a branch o the logarithm. In this paper, there appears the problem o inding the logarithm o a given unction, by dierent techniques. It is demonstrated that the above-mentioned problem is an inverse stable problem in Hadamard sense [3]. Thereore, the problem o existence and uniqueness is solved. In addition, it is demonstrated that it is stable, which transorms it into a well-posed inverse problem. It is realized the analysis o the solution o the inverse problem or when the domain o the inverse operator, it corresponds to the unctions that are not annulled and in addition they are holomorphic in some region [4]. It is demonstrated that when Ω is a simply connected domain, the solution o the inverse problem exists and is unique. In a similar way, when the space Y, it corresponds to the space o the continuous unctions that are not annulled on a region Ω X simply connected and where Z is a Hausdor s space and connected or paths, the solution o the inverse problem exists, is unique and is stable [5].. Exposition o 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 unction has logarithm. To answer to the previous question, an operator A: X Y is considered, i A = exp, and X, Y are spaces o unctions, it is possible

2 S. R. MORA ET AL. to pose the direct problem: From a unction on X, ind to a unction g Y g = exp. The corresponding inverse problem is given a unction g Y, to ind to a unction X, such that g = exp(. It s known that the inverse o the operator exp is the operator log, so the inverse problem can be seen as the search o the logarithm o a given unction. Notice that as is considered the operator A = exp, it has elt to take as space Y to the unctions that are not annulled and in addition be continuous or holomorphic on some region; nevertheless they must determine the conditions that the region must ulill to demonstrate the existence, uniqueness and stability o the inverse problem. 3. Holomorphic Logarithm such that In this section, it is considered that X corresponds to the holomorphic unctions and Y to the holomorphic unctions that are not annulled on any region. There s demonstrated that when Ω is a simply connected domain, the solution o the inverse problem exists and is unique when a branch o the logarithm is ixed. The problem consists o knowing i given a unction holomorphic can be a unction g Y, such that = exp( g. Deinition: Given a domain Ω, and be : Ω C and g: Ω C. It is said that g is a logarithm o i = exp( g. I the unctions and g are holomorphic in Ω, it is said that g is a holomorphic logarithm o [6]. Theorem : Let exp : H( Ω H Ω be, where Ω is a simply connected set o the plane, then the solution o the inverse problem exists and is unique. Proo: I it thinks that Ω it is a simply connected domain, by theorem o the Riemann application, the simply connected domains o the lat sound o two types: the conormal equivalent to the plane and the conormal equivalent to the unitary disc D. For such a motive, the existence o the solution o the inverse problem, are obtained as a consequence o the propositions, and 3. As or the uniqueness, it is essential to take a branch o the logarithm. Proposition : Every unction H ( D Proo: It is known that ined by: where admits a holomorphic logarithm. z or every z D. Let ψ w be a point such that exp( w ( z ( ξ ( ξ =. It s de- = w + dξ, ( γ z γ z is the curve that joins to z with z. it s holomorphic and is not annulled on D, then holomorphic on D. Then, ( ξ d ξ it s deinite as well. This way, ψ γ z ξ exp( ψ =, then h h z =. Consider exp( ψ ψ exp( ψ z z z z z = = ( ψ ( z z D. Then, h( z it is constant on D. So, h( z Then = exp( ψ. Let g = ψ w + log ( w be, then g ( w exp exp = = =. z w = exp and the proo it is complete. Proposition : Let Ω be a simply connected domain own o, admits a holomorphic logarithm. Proo: Consider H ( Ω, since Ω be a simply connected domain own o, there is an conormal unction g: D Ω, such that g H( D. Since H ( Ω, ollows that g H ( D. By the proposition, the unction g admits a holomorphic logarithm in D. That is, there exists h: D such that g( w = exp ( h( w. Hereby = exp( h g and thereore admits a holomorphic logarithm in Ω. Proposition 3: Every unction H ( admits a holomorphic logarithm. Proo: Since it is not annulled in and H(. Then H (, that is, there exist a z is

3 S. R. MORA ET AL. 3 unction F H Notice that such that is the primitive o, that is to say, F = or everything z. ( ( F ( F F ( F ( F ( F exp = exp exp = exp =. exp Then F = C, i C = exp( k it ollows that exp( k F 4. Continuous Logarithm = +. In this section, the inverse problem it s studied, when the condition weakens o being a holomorphic unction to being a continuous unction. Nevertheless, on having asked him only continuity to the unctions and having tried to solve the inverse problem, it is needed to do more restrictions on the domain o the above mentioned unctions. It is considered to be that X = C( Ω and the space Y = C ( Ω that correspond to the spaces o continuous and continuous that are not annulled unction respectively. There is demonstrated that when Ω Z it s a simply connected domain and Z is Hausdor space and connected by paths, the solution o the inverse problem exists, is unique i a branch o the logarithm is ixed adapted and is stable. Theorem Let exp : C( Ω C Ω be, where Ω a simply connected set is (not necessarily it is a subset o the plane content in Hausdor space and connected by paths; then the solution o the inverse problem exists, is unique and is stable. Proo: The existence o the solution o the inverse problem, it s demonstrated in the proposition 3 and the existence in the proposition 4. Since exp : C( Ω C ( Ω, where C ( Ω and C Ω are Banach spaces, in addition, the operator is continuous; then, by the open mapping theorem, the inverse operator is continuous and thereore, the solution o the inverse problem is stable. Deinition: Let X and Y be topological spaces. Consider to p: X Y a continuous unction. Is say that p is a covering map i y Y, it exists a neighborhood U or y, with ollowing properties: p ( U = U j with U k U l = k l. j J U j is a neighborhood or X such that p U is a homeomorphism on U, j J. j Proposition 4: The exp : unction is a covering map. Deinition: Let : X Y be a continuous unction and let p: X Y be a covering map. The g application is a liting or (in respect o p i ( p g =. Note that i it is known that the exp unction is a covering map, then to deine the liting, and under the conditions or the existence and uniqueness o the liting, it will be had that ( exp g = ; and thereore on having demonstrated the existence and uniqueness o the liting, there will be demonstrated the existence and uniqueness o the logarithm o a continuous unction. 4.. Uniqueness o the Liting For the uniqueness o the liting it is necessary that the X topological space be connected and Hausdor. Theorem 3 Let p: X Y be a local homeomorphism, let X be a connected and Hausdor space and let : X Y be a continuous application. Suppose that g and g are liting o. Then, i x X exists, such that g( x = g( x, it ollows that g = g in X. Proo: The set E = { x X : g( x = g( x } is deined. Note that x E and thereore E. I it is shown that E is open and closed it ollow that E = X. To show that E is a closed set, enough to prove that X \ E is an open set. Let y X \ E be, clearly ( y ( y and since X is a Hausdor space, exist neighborhoods V V or y ( y ( y and ( y, respectively such that V V y ( y =. Consider that ( y ( y W = V V, note that W isn t an empty set, then y W, so, W is open set since is a inite intersection o open sets. Let x W be, then ( x V ( y and ( x V ( y. What implies that ( x ( y. Thereore W X \ E and so, the X \ E open set and then the E set is closed. To show that E is an open set. Let x E a = x = x. By hypothesis, there is a neighborhood U or be and

4 4 S. R. MORA ET AL. = U is open set and p U is an homeomorphism. Since and are continuous unc V U and p( ( v = ( v = p( ( v. Since p U is an injective unction, it ollows that ( v = ( v so, V \ E it s a closed set and then the E set is open. Since Y is connected space and E is a non-empty, open and closed set, it ollows that E = Y. a such that pu tion, there is V an neighborhood or y such that \ i 4.. Existence o the Liting The study o the existence o the liting needs o the study o the undamental group. Let αk : I X be, where k =,, curves that begin and end in x, it is to say, they are closed curves. Is said that α it is related with α, α α, i it exists H : I I X a continuous unction such that (, = α, (, = α, (, (, H t t H t t H s = H s = x The H unction 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 [ α ] are ormed by the set o curves β that are homotopic to α. Intuitively it is possible to deine the operation join curved and the above mentioned operation gives a structure o group. Deinition: The irst group o homotopy o X is deined, with basis x X : ( X x {[ α ]} π, =, where α is a closed curve. We noted that α is the curve α ( t. The ollowing result gives necessary and suicient conditions or the existence o the liting. Theorem 4: Let p: ( X, x ( Y, y be a covering map and let : ( X, x Y, y be a continuous unction, with X a connected by paths set. Then they are equivalent: g: X, x X, x there exists or ; A liting ( π ( X, x p ( π ( X, x. Where g is the induced mapping o undamental groups. Proo: The demonstration obviously, because = p. There will be demonstrated that. Let x X be and let γ be a curve in X rom x to x. The curve γ in Y that it begins in y it has a unique liting that it begins x γ. ( x = γ ( is deined. It is demonstrated that it is deinite as well, independently o the choice o γ, let γ be another curve rom is a closed curve o h to y with: x to x. Then ( γ ( γ ( π, ( π (, [ ] h X x p X x. This means that there is a homotopy h t rom h to a closed curve h that gets up to a closed curve h in X based in x. Property homotopy covering is applied to h t to obtain a liting h t. Since h is a closed curve to x, h it s too. By the uniqueness o the lit curve, the irst hal o h is γ and the second hal is route the other way around, with the common midpoint γ γ = γ. This shows that γ it s deinite as well. Need to show that is a continuous unction. Let U Y be a neighborhood o ( x that it has a liting U X containing to ( x such that p : U U is an homeomorphism. A connected by path neighborhood V is chosen or x with ( V U. The curves rom x to x V points, it is possible to take a given ixed curve γ rom x to x ollow by curves η in V rom y to the points y. Then, the curves ( γ ( η in Y it has liting ( γ ( η where η = p η and p: U U is the inverse o p: U U V U and = p, thereore is continuous in x.. So, V REFERENCES [] C. W. Groetsch, Inverse Problems: Activities or Undergraduates, The Mathematical Association o America, Ohio, 999. [] A. Browder, Topology in the Complex Plane, The American Mathematical Monthly, Vol. 7 No., 6, pp [3] A. Hatcher, Algebraic Topology, Cambridge University Press, Cambridge, 9.

5 S. R. MORA ET AL. 5 [4] L. V. Ahlors, Complex Analysis, McGraw-Hill, New York, 979. [5] A. Kirsch, An Introduction to the Mathematical Theory o Inverse Problems, nd Edition, Springer, Berlin,. [6] E. Stein and R. Shakarchi, Complex Analysis, Princeton University Press, Princeton, 9.