Abstract

We study the properties of locally uniformly differentiable functions on , a non-Archimedean field extension of the real numbers that is real closed and Cauchy complete in the topology induced by the order. In particular, we show that locally uniformly differentiable functions are , they include all polynomial functions, and they are closed under addition, multiplication, and composition. Then we formulate and prove a version of the inverse function theorem as well as a local intermediate value theorem for these functions.

1. Introduction

We start this section by reviewing some basic terminology and facts about non-Archimedean fields. So let be an ordered non-Archimedean field extension of . We introduce the following terminology.

Definition 1.1 (~, , , , ). For , we say if there exist such that and , where denotes the usual absolute value on : For nonnegative , one says that is infinitely smaller than and write if for all , and we say that is infinitely small if and is finite if ; finally, we say that is approximately equal to and write if and . We also set , the class of under the equivalence relation ~.

The set of equivalence classes (under the relation ~) is naturally endowed with an addition via and an order via if (or ), both of which are readily checked to be well defined. It follows that is an ordered group, often referred to as the Hahn group or skeleton group, whose neutral element is [1], the class of 1. It follows from the previous part that the projection from to is a valuation.

The theorem of Hahn [2] provides a complete classification of non-Archimedean extensions of in terms of their skeleton groups. In fact, invoking the axiom of choice, it is shown that the elements of any such field can be written as formal power series over its skeleton group with real coefficients, and the set of appearing exponents forms a well-ordered subset of .

From general properties of formal power series fields [3, 4], it follows that if is divisible, then is real closed; that is, every positive element of is a square in and every polynomial of odd degree over has at least one root in . For a general overview of the algebraic properties of formal power series fields, we refer to the comprehensive overview by Ribenboim [5], and for an overview of the related valuation theory, the book by Krull [6]. A thorough and complete treatment of ordered structures can also be found in [7].

Throughout this paper, we will denote by any totally ordered non-Archimedean field extension of that is complete in the order topology and whose skeleton group is Archimedean, that is, a subgroup of . The coefficient of the th power in the Hahn representation of a given will be denoted by , and the number will be defined by and for . It is easy to check that, for , if and only if , and if and only if ; moreover, for all .

The smallest such field is the field of the formal Laurent series whose skeleton group is , and the smallest such field that is also real closed is the Levi-Civita field , first introduced in [8, 9]. In this latter case , and for any element , the set of exponents in the Hahn representation of is a left-finite subset of ; that is, below any rational bound there are only finitely many exponents. The Levi-Civita field is of particular interest because of its practical usefulness. Since the supports of the elements of (when viewed as maps from into ) are left-finite, it is possible to represent these numbers on a computer [1]. Having infinitely small numbers, the errors in classical numerical methods can be made infinitely small and hence irrelevant in all practical applications. One such application is the computation of derivatives of real functions representable on a computer [10], where both the accuracy of formula manipulators and the speed of classical numerical methods are achieved. For a review of the Levi-Civita field , see [11, 12] and the references therein.

In the wider context of valuation theory, it is interesting to note that the topology induced by the order on is the same as that introduced via the valuation , as shown in Remark 1.2. It follows therefore that the field is just a special case of the class of fields discussed in [13].

Remark 1.2. The mapping , given by , is an ultrametric distance (and hence a metric); the valuation topology it induces is equivalent to the order topology (we will use to denote either one of the two topologies in the rest of the paper). If is an open set in the order topology and , then there exists in such that, for all , . Let ; then we also have that, for all , , and hence is open with respect to the valuation topology. The other direction of the equivalence of the topologies follows analogously.

In this paper, we will study the properties of locally uniformly differentiable functions on , thus expanding on the work done in [14]. In particular, we will show that this class of functions is closed under addition, multiplication, and composition of functions. Then we will state and prove a more general version of the inverse function theorem than that proved in [14], as well as a local intermediate value theorem for -valued locally uniformly differentiable functions on . The stronger condition (local uniform differentiability) on the function than that of the real case is needed for the proof of both theorems because of the total disconnectedness of the field in the order topology.

2. Preliminaries

In this section we review some of the topological properties of the field which helps the reader understand the differences between and . We begin this with the following definition.

Definition 2.1. Let . Then we say that is compact in if every open cover of in has a finite subcover.

Remark 2.2. Since is induced by a metric on , namely, the valuation metric mentioned in the Introduction, it follows by the Borel-Lebesgue Theorem (see, e.g., [15, Section 9.2]) that is compact in if and only if is sequentially compact.

Theorem 2.3. is a totally disconnected topological space. It is Hausdorff and nowhere locally compact. There are no countable bases. The topology induced to is the discrete topology.

Proof. Let contain more than one point, and let in be given. Without loss of generality, we may assume that . Let Then and are disjoint and open in , and , and . This shows that any subset of containing more than one point is disconnected, and hence is totally disconnected. It follows that is Hausdorff. That is Hausdorff also follows from the fact that it is a metric space ([16, p. 66, Problem 7(a)]).
To prove that is nowhere locally compact, let be given and let be a neighborhood of . We show that the closure of is not compact. Let in be such that and consider the sets where is the infinitely small positive number defined in the introduction. Then it is easy to check that is open in for all , and ; in particular, . But it is impossible to select finitely many of the โs to cover because each of the infinitely many elements of , , is contained only in the set .
There cannot be any countable bases because the uncountably many open sets , with , are disjoint. The open sets induced on by the sets are just the singletons . Thus, in the induced topology, all sets are open and the induced topology is therefore discrete.

As an immediate consequence of the fact that is totally disconnected, it follows that, for any , the connected component of is . Moreover, there are sets that are both open and closed, as we will show hereinafter.

Definition 2.4. Let . Then we say that is clopen in if it is both open and closed.

Proposition 2.5. For any and for any in , the set is clopen in .

Proof. Let be given. For all , we have that Thus, and hence is open.
Now let . Then for all , we have that
so . Thus, and hence is open. That is, is closed.

Similarly we can show that the sets and are clopen for any and any in .

Proposition 2.6. Let be given and let be a neighborhood of . Then there is a clopen set such that .

Proof. Let in be such that . Let . Then is clopen by Proposition 2.5 and .

It follows that the clopen sets form a base for the order topology. Moreover, the quasi-component of any is .

As an immediate consequence of the fact that is nowhere locally compact, we obtain the following result.

Corollary 2.7. For all in , none of the intervals , , , or are compact in .

Since is induced on by the order, we define boundedness of a set in as follows.

Definition 2.8. Let . Then we say that is bounded in if there exists in such that for all .

Proposition 2.9. Let be compact in . Then is closed and bounded in . Moreover, has an empty interior in ; that is,

Proof. That is closed in follows from the fact that is a Hausdorff topological space and is compact in (see [17, p. 36]).
Now we show that is bounded in . For each , let . Then, for each , is open in . Moreover, since the skeleton group of is Archimedean it follows that . Since is compact in , we can choose a finite subcover; thus, there is and there exist in such that It follows that for all , and hence is bounded in .
Finally, we show that . Assume not. Then there exist in such that . Since is a closed subset of the compact set , it follows that is compact in , which contradicts Corollary 2.7.

The following examples show that there are (countably infinite) closed and bounded sets that are not compact while there are uncountable sets that are compact in .

Example 2.10. Let . Then clearly, is countably infinite and bounded in . We show that is closed in . Let be given and let . If , then there exists such that . Let and let . Then is open in and . Thus, is open, and hence is closed in .
Next we show that is not compact in . For each , let . Then is open in for each and , but we cannot select a finite subcover since each is contained only in .

Example 2.11. Let denote the Cantor-like set constructed in the same way as the standard real Cantor set ; but instead of deleting the middle third, we delete from the middle an open interval times the size of each of the closed subintervals of at each step of the construction. Then is compact in .
It turns out that if we view as an infinite dimensional vector space over , then is not a vector topology; that is, is not a linear topological space.

Theorem 2.12. is not a vector topology.

Proof. Assume to the contrary that is a vector topology. Then, by continuity of scalar multiplication, there exists an open set and there exists an open set such that for all and for all . Let and be given. Since is open, there exists in such that . Hence Thus, which contradicts the fact that , since both and are finite and is infinitely small.

Since any normed vector space, with the metric topology induced by its norm, is a linear topological space ([18, Proposition III.1.3]), we readily infer from Theorem 2.12 that there can be no norm on that would induce the same topology as on .

We finish this section with the following criterion for convergence for an infinite series, which does not hold for real numbers series.

Proposition 2.13. For each , let be an element of . Then the series converges if and only if the sequence converges to zero.

Proof. Assume that converges, and let denote the sequence of partial sums of the series: . Thus converges and hence it is Cauchy. Now let in be given. Then there exists such that for each we have that . It follows that for all or, equivalently, for all . Hence the sequence converges to zero.
Now assume that the sequence converges to zero. Let be the sequence of partial sums of the series and let be given in . Then there is an such that for all . Thus, for all , we have that . Thus is Cauchy, and hence converges since is complete.

3. Local Uniform Continuity

In this section we introduce the concept of local uniform continuity of a function from a subset of to and study properties of such functions that will be relevant to our discussion of locally uniformly differentiable functions later in Section 4. We start with the following definitions.

Definition 3.1. Let , let , and let be given. Then one says that is continuous at if for every there exists such that whenever and .

Definition 3.2. Let , let , and let be given. Then one says that locally uniformly continuous at if there is a neighborhood of in such that is uniformly continuous on . That is, for every , there exists such that whenever and .

Exactly as in real calculus, one can easily show that if are (locally uniformly) continuous at and if , then is (locally uniformly) continuous at . Moreover, if and if is (locally uniformly) continuous at and is (locally uniformly) continuous at , then is (locally uniformly) continuous at .

Lemma 3.3. Let be given and let be a neighborhood of . Then there exist sequences and as well as mutually disjoint clopen sets and and a continuous function such that(1)the set has no limit point;(2);(3);(4) for each ;(5) for any ;(6), .

Proof. Since is not compact, there is a sequence that has no limit point in . Without loss of generality, we can take since will still have no limit point. For each , let be such that and for any and . Since is a null sequence in , it follows that as . Assume that has a limit point in and let be such a limit point. Since has no limit point, there is an such that . There exists such that, for all , . Since is the limit point of , there must be such that . But then . This is a contradiction. Hence has no limit point.
Since has no limit point, there exist and such that , and . For each let in be such that and . Then and all of , , are mutually disjoint open sets. By Proposition 2.6, there are clopen neighborhoods of such that and a clopen neighborhood of such that . is open as it is the union of open sets, but it is also closed as we will show hereinafter. Let be given. Then there exists a sequence such that . has no limit points, so is separated from . Therefore, there exists such that . Moreover, there exists such that for every , . It follows that for all and for all . , for each ; it follows that . That is, which is a finite union of closed sets and hence is itself closed. Since and which is closed, it follows that . Thus, and hence is closed. Define on as follows: Then is continuous on because is clopen.

In the real case, any function that is continuous on a neighborhood is also locally uniformly continuous on that neighborhood. This property does not hold in non-Archimedean fields.

Theorem 3.4. Let be given and let be a neighborhood of . Then there is a continuous function that is not locally uniformly continuous at .

Proof. Let and apply Lemma 3.3 to and to get , , , and which correspond to , , , and , respectively, in that lemma. Let and apply Lemma 3.3 to and to get , , , and . Continuing inductively, we can apply Lemma 3.3 to and in order to get , , , and . The resulting โs satisfy for every and (the Kronecker delta) for every . Let , which converges (pointwise), by Proposition 2.13, since as for all .
To show that is continuous on , let be given. Let in be given and let be such that . For each , let be such that whenever and , which is possible since each is continuous at . Let . Then for all satisfying , we have that Thus, is continuous at , for all , and hence is continuous on .
Now we show that is not locally uniformly continuous at . Let be a neighborhood of , let be such that , and let . So . It follows that and are such that as , but Therefore is not locally uniformly continuous at .

One can in fact show that there are continuous functions which are not locally uniformly continuous without using the property of total disconnectedness of . By using the method prescribed previously, the derivatives of the constructed functions are calculated easily, and this will prove useful when dealing with local uniform differentiability in Section 4.

Example 3.5. In the following, we will provide an explicit example of a function which is continuous but not locally uniformly continuous. Imitating the proof of Theorem 3.4, we let and let Then is well defined on since the โs are mutually disjoint sets. We will show that is continuous on but is not locally uniformly continuous at 0. Let be given. We will distinguish the following three cases.

Case 1 (). In this case, the function is constant on the open set containing , and hence is continuous at .

Case 2 (). Let be such that . Then but . Let . Then is clopen, and is also clopen. and . Therefore is continuous on since is constant on disjoint open sets that cover . Hence is continuous at .

Case 3 (). Let in be given and let be such that . Let . Then, for , we have that for all and every . So . This shows that is continuous at .

Thus, is continuous at for all and hence is continuous on . To see that is not locally uniformly continuous at 0, let be any neighborhood of 0. Let be such that and let . Then the sequences in defined previously are such that as , but . Thus, for every neighborhood of 0, there exists such that for each , there are such that while , which shows that is not locally uniformly continuous at 0.

4. Local Uniform Differentiability

Definition 4.1. Let , let , and let be given. Then we say that is locally uniformly differentiable () at if there is a neighborhood of in such that for every , there exists such that whenever and .

Definition 4.2. Let and let . Then we say that is locally uniformly differentiable () on if is locally uniformly differentiable at for all .

Definition 4.3. Let , let , and let be given. Then we say that is if th derivative of , , exists and is on for each .

Proposition 4.4. Let and let be . Then is on for all and satisfying .

Proposition 4.5. Let be given, let , and let be such that is and is on . Then f is on .

A noteworthy property of local uniform differentiability is that it is not inherited by the function from its derivatives, nor passed from the function onto its derivatives. Indeed, the following is an explicit example of a function whose derivative is everywhere zero, but it is not itself locally uniformly differentiable.

Example 4.6. Let , , and be as in Example 3.5. Define by Then is well defined since the are mutually disjoint. Let be given. If for some , then and hence . Also, if for any , then . Therefore, for all .
Note that is locally constant on and hence on . We will show that is differentiable at 0 with too. So let in be given. Let . Then for , we have that This shows that is differentiable at 0 with . Altogether, on . Therefore is on with for all .
Now we show that is not at 0. Consider the sequences and , where and . Both of these sequences converge to zero. Moreover, but Thus, for any neighborhood of 0, , and for any , there are such that , but . This shows that is not at 0.

Example 4.6 shows that the property is not necessarily inherited from the derivatives of the function, since is . Here is another example that shows that the property is not passed on from a function to its derivatives.

Example 4.7. Let be given by We will show that is on with derivative and then we will show that is not by showing that it is not at 0. So let be given. Let which is an open (clopen) neighborhood of . Then for all , we have that This shows that is locally uniformly differentiable at with derivative .
For , let in be given. Let . Then for all in , we will show thatCase 1 (). Then and . It follows that because .Case 2 (). Then and . It follows that Case 3 (). Then Thus, is locally uniformly differentiable at 0 with derivative . Altogether, it follows that is on with derivative
Next we show that is not differentiable at 0. Take the sequence and the sequence . Then both sequences converge to 0. But If were differentiable at 0, then both limits in (4.13) would be equal to . Since the two limits are different, we conclude that is not differentiable at 0, and hence is not on .
The previous two examples illustrate that the most natural definition for is given in Definition 4.3.

Proposition 4.8. Let be at . Then is at .

Proof. Let be a neighborhood of in such that is uniformly differentiable on and let be such that . Let in be given. Then there is , , such that for all satisfying we have that It follows that for ,

Remark 4.9. Proposition 4.8 shows that the class of locally uniformly differentiable functions is a subset of the class of functions. However, this is still large enough to include all polynomial functions as Corollary 4.14 will show.

Proposition 4.10. Let be locally uniformly differentiable at . Then is locally uniformly continuous at .

Proof. Let be a neighborhood of in such that is uniformly differentiable on . By Proposition 4.8, is continuous at . Let be a neighborhood of such that for every , . Since is uniformly differentiable on , there exists such that for all satisfying , we have that Let in be given. Let . Then for all satisfying , we have that

Proposition 4.11. Let , let be given, let be given, and let be at . Then is at , with derivative

Proof. Without loss of generality, we may assume that . Let in be given. Then there exists in such that , whenever . It follows that, for all , we have that

Proposition 4.12 (Chain Rule). Let , let be given, and let and be such that is at and is at . Then is at with derivative .

Proof. Let be a neighborhood of in such that is uniformly differentiable on , and let be a neighborhood of in such that and is uniformly differentiable and uniformly continuous on (Proposition 4.10). The condition that is always possible because is continuous. Since is continuous at and is continuous at (Proposition 4.8), it follows that is continuous at . So there exists in such that and for all we have that Also, since is continuous at , there is a in such that , and for all , we have that Since is uniformly differentiable on , there is a in such that , and for all satisfying , we have that Let in be given. Then, since is uniformly differentiable on , there exists an such that whenever and , it follows that Since is uniformly differentiable on , there exists in such that , and for all satisfying , we have that Finally, is uniformly continuous on . Thus, there exists in such that , and for all satisfying , we have that Let Then , and for all , we have that and ; and hence

Proposition 4.13 (Product Rule). Let , let be given, and let be at . Then is at with .

Proof. Let be a neighborhood of in such that and are uniformly differentiable on and is uniformly continuous on (Proposition 4.10). Since , , and are continuous at , there exists in such that , and for all , we have that Let in be given. Since and are uniformly differentiable on , and is uniformly continuous on , there exists such that whenever and , we have that Let be such that . Then,

Since the function is on , as can be easily checked, it follows from Propositions 4.11 and 4.13 that any polynomial function is on .

Corollary 4.14. Let be a polynomial function. Then is on .

Since the derivative of a polynomial function is again a polynomial function, we readily obtains the following result.

Corollary 4.15. Let be a polynomial function. Then is on .

Proposition 4.16. The function defined as is .

Proof. First we note that is infinitely often differentiable on , with derivatives Now we prove that is on . So let and let in be given. Let Then is a neighborhood of and for all . Moreover, for all , we have that since so that Thus, for all , is at , and hence is on .
Applying Propositions 4.11 and 4.13, it then follows that is on for all , and hence is on .

5. Inverse Function Theorem

The following version of the inverse function theorem for functions was proven in [14].

Theorem 5.1 (inverse function theorem). Let