On Locally Uniformly Differentiable Functions on a Complete Non-Archimedean Ordered Field Extension of the Real Numbers

Abstract

We study the properties of locally uniformly differentiable functions on 𝒩, 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 𝐶¹, 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 𝐹 be an ordered non-Archimedean field extension of ℝ. We introduce the following terminology.

Definition 1.1 (~, ≈, ≪, 𝑆_𝐹, 𝜆). For 𝑥,𝑦∈𝐹^∗∶=𝐹⧵{0}, we say 𝑥∼𝑦 if there exist 𝑛,𝑚∈ℕ such that 𝑛|𝑥|>|𝑦| and 𝑚|𝑦|>|𝑥|, where |⋅| denotes the usual absolute value on 𝐹: |𝑥|=𝑥−𝑥iiff𝑥𝑥≥<00,.(1.1) For nonnegative 𝑥,𝑦∈𝐹, one says that 𝑥 is infinitely smaller than 𝑦 and write 𝑥≪𝑦 if 𝑛𝑥<𝑦 for all 𝑛∈ℕ, and we say that 𝑥 is infinitely small if 𝑥≪1 and 𝑥 is finite if 𝑥∼1; finally, we say that 𝑥 is approximately equal to 𝑦 and write 𝑥≈𝑦 if 𝑥∼𝑦 and |𝑥−𝑦|≪|𝑥|. We also set 𝜆(𝑥)=[𝑥], the class of 𝑥 under the equivalence relation ~.

The set of equivalence classes 𝑆_𝐹 (under the relation ~) is naturally endowed with an addition via [𝑥]+[𝑦]=[𝑥⋅𝑦] and an order via [𝑥]<[𝑦] if |𝑦|≪|𝑥| (or |𝑥|≫|𝑦|), both of which are readily checked to be well defined. It follows that (𝑆_𝐹,+,<) 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 𝜆 from 𝐹^∗ to 𝑆_𝐹 is a valuation.

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

From general properties of formal power series fields [3, 4], it follows that if 𝑆_𝐹 is divisible, then 𝐹 is real closed; that is, every positive element of 𝐹 is a square in 𝐹 and every polynomial of odd degree over 𝐹 has at least one root in 𝐹. 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 𝒩 any totally ordered non-Archimedean field extension of ℝ that is complete in the order topology and whose skeleton group 𝑆_𝒩 is Archimedean, that is, a subgroup of ℝ. The coefficient of the 𝑞th power in the Hahn representation of a given 𝑥 will be denoted by 𝑥[𝑞], and the number 𝑑 will be defined by 𝑑[1]=1 and 𝑑[𝑞]=0 for 𝑞≠1. It is easy to check that, for 𝑞∈𝑆_𝒩, 0<𝑑^𝑞≪1 if and only if 𝑞>0, and 𝑑^𝑞≫1 if and only if 𝑞<0; moreover, 𝑥≈𝑥[𝜆(𝑥)]𝑑^𝜆(𝑥) for all 𝑥≠0.

The smallest such field 𝒩 is the field 𝐿 of the formal Laurent series whose skeleton group is 𝑆_𝐿=ℤ, and the smallest such field that is also real closed is the Levi-Civita field ℛ, first introduced in [8, 9]. In this latter case 𝑆_ℛ=ℚ, and for any element 𝑥∈ℛ, the set of exponents in the Hahn representation of 𝑥 is a left-finite subset of ℚ; that is, below any rational bound 𝑟 there are only finitely many exponents. The Levi-Civita field ℛ is of particular interest because of its practical usefulness. Since the supports of the elements of ℛ (when viewed as maps from 𝑆_ℛ=ℚ into ℝ) 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 ℛ, 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 𝒩 is the same as that introduced via the valuation 𝜆, as shown in Remark 1.2. It follows therefore that the field 𝒩 is just a special case of the class of fields discussed in [13].

Remark 1.2. The mapping Λ∶𝒩×𝒩→ℝ, given by Λ(𝑥,𝑦)=exp(−𝜆(𝑥−𝑦)), is an ultrametric distance (and hence a metric); the valuation topology it induces is equivalent to the order topology (we will use 𝜏_𝑣 to denote either one of the two topologies in the rest of the paper). If 𝐴 is an open set in the order topology and 𝑎∈𝐴, then there exists 𝑟>0 in 𝒩 such that, for all 𝑥∈𝒩, |𝑥−𝑎|<𝑟⇒𝑥∈𝐴. Let 𝑙=exp(−𝜆(𝑟)); then we also have that, for all 𝑥∈𝒩, Λ(𝑥,𝑎)<𝑙⇒𝑥∈𝐴, and hence 𝐴 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 𝒩, 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 𝒩-valued locally uniformly differentiable functions on 𝒩. 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 𝒩 in the order topology.

2. Preliminaries

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

Definition 2.1. Let 𝐴⊂𝒩. Then we say that 𝐴 is compact in (𝒩,𝜏_𝑣) if every open cover of 𝐴 in (𝒩,𝜏_𝑣) has a finite subcover.

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

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

Proof. Let 𝐴⊂𝒩 contain more than one point, and let 𝑎≠𝑏 in 𝐴 be given. Without loss of generality, we may assume that 𝑎<𝑏. Let 𝐺₁={𝑥∈𝒩∶|𝑥−𝑎|≪𝑏−𝑎},𝐺₂=𝒩⧵𝐺₁.(2.1) Then 𝐺₁ and 𝐺₂ are disjoint and open in (𝒩,𝜏_𝑣), 𝑎∈𝐺₁∩𝐴 and 𝑏∈𝐺₂∩𝐴, and 𝐴⊂𝐺₁∪𝐺₂=𝒩. This shows that any subset of (𝒩,𝜏_𝑣) containing more than one point is disconnected, and hence (𝒩,𝜏_𝑣) is totally disconnected. It follows that (𝒩,𝜏_𝑣) is Hausdorff. That (𝒩,𝜏_𝑣) is Hausdorff also follows from the fact that it is a metric space ([16, p. 66, Problem 7(a)]).
To prove that (𝒩,𝜏_𝑣) is nowhere locally compact, let 𝑥∈𝒩 be given and let 𝑈 be a neighborhood of 𝑥. We show that the closure 𝑈 of 𝑈 is not compact. Let 𝜖>0 in 𝒩 be such that (𝑥−𝜖,𝑥+𝜖)⊂𝑈 and consider the sets 𝑀𝑀₋₁={𝑦∈𝒩∶𝑦<𝑥or𝑦−𝑥≫𝑑⋅𝜖},_𝑛=(𝑥+(𝑛−1)𝑑⋅𝜖,𝑥+(𝑛+1)𝑑⋅𝜖)for𝑛=0,1,2,…,(2.2) where 𝑑 is the infinitely small positive number defined in the introduction. Then it is easy to check that 𝑀_𝑛 is open in (𝒩,𝜏_𝑣) for all 𝑛≥−1, and ⋃^∞𝑛=−1𝑀_𝑛=𝒩; in particular, 𝑈⊂⋃^∞𝑛=−1𝑀_𝑛. But it is impossible to select finitely many of the 𝑀_𝑛’s to cover 𝑈 because each of the infinitely many elements 𝑥+𝑛𝑑⋅𝜖 of 𝑈, 𝑛=−1,0,1,2,…, is contained only in the set 𝑀_𝑛.
There cannot be any countable bases because the uncountably many open sets 𝑀_𝑋=(𝑋−𝑑,𝑋+𝑑), with 𝑋∈ℝ, are disjoint. The open sets induced on ℝ by the sets 𝑀_𝑋 are just the singletons {𝑋}. Thus, in the induced topology, all sets are open and the induced topology is therefore discrete.

As an immediate consequence of the fact that (𝒩,𝜏_𝑣) is totally disconnected, it follows that, for any 𝑥₀∈𝒩, the connected component of 𝑥₀ is {𝑥₀}. Moreover, there are sets that are both open and closed, as we will show hereinafter.

Definition 2.4. Let Ω⊂𝒩. Then we say that Ω is clopen in (𝒩,𝜏_𝑣) if it is both open and closed.

Proposition 2.5. For any 𝑥₀∈𝒩 and for any 𝑎>0 in 𝒩, the set Ω={𝑥∶|𝑥−𝑥₀|≪𝑎} is clopen in (𝒩,𝜏_𝑣).

Proof. Let 𝑥∈Ω be given. For all 𝑦∈(𝑥−𝑎𝑑,𝑥+𝑎𝑑), we have that ||𝑦−𝑥₀||≤||𝑦−𝑥||+||𝑥−𝑥₀||<𝑎𝑑+||𝑥−𝑥₀||≪𝑎.(2.3) Thus, (𝑥−𝑎𝑑,𝑥+𝑎𝑑)⊂Ω and hence Ω is open.
Now let 𝑥∈𝒩⧵Ω. Then for all 𝑦∈(𝑥−𝑎𝑑,𝑥+𝑎𝑑), we have that ||𝑦−𝑥₀||≥||||𝑦−𝑥||−||𝑥−𝑥₀||||≈||𝑥−𝑥₀||≪_¨¨𝑎;(2.4)
so 𝑦∈𝒩⧵Ω. Thus, (𝑥−𝑎𝑑,𝑥+𝑎𝑑)⊂𝒩⧵Ω and hence 𝒩⧵Ω is open. That is, Ω is closed.

Similarly we can show that the sets {𝑥∶|𝑥−𝑥₀|∼𝑎} and {𝑥∶|𝑥−𝑥₀|≈𝑎} are clopen for any 𝑥₀∈𝒩 and any 𝑎>0 in 𝒩.

Proposition 2.6. Let 𝑥₀∈𝒩 be given and let Ω be a neighborhood of 𝑥₀. Then there is a clopen set 𝐿 such that 𝑥₀∈𝐿⊂Ω.

Proof. Let 𝜖>0 in 𝒩 be such that (𝑥₀−𝜖,𝑥₀+𝜖)⊂Ω. Let 𝐿={𝑥∈𝒩∶|𝑥−𝑥₀|≪𝜖}. Then 𝐿 is clopen by Proposition 2.5 and 𝑥₀∈𝐿⊂(𝑥₀−𝜖,𝑥₀+𝜖)⊂Ω.

It follows that the clopen sets form a base for the order topology. Moreover, the quasi-component of any 𝑥₀∈𝒩 is {𝑥₀}.

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

Corollary 2.7. For all 𝑎<𝑏 in 𝒩, none of the intervals (𝑎,𝑏), (𝑎,𝑏], [𝑎,𝑏), or [𝑎,𝑏] are compact in (𝒩,𝜏_𝑣).

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

Definition 2.8. Let 𝐴⊂𝒩. Then we say that 𝐴 is bounded in (𝒩,𝜏_𝑣) if there exists 𝑀>0 in 𝒩 such that |𝑥|≤𝑀 for all 𝑥∈𝐴.

Proposition 2.9. Let 𝐴 be compact in (𝒩,𝜏_𝑣). Then 𝐴 is closed and bounded in (𝒩,𝜏_𝑣). Moreover, 𝐴 has an empty interior in (𝒩,𝜏_𝑣); that is, int(𝐴)∶={𝑎∈𝐴∶∃𝑟>0in𝒩∋(𝑎−𝑟,𝑎+𝑟)⊂𝐴}=∅.(2.5)

Proof. That 𝐴 is closed in (𝒩,𝜏_𝑣) follows from the fact that (𝒩,𝜏_𝑣) is a Hausdorff topological space and 𝐴 is compact in (𝒩,𝜏_𝑣) (see [17, p. 36]).
Now we show that 𝐴 is bounded in (𝒩,𝜏_𝑣). For each 𝑛∈ℕ, let 𝐺_𝑛=(−𝑑^−𝑛,𝑑^−𝑛). Then, for each 𝑛∈ℕ, 𝐺_𝑛 is open in (𝒩,𝜏_𝑣). Moreover, since the skeleton group of 𝒩 is Archimedean it follows that 𝐴⊂⋃_𝑛∈ℕ𝐺_𝑛=𝒩. Since 𝐴 is compact in (𝒩,𝜏_𝑣), we can choose a finite subcover; thus, there is 𝑚∈ℕ and there exist 𝑗₁<𝑗₂<⋯<𝑗_𝑚 in ℕ such that𝐴⊂⋃^𝑚_𝑙=1𝐺_𝑗𝑙=𝐺_𝑗𝑚=−𝑑^−𝑗𝑚,𝑑^−𝑗𝑚.(2.6) It follows that |𝑥|<𝑑^−𝑗𝑚 for all 𝑥∈𝐴, and hence 𝐴 is bounded in (𝒩,𝜏_𝑣).
Finally, we show that int(𝐴)=∅. Assume not. Then there exist 𝑎<𝑏 in 𝐴 such that [𝑎,𝑏]⊂𝐴. Since [𝑎,𝑏] is a closed subset of the compact set 𝐴, it follows that [𝑎,𝑏] is compact in (𝒩,𝜏_𝑣), 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 (𝒩,𝜏_𝑣).

Example 2.10. Let 𝐴=[0,1]∩ℚ. Then clearly, 𝐴 is countably infinite and bounded in (𝒩,𝜏_𝑣). We show that 𝐴 is closed in (𝒩,𝜏_𝑣). Let 𝑥∈𝒩⧵𝐴 be given and let 𝐺₀=(𝑥−𝑑,𝑥+𝑑). If 𝐺₀∩𝐴≠∅, then there exists 𝑞∈𝐴 such that 𝐺₀∩𝐴={𝑞}. Let 𝑟=|𝑞−𝑥| and let 𝐺=(𝑥−𝑟,𝑥+𝑟). Then 𝐺 is open in (𝒩,𝜏_𝑣) and 𝐺∩𝐴=∅. Thus, 𝒩⧵𝐴 is open, and hence 𝐴 is closed in (𝒩,𝜏_𝑣).
Next we show that 𝐴 is not compact in (𝒩,𝜏_𝑣). For each 𝑞∈𝐴, let 𝐺_𝑞=(𝑞−𝑑,𝑞+𝑑). Then 𝐺_𝑞 is open in (𝒩,𝜏_𝑣) for each 𝑞 and 𝐴⊂⋃_𝑞∈𝐴𝐺_𝑞, but we cannot select a finite subcover since each 𝑡∈𝐴 is contained only in 𝐺_𝑡.

Example 2.11. Let 𝐶_𝒩 denote the Cantor-like set constructed in the same way as the standard real Cantor set 𝐶; but instead of deleting the middle third, we delete from the middle an open interval (1−2𝑑) times the size of each of the closed subintervals of [0,1] at each step of the construction. Then 𝐶_𝒩 is compact in (𝒩,𝜏_𝑣).
It turns out that if we view 𝒩 as an infinite dimensional vector space over ℝ, then 𝜏_𝑣 is not a vector topology; that is, (𝒩,𝜏_𝑣) is not a linear topological space.

Theorem 2.12. 𝜏_𝑣 is not a vector topology.

Proof. Assume to the contrary that (𝒩,𝜏_𝑣) is a vector topology. Then, by continuity of scalar multiplication, there exists an open set 𝑂_ℝ⊂ℝ and there exists an open set 𝑂_𝒩⊂𝒩 such that 𝛼𝑥∈(1−𝑑,1+𝑑) for all 𝛼∈𝑂_ℝ and for all 𝑥∈𝑂_𝒩. Let 𝛼₀∈𝑂_ℝ and 𝑥₀∈𝑂_𝒩 be given. Since 𝑂_ℝ is open, there exists 𝑟>0 in ℝ such that (𝛼₀−2𝑟,𝛼₀+2𝑟)⊂𝑂_ℝ. Hence 𝛼₀𝑥₀∈(1−𝑑,1+𝑑),𝛼₀+𝑟𝑥₀∈(1−𝑑,1+𝑑).(2.7) Thus, 𝑟||𝑥₀||=||𝛼₀+𝑟𝑥₀−𝛼₀𝑥₀||<2𝑑,(2.8) which contradicts the fact that 𝑟|𝑥₀|≫2𝑑, since both 𝑟 and |𝑥₀| are finite and 𝑑 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 𝒩 that would induce the same topology as 𝜏_𝑣 on 𝒩.

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 𝑛∈ℕ, let 𝑥_𝑛 be an element of 𝒩. Then the series ∑^∞𝑛=1𝑥_𝑛 converges if and only if the sequence (𝑥_𝑛) converges to zero.

Proof. Assume that ∑^∞𝑛=1𝑥_𝑛 converges, and let (𝑦_𝑛) denote the sequence of partial sums of the series: 𝑦_𝑛=∑^𝑛𝑖=1𝑥_𝑖. Thus (𝑦_𝑛) converges and hence it is Cauchy. Now let 𝜖>0 in 𝒩 be given. Then there exists 𝑁∈ℕ such that for each 𝑛,𝑚>𝑁 we have that |𝑦_𝑛−𝑦_𝑚|<𝜖. It follows that |𝑥_𝑛+1|=|𝑦_𝑛+1−𝑦_𝑛|<𝜖 for all 𝑛>𝑁 or, equivalently, |𝑥_𝑛|<𝜖 for all 𝑛>𝑁+1. Hence the sequence (𝑥_𝑛) converges to zero.
Now assume that the sequence (𝑥_𝑛) converges to zero. Let (𝑦_𝑛) be the sequence of partial sums of the series ∑^∞𝑛=1𝑥_𝑛 and let 𝜖>0 be given in 𝒩. Then there is an 𝑁∈ℕ such that |𝑥_𝑛|<𝑑𝜖 for all 𝑛>𝑁. Thus, for all 𝑛>𝑚>𝑁, we have that |𝑦_𝑛−𝑦_𝑚|=|∑^{𝑛𝑖=𝑚+1}𝑥_𝑖|<(𝑚−𝑛)𝑑𝜖≪𝜖. Thus (𝑦_𝑛) is Cauchy, and hence ∑^∞𝑛=1𝑥_𝑛 converges since (𝒩,𝜏_𝑣) is complete.

3. Local Uniform Continuity

In this section we introduce the concept of local uniform continuity of a function from a subset of 𝒩 to 𝒩 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 𝐴⊂𝒩, let 𝑓∶𝐴→𝒩, and let 𝑥₀∈𝐴 be given. Then one says that 𝑓 is continuous at 𝑥₀ if for every 𝜖>0 there exists 𝛿>0 such that |𝑓(𝑥)−𝑓(𝑥₀)|<𝜖 whenever 𝑥∈𝐴 and |𝑥−𝑥₀|<𝛿.

Definition 3.2. Let 𝐴⊂𝒩, let 𝑓∶𝐴→𝒩, and let 𝑥₀∈𝐴 be given. Then one says that 𝑓 locally uniformly continuous at 𝑥₀ if there is a neighborhood Ω of 𝑥₀ in 𝐴 such that 𝑓 is uniformly continuous on Ω. That is, for every 𝜖>0, there exists 𝛿>0 such that |𝑓(𝑦)−𝑓(𝑥)|<𝜖 whenever 𝑥,𝑦∈Ω and |𝑦−𝑥|<𝛿.

Exactly as in real calculus, one can easily show that if 𝑓,𝑔∶𝐴→𝒩 are (locally uniformly) continuous at 𝑥₀∈𝐴 and if 𝛼∈𝒩, then 𝑓+𝛼𝑔 is (locally uniformly) continuous at 𝑥₀. Moreover, if 𝐴,𝐵⊂𝒩 and if 𝑓∶𝐴→𝐵 is (locally uniformly) continuous at 𝑥₀∈𝐴 and 𝑔∶𝐵→𝒩 is (locally uniformly) continuous at 𝑓(𝑥₀), then 𝑔∘𝑓∶𝐴→𝒩 is (locally uniformly) continuous at 𝑥₀.

Lemma 3.3. Let 𝑥₀∈𝒩 be given and let Ω be a neighborhood of 𝑥₀. Then there exist sequences (𝑥_𝑛) and (𝑦_𝑛) as well as mutually disjoint clopen sets 𝑈_𝑛 and 𝑈₀ and a continuous function 𝑓 such that(1)the set {𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ} has no limit point;(2)lim_𝑛→∞|𝑥_𝑛−𝑦_𝑛|=0;(3)𝑥₀∉{𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ};(4)𝑥_𝑛∈𝑈_𝑛 for each 𝑛∈ℕ∪{0};(5)𝑦_𝑛∉𝑈_𝑚 for any 𝑛,𝑚∈ℕ;(6)𝑓(∪_𝑛∈ℕ𝑈_𝑛)={1}, 𝑓(𝒩⧵∪_𝑛∈ℕ𝑈_𝑛)={0}.

Proof. Since 𝒩 is not compact, there is a sequence (𝑥_𝑛)⊂Ω that has no limit point in 𝒩. Without loss of generality, we can take 𝑥₀∉{𝑥_𝑛∶𝑛∈ℕ} since {𝑥_𝑛∶𝑛∈ℕ}⧵{𝑥₀} will still have no limit point. For each 𝑛∈ℕ, let 𝑦_𝑛∈𝒩 be such that |𝑦_𝑛−𝑥_𝑛|<𝑑^𝑛 and 𝑦_𝑛≠𝑥_𝑚 for any 𝑛∈ℕ and 𝑚∈ℕ∪{0}. Since (𝑑^𝑛) is a null sequence in 𝒩, it follows that |𝑦_𝑛−𝑥_𝑛|→0 as 𝑛→∞. Assume that {𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ} has a limit point in 𝒩 and let 𝑐 be such a limit point. Since (𝑥_𝑛) has no limit point, there is an 𝜖>0 such that (𝑐−𝜖,𝑐+𝜖)∩{𝑥_𝑛∶𝑛∈ℕ}=∅. There exists 𝑁∈ℕ such that, for all 𝑛≥𝑁, |𝑦_𝑛−𝑥_𝑛|<𝜖/2. Since 𝑐 is the limit point of {𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ}, there must be 𝑀>𝑁 such that 𝑦_𝑀∈(𝑐−𝜖/2,𝑐+𝜖/2). But then |𝑥_𝑀−𝑐|≤|𝑥_𝑀−𝑦_𝑀|+|𝑦_𝑀−𝑐|<𝜖/2+𝜖/2=𝜖. This is a contradiction. Hence {𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ} has no limit point.
Since {𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ} has no limit point, there exist 𝑈^′_𝑛 and 𝑈^0 such that 𝑥₀∈𝑈^0, {𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ}∩𝑈^𝑛={𝑥_𝑛} and {𝑥_𝑛,𝑦_𝑛∶𝑛∈ℕ}∩𝑈^0=∅. For each 𝑛∈ℕ∪{0} let 𝜖_𝑛>0 in 𝒩 be such that 𝜖_𝑛≤𝑑^𝑛 and (𝑥_𝑛−𝜖_𝑛,𝑥_𝑛+𝜖_𝑛)⊂𝑈^𝑛. Then (𝑥₀−𝜖₀,𝑥₀+𝜖₀) and all of (𝑥_𝑛−𝜖_𝑛/2,𝑥_𝑛+𝜖_𝑛/2), 𝑛∈ℕ, are mutually disjoint open sets. By Proposition 2.6, there are clopen neighborhoods 𝑈_𝑛 of 𝑥_𝑛 such that 𝑈_𝑛⊂(𝑥_𝑛−𝜖_𝑛/2,𝑥_𝑛+𝜖_𝑛/2) and a clopen neighborhood 𝑈₀ of 𝑥₀ such that 𝑈₀⊂(𝑥₀−𝜖₀/2,𝑥₀+𝜖₀/2). ⋃^∞𝑛=1𝑈_𝑛 is open as it is the union of open sets, but it is also closed as we will show hereinafter. Let 𝑥∈cl(⋃^∞𝑛=1𝑈_𝑛) be given. Then there exists a sequence (𝑧_𝑚)⊂⋃^∞𝑛=1𝑈_𝑛 such that 𝑧_𝑛→𝑥. (𝑥_𝑛) has no limit points, so 𝑥 is separated from (𝑥_𝑛). Therefore, there exists 𝑁∈ℕ such that (𝑥−𝑑^𝑁,𝑥+𝑑^𝑁)∩{𝑥_𝑛∶𝑛∈ℕ}=∅. Moreover, there exists 𝑀∈ℕ such that for every 𝑚≥𝑀, |𝑧_𝑚−𝑥|<(1/2)𝑑^𝑁. It follows that |𝑧_𝑚−𝑥_𝑘|≥||𝑥_𝑘−𝑥|−|𝑧_𝑚−𝑥||>𝑑^𝑁−(1/2)𝑑^𝑁=(1/2)𝑑^𝑁≫𝑑^𝑘 for all 𝑘>𝑁 and for all 𝑚≥𝑀. 𝑈_𝑘⊂(𝑥_𝑘−𝑑^𝑘,𝑥_𝑘+𝑑^𝑘), for each 𝑘∈ℕ; it follows that {𝑧_𝑚∶𝑚≥𝑀}∩⋃^{∞𝑘=𝑁+1}𝑈_𝑘=∅. That is, {𝑧_𝑚∶𝑚≥𝑀}⊂⋃^𝑁𝑘=1𝑈_𝑘 which is a finite union of closed sets and hence is itself closed. Since 𝑧_𝑚→𝑥 and {𝑧_𝑚∶𝑚≥𝑀}⊂⋃^𝑁𝑘=1𝑈_𝑘 which is closed, it follows that 𝑥∈⋃^𝑁𝑘=1𝑈_𝑘⊂⋃^∞𝑘=1𝑈_𝑘. Thus, ⋃^∞𝑘=1U_𝑘=cl(⋃^∞𝑘=1𝑈_𝑘) and hence ⋃^∞𝑘=1𝑈_𝑘 is closed. Define 𝑓 on 𝒩 as follows:𝑓(𝑥)=⎧⎪⎨⎪⎩1if𝑥∈^∞_𝑛=10otherwise.𝑈_𝑛,(3.1) Then 𝑓 is continuous on 𝒩 because ⋃^∞𝑛=1𝑈_𝑛 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 𝑥₀∈𝒩 be given and let Ω be a neighborhood of 𝑥₀. Then there is a continuous function 𝑓∶Ω→𝒩 that is not locally uniformly continuous at 𝑥₀.

Proof. Let Ω₁=Ω∩(𝑥₀−𝑑,𝑥₀+𝑑) and apply Lemma 3.3 to 𝑥₀ and Ω₁ to get 𝑓₁, 𝑈₁, (𝑥_1,𝑛), and (𝑦_1,𝑛) which correspond to 𝑓, 𝑈₀, (𝑥_𝑛), and (𝑦_𝑛), respectively, in that lemma. Let Ω₂=𝑈₁∩(𝑥₀−𝑑²,𝑥₀+𝑑²) and apply Lemma 3.3 to 𝑥₀ and Ω₂ to get 𝑓₂, 𝑈₂, (𝑥_2,𝑛), and (𝑦_2,𝑛). Continuing inductively, we can apply Lemma 3.3 to 𝑥₀ and Ω_𝑘+1=𝑈_𝑘∩(𝑥₀−𝑑^𝑘+1,𝑥₀+𝑑^𝑘+1) in order to get 𝑈_𝑘+1, 𝑓_𝑘+1, (𝑥_𝑘+1,𝑛), and (𝑦_𝑘+1,𝑛). The resulting 𝑓_𝑘’s satisfy 𝑓_𝑘(𝑦_𝑙,𝑛)=0 for every 𝑘,𝑙,𝑛∈ℕ and 𝑓_𝑘(𝑥_𝑙,𝑛)=𝛿_𝑘,𝑙 (the Kronecker delta) for every 𝑘,𝑙,𝑛∈ℕ. Let 𝑓=∑^∞𝑘=1𝑑^𝑘𝑓_𝑘, which converges (pointwise), by Proposition 2.13, since |𝑑^𝑘𝑓_𝑘(𝑥)|≤𝑑^𝑘→0 as 𝑘→∞ for all 𝑥∈Ω.
To show that 𝑓 is continuous on Ω, let 𝑡∈Ω be given. Let 𝜖>0 in 𝒩 be given and let 𝑁∈ℕ be such that 𝑑^𝑁≪𝜖. For each 𝑛<𝑁, let 𝛿_𝑛>0 be such that |𝑓_𝑛(𝑥)−𝑓_𝑛(𝑡)|<𝑑^𝑁 whenever 𝑥∈Ω and |𝑥−𝑡|<𝛿_𝑛, which is possible since each 𝑓_𝑛 is continuous at 𝑡. Let 𝛿=min{𝛿_𝑛∶𝑛<𝑁}. Then for all 𝑥∈Ω satisfying |𝑥−𝑡|<𝛿, we have that||𝑓(𝑥)−𝑓(𝑡)||=|||||^∞_𝑛=1𝑑^𝑛𝑓_𝑛(𝑥)−𝑓_𝑛(𝑡)|||||≤|||||^𝑁_𝑛=1+|||||𝑑^𝑛𝑓_𝑛(𝑥)−𝑓_𝑛(𝑡)_∞|||||_𝑛=_𝑁+1𝑑^𝑛𝑓_𝑛(𝑥)|||||+|||||^∞_𝑛=_𝑁+1𝑑^𝑛𝑓_𝑛(𝑡)|||||≤^𝑁_𝑛=1+|||||_∞𝑑^𝑛||𝑑^𝑁||_𝑛=_𝑁+1𝑑^𝑛|||||+|||||_∞𝑛=_𝑁+1𝑑^𝑛|||||<𝑑_𝑁+𝑑^𝑁+𝑑^𝑁≪𝜖.(3.2) Thus, 𝑓 is continuous at 𝑡, for all 𝑡∈Ω, and hence 𝑓 is continuous on Ω.
Now we show that 𝑓 is not locally uniformly continuous at 𝑥₀. Let Δ be a neighborhood of 𝑥₀, let 𝑀∈ℕ be such that (𝑥₀−𝑑^𝑀,𝑥₀+𝑑^𝑀)⊂Δ, and let 𝜖=(1/2)𝑑^𝑀. So Ω_𝑀⊂(𝑥₀−𝑑^𝑀,𝑥₀+𝑑^𝑀)⊂Δ. It follows that 𝑥_𝑀,𝑛∈Ω_𝑀⊂Δ and 𝑦_𝑀,𝑛∈Ω_𝑀⊂Δ are such that |𝑦_𝑀,𝑛−𝑥_𝑀,𝑛|→0 as 𝑛→∞, but||𝑓𝑦_𝑀,𝑛−𝑓𝑥_𝑀,𝑛||=|||||^∞_𝑘=1𝑑^𝑘𝑓_𝑘𝑦_𝑀,𝑛−𝑓_𝑘𝑥_𝑀,𝑛|||||=𝑑^𝑀>𝜖.(3.3) Therefore 𝑓 is not locally uniformly continuous at 𝑥₀.

One can in fact show that there are continuous functions which are not locally uniformly continuous without using the property of total disconnectedness of 𝒩. 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 𝑥𝑦_{𝑚,𝑛𝑚,𝑛}=𝑛𝑑^𝑚,𝑈_𝑚,𝑛=𝑛𝑑^𝑚+𝑑^𝑛+𝑚,=𝑥∈𝒩∶||𝑥−𝑛𝑑^𝑚||≪_∼𝑑^𝑚+𝑚+1𝑈_𝑚,𝑜,=𝑥∈𝒩∶|𝑥|≪_∼𝑑^𝑚+1,(3.4) and let 𝑓(𝑥)=𝑑^𝑚0oifth𝑥er∈w𝑈ise._𝑚,𝑛forsome𝑚,𝑛∈ℕ,(3.5) Then 𝑓 is well defined on 𝒩 since the 𝑈_𝑚,𝑛’s are mutually disjoint sets. We will show that 𝑓 is continuous on 𝒩 but 𝑓 is not locally uniformly continuous at 0. Let 𝑡∈𝒩 be given. We will distinguish the following three cases.

Case 1 (𝑡<0). In this case, the function 𝑓 is constant on the open set (−∞,0) containing 𝑡, and hence 𝑓 is continuous at 𝑡.

Case 2 (𝑡>0). Let 𝑀∈ℕ be such that 𝑑^𝑀+1≪𝑡≪_∼𝑑^𝑀. Then 𝑡∈𝑈_𝑀−1,0 but 𝑡∉𝑈_𝑀,0. Let Δ=𝑈_𝑀−1,0⧵𝑈_𝑀,0. Then Δ is clopen, and ⋃^∞𝑛=1𝑈_𝑀,𝑛⊂Δ is also clopen. 𝑓(⋃^∞𝑛=1𝑈_𝑀,𝑛)={𝑑^𝑀} and 𝑓(Δ⧵⋃^∞𝑛=1𝑈_𝑀,𝑛)={0}. Therefore 𝑓 is continuous on Δ since 𝑓 is constant on disjoint open sets that cover Δ. Hence 𝑓 is continuous at 𝑡∈Δ.

Case 3 (𝑡=0). Let 𝜖>0 in 𝒩 be given and let 𝑀∈ℕ be such that 𝑑^𝑀≪𝜖. Let 𝛿=𝑑^𝑀+1. Then, for |𝑥|<𝛿, we have that 𝑥∉𝑈_𝑘,𝑛 for all 𝑘<𝑀 and every 𝑛∈ℕ. So |𝑓(𝑥)−𝑓(0)|≤𝑑^𝑀≪𝜖. This shows that 𝑓 is continuous at 𝑡=0.

Thus, 𝑓 is continuous at 𝑡 for all 𝑡∈𝒩 and hence 𝑓 is continuous on 𝒩. To see that 𝑓 is not locally uniformly continuous at 0, let Ω be any neighborhood of 0. Let 𝑀∈ℕ be such that (−𝑑^𝑀,𝑑^𝑀)⊂Ω and let 𝜖=(1/2)𝑑^𝑀. Then the sequences (𝑥_𝑀,𝑛)and(𝑦_𝑀,𝑛) in Ω defined previously are such that |𝑦_𝑀,𝑛−𝑥_𝑀,𝑛|=𝑑^𝑛+𝑀→0 as 𝑛→∞, but |𝑓(𝑦_𝑀,𝑛)−𝑓(𝑥_𝑀,𝑛)|=𝑑^𝑀>𝜖. Thus, for every neighborhood Ω of 0, there exists 𝜖₀>0 such that for each 𝛿>0, there are 𝑥,𝑦∈Ω such that |𝑦−𝑥|<𝛿 while |𝑓(𝑦)−𝑓(𝑥)|≥𝜖₀, which shows that 𝑓 is not locally uniformly continuous at 0.

4. Local Uniform Differentiability

Definition 4.1. Let 𝐴⊂𝒩, let 𝑓∶𝐴→𝒩, and let 𝑥₀∈𝐴 be given. Then we say that 𝑓 is locally uniformly differentiable (LUD) at 𝑥₀ if there is a neighborhood Ω of 𝑥₀ in 𝐴 such that for every 𝜖>0, there exists 𝛿>0 such that |𝑓(𝑦)−𝑓(𝑥)−𝑓^′(𝑥)(𝑦−𝑥)|<𝜖|𝑦−𝑥| whenever 𝑥,𝑦∈Ω and |𝑦−𝑥|<𝛿.

Definition 4.2. Let 𝐴⊂𝒩 and let 𝑓∶𝐴→𝒩. Then we say that 𝑓 is locally uniformly differentiable (LUD) on 𝐴 if 𝑓 is locally uniformly differentiable at 𝑥 for all 𝑥∈𝐴.

Definition 4.3. Let 𝐴⊂𝒩, let 𝑓∶𝐴→𝒩, and let 𝑘∈ℕ be given. Then we say that 𝑓 is LUD^𝑘 if 𝑘th derivative of 𝑓, 𝑓^(𝑘), exists and 𝑓^(𝑙) is LUD on 𝐴 for each 𝑙∈{0,1,…,𝑘−1}.
The following two results follow readily from Definition 4.3.

Proposition 4.4. Let 𝐴⊂𝒩 and let 𝑓∶𝐴→𝒩 be LUD^𝑘. Then 𝑓^(𝑛) is LUD^𝑙 on 𝐴 for all 𝑙≥1 and 𝑛≥0 satisfying 𝑙+𝑛≤𝑘.

Proposition 4.5. Let 𝑙∈ℕ be given, let 𝐴⊂𝒩, and let 𝑓∶𝐴→𝒩 be such that 𝑓 is LUD^𝑙 and 𝑓^(𝑙) is LUD on 𝐴. Then f is LUD^𝑙+1 on 𝐴.

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 𝑥_𝑚,𝑛, 𝑦_𝑚,𝑛, and 𝑈_𝑚,𝑛 be as in Example 3.5. Define 𝑔∶𝒩→𝒩 by 𝑔(𝑥)=(𝑛−1)²0𝑑^2𝑚iofth𝑥er∈w𝑈ise._𝑚,𝑛forsome𝑚,𝑛∈ℕ,(4.1) Then 𝑔 is well defined since the 𝑈_𝑚,𝑛 are mutually disjoint. Let 𝑥∈𝒩 be given. If 𝑥∈𝑈_𝑚,𝑛 for some 𝑚,𝑛∈ℕ, then 𝑥≈𝑛𝑑^𝑚 and hence |𝑔(𝑥)|=(𝑛−1)²𝑑^2𝑚<|𝑥|². Also, if 𝑥∉𝑈_𝑚,𝑛 for any 𝑚,𝑛∈ℕ, then |𝑔(𝑥)|=0<|𝑥|². Therefore, |𝑔(𝑥)|<|𝑥|² for all 𝑥∈𝒩.
Note that 𝑔 is locally constant on 𝒩⧵{0} and hence 𝑔^′=0 on 𝒩⧵{0}. We will show that 𝑔 is differentiable at 0 with 𝑔^(0)=0 too. So let 𝜖>0 in 𝒩 be given. Let 𝛿=𝜖. Then for 0<|𝑥|<𝛿, we have that|||𝑔(𝑥)−𝑔(0)𝑥|||=|||𝑔𝑥(𝑥)|||<|𝑥|<𝛿=𝜖.(4.2) This shows that 𝑔 is differentiable at 0 with 𝑔^(0)=0. Altogether, 𝑔^=0 on 𝒩. Therefore 𝑔 is 𝐶^∞ on 𝒩 with 𝑔^(𝑘)=0 for all 𝑘∈ℕ.
Now we show that 𝑔 is not LUD at 0. Consider the sequences (𝑥_𝑛,𝑛)_𝑛∈ℕ and (𝑦_𝑛,𝑛)_𝑛∈ℕ, where 𝑥_𝑛,𝑛=𝑛𝑑^𝑛 and 𝑦_𝑛,𝑛=𝑛𝑑^𝑛+𝑑^2𝑛. Both of these sequences converge to zero. Moreover,||𝑦_𝑛,𝑛−𝑥_𝑛,𝑛||=𝑑^2𝑛⟶0as𝑛⟶∞,(4.3) but ||𝑔𝑦_𝑛,𝑛−𝑔𝑥_𝑛,𝑛||=(𝑛−1)²𝑑^2𝑛≥𝑑^2𝑛=||𝑦_𝑛,𝑛−𝑥_𝑛,𝑛||for𝑛≥2.(4.4) Thus, for any neighborhood Ω of 0, 𝜖₀=1, and for any 𝛿>0, there are 𝑦,𝑥∈Ω such that |𝑦−𝑥|<𝛿, but |𝑔(𝑦)−𝑔(𝑥)−𝑔^′(𝑥)(𝑦−𝑥)|≥|𝑦−𝑥|. This shows that 𝑔 is not LUD at 0.

Example 4.6 shows that the property LUD is not necessarily inherited from the derivatives of the function, since 𝑔^′=0 is LUD^∞. Here is another example that shows that the LUD property is not passed on from a function to its derivatives.

Example 4.7. Let 𝑓∶𝒩→𝒩 be given by 𝑓(𝑥)=𝑥𝑑^𝜆(𝑥)0iiff𝑥𝑥≠=00,.(4.5) We will show that 𝑓 is LUD on 𝒩 with derivative 𝑓^′(𝑥)=𝑔(𝑥)=𝑑^𝜆(𝑥)0iiff𝑥𝑥≠=00,,(4.6) and then we will show that 𝑓^′ is not LUD by showing that it is not LUD at 0. So let 𝑥₀≠0 be given. Let Ω={𝑥∈𝒩∶𝑥∼𝑥₀} which is an open (clopen) neighborhood of 𝑥₀. Then for all 𝑥,𝑦∈Ω, we have that ||𝑓(𝑦)−𝑓(𝑥)−𝑔(𝑥)(𝑦−𝑥)||==||||𝑦𝑦𝑑𝑑^𝜆(𝑦)_𝜆(𝑥0)−𝑥𝑑^𝜆(𝑥)−𝑑^𝜆(𝑥)(𝑦−𝑥)||−𝑥𝑑^𝜆(𝑥0)−𝑑^𝜆(𝑥0)(𝑦−𝑥)||=0.(4.7) This shows that 𝑓 is locally uniformly differentiable at 𝑥₀ with derivative 𝑓^′(𝑥₀)=𝑔(𝑥₀)=𝑑^𝜆(𝑥0).
For 𝑥₀=0, let 𝜖>0 in 𝒩 be given. Let 𝛿=𝜖𝑑. Then for all 𝑥≠𝑦 in (−𝛿,𝛿), we will show that||𝑓(𝑦)−𝑓(𝑥)−𝑔(𝑥)(𝑦−𝑥)||<𝜖||𝑦−𝑥||.(4.8)Case 1 (𝑥=0). Then 𝑓(𝑥)=𝑔(𝑥)=0 and 𝑦≠0. It follows that ||𝑓(𝑦)−𝑓(𝑥)−𝑔(𝑥)(𝑦−𝑥)||=||𝑓(𝑦)||=𝑑^𝜆(𝑦)||𝑦||=𝑑^𝜆(𝑦)||𝑦−𝑥||<𝜖||𝑦−𝑥||.(4.9) because 𝑑^𝜆(𝑦)∼|𝑦|<𝛿≪𝜖.Case 2 (𝑦=0). Then 𝑓(𝑦)=0 and 𝑥≠0. It follows that ||𝑓(𝑦)−𝑓(𝑥)−𝑔(𝑥)(𝑦−𝑥)||==0||−<𝑓(𝜖𝑥||𝑦)+−𝑔𝑥||(𝑥.)𝑥||=||−𝑑^𝜆(𝑥)𝑥+𝑑^𝜆(𝑥)𝑥||(4.10)Case 3 (𝑥≠0≠𝑦). Then||𝑓(𝑦)−𝑓(𝑥)−𝑔(𝑥)(𝑦−𝑥)||==||||𝑑𝑑^𝜆(𝑦)𝑦−𝑑^𝜆(𝑥)𝑥−𝑑^𝜆(𝑥)(𝑦−𝑥)_𝜆(𝑦)||<𝑑₋₁−𝑑^𝜆(𝑥)||||𝑦||<𝑑⁻¹||𝑦−𝑥||||𝑦||𝛿||𝑦−𝑥||=𝜖||𝑦−𝑥||.(4.11) Thus, 𝑓 is locally uniformly differentiable at 0 with derivative 𝑓^(0)=𝑔(0)=0. Altogether, it follows that 𝑓 is LUD on 𝒩 with derivative 𝑓^′(𝑥)=𝑔(𝑥)=𝑑^𝜆(𝑥)0iiff𝑥𝑥≠=00,.(4.12)
Next we show that 𝑓^′ is not differentiable at 0. Take the sequence (𝑥_𝑛)=𝑑^𝑛 and the sequence (𝑦_𝑛)=2𝑑^𝑛. Then both sequences converge to 0. Butlim_𝑛⟶∞𝑓^𝑥_𝑛−𝑓^(0)𝑥_𝑛=lim_𝑛⟶∞𝑑^𝑛𝑑^𝑛lim_𝑛⟶∞=1,𝑓^𝑦_𝑛−𝑓^(0)𝑦_𝑛=lim_𝑛⟶∞𝑑^𝑛2𝑑^𝑛=12.(4.13) If 𝑓^(𝑥) were differentiable at 0, then both limits in (4.13) would be equal to 𝑓^(0). Since the two limits are different, we conclude that 𝑓^ is not differentiable at 0, and hence 𝑓^ is not LUD on 𝒩.
The previous two examples illustrate that the most natural definition for LUD^𝑘 is given in Definition 4.3.

Proposition 4.8. Let 𝑓∶𝐴→𝒩 be LUD at 𝑥₀∈𝐴. Then 𝑓 is 𝐶¹ at 𝑥₀.

Proof. Let Ω be a neighborhood of 𝑥₀ in 𝐴 such that 𝑓 is uniformly differentiable on Ω and let 𝛿₀>0 be such that (𝑥₀−𝛿₀,𝑥₀+𝛿₀)⊂Ω. Let 𝜖>0 in 𝒩 be given. Then there is 𝛿>0, 𝛿≤𝛿₀, such that for all 𝑥,𝑦∈Ω satisfying |𝑦−𝑥|<𝛿 we have that |||𝑓(𝑦)−𝑓(𝑥)−𝑓^′(𝑥)(𝑦−𝑥)|||<𝜖2||𝑦−𝑥||.(4.14) It follows that for 0<|𝑥−𝑥₀|<𝛿, ||𝑓^(𝑥)−𝑓^𝑥₀||≤||||𝑓𝑥⁰−𝑓(𝑥)𝑥₀−𝑥−𝑓^(𝑥)||||+||||𝑓(𝑥)−𝑓𝑥⁰𝑥−𝑥₀−𝑓^𝑥₀||||<𝜖2+2𝜖=𝜖.(4.15)

Remark 4.9. Proposition 4.8 shows that the class of locally uniformly differentiable functions is a subset of the class of 𝐶¹ functions. However, this is still large enough to include all polynomial functions as Corollary 4.14 will show.

Proposition 4.10. Let 𝑓∶𝐴→𝒩 be locally uniformly differentiable at 𝑥₀∈𝐴. Then 𝑓 is locally uniformly continuous at 𝑥₀.

Proof. Let Δ be a neighborhood of 𝑥₀ in 𝐴 such that 𝑓 is uniformly differentiable on Δ. By Proposition 4.8, 𝑓^ is continuous at 𝑥₀. Let Ω⊂Δ be a neighborhood of 𝑥₀ such that for every 𝑥∈Ω, |𝑓^(𝑥)|<1+|𝑓^(𝑥₀)|. Since 𝑓 is uniformly differentiable on Δ, there exists 𝛿₁>0 such that for all 𝑥,𝑦∈Ω⊂Δ satisfying |𝑦−𝑥|<𝛿₁, we have that ||𝑓(𝑦)−𝑓(𝑥)−𝑓^(𝑥)(𝑦−𝑥)||<||𝑦−𝑥||.(4.16) Let 𝜖>0 in 𝒩 be given. Let 𝛿=min{𝛿₁,𝜖/(2+|𝑓^(𝑥₀)|)}. Then for all 𝑥,𝑦∈Ω⊂Δ satisfying |𝑦−𝑥|<𝛿, we have that ||𝑓(𝑦)−𝑓(𝑥)||<<𝛿||𝑦−2𝑥+||||+𝑓_||𝑓^(𝑥)||||𝑦−𝑥||<||𝑦−𝑥||2+||𝑓^𝑥₀||𝑥₀||<𝜖.(4.17)

Proposition 4.11. Let 𝐴⊂𝒩, let 𝑥₀∈𝐴 be given, let 𝛼∈𝒩 be given, and let 𝑓,𝑔∶𝐴→𝒩 be LUD at 𝑥₀. Then 𝑓+𝛼𝑔 is LUD at 𝑥₀, with derivative (𝑓+𝛼𝑔)^′𝑥₀=𝑓^𝑥₀+𝛼𝑔^𝑥₀.(4.18)

Proof. Without loss of generality, we may assume that 𝛼≠0. Let 𝜖>0 in 𝒩 be given. Then there exists 𝛿>0 in 𝒩 such that (𝑥₀−𝛿,𝑥₀+𝛿)⊂𝐴, ||𝑓(𝑦)−𝑓(𝑥)−𝑓^(𝑥)(𝑦−𝑥)||<𝜖||𝑔(𝑦)−𝑔(𝑥)−𝑔_2||𝑦−𝑥||,(𝑥)(𝑦−𝑥)||<𝜖2|𝛼|||𝑦−𝑥||,(4.19) whenever 𝑥,𝑦∈(𝑥₀−𝛿,𝑥₀+𝛿). It follows that, for all 𝑥,𝑦∈(𝑥₀−𝛿,𝑥₀+𝛿), we have that ||(𝑓=+||𝛼𝑓𝑔()𝑦()𝑦)−−𝑓((𝑓𝑥+)−𝛼𝑔𝑓)_(𝑥)−𝑓^(𝑥)(𝑦−𝑥)−𝛼𝑔^(𝑥)(𝑦−𝑥)||≤||𝑓(𝑦)−𝑓(𝑥)−𝑓_(𝑥)(𝑦−𝑥)+𝛼𝑔(𝑦)−𝑔(𝑥)−𝑔^(𝑥)(𝑦−𝑥)||<𝜖(𝑥)(𝑦−𝑥)||+|𝛼|||𝑔(𝑦)−𝑔(𝑥)−𝑔^(𝑥)(𝑦−𝑥)||2||𝑦−𝑥||+|𝛼|2𝜖|𝛼|||𝑦−𝑥||=𝜖||𝑦−𝑥||.(4.20)

Proposition 4.12 (Chain Rule). Let 𝐴,𝐵⊂𝒩, let 𝑥₀∈𝐴 be given, and let 𝑔∶𝐴→𝐵 and 𝑓∶𝐵→𝒩 be such that 𝑔 is LUD at 𝑥₀ and 𝑓 is LUD at 𝑔(𝑥₀). Then 𝑓∘𝑔 is LUD at 𝑥₀ with derivative (𝑓∘𝑔)^′(𝑥₀)=(𝑓^∘𝑔(𝑥₀))⋅𝑔^(𝑥₀).

Proof. Let Δ be a neighborhood of 𝑔(𝑥₀) in 𝐵 such that 𝑓 is uniformly differentiable on Δ, and let Ω be a neighborhood of 𝑥₀ in 𝐴 such that 𝑔(Ω)⊂Δ and 𝑔 is uniformly differentiable and uniformly continuous on Ω (Proposition 4.10). The condition that 𝑔(Ω)⊂Δ is always possible because 𝑔 is continuous. Since 𝑔 is continuous at 𝑥₀ and 𝑓^ is continuous at 𝑔(𝑥₀) (Proposition 4.8), it follows that 𝑓^∘𝑔 is continuous at 𝑥₀. So there exists 𝛿₁>0 in 𝒩 such that (𝑥₀−𝛿₁,𝑥₀+𝛿₁)⊂Ω and for all 𝑥∈(𝑥₀−𝛿₁,𝑥₀+𝛿₁) we have that ||𝑓^∘𝑔(𝑥)||<||𝑓^∘𝑔𝑥₀||+1.(4.21) Also, since 𝑔^ is continuous at 𝑥₀, there is a 𝛿₂>0 in 𝒩 such that (𝑥₀−𝛿₂,𝑥₀+𝛿₂)⊂Ω, and for all 𝑥∈(𝑥₀−𝛿₂,𝑥₀+𝛿₂), we have that ||𝑔^(𝑥)||<||𝑔^𝑥₀||+12.(4.22) Since 𝑔 is uniformly differentiable on Ω, there is a 𝛿₃>0 in 𝒩 such that (𝑥₀−𝛿₃,𝑥₀+𝛿₃)⊂Ω, and for all 𝑥,𝑦∈(𝑥₀−𝛿₃,𝑥₀+𝛿₃) satisfying |𝑦−𝑥|<𝛿₃, we have that ||𝑔(𝑦)−𝑔(𝑥)||<||𝑔^(𝑥)||+12||𝑦−𝑥||≤||𝑔^𝑥₀||+1||𝑦−𝑥||.(4.23) Let 𝜖>0 in 𝒩 be given. Then, since 𝑓 is uniformly differentiable on Δ, there exists an 𝜂>0 such that whenever 𝑢,𝑣∈Δ and |𝑢−𝑣|<𝜂, it follows that ||𝑓(𝑣)−𝑓(𝑢)−𝑓^(𝑢)(𝑣−𝑢)||<𝜖2||𝑔^𝑥₀||+1|𝑣−𝑢|.(4.24) Since 𝑔 is uniformly differentiable on Ω, there exists 𝛿₄>0 in 𝒩 such that (𝑥₀−𝛿₄,𝑥₀+𝛿₄)⊂Ω, and for all 𝑥,𝑦∈(𝑥₀−𝛿₄,𝑥₀+𝛿₄) satisfying |𝑦−𝑥|<𝛿₄, we have that ||𝑔(𝑦)−𝑔(𝑥)−𝑔^(𝑥)(𝑦−𝑥)||<𝜖2||𝑓^∘𝑔𝑥₀||+1||𝑦−𝑥||.(4.25) Finally, 𝑔 is uniformly continuous on Ω. Thus, there exists 𝛿₅>0 in 𝒩 such that (𝑥₀−𝛿₅,𝑥₀+𝛿₅)⊂Ω, and for all 𝑥,𝑦∈(𝑥₀−𝛿₅,𝑥₀+𝛿₅) satisfying |𝑦−𝑥|<𝛿₅, we have that ||𝑔(𝑦)−𝑔(𝑥)||<𝜂.(4.26) Let 𝛿=min𝛿₁,𝛿₂,𝛿₃,𝛿₄,𝛿₅.(4.27) Then (𝑥₀−𝛿,𝑥₀+𝛿)⊂Ω, and for all 𝑥,𝑦∈(𝑥₀−𝛿/2,𝑥₀+𝛿/2), we have that |𝑦−𝑥|<𝛿 and |𝑔(𝑦)−𝑔(𝑥)|<𝜂; 𝑔(𝑦),𝑔(𝑥)∈Δ and hence ||𝑓∘𝑔≤(𝑦||𝑓)−∘𝑔𝑓(∘𝑦𝑔)−(𝑥𝑓)−∘𝑔𝑓(^𝑥∘)𝑔−(𝑥𝑓)_𝑔^(𝑥)(𝑦−𝑥)||+||𝑓∘𝑔(𝑥)(𝑔(𝑦)−𝑔(𝑥))_||=||𝑓(𝑔∘(𝑦𝑔))(𝑥−)𝑓𝑔((𝑦𝑔)(−𝑥)𝑓)−^∘𝑓𝑔_(𝑥)𝑔(𝑥)−𝑓^∘𝑔(𝑥)𝑔^(𝑥)(𝑦−𝑥)||+||𝑓(𝑔(𝑥))(𝑔(𝑦)−𝑔(𝑥))_||<∘𝑔𝜖(𝑥)||||𝑔(𝑦)−𝑔(𝑥)−𝑔^(𝑥)(𝑦−𝑥)||2||𝑔^𝑥₀||+1||𝑔(𝑦)−𝑔(𝑥)||+2||𝑓^∘𝑔𝜖𝑥₀||+1||𝑓^<𝜖∘𝑔𝑥₀||+1||𝑦−𝑥||2||𝑦−𝑥||+2𝜖||𝑦−𝑥||=𝜖||𝑦−𝑥||.(4.28)

Proposition 4.13 (Product Rule). Let 𝐴⊂𝒩, let 𝑥₀∈𝐴 be given, and let 𝑓,𝑔∶𝐴→𝒩 be LUD at 𝑥₀. Then 𝑓𝑔 is LUD at 𝑥₀ with (𝑓𝑔)^′(𝑥₀)=𝑓^(𝑥₀)𝑔(𝑥₀)+𝑓(𝑥₀)𝑔^(𝑥₀).

Proof. Let Ω be a neighborhood of 𝑥₀ in 𝐴 such that 𝑓 and 𝑔 are uniformly differentiable on Ω and 𝑔 is uniformly continuous on Ω (Proposition 4.10). Since 𝑓, 𝑔, and 𝑓^ are continuous at 𝑥₀, there exists 𝛿₁>0 in 𝒩 such that (𝑥₀−𝛿₁,𝑥₀+𝛿₁)⊂Ω, and for all 𝑥∈(𝑥₀−𝛿₁,𝑥₀+𝛿₁), we have that ||𝑓(𝑥)||<||𝑓𝑥₀||+1,||𝑔(𝑥)||<||𝑔𝑥₀||+1,||𝑓^(𝑥)||<||𝑓^𝑥₀||+1.(4.29) Let 𝜖>0 in 𝒩 be given. Since 𝑓 and 𝑔 are uniformly differentiable on Ω, and 𝑔 is uniformly continuous on Ω, there exists 𝛿>0 such that whenever |𝑦−𝑥|<𝛿 and 𝑥,𝑦∈Ω, we have that ||𝑔(𝑦)−𝑔(𝑥)||<𝜖||𝑓(𝑦)−𝑓(𝑥)−𝑓_3||𝑓^𝑥₀||+1,(𝑥)(𝑦−𝑥)||<𝜖||𝑔(𝑦)−𝑔(𝑥)−𝑔_3||𝑔𝑥₀||+1||𝑦−𝑥||,(𝑥)(𝑦−𝑥)||<𝜖3||𝑓𝑥₀||+1||𝑦−𝑥||.(4.30) Let 𝑥,𝑦∈(𝑥₀−𝛿₁,𝑥₀+𝛿₁) be such that |𝑦−𝑥|<𝛿. Then, ||𝑓(𝑦≤)𝑔||𝑓(𝑦()𝑦−)𝑔𝑓((𝑦𝑥))−𝑔𝑓(𝑥()𝑥−)𝑔𝑓(𝑦^)(−𝑥)𝑓𝑔_(𝑥)+𝑓(𝑥)𝑔^(𝑥)(𝑦−𝑥)||+||𝑓(𝑥)𝑔(𝑦)−𝑓(𝑥)𝑔(𝑥)−𝑓(𝑥(𝑥))𝑔𝑔(_𝑥)(𝑦−𝑥)||≤||𝑓(𝑦)𝑔(𝑦)−𝑓(𝑥)𝑔(𝑦)−𝑓_(𝑥)(𝑦−𝑥)||+||𝑓(𝑥)𝑔(𝑦)(𝑦−𝑥)_||+||𝑓((𝑥𝑥))𝑔(𝑦(𝑦−)−𝑥)𝑓(𝑔(𝑥(𝑦))𝑔−(𝑥𝑔)(−𝑥)𝑓)||(𝑥)𝑔_=||𝑔(𝑦)||||𝑓(𝑦)−𝑓(𝑥)−𝑓_(𝑥)(𝑦−𝑥)||+||𝑓(𝑥)(𝑦−𝑥)_||+||𝑓((𝑥𝑥))||||||𝑔||𝑔((𝑦𝑦))−−𝑔𝑔((𝑥𝑥))−||||𝑦𝑔−_𝑥||<||𝑔𝑥(𝑥)(𝑦−𝑥)₀||||+1𝜖+||𝑓3||𝑔𝑥₀||+1||𝑦−𝑥_||𝑥₀||+1𝜖+||𝑓𝑥3||𝑓^𝑥₀||+1||𝑦−𝑥₀||||+1𝜖=𝜖3||𝑓𝑥₀||+1||𝑦−𝑥||3+3𝜖+3𝜖||𝑦−𝑥||=𝜖||𝑦−𝑥||.(4.31)

Since the function 𝑓(𝑥)=𝑥 is LUD on 𝒩, as can be easily checked, it follows from Propositions 4.11 and 4.13 that any polynomial function is LUD on 𝒩.

Corollary 4.14. Let 𝑓∶𝒩→𝒩 be a polynomial function. Then 𝑓 is LUD on 𝒩.

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

Corollary 4.15. Let 𝑓∶𝒩→𝒩 be a polynomial function. Then 𝑓 is LUD^∞ on 𝒩.

Proposition 4.16. The function ℎ∶𝒩⧵{0}→𝒩⧵{0} defined as ℎ(𝑥)=1/𝑥 is LUD^∞.

Proof. First we note that ℎ is infinitely often differentiable on 𝒩⧵{0}, with derivatives ℎ^(𝑙)(𝑥)=(−1)^𝑙𝑙!𝑥^𝑙+1foreach𝑙∈ℕ.(4.32) Now we prove that ℎ is LUD on 𝒩⧵{0}. So let 𝑥₀∈𝒩⧵{0} and let 𝜖>0 in 𝒩 be given. Let Ω=𝑥∈𝒩∶||𝑥−𝑥₀||<min||𝑥₀||𝑑,𝜖𝑑||𝑥₀||³.(4.33) Then Ω is a neighborhood of 𝑥₀ and 𝑥≈𝑥₀ for all 𝑥∈Ω. Moreover, for all 𝑥,𝑦∈Ω, we have that |||ℎ(𝑦)−ℎ(𝑥)+1𝑥²(𝑦−𝑥)|||=||||1𝑦−𝑥1+𝑥1²(𝑦−𝑥)||||=||||1𝑥²−1=||𝑦−𝑥||𝑥𝑦||||||𝑦−𝑥||𝑥²||𝑦||||𝑦−𝑥||<𝜖||𝑦−𝑥||,(4.34) since 𝑥²||𝑦||≈||𝑥₀||³,||𝑦−𝑥||≤||𝑦−𝑥₀||+||𝑥−𝑥₀||<2𝜖𝑑||𝑥₀||³≪𝜖||𝑥₀||³,(4.35) so that ||𝑦−𝑥||𝑥²||𝑦||<𝜖.(4.36) Thus, for all 𝑥₀∈𝒩⧵{0}, ℎ is LUD at 𝑥₀, and hence ℎ is LUD on 𝒩⧵{0}.
Applying Propositions 4.11 and 4.13, it then follows that ℎ^(𝑙)(𝑥)=(−1)^𝑙𝑙!/𝑥^𝑙+1 is LUD on 𝒩⧵{0} for all 𝑙∈ℕ, and hence ℎ is LUD^∞ on 𝒩⧵{0}.

5. Inverse Function Theorem

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

Theorem 5.1 (inverse function theorem). Let 𝐴⊂𝒩 be open, let 𝑓∶𝐴→𝒩 be locally uniformly differentiable on 𝐴, and let 𝑥₀∈𝐴 be such that 𝑓^′(𝑥₀)≠0. Then there is a neighborhood Ω of 𝑥₀ in 𝐴 and a function 𝑔∶𝑓(Ω)→𝒩, such that(i)𝑔=𝑓|^Ω−1;(ii)𝑓|_Ω is injective;(iii)𝑓(Ω) is open;(iv)𝑔 is locally uniformly differentiable on 𝑓(Ω), with 𝑔^=1/𝑓^∘𝑔.

In this paper, we state and prove a more general version of the inverse function theorem for functions from (a subset of) 𝒩 to 𝒩.

Theorem 5.2 (General Inverse Function Theorem for One-Variable Functions). Let 𝐴⊂𝒩 be open, let 𝑓∶𝐴→𝒩 be LUD^𝑙 on 𝐴, and let 𝑥₀∈𝐴 be such that 𝑓^′(𝑥₀)≠0. Then there is a neighborhood Ω of 𝑥₀ in 𝐴 and a function 𝑔∶𝑓(Ω)→𝒩, such that(i)𝑔=𝑓|^Ω−1;(ii)𝑓|_Ω is injective;(iii)𝑓(Ω) is open;(iv)𝑔 is LUD^𝑙 on 𝑓(Ω); (v)𝑔^′=1/𝑓^∘𝑔.

Proof. (i), (ii), (iii), and (v) are proven in [14]. To prove (iv), first recall that the function ℎ∶𝒩⧵{0}→𝒩⧵{0} given by ℎ(𝑥)=1/𝑥 is LUD^∞ on 𝒩⧵{0} (by Proposition 4.16). Let 𝑦∈𝑓(Ω) be given; then there exists 𝜉∈Ω such that 𝑦=𝑓(𝜉). We show by induction on 𝑛 that 𝑔 is LUD^𝑛 at 𝑓(𝜉) for all 𝑛≤𝑙. We know that 𝑔 is LUD¹ at 𝑓(𝜉) from [14]. Now assume that 𝑔 is LUD^𝑘 (𝑘<𝑙) at 𝑓(𝜉). Then 𝑔^(𝑘)=(𝑔^′)^(𝑘−1)=(ℎ∘𝑓^∘𝑔)^(𝑘−1) (by (v)). We have that ℎ is LUD^𝑘 on 𝒩⧵{0} (by Proposition 4.16), 𝑓^′ is LUD^𝑘 at 𝜉 (by Proposition 4.4), and 𝑔 is LUD^𝑘 at 𝑓(𝜉). Thus, by the Chain Rule (Proposition 4.12), it follows that ℎ∘𝑓^′∘𝑔 is LUD^𝑘 at 𝑓(𝜉). It follows that 𝑔^(𝑘)=(ℎ∘𝑓^′∘𝑔)^(𝑘−1) is LUD at 𝑓(𝜉) (by Proposition 4.4) and hence 𝑔 is LUD^𝑘+1 at 𝑓(𝜉) (by Proposition 4.5). This completes the induction and shows that 𝑔 is LUD^𝑙 at 𝑓(𝜉).

6. Intermediate Value Theorem

The intermediate value theorem is an important key result in real analysis. However, while all continuous real-valued functions on ℝ have the intermediate value property, this is not the case for 𝒩-valued functions on 𝒩. In fact, since 𝒩 is not connected, any function which takes on two distinct constant values on a separation of the field will be continuous but will not attain any value between the constants. The following example illustrates this.

Example 6.1. Let 𝑓∶[0,1]→𝒩 be given by 𝑓(𝑥)=10iiff𝑥𝑥∼≪11,.(6.1) Then 𝑓 is LUD^∞ on [0,1] as 𝑓 is locally constant everywhere. But 𝑓(𝑥) does not attain on [0,1] any values between 𝑓(0)=0 and 𝑓(1)=1. So even the property of LUD^∞ is not strong enough to ensure an intermediate value property for the function.

The next question is whether any kind of local intermediate value property can be assured. That is, can we find sufficient conditions for a function to have the intermediate value property on some neighborhood of a point? The answer is yes as we will see in Theorem 6.3, but first we present the following example which shows that even the LUD^∞ property is not quite sufficient to ensure the local intermediate value property, and it demonstrates the need for the added hypothesis to Theorem 6.3.

Example 6.2. Let 𝑓∶𝒩→𝒩 be given by 𝑓(𝑥)=𝑑^2𝜆(𝑥)0iiff𝑥𝑥≠=00,.(6.2) We will show that 𝑓 is LUD^∞ on 𝒩. First, note that 𝑓 is locally constant everywhere but at 0. Hence 𝑓 is trivially LUD^∞ on 𝒩⧵{0} with 𝑓^′=0. It remains to show that 𝑓 is LUD at 0 with 𝑓^′(0)=0. Let Ω=(−1,1) and let 𝜖>0 in 𝒩 be given. Let 𝛿=𝜖𝑑. Let 𝑥,𝑦∈Ω be such that |𝑦−𝑥|<𝛿. Without loss of generality, we may assume that |𝑥|<|𝑦|. We distinguish two possible cases.Case 1 (|𝑦|∼|𝑥|). Then |𝑓(𝑦)−𝑓(𝑥)|=0<𝜖|𝑦−𝑥|.Case 2 (|𝑦|≫|𝑥|). Then ||𝑓(𝑦)−𝑓(𝑥)||≈𝑑^2𝜆(𝑦)∼||𝑦||²<𝛿||𝑦||≪𝜖||𝑦−𝑥||,(6.3) since 𝛿≪𝜖 and |𝑦|≈|𝑦−𝑥|, and this shows that |𝑓(𝑦)−𝑓(𝑥)|≪𝜖|𝑦−𝑥|. So, 𝑓 is LUD^∞ on 𝒩 (with all derivatives equal to 0 everywhere); however, clearly 𝑓 does not have the intermediate value property in any neighborhood of 0.

Theorem 6.3 (Local intermediate value theorem). Let 𝐴⊂𝒩 and let 𝑓∶𝐴→𝒩 be LUD at 𝑥₀∈𝐴 with 𝑓^′(𝑥₀)≠0. Then there is a neighborhood Ω of 𝑥₀ such that for any 𝑎<𝑏 in 𝑓(Ω) and for any 𝑐∈(𝑎,𝑏), there is an 𝑥∈Ω such that 𝑓(𝑥)=𝑐. Moreover, 𝑥 is strictly between 𝑓⁽⁻¹⁾(𝑎) and 𝑓⁽⁻¹⁾(𝑏).

Proof. Without loss of generality, we may assume that 𝑓^(𝑥₀)>0, since if 𝑓^(𝑥₀)<0, we could then apply this proof to (−𝑓) and get the desired result. Since 𝑓 is LUD at 𝑥₀, there exists a neighborhood Δ of 𝑥₀ in 𝐴 such that 𝑓 is uniformly differentiable on Δ. Since 𝑓^′ is continuous at 𝑥₀ (by Proposition 4.8), there is 𝛿₁>0 such that (𝑥₀−𝛿₁,𝑥₀+𝛿₁)⊂Δ, and for any 𝑥∈(𝑥₀−𝛿₁,𝑥₀+𝛿₁), we have that 𝑓^(𝑥)>23𝑓^𝑥₀.(6.4) Since 𝑓 is uniformly differentiable on Δ, there exists 𝛿<𝛿₁ such that for all 𝑥,𝑦∈(𝑥₀−𝛿₁,𝑥₀+𝛿₁) satisfying |𝑦−𝑥|<𝛿 we have that ||𝑓(𝑦)−𝑓(𝑥)−𝑓^(𝑥)(𝑦−𝑥)||<𝑓^𝑥₀3||𝑦−𝑥||.(6.5) It follows that for 𝑦,𝑥∈(𝑥₀−𝛿/2,𝑥₀+𝛿/2) we have that 𝑓(𝑦)−𝑓(𝑥)𝑦−𝑥>𝑓^(𝑥)−𝑓^𝑥₀3>13𝑓^𝑥₀>0.(6.6) Hence 𝑓 is strictly increasing on Ω₁=(𝑥₀−𝛿/2,𝑥₀+𝛿/2). Applying Theorem 5.2 to 𝑓 yields a neighborhood Ω₂⊂Ω₁ of 𝑥₀ such that 𝑓(Ω₂) is open. Let 𝜖>0 be such that (𝑓(𝑥₀)−𝜖,𝑓(𝑥₀)+𝜖)⊂𝑓(Ω₂) and let Ω=𝑓⁻¹((𝑓(𝑥₀)−𝜖,𝑓(𝑥₀)+𝜖)) which is an open neighborhood of 𝑥₀. Now let 𝑎,𝑏∈𝑓(Ω) be such that 𝑎<𝑏, and let 𝑐∈(𝑎,𝑏) be given. Then 𝑐∈(𝑓(𝑥₀)−𝜖,𝑓(𝑥₀)+𝜖)⊂𝑓(Ω₂) since 𝑎,𝑏∈(𝑓(𝑥₀)−𝜖,𝑓(𝑥₀)+𝜖) and (𝑓(𝑥₀)−𝜖,𝑓(𝑥₀)+𝜖) is a convex set. So there is 𝑥∈Ω₂ such that 𝑓(𝑥)=𝑐. It follows that 𝑥∈Ω because 𝑓(𝑥)=𝑐∈(𝑓(𝑥₀)−𝜖,𝑓(𝑥₀)+𝜖). It is also true that 𝑎<𝑐=𝑓(𝑥)<𝑏; and since 𝑓 is increasing on Ω, it follows that 𝑓⁻¹(𝑎)<𝑥<𝑓⁻¹(𝑏).

Acknowledgment

This research was supported by a University of Manitoba start-up fund for K. Shamseddine.