Abstract

This article characterizes the isometries between spaces of all differentiable functions from a compact interval of the real line into a strictly convex Banach space.

1. Introduction

The main purpose of this article is to characterize the isometries of the space of all (continuously) differentiable functions from a compact interval of the real line into a strictly convex Banach space.

Cambern ([1], Theorem 6.5.5) and Pathak [2] investigated the surjective linear isometries over spaces of complex-valued differentiable functions on the interval and gave a representation for such operators, and Jarosz and Pathak [3] studied those operators over differentiable function spaces defined on te compact subsets of the real line (without isolated points). Also, Wang [4] studied the isometries between spaces of scalar-valued (real case and complex case) differentiable functions (and vanish at infinity) on the locally compact subsets of the Euclidean spaces (without isolated points) and gave a representation for such operators.

On the other hand, Botelho and Jamison [5] extended these results to the surjective linear isometries on spaces of -time differentiable functions on with values in a finite-dimensional Hilbert space. In [5], the main result is valid whenever the (real) dimension of the Hilbert space is bigger than one. Compare with the assumptions in ([6], Theorem 2.13). In [6], the authors investigated the isometries between spaces of -times differentiable functions (and vanish at infinity) on an open subset of the real line with values in a strictly convex Banach space. Also, Li and Wang [7] studied the isometries between spaces of -times differentiable functions (and vanish at infinity) on an open subset of the Euclidean space with values in a reflexive and strictly convex Banach space. In [7], there is a gap in the proof of Theorem 3.5 (page 553); in the proof of Claim 1, why must have at most one nonzero term? Compare with the proofs in ([4], Lemma 2.5) and ([2], Lemma 2.3).

Suppose that and are Banach spaces (on the real line). Denote the space of all -functions by , and on this space, we consider the following norm:

Let be a surjective linear isometry. In [5], it is shown when is a finite dimensional Hilbert space, there exists a linear isometry such that either or , for all and . In this article, we extend the above result to a surjective isometry , whenever Banach spaces and are strictly convex.

In [6], the authors described the surjective isometries between spaces of -times differentiable functions (and vanish at infinity) on an open subset of the real line with values in a strictly convex Banach space (with dimension greater than one). Such a representation is as , where is a surjective linear isomorphism (for all ) and is a diffeomorphism (over ), and its proof is based on the main result of [8]. By using the proof of ([6], Lemma 2.12), we recover their representation in our setting. Therefore, our description is better and its proof is somewhat shorter and more elementary than the one in [6]. Also, in the appendix, we provide a proof for Theorem 13 for special case . In a forthcoming article, we extend our result to the class of -times differentiable functions.

2. Characterization of -Set in

Suppose that is a Banach space on the real line. For and , we define where .

A -set in a Banach space is a subset of with the property that for any finite collection , , and such that is maximal with respect to this property.

In this section, we show that every -set of has a unique representation by some , up to and , whenever is strictly convex (compare with ([9], Lemma 7.2.2)). We start this section with two elementary Lemmas.

Lemma 1. For any and such that , there exists in such a way that , and for all . Moreover, when or , we can state and prove a similar result.

Proof. For , define as follows:

Then, it is easy to verify that and for all . It is clear that can be extended as a -function on with support in , for any . Denote this extension of by . Now, by choosing small enough (fixing and ), we have for all . Now, one can easily construct the desired function as in Lemma 1.

For , the function denotes the constant function with value .

Lemma 2. Let be a Banach space, and . For any , there exists in such a way that , and for all .

Proof. Let be a function with the properties as in Lemma 1. Then, is the desired function.

Theorem 3. Let be a strictly convex Banach space. Then, every -set of is as , for some and , and vice versa is a -set of .

Proof. First, we show that is a -set of , for any and . To do this, it is enough to show the maximality property for . Suppose that such that , for any , we show that . Suppose that . So, there exists an open neighborhood of in , say , such that , for all . By Lemma 1, there exists such that , and , and also for all . Then, we have for all . This is a contradiction. Therefore, we get .

Next, we prove that and . We know that (see the proof of Lemma 2) and for all . Also, we have

By Equation (10) and the assumptions, we see that all inequalities in Equation (11) are equalities. This implies that

Since is strictly convex, this implies that and and then and .

To prove the converse, suppose that is a -set in . For , define

It is clear that is a compact nonempty subset of . Also, for any finite collection of elements , we have . By contradiction, suppose that . This implies that for any , there exists some , for , such that . Then, we obtain for all . So, we get for all . This is a contradiction. So, . This implies that . Let . Finally, by the maximality property of , since is strictly convex, we see that , for some .

Next, we state a few simple facts about -sets in , whenever is a strictly convex Banach space.

Proposition 4. Let be a strictly convex Banach space. Consider the space , then we have (i) is not trivial, i.e., (ii) is not trivial, whenever (iii)If , or , or , then (iv)If , then(v)If , then(vi)If and , then

Proof. It is straightforward (using Lemma 1 and Lemma 2).

3. Main Results

In this section, by using results of the previous section about -sets in , we obtain a few important properties of a given isometry in order to characterize such isometry.

Proposition 5. Let be a surjective linear isometry, where and are two strictly convex Banach spaces. Then, (i) maps a -set in to a -set in . In particular, for any and , there exist and such that (ii)If and , then

Proof. (i) Since is an isometry, by definition, it is easy to see that maps a -set to a -set. Now, by Theorem 3, the proof is complete. (ii) By contradiction, suppose that .

First, we assume that . By Proposition 4 (vi), there exist such that

Therefore, we obtain

On the other hand, by part (i), we know that , for some and . Also, either or . This is a contradiction, by Proposition 4 (ii).

Next, we assume that and . Then, there exist and such that , for some . Now, by considering and since and , we obtain a contradiction as before (note that ).

Finally, we assume that (see [3], page 202). Also, we present a proof for this in the appendix.

Corollary 6. Let be a surjective linear isometry, where and are two strictly convex Banach spaces. Then, there exists a bijection such that , does not depend on (and ).

Proof. It is an immediate consequence of Proposition 5.

Theorem 7. Let be a surjective linear isometry, where and are two strictly convex Banach spaces. Then, maps constant functions to constant functions. In particular, induces a linear isometry from onto .

Proof. Suppose that and the function is as in Corollary 6. It is clear that , for all and . Therefore, by Proposition 4 (iv) and (v) and Proposition 5 (ii), for any , we see that either or . Now, the theorem is an immediate consequence of the following simple fact from real analysis: (i)Let be a differentiable function such that for any , either or . Then, is a constant function

Proposition 8. Let be a surjective linear isometry, where and are two strictly convex Banach spaces. Consider and and suppose that , for some and . Then, (i)If , then,for some (ii)If , then,for some

Proof. (i) By Proposition 5 (ii), we know that , for some . On the other hand, the constant function belongs to , for all . Therefore, by Theorem 7, is a constant function with value in . This implies that (note that ). So, this completes the proof. (ii) The proof is an immediate consequence of part (i) applied to (by contradiction and using Proposition 4 (vi)).

Lemma 9. Suppose that , i.e., is continuously twice differentiable, such that and , for some . For any such that , there exists such that , , and for all . Moreover, when or , we can state and prove a similar result.

Proof. Since , there is a positive constant such that for all . By using this fact, the rest of proof is similar to the proof of Lemma 1, with a slight modification.

Remark 10. Lemma 9 is meaningful when . In fact, for any , there exists satisfying the conditions in Lemma 9, except, replacing the condition of with .

Theorem 11. Let be a surjective linear isometry, where and are two strictly convex Banach spaces. Suppose that is as in Corollary 6 and consider . Suppose that satisfies the condition , then .

Proof. By Lemma 9 and Remark 10, there exists a function in such a way that : for all . This implies that , for all and some . Now, by Proposition 8 (ii), we see that , for all and some . So, by Proposition 4 (iv), we have . Similarly, we can show that . This completes the proof of theorem.

Corollary 12. In Theorem 11, we can replace the condition with the weaker assumption .

Proof. It is an immediate consequence of the density of in (with norm).

Theorem 13. Let be a surjective linear isometry, where and are two strictly convex Banach spaces. Then, there exists a surjective linear isometry , such that either or , for all and .

Proof. By Theorem 7, we know that maps the constant function with value to the constant function , for some . So, induces a surjective linear isometry . Now, let be an arbitrary element of and let be an arbitrary element of . Define . It is clear that , so by Corollary 12, we obtain . This implies that , for all and . Finally, by a standard argument, we can show that is differentiable on and the absolute value of its derivative is a constant function with value . This completes the proof of theorem.

Remark 14. Theorem 13 remains meaningful and valid for a surjective linear isometry as well, and as a consequence, we see that the intervals and should have equal length.

Appendix

In this appendix, we provide a proof for Proposition 5 (ii) whenever . To do this, we need to show that .

Without loss of generality, we may assume that and and also and , for .

On the other hand, by Proposition 1.3 (or Equation (1.2)) in [5] (see also ([1], 6.5)), for any , and , there exist , , and , such that for all , where . In particular, there exist , , and , for , such that for all , where .

Now, we show that , , , and . To do this, by Lemma 1, there exists a function in such a way that , , and also for all . Since and

This implies that and . Similarly, we can show that and .

Next, we show that maps a constant function to a constant function. Suppose that the image of under is the constant function with value . Then, by Equation (A.1), we have for all , where and . Then, we obtain for all . Since is continuous and is connected, we see that should be constant with value .

By a simple argument similar to the proof of ([2], Lemma 1.4), we can show that the map is a well-defined homeomorphism from onto itself (for fixed values and in Equation (A.1)); we denote this homeomorphism by .

Let denote the set of all such that for all , where and . One can easily show that is a closed and open set in , and since , we obtain . Similarly, we can define and show that . Therefore, we have for all and all , where (note that , and ). If , by considering and the corresponding homeomorphism , we obtain a contradiction.

So, we have . Now, by choosing in Equation (A.8), we see that satisfies for all . From the second equation, we see that , for some constant . By the first equation, we have for all . In particular, for , we get and . On the other hand, since is a homeomorphism on , is equal to or .

If , we obtain and then we get for all . This implies that and , for all . Then, since , we obtain .

If , we obtain and then we get for all . This implies that and , for all . Then, since , we obtain .

Therefore, either , whenever or , whenever . Similarly, by using and , we can show that either , whenever or , whenever . Finally, we obtain , as desired.

Remark 15. By the above constructions, one can provide a direct proof for Theorem 13, whenever . Also, in ([1], 6.5), one can find a proof for Theorem 13, whenever is the complex plane (note that its dimension over the real line is ).

Data Availability

The data used to support the findings of this study are available from the author upon request.

Conflicts of Interest

The author declares that there are no conflicts of interest.

Acknowledgments

The author would like to thank the Research Council of Sharif University of Technology for the support.