THE TINGLEY PROBLEM ON THE UNIT SPHERE OF COMPLEX L p [ 0 , 1 ] SPACE

. In this paper, we investigate the problem of extending isometric operators from unit sphere of complex L p spaces (1 < p < ∞ , p (cid:54) = 2) to general complex Banach spaces. By studying the isometric operators, we prove the Tingley problem on complex L p spaces and provide a positive answer under some conditions. That is, it is proved that for a surjective isometry V 0 on any complex L p [ 0 , 1 ] unit sphere to any general complex Banach space E unit sphere, Under some conditions, V 0 can be extended to a linear isometry from the entire space L p [ 0 , 1 ] to E .


INTRODUCTION
The classical Mazur-Ulam theorem [1] states that if X and Y are real normed spaces and T : X → Y is a surjective isometry, then T is affine.In particular, if T (0) = 0, then T is a linear isometry.However, local surjective isometries often lack desirable properties, leading to the following question raised by Tingley in 1987.
Tingley Problem [2]: Let X and Y be two normed spaces, and T 0 be a surjective isometry between the unit spheres S(X) and S(Y ).Is there a linear isometry T : X → Y defined on the entire space such that T | S(X) = T 0 ?
The formulation of the Tingley problem has been a milestone in the study of isometries.
Its significance lies in the fact that if the conclusion holds, then the geometric properties of a mapping on the unit sphere in a spatial context will determine its properties throughout the entire space.In 2011, L. Cheng and Y. Dong introduced the concept of Mazur-Ulam property related to the Tingley problem and the Mazur-Ulam theorem [3], making significant contributions to the study of isometric problems.The Mazur-Ulam property states that if a surjective isometry from the unit sphere of a normed space X to the unit sphere of an arbitrary normed space Y can be extended to a real linear isometry on the entire space, then the normed space X is said to have the Mazur-Ulam property [3] (MUP).
In recent years, researchers have primarily focused on the study of isometric extension problems in similar or different types of classical Banach spaces.Ding Guanggui and other scholars have summarized the conclusions regarding isometric problems in these spaces.In the case of similar type spaces, Ding Guanggui [4] proved in 2007 that two surjective isometries between the unit spheres of l ∞ (Γ)-type spaces can be extended to real linear isometries on the entire spaces.In 2011 and 2012, Tan Dongni provided affirmative answers to the Tingley problem on the F-spaces (L p (v), 0 < p < 1) and Tsirelson space, James space (see [5,6]), respectively.In the case of different type spaces [7], Ding Guanggui initially proved that a surjective isometry from the unit sphere S(E) of an arbitrary Banach space E to the unit sphere S(C[0, 1]) of the space of continuous functions on [0, 1] can be extended to a linear isometry on the entire space under certain conditions.Later, Wang Jianhua and Fang Xinian [8] made improvements by removing the additional conditions.Subsequently, Liu Rui [9] proved that the surjective isometry from the unit sphere of L ∞ (µ)-type spaces to the unit sphere of an arbitrary Banach space, as well as Tan Dongni [10], proved that the isometry between the unit spheres of Banach spaces is linearly extendable when the Banach space is a locally GL space, thereby providing more general results.In 2012, Tan Dongni [11] studied L p (µ, H) spaces (where H is a Hilbert space, 1 < p < ∞, and p = 2) and also provided exact conclusions.
Most of the research on the Tingley problem in real Banach spaces has been conducted by scholars (for further references, see [12,13,14,15,16,17,18]).While it is not always possible to extend a surjective isometry between the unit spheres of arbitrary complex Banach spaces to complex linear or conjugate linear operators on the entire space, this problem holds true for certain classical complex Banach spaces.In 2014, Yi Jijin, Wang Ruidong, and Wang Xiaoxiao [19] established that any surjective isometry from the unit sphere of a complex l p (Γ) space (where 1 < p < ∞ and p = 2) to the unit sphere of a complex l p (∆) space can be extended to a real linear isometry on the entire space.
Building upon the inspiration from the aforementioned papers, this study considers the complex L p [0, 1] spaces and aims to prove that any surjective isometry from the unit sphere of a complex L p [0, 1] space (where 1 < p < ∞ and p = 2) to the unit sphere of an arbitrary complex Banach space can always be linearly extended to the entire space.

PRELIMINARY KNOWLEDGE
Definition 2.1.Introduces the definition of the complex L p [0, 1]space as follows: Here, p ∈ (1, ∞) and p = 2.The norm on the space is defined as: States that for normed spaces X and Y, an operator V : X → Y is said to be a

RELATED LEMMAS AND RESULTS
Theorem 3.1.Let E be a complex Banach space, V 0 be a surjective isometry from complex Then V 0 can be linearly and isometrically extended to the whole space L p [0, 1].
In order to prove this theorem, we need to introduce some notations and symbols, which will be used throughout this section. Notations: (1) The mapping V 0 : complex S (L p [0, 1]) → complex S(E) is a surjective isometry.
(2) For positive real numbers α and β , we have we denote f ∧ g = 0.And denote (3) We use the expressions to represent elements in complex S (L p [0, 1]).
The representation in (3) is justified because we have the following Lemma 3.2, which guarantees: The proof is mainly divided into two cases: 1 < p < 2 and 2 < p < ∞.For the case of 2 < p < ∞,the following inequality in L p [0, 1] space, with 1 < p < ∞ and p = 2, is known as the Clarkson inequality and is an important inequality in L p [0, 1] space with 1 < p < ∞ and p = 2.
Lemma 3.2.For S (L p [0, 1]) (1 < p < ∞, and p = 2), for any two elements f and g, the following hold: Moreover, equality holds in (i) and (ii) if and only if f ⊥ g.
This certification is detailed in the reference [12].
Lemma 3.3.Suppose X and Y are complex Banach spaces, and let V 0 : S(X) → S(Y ) be a surjective isometry.Then, X is strictly convex if and only if Y is strictly convex.Moreover, for every x ∈ S(X) and y ∈ S(Y ), we have This certification is detailed in the reference [20].It will not be proved here.From this theorem, we can deduce that in a strictly convex space, for any x, y, we have x + y = V 0 (x) + V 0 (y) holds.Thus, in notation (3), we have which leads to: The following proof comes from the references [11] slightly modified.For completeness, we give the proof.
Lemma 3.4.Let E be a complex Banach space, and V 0 : S (L p [0, 1]) → S(E) be a surjective isometry.Let f , g ∈ S (L p [0, 1]) be complex-valued functions satisfying f ∧ g = 0. Let α and β be positive real numbers such that ) with a ∧ x = 0 and b ∧ x = 0.Then, the following hold: , where λ ∈ T and T is the unit sphere in the complex plane.
Furthermore, since the complex L p [0, 1] space is strictly convex, from Lemma 3.3, we can conclude that the space E is also strictly convex.Moreover, based on equations (3.2) and (3.3), we have: Apologies for the repeated content.It seems there was an issue with the response.Let's continue with the proof: By adding and subtracting the previous two equations, we have: From these equations, we can deduce: From equation (3.4), we observe the following relation: Based on Lemma 3.2 (i), this equation holds with equality if and only if The above proof also applies to the case when β ≤ α.Therefore, the result holds.
Case 2: When 1 < p < 2 and assuming β ≤ α, we can use the same approach as in Case 1 to obtain: Similar to the steps in Case 1, we can verify statements (i) and (ii).
To prove statement (iii), we want to show that Based on the definition of the operator H, we can rewrite this as: and f ∧ g = 0, we can apply the properties of V 0 to obtain: Hence, equation (3.5) is proven.
), then T 0 can be extended to a linear isometry on the entire space.
Proof: The theorem will be proven in three steps.
Step 1: Let ∀ f ∈ S(L p [0, 1]), where f = ∑ n i=1 a i f i , with f i ∧ f j = 0 for i = j, a i ∈ T, and The goal is to prove that for ∀1 ≤ i ≤ n, we have: Given that T 0 ( f i ) ∧ T 0 ( f j ) = 0 for i = j, by definition, we have supp(T 0 ( f i )) ∩ supp(T 0 ( f j )) = / 0 holds.In other words, we have A i ∩ A j = / 0 for i = j.It can be deduced that: Step 2: Next, we will mainly prove that T 0 ( f Let a i ∈ T, and we can write a i = |a i | e iθ i , where sign (a i ) = e iθ i .Since T 0 is 1-Lipschitz and we have the following: (ii) On the other hand, we have Here, T 0 −e iθ i f i = 1.Therefore, we have: Combining (i) and (ii), we obtain: By observing that the form of equation ( 3 Furthermore, from the conclusion obtained in Step 1, we have Therefore, we can easily obtain: Here, we have 1 = T 0 e iθ i f i = e iθ i T 0 ( f i ) .Therefore, Given that the complex space L p [0, 1] is strictly convex, equation (3.7) holds if and only if α > 0 and T 0 ( f )| A i = αe iθ i T 0 ( f i ).We will prove that αe iθ i = a i . Since we have: If αe iθ i = −a i , then it contradicts the given condition.Hence, we conclude that αe iθ i = a i .
In conclusion, we have shown that T 0 ( f Step 3: Let X denote the space of all simple functions in L p [0, 1].We construct an operator on the space X as follows: Let L p [0, 1] be a complex function space.Consider any simple function in this space: and a i x i = 0. (i) First, we prove that the operator T is isometric: (ii) We prove that the operator T is linear on the space X, using the definition of T and the properties of T 0 from Step 2: Thus, T is a linear isometry on the space X.Furthermore, since X is dense in L p [0, 1], T being an isometry on the space X can be linearly extended to the entire space L p [0, 1].
Lemma 3.5 is an important lemma in complex L p [0, 1], and as a consequence, we obtain the following.
Corollary 3.6.Let T 0 be a 1-Lipschitz mapping from then T 0 can be isometrically extended to the entire complex L p [0, 1] space.
Proof: Given the assumption on T 0 in the condition, for any Since T 0 is a 1-Lipschitz mapping, we have: From T 0 ( f ) = −T 0 ( f ), we can conclude that T 0 is an odd function, i.e., T 0 (− f ) = −T 0 ( f ) holds.
Then, for any f , g ∈ S (L p [0, 1]) (where f ∧ g = 0) satisfying condition (i) in Lemma 3.2, we examine the following expression: According to the equality condition in Lemma 3.2, we can see that equality holds if and only if Therefore, by utilizing the conclusion in Lemma 3.5, we can deduce that T 0 can be uniquely linearly extended to the entire complex L p [0, 1] space.
Theorem 3.7.Let E be a complex Banach space, and let V 0 be a surjective isometry from x ∈ S (L p [0, 1]) with a ∧ x = 0 and b ∧ x = 0.Suppose there exist positive real numbers α and β such that α p + β p = 1.Let f be a fixed element on the unit sphere of S (L p [0, 1]).Then, for any element g on the unit sphere that is orthogonal to f , we define the mapping: Then, (i) the mapping Φ f (g) is a linear isometry.(ii) For all elements g 1 and g 2 in S (L p [0, 1]) Proof: From the conditions and definitions, it is straightforward to see that Now, let A 0 = supp f .In fact, the defined mapping Φ f (g) can be seen as a mapping from the unit sphere S (L p [0, 1]\A 0 ) to the unit sphere S (L p [0, 1]).According to Corollary 3.6, we only need to prove that this mapping is 1-Lipschitz and its range is symmetric.
Next, we prove that the range of Φ f (g) is symmetric:From the definition of H g (α f , β g) and Lemma 3.3, we know that H −g (α f , β (−g)) = −H g (α f , β g).Therefore, by Corollary 3.6, Φ f (g) is a linear isometry.Now, we prove (ii).Since Φ f (g) is an isometry, Equation (3.8) becomes an equality.Hence, we obtain: Using Lemma 3.3, we have: This implies that . Thus, we have proved (ii).
Theorem 3.8.Let E be a complex Banach space, and let V 0 be a surjective isometry from a ∧ x = 0 and b ∧ x = 0. Suppose there exist positive real numbers α and β such that α p + β p = 1.Take a fixed element g on the unit sphere of S (L p [0, 1]).Then, for any element f on the unit sphere that is orthogonal to g, we define the mapping: Then, (i) the mapping Φ g ( f ) is a linear isometry.(ii) For all elements f 1 and f 2 in S (L p [0, 1]) Proof: The proof of Theorem 3.8 is similar to that of Theorem 3.7 and is omitted.
To establish an important result in this section, we first introduce an important lemma that has been mentioned and proven in references [21,22].
Theorem 3.10.Let E be a complex Banach space, and let V 0 be a surjective isometry from a ∧ x = 0 and b ∧ x = 0. Suppose there exist positive real numbers α and β such that Then, the following equations hold:

Proof:
According to Lemma 3.4(iii), we have and similarly, it suffices to prove that Step 1: Let us first prove that for ∀ f , g ∈ S(L p [0, 1]) such that f ∧ g = 0, the following statements hold: Indeed, without loss of generality, let's assume α ≥ 2 −1/p .Since V 0 is an isometry, we can examine the following: Step 2: We will prove that for any measurable set A ⊂ [0, 1] and any function f ∈ S(L p [0, 1]) satisfying 0 ≤ µ(A) ≤ 1 and µ(supp f ∩ A) = 0, the following statements hold: Proof by contradiction: Let's assume that µ(supp Then there exists a measurable set B, with 0 < µ(B) < 1, such that

With the chosen element χ
Therefore, by applying Theorem 3.7(ii), we obtain ).On the other hand, from Step 1, we know that However, according to Lemma 3.4 (ii), we know that µ(B) 1/p = 0. Hence, we have: This contradicts equation (3.11), and thus our assumption is not valid.Therefore, we conclude that µ(supp Similarly, we can prove that µ(supp Next, we will prove that the following situation cannot occur: Proof by contradiction: Assume that µ(supp µ(A 2 ) 1/p .According to Theorem 3.7 (i), the operator H g (α f , β g) is a linear isometry.Therefore, we have: (3.12) Moreover, from the conclusion in Step 1, we have µ(supp holds.Thus, we can deduce from equation (3.12): This contradicts the definition of A 2 as A \ A 1 .Hence, our assumption is not valid.
Step 3: Finally, we prove that By using Lemma 3.9 (Lamperti's theorem), for all linear isometries U on the space L p [0, 1] → L p [0, 1], there exists a unified representation: where f ∈ L p [0, 1] and T 1 is a linear transformation induced by T .
According to Theorem 3.7 (i), we know that ) is a linear isometry.By Lemma 3.9, let T be a regular set isomorphism associated with Φ f , and T 1 be the linear transformation induced by T .Note that for any A ⊂ [0, 1] with 0 ≤ µ(A) ≤ 1, we have T 1 (χ A ) = χ T (A) and T (A) = A due to the properties of the linear transformation T 1 .Then, we have Here, h satisfies the following equation: Hence, we have |h| = 1 a.e.
Proof: The proof is divided into two steps. Step Step 2: Let X denote the space of all simple functions in L p [0, 1].Consider any simple function in the complex L p [0, 1] space: where {λ i } n i=1 ⊂ C and {A i } n i=1 is a sequence of measurable sets satisfying 0 ≤ µ (A i ) < 1 for 1 ≤ i ≤ n, and µ A i ∩ A j = 0 for i = j.We construct an operator V on the space X as follows: We will prove that the operator V is isometric: Hence, V is a linear isometry on the space X.Since X is dense in L p [0, 1] and both L p [0, 1] and E are complete, V 0 can be linearly extended to the entire space L p [0, 1].
Next, we will consider the case of1 < p < 2,The following Lemmas 3.12 and 3.13 are similar to the case of 2 < p < ∞ in the first part and their proof processes are omitted.