Abstract
Nirenberg proposed a problem as to whether or not a continuous and expansive operator (where is a Hilbert space) is surjective if . I shall give a positive answer for the problem provided that is unbounded. For contents related to this paper, the reader is referred to the remarks and the study of Asfaw (2021). The present paper gives a complete answer for the problem that has been open for about 47 years.
1. Introduction and Preliminaries
In what follows, denotes a real Hilbert space with an inner product , and the symbol denotes the norm in . An operator is called “-expansive” if there exists such that for all and . The operator is called expansive if . In 1974, Nirenberg [1] proposed a problem as to whether or not a continuous and expansive operator is surjective provided that .
The main goal of the present paper is to give a positive answer for the problem provided that is unbounded. Clearly, cannot be surjective if is bounded.
Remark 1. Recently, I published a paper in [2] planning to give a proof of the problem. A minor error was found after publication. The editors decided a retraction based on their belief that the error is profound and not fixable. However, the error was minor, not profound and fixable. I submitted a corrigendum about the published paper but the editors decided to retract it based on the error in the original paper. Thus, the present paper reuses contents from [2]. The differences between the current and the retracted papers are given below. (i)In Theorem 7, it is assumed that has unbounded interior which was not assumed in [2]. It is obvious to see that cannot be surjective if is bounded(ii)The sequences and used in Step 1 in [2] are not required (in the current paper) to show the property of the operator . Instead, the property of is achieved based on Kirszbraun-Valentine [3, 4] extension theorem (c.f., on page 5 of the current paper). It is essential to mention that Valentine’s [4] result is a generalization of Kirszbraun [3] extension theorem
Morel and Steinlein [5] constructed an example of such an operator defined from into satisfying the stated conditions but fails to be surjective. In a Hilbert space setting (i.e., if ), Szczepanski [6] constructed a family of nonsurjective continuous maps such that contains and is expansive on each sphere and for all or , where . In addition, Szczepanski [7] gave a negative answer for the problem in by constructing a continuous and nonsurjective operator satisfying weak expansive condition. Chang and Shujie [8], Theorem 7, gave a positive answer for the problem provided that is Frèchet differentiable on (where is a real Banach space and is reflexive), and its derivative is locally bounded. Recently, Kartsatos [9], Theorem 6 (with and ), and Xiang [10], Theorem 2.12 (with ), gave surjectivity results if the condition is replaced by strong monotonicity of .
The following remarks on the recent result due to Ives and Preiss [11], are essential.
Remark 2. Ives and Preiss [11], Theorem 4, constructed an operator (where ) satisfying the following properties. (i) for all (ii)The image of is nowhere dense in (iii) is not surjectiveWe shall show that (i) and (ii) imply the boundedness of . Indeed, let . The continuity and expansiveness of and the closedness of show that is closed. Since is nowhere dense, it follows that , i.e., and . Let . Then, for each , there exists a sequence in such that as . Since is nonempty and open containing ( is a neighbourhood of ), there exists a positive integer such that for all . In addition, we see that i.e., we have Since for all (i.e., for all ) and and are disjoint, we conclude that for all . This shows that , i.e., . Thus, the set is bounded, i.e., is not surjective.
Remark 3. Remark 2 shows that conditions (i) through (v) in Theorem 4 due to Ives and Preiss [11] imply that has a bounded interior, i.e., is nonsurjective. It is essential to mention that the result in [11] did not give a negative answer for Nirenberg’s problem if is unbounded. In addition, we point out that the case of boundedness of can be ignored from the statement of Nirenberg’s problem.
The following definition is needed in the squeal.
Definition 4. . Let be a real reflexive Banach space and be its dual space. Let be a nonempty, bounded, and open subset of and be bounded on finite dimensional subsets of . Then, is said to be “of type ” if a sequence in satisfies and ; then, and as .
Browder [12], Definition 5, introduced the following definition.
Definition 5. . Let be a nonempty, bounded and open subset of a reflexive Banach space . A family of operators , defined from into , is called a “homotopy of class ” if for sequences in and in such that , and , we have and as .
The operator defined by is called the normalized duality mapping. It is well-known that is single-valued, bounded, monotone, bicontinuous, surjective, and of type if and are locally uniformly convex and real reflexive Banach spaces. The operator (where is the identity operator on ) if is a real Hilbert space. Browder [12], Theorems 3 and 4, constructed a degree mapping for the class of operators of type that satisfies the basic properties listed in Lemma 6.
Lemma 6. Let be a nonempty, bounded, and open subset of a real reflexive Banach space . Let . Then, the following basic properties hold. (i)Normalization. if and if .(ii)Existence. Let be a bounded operator of type such that and . Then, .(iii)Homotopy Invariance. Let be a homotopy of class defined from into such that for all . Then, is independent of .(iv)Invariance under Affine Homotopy. Let () be a bounded operator of type and for . Then, is a homotopy of class , and is independent of provided that for all
2. Main Result
Theorem 7 gives a proof of the Nirenberg problem.
Theorem 7. Let be a real Hilbert space and be a continuous and -expansive operator such that is unbounded. Then, is surjective.
Proof. Choose . Then, there exists such that . Let . Then, the operator is -expansive and continuous such that . It follows that is unbounded if and only if is unbounded. Clearly, the set is open and , i.e., . In addition, we see that is surjective if and only if is surjective. Thus, we assume without loss of generality that . Let and . The continuity and -expansive conditions imply that is closed and
for all and . Then, it follows that
for all and , i.e., is a continuous and monotone operator. It holds that because is closed and . Let be the restriction of on . Let , and for each and , let . Let
The monotonicity of implies
for all and . The monotonicity of and the condition yield
for all . Since is unbounded, the right side of (7) approaches infinity as . Then, there exists such that
for all . Let We shall show that the operator is of type . We shall apply Kirszbraun-Valentine [3, 4] extension theorem. Indeed, the Lipschitz operator (with Lipschitz constant ) can be extended to an operator with the same Lipschitz constant. Thus, the operator is continuous and uniformly monotone, i.e., is of type . Let be the restriction of on . We shall show that is of type . Indeed, let for all such that as and
It is enough to show that . Since on , it follows that
The uniform monotonicity of with domain (i.e., using (5), (6) and on ) shows that
for all . Thus, (9)–(11) yield
Since as (because ), (12) shows that
i.e., as . Clearly because for all . The continuity of implies that as . Thus, the operator is bounded, continuous, and of type .
Clearly, the identity operator is bounded and of type . For each , let be given by
Then, (iv) of Lemma 6 shows that the family is an affine homotopy of class . Clearly, for all . Thus, exactly one of the following holds.
(i)There exists such that for each , there exist and such that for all (ii)For each there exists such thatSuppose (i) holds, i.e., there exists such that for each , this corresponds , , and such that and for all , i.e., for each , we have
for all . Replacing by , by , and by in (16), for each , we get
for all . Next we choose a sequence of natural numbers such that for all . Let for all . Clearly, and for all (i.e., for all ). Since for all , (17) implies
for all . Clearly, for all . Dividing (18) by for all , we arrive at
for all , where . The boundedness of and shows that is bounded. Assume without loss of generality that as . Suppose . The sequence cannot converge to because for all . Then, we consider either is unbounded or bounded. Suppose is unbounded, i.e., there exists a subsequence of , denoted by , such that as . Since , and are bounded, letting in (18) implies
as , i.e., as , where for all . Since as and is bounded, we conclude that as . Letting gives .
Next we assume that is bounded. Assume without loss of generality that there exists a subsequence, denoted again by , such that as . Then, and
as . Since , letting gives .
Next we consider the case . Then, we have and
for all , where . Since as and , it follows that as . Letting in (19) yields . Letting shows that . In all cases, each satisfying (i) lies in .
Suppose (ii) holds. Then, there exists a subsequence of , denoted again by , such that for all and . Then, (iv) of Lemma 6. shows that the family is a homotopy of class defined from into such that for all and . Let stand for Browder degree mapping. Applying (iii) of Lemma 6., we conclude that is independent of , i.e., (i) and (iii) of Lemma 6. give
for all . In addition, (ii) of Lemma 6. confirms that for all . Thus, for each positive integer , there exist and such that for all and
for all . Since and are bounded, it follows that is bounded. In addition, (24) gives
as because and , where . Thus, letting shows that . Thus, each satisfying (i) or (ii) lies in . This shows that is surjective. The proof is completed.
Data Availability
No data were used to support this study.
Conflicts of Interest
The author declares that there are no conflicts of interest regarding the publication of the paper.