# A Positive Answer on Nirenberg’s Problem on Expansive Mappings in Hilbert Spaces

Authors: Teffera M. Asfaw
Journal: Abstract and Applied Analysis
 (2022)

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## Abstract

Nirenberg proposed a problem as to whether or not a continuous and expansive operatorT:X⟶X (where X is a Hilbert space) is surjective if RT∘≠∅. I shall give a positive answer for the problem provided that RT∘ 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.


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## Body

## 1. Introduction and Preliminaries

In what follows,X denotes a real Hilbert space with an inner product <·>, and the symbol ∥.∥ denotes the norm in X. An operator T:X⟶X is called “α-expansive” if there exists α>0 such that ∥Tx−Ty∥≥α∥x−y∥ for all x∈X and y∈X. The operator is called expansive if α=1. In 1974, Nirenberg [1] proposed a problem as to whether or not a continuous and expansive operator is surjective provided that RT∘≠∅.The main goal of the present paper is to give a positive answer for the problem provided thatRT∘ is unbounded. Clearly, T cannot be surjective if RT∘ 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 Theorem7, it is assumed that RT has unbounded interior which was not assumed in [2]. It is obvious to see that T cannot be surjective if RT∘ is bounded(ii)
The sequencespn,m and qn,m used in Step 1 in [2] are not required (in the current paper) to show the S+ property of the operator S+λI. Instead, the S+ property of S+λI is achieved based on Kirszbraun-Valentine [3, 4] extension theorem (c.f., L10−L27 on page 5 of the current paper). It is essential to mention that Valentine’s [4] result is a generalization of Kirszbraun [3] extension theoremMorel and Steinlein [5] constructed an example of such an operator T defined from ℓ1N into ℓ1N satisfying the stated conditions but fails to be surjective. In a Hilbert space setting (i.e., if H=ℓ2N), Szczepanski [6] constructed a family of nonsurjective continuous maps Fε:H⟶Hε>0 such that RFε contains B10 and Fε is expansive on each sphere and ∥Fεx−Fεy∥≥∥x−y∥ for all x∈Pε or y∈Pε, where Pε=x∈H:1<∥x∥<1+ε. In addition, Szczepanski [7] gave a negative answer for the problem in X=ℓ2N by constructing a continuous and nonsurjective operator T satisfying weak expansive condition. Chang and Shujie [8], Theorem 7, gave a positive answer for the problem provided that T:X⟶Y is Frèchet differentiable on X (where X is a real Banach space and Y is reflexive), and its derivative is locally bounded. Recently, Kartsatos [9], Theorem 6 (with α=1 and c∈0,1), and Xiang [10], Theorem 2.12 (with c∈−∞,α), gave surjectivity results if the condition RT∘≠∅ is replaced by strong monotonicity of T+αI.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 P:X⟶X (where X=L20,∞) satisfying the following properties.
(i)
Px=x for all x∈B¯10(ii)
TheP image of x∈X:∥x∥≥1 is nowhere dense in X(iii)
P is not surjective
We shall show that (i) and (ii) imply the boundedness ofRP∘. Indeed, let F=PX\B10. The continuity and expansiveness of P and the closedness of X\B10 show that F is closed. Since F is nowhere dense, it follows that F∘=∅, i.e., F=∂F and X\F¯=X. Let V=RP∘. Then, for each x∈V, there exists a sequence xn in X\F such that xn⟶x as n⟶∞. Since V is nonempty and open containing x (V is a neighbourhood of x), there exists a positive integer N such that xn∈V for all n≥N. In addition, we see that
(1)RP=PB10∪PX\B10=B10∪F,
i.e., we haveV=V∩B10∪F=B10∪F∩V. Since xn∈F for all n (i.e., xn∈F∩V for all n) and B10 and F∩V are disjoint, we conclude that xn∈B10 for all n≥N. This shows that x∈B¯10, i.e., V⊆B¯10. Thus, the set V is bounded, i.e., P is not surjective.Remark 3.
Remark2 shows that conditions (i) through (v) in Theorem 4 due to Ives and Preiss [11] imply that RT has a bounded interior, i.e., T is nonsurjective. It is essential to mention that the result in [11] did not give a negative answer for Nirenberg’s problem if RT∘ is unbounded. In addition, we point out that the case of boundedness of RT∘ can be ignored from the statement of Nirenberg’s problem.The following definition is needed in the squeal.Definition 4.
. LetY be a real reflexive Banach space and Y∗ be its dual space. Let G be a nonempty, bounded, and open subset of Y and S:G¯⟶Y∗ be bounded on finite dimensional subsets of Y. Then, S is said to be “of type S+” if a sequence xn in G¯ satisfies xn⇀x0 and limsupn⟶∞Sxn,xn−x0≤0; then, xn⟶x0 and Sxn⇀Sx0 as n⟶∞.
Browder [12], Definition 5, introduced the following definition.Definition 5.
. LetG be a nonempty, bounded and open subset of a reflexive Banach space Y. A family of operators Stt∈0,1, defined from G¯ into Y∗, is called a “homotopy of class S+” if for sequences xn in G¯ and tn in 0,1 such that xn⇀x0, tn⟶t0 and limsupn⟶∞Stnxn,xn−x0≤0, we have xn⟶x0 and Stnxn⇀St0x0 as n⟶∞.The operatorJ:Y⟶2Y∗ defined by
(2)Jy=y∗∈Y∗:y∗,y=∥y∥2,∥y∥=∥y∗∥,is called the normalized duality mapping. It is well-known that J is single-valued, bounded, monotone, bicontinuous, surjective, and of type S+ if Y and Y∗ are locally uniformly convex and real reflexive Banach spaces. The operator J=I (where I is the identity operator on Y) if Y is a real Hilbert space. Browder [12], Theorems 3 and 4, constructed a degree mapping d for the class of operators of type S+ that satisfies the basic properties listed in Lemma 6.Lemma 6.
LetG be a nonempty, bounded, and open subset of a real reflexive Banach space Y. Let f∗∈Y∗. Then, the following basic properties hold.
(i)
Normalization. dJ,G,0=1 if 0∈G and dJ,G,0=0 if 0∈G¯.(ii)
Existence. Let S:G¯⟶Y∗ be a bounded operator of type S+ such that f∗∈S∂G and dS,G,f∗≠0. Then, f∗∈SG.(iii)
Homotopy Invariance. Let Stt∈0,1 be a homotopy of class S+ defined from G¯ into Y∗ such that f∗∈St∂G for all t∈0,1. Then, dSt,G,f∗ is independent of t∈0,1.(iv)
Invariance under Affine Homotopy. Let Si:G¯⟶Y∗ (i=1,2) be a bounded operator of type S+ and Kt=tS1+1−tS2 for t∈0,1. Then, Ktt∈0,1 is a homotopy of class S+, and dKt,G,f∗ is independent of t∈0,1 provided that f∗∈Kt∂G for all t∈0,1


## 2. Main Result

Theorem7 gives a proof of the Nirenberg problem.Theorem 7.
LetX be a real Hilbert space and T:X⟶X be a continuous and α-expansive operator such that RT∘ is unbounded. Then, T is surjective.Proof.
Choosex0∈RT∘. Then, there exists r>0 such that Brx0⊆RT. Let T~x=Tx−x0. Then, the operator T~:X⟶X is α-expansive and continuous such that RT−x0=RT~. It follows that RT∘ is unbounded if and only if RT~∘ is unbounded. Clearly, the set V=Brx0−x0 is open and 0∈V⊆RT~, i.e., 0∈RT~∘. In addition, we see that T is surjective if and only if T~ is surjective. Thus, we assume without loss of generality that 0∈RT∘. Let U=RT∘ and f∈X. The continuity and α-expansive conditions imply that RT is closed and
(3)∥T−1x−T−1y∥≤α−1∥x−y∥,for all x∈RT and y∈RT. Then, it follows that
(4)T−1x+α−1x−T−1y+α−1y,x−y=T−1x−T−1y,x−y+α−1∥x−y∥2≥−∥T−1x−T−1y∥∥x−y∥+α−1∥x−y∥2≥−α−1∥x−y∥2+α−1∥x−y∥2=0,
for allx∈RT and y∈RT, i.e., T−1+α−1I:RT⟶X is a continuous and monotone operator. It holds that U¯⊆RT because RT is closed and U⊆RT. Let S:U¯⟶X be the restriction of T−1+α−1I on U¯. Let λ>0, and for each f∈X and μ∈0,1, let fμ=μ−1f. Let
(5)Sλfx=Sx+λx−fμ,x∈U¯.
The monotonicity ofS implies
(6)Sλfx−Sλfy,x−y≥λ∥x−y∥2,for all x∈U¯ and y∈U¯. The monotonicity of S and the condition 0∈U¯ yield
(7)Sλfx,x=Sx−S0+S0+λx−fμ,x≥λ∥x∥2−λ∥fμ∥+∥S0∥∥x∥,for all x∈U¯. Since U¯ is unbounded, the right side of (7) approaches infinity as ∥x∥⟶∞. Then, there exists R=Rfμ>0 such that
(8)Sx+λx−fμ,x>0,for all x∈U¯∩∂BR0. Let Ω=U∩BR0. We shall show that the operator S+λI−fμ:Ω¯⟶X is of type S+. We shall apply Kirszbraun-Valentine [3, 4] extension theorem. Indeed, the Lipschitz operator T−1 (with Lipschitz constant α−1) can be extended to an operator P:X⟶X with the same Lipschitz constant. Thus, the operator S~=P+α−1I+λI−fμ:X⟶X is continuous and uniformly monotone, i.e., S~ is of type S+. Let S~Ω¯ be the restriction of S~ on Ω¯. We shall show that S~Ω¯ is of type S+. Indeed, let xn∈Ω¯ for all n such that xn⇀x0∈X as n⟶∞ and
(9)limsupn⟶∞S~Ω¯xn,xn−x0≤0.
It is enough to show thatxn⟶x0∈Ω¯. Since S~=S~Ω¯ on Ω¯, it follows that
(10)limsupn⟶∞S~xn,xn−x0≤0.
The uniform monotonicity ofS~ with domain X (i.e., using (5), (6) and S~=Sλf on Ω¯) shows that
(11)S~xn−S~x0+S~x0,xn−x0=S~xn−S~x0,xn−x0+S~x0,xn−x0≥λ∥xn−x0∥2+S~x0,xn−x0,for all n. Thus, (9)–(11) yield
(12)limsupn⟶∞λ∥xn−x0∥2+S~x0,xn−x0≤limsupn⟶∞S~xn,xn−x0≤0.
SinceS~x0,xn−x0⟶0 as n⟶∞ (because xn⇀x0), (12) shows that
(13)limsupn⟶∞λ∥xn−x0∥2≤0,
i.e.,xn⟶x0 as n⟶∞. Clearly x0∈Ω¯ because xn∈Ω¯ for all n. The continuity of S~Ω¯ implies that S~Ωxn⟶S~Ωx0 as n⟶∞. Thus, the operator S~Ω¯=S+λI−fμ:Ω¯⟶X is bounded, continuous, and of type S+.
Clearly, the identity operatorI:Ω¯⟶X is bounded and of type S+. For each λ>0, let Hλ:0,1×Ω¯⟶X be given by
(14)Hλt,x=tSx+λx−λfμ+1−tx,t,x∈0,1×Ω¯.
Then, (iv) of Lemma6 shows that the family Hλt,.t∈0,1 is an affine homotopy of class S+. Clearly, 0≠Hλ0,x for all x∈∂Ω. Thus, exactly one of the following holds.
(i)
There existsλ>0 such that for each λ′≥λ, there exist tλ′∈0,1 and xλ′∈∂Ω such that Hλ′′tλ′,xλ′=0 for all λ′′≥λ′(ii)
For eachλ>0 there exists λ′≥λ such that(15)Hλ′′t,x≠0forallt,x∈0,1×∂Ωandsomeλ′′≥λ′.
Suppose (i) holds, i.e., there existsn>0 such that for each n′≥n, this corresponds tn′∈0,1, xn′∈∂Ω⊆Ω¯⊆U¯⊆RT, and yn′∈X such that xn′=Tyn′ and Hn′′tn′,xn′=0 for all n′′≥n′, i.e., for each n′≥n, we have
(16)tn′yn′+α−1Tyn′+n′′Tyn′−n′′fμ+1−tn′Tyn′=0,for all n′′≥n′. Replacing n by n0, n′ by n, and n′′ by n′ in (16), for each n≥n0, we get
(17)tnyn+α−1Tyn+n′Tyn−n′fμ+1−tnTyn=0,for all n′≥n. Next we choose a sequence of natural numbers Nn such that tn∈1/Nn,1 for all n. Let λn=Nn+n−1 for all n. Clearly, λn≥n and tn∈1/λn,1 for all n (i.e., tnλn>1 for all n). Since λn≥n for all n, (17) implies
(18)tnyn+α−1Tyn+λnTyn−λnfμ+1−tnTyn=0,for all n. Clearly, tn≠0 for all n. Dividing (18) by tn for all n, we arrive at
(19)γn−1yn+Tyn=λnγn−1fμ,for all n, where γn=α−1+λn+1−tntn−1. The boundedness of xn and T−1 shows that yn is bounded. Assume without loss of generality that tn⟶t0∈0,1 as n⟶∞. Suppose t0=0. The sequence tnλn cannot converge to 0 because tnλn>1 for all n. Then, we consider either tnλn is unbounded or bounded. Suppose tnλn is unbounded, i.e., there exists a subsequence of tnλn, denoted by tnλn, such that tnλn⟶∞ as n⟶∞. Since t0=0, yn and Tyn are bounded, letting n⟶∞ in (18) implies
(20)tnλnTyn−tnλnfμ+Tyn⟶0,as n⟶∞, i.e., cnTyn⟶fμ as n⟶∞, where cn=1+tnλn/tnλn for all n. Since cn⟶1+ as n⟶∞ and Tyn is bounded, we conclude that xn=Tyn⟶fμ∈RT¯=RT as n⟶∞. Letting μ⟶1− gives f∈RT.
Next we assume thattnλn is bounded. Assume without loss of generality that there exists a subsequence, denoted again by tnλn, such that tnλn⟶a∈1,∞ as n⟶∞. Then, cn⟶a+1a−1 and
(21)xn=Tyn⟶aa+1−1fμ∈RT¯=RT,as n⟶∞. Since d=aa+1−1∈0,1, letting μ⟶d gives f∈RT.
Next we consider the caset0∈0,1. Then, we have μn=α−1+1−tntn−1⟶μ0=α−1+1−t0t0−1∈ℝ and
(22)λnγn−1=λnα−1+λn+1−tntn−1=Nn+n−1μn+Nn+n−1=11+ξnμn,
for alln, where ξn=Nn+n−1−1. Since ξn⟶0 as n⟶∞ and μ0∈ℝ, it follows that λnγn−1⟶1 as n⟶∞. Letting n⟶∞ in (19) yields xn=Tyn⟶fμ∈RT¯=RT. Letting μ⟶1− shows that f∈RT. In all cases, each f∈X satisfying (i) lies in RT.
Suppose (ii) holds. Then, there exists a subsequence ofn, denoted again by n, such that 0∈Hnt,∂Ω for all t∈0,1 and n. Then, (iv) of Lemma 6. shows that the family Hnt,.t∈0,1 is a homotopy of class S+ defined from Ω¯ into X such that 0∈Hnt,∂Ω for all t∈0,1 and n. Let d stand for Browder degree mapping. Applying (iii) of Lemma 6., we conclude that dHnt,.,Ω,0 is independent of t∈0,1, i.e., (i) and (iii) of Lemma 6. give
(23)dS+nI−nfμ,Ω,0=dI,Ω,0=1
for alln. In addition, (ii) of Lemma 6. confirms that nfμ∈S+nIΩ for all n. Thus, for each positive integer n, there exist xn∈Ω and yn∈X such that xn=Tyn for all n and
(24)nfμ=Sxn+nxn=T−1xn+α−1xn+nxn=n+α−1Tyn+yn,
for alln. Since xn and T−1 are bounded, it follows that yn is bounded. In addition, (24) gives
(25)xn=Tyn=nτnfμ−τnyn⟶fμ∈Ω¯⊆U¯⊆RT,as n⟶∞ because τn⟶0 and nτn⟶1, where τn=n+α−1−1. Thus, letting μ⟶1− shows that f∈RT. Thus, each f∈X satisfying (i) or (ii) lies in RT. This shows that T is surjective. The proof is completed.



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*Source: 9487405-2022-03-17.xml*