Rev. Anal. Num´er. Th´eor. Approx., vol. 30 (2001) no. 1, pp. 9–14 ictp.acad.ro/jnaat
COMPACTNESS IN SPACES OF LIPSCHITZ FUNCTIONS
S¸TEFAN COBZAS¸
Dedicated to the memory of Acad. Tiberiu Popoviciu
Abstract. The aim of this paper is to prove a compactness criterium in spaces of Lipschitz and Fr´echet differentiable mappings.
MSC 2000. 46E15.
1. INTRODUCTION
In the last years there have been an increasing interest in the study of Lipschitz functions and of spaces of Lipschitz functions, as a first step to extend to the nonlinear setting results from linear functional analysis. For instance, in the attempt to build a spectral theory for nonlinear operators, a special attention was paid to spectra of Lipschitz operators (see, e.g., [9], [2], [4]).
Lipschitz duals, meaning spaces of Lipschitz functions on a metric linear space, were used to study best approximation problems in such spaces (see [10]). A good account on Banach spaces and Banach algebras of Lipschitz functions is given in the monograph [11]. The monograph [6] contains a comprehensive study of Lipschitz functions on Banach spaces and their applications to the geometry of Banach spaces (e.g. the Lipschitz classification of Banach spaces).
As asserts Appell [1], apparently there is no compactness criterium in spaces of H¨older functions, and some criteria given in the literature turned to be false (e.g. that in [7]). The aim of this Note is to prove such a criterium (a true one, I hope) for families of Lipschitz and Fr´echet differentiable mappings. The paper by J. Batt [5] contains a detailed study of compactness for nonlinear mappings and their adjoints, including Schauder type theorems. A Schauder type theorem for differentiable mappings was proved also by Yamamuro [12].
“Babe¸s-Bolyai” University, Faculty of Mathematics and Computer Science, RO-3400 Cluj–Napoca, Romania, e-mail: [email protected].
2. THE RESULT
Let X, Y be real or complex normed linear spaces, and Ω a subset of X.
Denote by Lip(Ω, Y) the space of all Lipschitz mappings from Ω to Y, i.e.
those mappings f : Ω→Y for which the number
(1) L(f) := sup{kf(x)−f(y)k/kx−yk:x, y∈Ω, x6=y}
is finite. The number L(f) defined by (1) is called the Lipschitz norm of the mapping f, and it is the smallest Lipschitz constant for f. The function L(·) is a seminorm on Lip(Ω, Y), so that (Lip(Ω, Y), L) is a seminormed space which is complete if Y is a Banach space. (The operations of addition and multiplication by scalars are defined pointwisely)
If Ω is an open subset ofX, denote byC1(Ω, Y) the space of all continuously Fr´echet differentiable mappings from Ω toY, and for K ⊂Ω put
C1Lip(K, Y) :={f ∈Lip(K, Y) :∃F ∈C1(Ω, Y) such that F|K=f}.
Let also L(X, Y) denote the space of all continuous linear operators from X toY equipped with the uniform norm.
The compactness result we shall prove is the following:
Theorem 1. Let X, Y be normed spaces, Ω an open subset of X and K a compact convex subset of Ω.
Suppose that Z is a subset ofC1Lip(K, Y) such that
(i) for every x∈K the set{f0(x) :f ∈Z} is totally bounded in L(X, Y);
(ii) for everyx∈K and every >0 there existsδ =δ(x, )>0 such that
∀x0 ∈B(x, δ)⊂Ω, ∀f ∈Z kf0(x)−f0(x0)k ≤. Then the set Z is totally bounded inLip(K, Y).
Conversely, if the setZ ⊂C1Lip(Ω, Y)is totally bounded inLip(Ω, Y)then Z satisfies the conditions (i) and (ii).
As consequence, one obtains the following corollary.
Corollary 1. IfY is a Banach space andZ ⊂C1Lip(K, Y) is closed and satisfies the conditions (i) and (ii) from Theorem 1then the set Z is compact in Lip(K, Y).
The proof of Theorem 1 will be based on the following lemma:
Lemma 1. Let X, Y be normed spaces and Ω an open subset of X. If g: Ω→Y satisfies
(2) kg(x0)−g(x)k ≤λkx0−xk,
for every x in a neighborhood U ⊂Ω of x0 and g is Fr´echet differentiable at x0, then
(3) kg0(x0)k ≤λ.
Conversely, if g is Fr´echet differentiable on an open convex neighborhood U ⊂Ω of x0 and
(4) kg0(x)k ≤λ, ∀x∈U,
then
(5) kg(x)−g(y)k ≤λkx−yk, ∀x, y∈U.
Proof of Lemma 1. Suppose that g: Ω → Y satisfies (2). The differentia- bility ofg atx0 implies the existence ofg0(x0)∈L(X, Y) such that
(6) g(x0+h)−g(x0) =g0(x0)h+khkα(h),
where limh→0α(h) = 0. Forn∈Nchoose δn>0 such that B(x0, δn)⊂Ω and kα(h)k ≤1/n, ∀h∈B(0, δn).
Then, from (6),
kg0(x0)k ≤ kg(x0+h)−g(x0)k+khkkα(h)k
≤(λ+ 1n)khk.
The inequality
kg0(x0)hk ≤(λ+n1)khk, ∀h, khk ≤δn, implies kg0(x0)k ≤λ+ 1/n, ∀n∈N,so thatkg0(x0)k ≤λ.
Conversely, suppose thatgis Fr´echet differentiable on an open convex neigh- borhoodU ⊂Ω of x0, and satisfies (4).
By the mean value theorem
kg(x)−g(y)k ≤ kx−yksup{kg0(ξ)k:ξ∈[x, y]} ≤λkx−yk, for all x, y∈U.
Lemma 1 is proved.
Proof of Theorem 1.
Suppose that the setZ ⊂C1Lip(K, Y) satisfies the conditions (i) and (ii), and let >0 be given.
By (ii), for every x∈K there existsδx>0 such that
(7) ∀f ∈Z and ∀x0 ∈B(x, δx)∩K kf0(x)−f0(x0)k ≤. Since the set K is compact, there exists x1, . . . , xp inK such that
(8) K⊂
p
[
k=1
B(xk, δk), where δk=δxk.
By (i), the set Yk = {f0(xk) : f ∈ Z} is totally bounded in L(X, Y), for k= 1,2, . . . , p.It follows that the set
W =Y1× · · · ×Yk
is totally bounded in (L(X, Y))p with respect to the norm k(A1, . . . , Ap)k= max{kA1k, . . . ,kApk}, as well as the set
H={(f0(x1), ..., f0(xp)) :f ∈Z} ⊂W.
Therefore we can findf1, . . . , fn inZ such that
(9) ∀f ∈Z ∃j∈ {1, . . . , n} such that kf0(xk)−fj0(xk)k ≤, fork= 1, . . . , p.
We shall show that{f1, . . . , fn}is a 3-net for the setZ with respect to the Lipschitz norm (1) on Lip(K, Y).
Letf ∈Z. By (9) there is j∈ {1, . . . , n} such that (10) kf0(xk)−fj0(xk)k ≤, for k= 1, . . . , n.
By the mean value theorem, we have for x, y∈K
(11) k(f −fj)(x)−(f −fj)(y)k ≤ kx−yksup{k(f0−fj0)(ξ)k:ξ ∈[x, y]}.
Since ξ ∈ [x, y] ⊂ K, by (8) there exists k ∈ {1, . . . , p} such that ξ ∈ B(xk, δk). But then
(12) k(f0−fj0)(ξ)k ≤ kf0(ξ)−f0(xk)k+k(f0−fj0)(xk)k+kfj0(xk)−fj0(ξ)k ≤3.
(We have applied (7) to the first and the last term, and (10) to the middle one).
By (11) and (12)
k(f−fj)(x)−(f−fj)(y)k ≤3kx−yk for all x, y∈K or, equivalently,
L(f −fj)≤3.
To prove the converse implication, suppose that Z ⊂Lip(Ω, Y)∩C1(Ω, Y) is totally bounded in Lip(Ω, Y), and let > 0 be given. Choose an -net {f1, . . . , fn} ⊂Z, i.e. ∀f ∈Z∃j∈ {1, . . . , n}such that ∀x, y∈Ω:
k(f −fj)(x)−(f−fj)(y)k ≤kx−yk.
Taking into account Lemma 1, one obtains kf0(x)−fj0(x)k ≤,
for all x ∈Ω, showing that {f10(x), . . . , fn0(x)} is an -net for the set{f0(x) : f ∈Z}. Therefore (i) holds.
To prove (ii), let > 0 and x ∈ Ω be fixed. Choose again an -net {f1, . . . , fn} for the set Z. Since the mappings fi are of class C1 there ex- istsδ >0 such that
(13) ∀x0 ∈B(x, δ)⊂Ω and ∀i∈ {1, . . . , n} kfi0(x)−fi0(x0)k ≤. Forf ∈Z choose j∈ {1, . . . , n} such that
(14) L(f −fj)≤.
By Lemma 1 this implies
(15) ∀y∈Ω k(f0−fj0)(y)k ≤. Taking into account (13) and (15), we obtain
kf0(x)−f0(x0)k ≤ kf0(x)−fj0(x)k+kfj0(x)−fj0(x0)k+kfj0(x0)−f0(x0)k ≤3 for all x0 ∈B(x, δ), which shows that (ii) holds too.
Theorem 1 is completely proved.
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Received February 7, 2000.
