View of Compactness in spaces of Lipschitz functions

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Rev. Anal. Num´er. Th´eor. Approx., vol. 30 (2001) no. 1, pp. 9–14



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.


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]



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,


(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⊂




B(xk, δk), where δkxk.

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.




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