Smoothness via directional smoothness and Marchaud's theorem in Banach spaces

Accepted Manuscript

Smoothness via directional smoothness and Marchaud’s theorem inBanach spaces

Michal Johanis, Ludek Zajícek

PII: S0022-247X(14)00901-9DOI: 10.1016/j.jmaa.2014.09.068Reference: YJMAA 18884

To appear in: Journal of Mathematical Analysis and Applications

Received date: 11 July 2014

Please cite this article in press as: M. Johanis, L. Zajícek, Smoothness via directional smoothnessand Marchaud’s theorem in Banach spaces, J. Math. Anal. Appl. (2015),http://dx.doi.org/10.1016/j.jmaa.2014.09.068

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http://dx.doi.org/10.1016/j.jmaa.2014.09.068

SMOOTHNESS VIA DIRECTIONAL SMOOTHNESS AND MARCHAUD’S THEOREM IN

BANACH SPACES

MICHAL JOHANIS AND LUDEK ZAJICEK

Abstract. Classical Marchaud’s theorem (1927) asserts that if f is a bounded function on [a, b], k ∈ N, and the (k+1)th

modulus of smoothness ωk+1(f ; t) is so small that η(t) =∫ t0

ωk+1(f ;s)

sk+1 ds < +∞ for t > 0, then f ∈ Ck((a, b)

)and f (k)

is uniformly continuous with modulus cη for some c > 0 (i.e. in our terminology f is Ck,cη-smooth). Using a known

version of the converse of Taylor theorem we easily deduce Marchaud’s theorem for functions on certain open connected

subsets of Banach spaces from the classical one-dimensional version. In the case of a bounded subset of Rn our result

is more general than that of H. Johnen and K. Scherer (1973), which was proved by quite a different method. We also

prove that if a locally bounded mapping between Banach spaces is Ck,ω-smooth on every line, then it is Ck,cω-smooth

for some c > 0.

1. Introduction

Classical Marchaud’s theorem (1927) asserts that if f is a bounded function on [a, b], k ∈ N, and the (k + 1)th

modulus of smoothness ωk+1(f ; t) is so small that η(t) =∫ t

0ωk+1(f ;s)

sk+1 ds < +∞ for t > 0, then f ∈ Ck((a, b)

)and f (k)

is uniformly continuous with modulus cη for some c > 0. Marchaud’s theorem was generalised to functions defined onbounded LG domains in R

n in [JS] (it is an easy consequence of Theorems 1 and 2 of [JS]).We will generalise Marchaud’s theorem to mappings on certain open connected subsets of Banach spaces. Namely

our proof works if the domain has the “uniform convex chain (UCC) property” (in particular, if it is convex andbounded or if it is convex and contains an unbounded cone). We show that our version of Marchaud’s theorem (in R

n)is more general than that of [JS] (which works with LG domains), see Proposition 21.

We deduce our version of Marchaud’s theorem rather easily from the classical one-dimensional version and a recent([J]) quantitative version of the Converse Taylor theorem. Both our proof and the proof of the result of [J] use only alittle of analysis; they are essentially based on several non-trivial but well-known properties of polynomials in Banachspaces.

The rough strategy of our proof is the following: If f satisfies the assumptions of Marchaud’s theorem in U ⊂ X,then its one-dimensional version implies that f is Ck,c1η-smooth on all segments in U . For convex U we then showthat this fact implies (see Proposition 12) that f is Ck,c2η-smooth on U . This is done by verifying (using Theorem 9)that near each a ∈ U the mapping f is well approximated by a polynomial of degree at most k and so the quantitativeversion of the Converse Taylor theorem of [J] implies that f is Ck,c2η-smooth on U .

The above method gives indeed very easily Marchaud’s theorem in higher dimensions under the assumption ofcontinuity of f on convex domains, see Proposition 13. We believe that this simple proof may be interesting also forpeople working in the function spaces theory in R

n. Moreover, it generalises to the domains with the UCC propertyquite easily, see Remark 14.

In the main section (Section 4) we prove a more general version (for vector-valued mappings on UCC domains)of Marchaud’s theorem (Theorem 20) under much weaker assumptions (e.g. for locally bounded mappings). Thisgeneralisation needs some additional technical results, nevertheless we believe it is interesting from the point of viewof differentiation theory in Banach spaces.

Theorem 9 also immediately implies Theorem 10 which is a new “directional version” of the Converse Taylor theoremand can be of some independent interest. Another new interesting result is a characterisation of Ck,cω-smoothnessvia the smoothness on one-dimensional affine subspaces in the domain (Theorem 19), which is analogous to thecharacterisation of polynomials or holomorphic mappings.

2. Preliminaries

We set N0 = N ∪ {0}. For x ∈ R we denote by �x� the ceiling of x, i.e. the unique number k ∈ Z satisfyingk − 1 < x ≤ k. All vector spaces considered are real. We denote by B(x, r), resp. U(x, r) the closed, resp. open ballin a normed linear space centred at x with radius r > 0. By BX we denote the closed unit ball of a normed linearspace X, i.e. BX = B(0, 1). By SX we denote the unit sphere of a normed linear space X. Let X, Y be normedlinear spaces and n ∈ N. By L(nX;Y ) we denote the space of continuous n-linear mappings from Xn to Y with the

Date: April 2014.

2010 Mathematics Subject Classification. 46G05, 46T20, 26B05.

Key words and phrases. Ck,ω-smoothness, Marchaud’s theorem, Converse Taylor theorem.

Supported by the grants GACR P201/11/0345 (M. Johanis) and GACR P201/12/0436 (L. Zajıcek).

1

2 MICHAL JOHANIS AND LUDEK ZAJICEK

norm ‖M‖ = supx1,...,xn∈BX‖M(x1, . . . , xn)‖. Recall that a continuous n-homogeneous polynomial from X to Y is a

mapping P that is given by P (x) = M(x, . . . , x), x ∈ X, for some M ∈ L(nX;Y ). By P(nX;Y ) we denote the space ofcontinuous n-homogeneous polynomials from X to Y with the norm ‖P‖ = supx∈BX

‖P (x)‖. By Pn(X;Y ) we denotethe space of continuous polynomials of degree at most n from X to Y with the norm ‖P‖ = supx∈BX

‖P (x)‖. Of course,

P0(X;Y ) = P(0X;Y ) is the space of constant mappings. The Polarisation formula (see e.g. [MO1, Statement 8, p. 62]or [HJ, Proposition 1.13]) implies the following fact:

Fact 1. Let X, Y be normed linear spaces and let n ∈ N. There is a mapping I : P(nX;Y ) → L(nX;Y ) that satisfies

P (h) = I(P )(h, . . . , h) which is a linear isomorphism into and ‖P‖ ≤ ‖I(P )‖ ≤ nn

n! ‖P‖.The mapping f : X → Y between topological spaces X and Y is said to be Baire measurable (or to have the Baire

property) if f−1(G) ⊂ X has the Baire property for every G ⊂ Y open, so each Borel measurable mapping is alsoBaire measurable.

Let X, Y be vector spaces, U ⊂ X, and f : U → Y . We define the first difference by

Δ1f(x;h) = f(x+ h)− f(x)

for all x ∈ U and h ∈ X such that the right-hand side is defined. Further, we inductively define the differences ofhigher order by

Δnf(x;h1, . . . , hn) = Δn−1f(x+ h1;h2, . . . , hn)−Δn−1f(x;h2, . . . , hn)

for all x ∈ U and h1, . . . , hn ∈ X such that the right-hand side is defined. We are also going to use the classicalnotation Δn

hf(x) = Δnf(x;h, . . . , h). It is easy to check that

Δnhf(x) =

n∑k=0

(−1)n−k

(n

k

)f(x+ kh). (1)

We will rely on the following fundamental result; it is a combination of [MO1, Satz I] and [MO2, Statement 12,p. 182], see also [HJ, Corollary 1.55]:

Theorem 2 ([MO1], [MO2]). Let X be a Banach space, Y a normed linear space, n ∈ N0, and let P : X → Y be suchthat φ ◦ P is Baire measurable for each φ ∈ Y ∗. Then P ∈ Pn(X;Y ) if and only if P satisfies the (Frechet) formula

Δn+1h P (x) =

n+1∑k=0

(−1)n+1−k

(n+ 1

k

)P (x+ kh) = 0 (2)

for all x, h ∈ X.

We remark that the above characterisation holds also in incomplete spaces X provided that we assume that φ ◦ Pis continuous ([MO1, Satz I*], see also [HJ, Theorem 2.49, Fact 1.49]).

The following characterisation is an immediate consequence: A Baire measurable mapping P : X → Y is a continuouspolynomial of degree at most n if and only if the restriction of P to each one-dimensional affine subspace of X is apolynomial of degree at most n; moreover, it suffices to test only the polynomiality of φ ◦ P , φ ∈ Y ∗. We note that asimilar characterisation holds also for holomorphic mappings (of course in the case of complex Banach spaces). Oneof our main results below is that Ck,cω-smoothness can be characterised in the same way (Theorem 19, Theorem 17).

Let X, Y be normed linear spaces, U ⊂ X open, f : U → Y , and x ∈ U . By Df(x) we denote the Frechet derivativeof f at x, and by Df(x)[h] we denote the evaluation of this derivative in h ∈ X. Similarly we denote by Dkf(x) thekth Frechet derivative of f at x. By dkf(x) we denote the k-homogeneous polynomial corresponding to the symmetrick-linear mapping Dkf(x), so dkf(x)[h] = Dkf(x)[h, . . . , h]. For convenience we put d0f = f . Recall that we identifyDkf(x) and D(Dk−1f)(x) using the formula Dkf(x)[h1, . . . , hk] =

(D(Dk−1f)(x)[h1]

)(h2, . . . , hk).

We recall the following well-known easy facts:

Fact 3. Let X be a normed linear space, U ⊂ X open, f : U → R, x ∈ U , h ∈ X, and k ∈ N. Further, letg(t) = f(x+ th). Then dkf(x+ th)[h] = g(k)(t) if the left-hand side exists.

Fact 4. Let X, Y , Z be normed linear spaces, L ∈ L(Y ;Z), U ⊂ X open, and k ∈ N. Let f : U → Y be k-timesFrechet differentiable at a ∈ U . Then Dk(L ◦ f)(a) = L ◦Dkf(a) and consequently also dk(L ◦ f)(a) = L ◦ dkf(a).

We say that f is Ck-smooth if Dkf (i.e. the mapping x �→ Dkf(x)) is continuous in the domain. Note that f isCk-smooth if and only if dkf is continuous in the domain by Fact 1. We denote by Ck(U ;Y ) the vector space ofall Ck-smooth mappings from U into Y . Further, C0(U ;Y ) = C(U ;Y ) is the space of continuous mappings. We setCk(U) = Ck(U ;R), k ∈ N0.

We also recall the following well-known corollary of the Taylor formula (it follows easily from [D, Theorem 8.14.2],or see [HJ, Corollary 1.108]).

SMOOTHNESS VIA DIRECTIONAL SMOOTHNESS AND MARCHAUD’S THEOREM IN BANACH SPACES 3

Theorem 5 (see e.g. [D]). Let X, Y be normed linear spaces, U ⊂ X an open convex set, k ∈ N, and f ∈ Ck(U ;Y ).Then for any x ∈ U and h ∈ X satisfying x+ h ∈ U we have∥∥∥∥∥f(x+ h)−

k∑j=0

1

j!djf(x)[h]

∥∥∥∥∥ ≤ 1

k!

(sup

t∈[0,1]

∥∥dkf(x+ th)− dkf(x)∥∥) · ‖h‖k.

Let (P, ρ), (Q, σ) be metric spaces. The minimal modulus of continuity of a uniformly continuous mapping f : P → Qis defined as ωf (δ) = sup {σ(f(x), f(y)); x, y ∈ P, ρ(x, y) ≤ δ} for δ ∈ [0,+∞). Clearly, ωf is continuous at 0.

A modulus is a non-decreasing function ω : [0,+∞) → [0,+∞] continuous at 0 with ω(0) = 0. The set of allmoduli will be denoted by M. (Notice that a modulus by our definition can be infinite on some interval.) We say thatf : P → Q is uniformly continuous with modulus of continuity ω ∈ M if ωf ≤ ω.

Let X, Y be normed linear spaces, U ⊂ X an open set, k ∈ N, f ∈ Ck(U ;Y ), and let ω ∈ M. We say that f isCk,ω-smooth on U (or f ∈ Ck,ω(U ;Y )) if dkf is uniformly continuous on U with modulus ω. Note that Fact 1 implies

that if f is Ck,ω-smooth, then Dkf is uniformly continuous on U with modulus kk

k! ω, and conversely if f is Ck-smoothand Dkf is uniformly continuous on U with modulus ω, then f is Ck,ω-smooth.

Further, we define the kth modulus of smoothness of f : U → Y , U ⊂ X any set, by

ωk(f ; t) = sup‖h‖≤t

[x,x+kh]⊂U

‖Δkhf(x)‖, t ∈ [0,+∞),

where [x, x+kh] denotes the segment with endpoints x and x+kh. We remark that unlike in the definition of modulusof continuity, in the definition of modulus of smoothness we are not allowed to “jump over the gaps in the domain”.

We will use the following version of Marchaud’s theorem:

Theorem 6 ([M], [DL]). Let k ∈ N. There is a constant Ak > 0 such that if a, b ∈ R, a < b, and f : [a, b] → R is abounded function, then f is k-times differentiable on (a, b) and for each t ≥ 0

ωf(k)(t) = ω1(f(k); t) ≤ Ak

∫ t

0

ωk+1(f ; s)

sk+1ds,

provided that the integral on the right-hand side is finite for some t > 0.

For a continuous f it is proved e.g. in [DL, Theorem 6.3.1] (where we put A = [a, b], p = ∞, and r = k+1); further,the original Marchaud’s version [M, § 29, cf. § 22] shows that the finiteness of the integral already gives the continuityof f .

3. Converse Taylor theorems and simple consequences

Let X, Y be normed linear spaces, U ⊂ X open, f : U → Y , and k ∈ N0. We say that f is T k-smooth at x ∈ U ifthere exists a polynomial P x ∈ Pk(X;Y ) satisfying P x(0) = f(x) and

f(x+ h)− P x(h) = o(‖h‖k), h → 0.

We say that f is T k-smooth on U if it is T k-smooth at every point x ∈ U . Recall that by Peano’s form of Taylor’s

theorem a Ck-smooth mapping is also T k-smooth and the approximating polynomial is given by∑k

j=01j!d

jf(x). A

converse statement is contained in the following theorem, see e.g. [LS], [AD], cf. also [J] or [HJ, Theorem 1.110].

Theorem 7 (Converse Taylor theorem; [LS], [AD]). Let X, Y be normed linear spaces, U ⊂ X an open set, f : U → Y ,and k ∈ N0. Then f ∈ Ck(U ;Y ) if and only if f is a T k-smooth mapping satisfying

lim(y,h)→(x,0)

h �=0

‖R(y, h)‖‖h‖k = 0

for every x ∈ U , where R(x, h) = f(x + h) − P x(h) and the polynomials P x ∈ Pk(X;Y ) come from the definition ofT k-smoothness of f at x.

We will rely on a quantitative version of the above theorem. Let X, Y be normed linear spaces, U ⊂ X an open set,f : U → Y , and k ∈ N0. We say that f is UT k-smooth on U with modulus ω if for each x ∈ U there is a polynomialP x ∈ Pk(X;Y ) satisfying

‖f(x+ h)− P x(h)‖ ≤ ω(‖h‖)‖h‖k for x+ h ∈ U .

For a convex bounded subset U of a normed linear space we define its “ellipticity” eU = diamUsup {r; ∃a∈U : B(a,r)⊂U} .

Theorem 8 ([J], see also [HJ, Theorem 1.125]). Let X, Y be normed linear spaces, U ⊂ X an open convex boundedset, f : U → Y , and k ∈ N. Suppose that f is UT k-smooth on U with modulus ω. Then f is Ck,mω-smooth on U withm = cke

kU , where ck > 0 is a constant depending only on k.


We note that the quantity eU is defined slightly differently in [J], but it is clear that it is not larger than eU asdefined here.

Now we are ready to present our main tool. Let X, Y be normed linear spaces, A ⊂ X, f : A → Y , and k ∈ N0.We say that f is weakly T k-smooth at x ∈ A if there exists a polynomial P x ∈ Pk(X;Y ) satisfying P x(0) = f(x) andf(x+ th)−P x(th) = o(tk), t → 0 for each h ∈ X. We say that f is weakly T k-smooth on A if it is weakly T k-smoothat every x ∈ A.

We say that f is directionally T k-smooth at x ∈ A if the mapping t �→ f(x+th) is T k-smooth at 0 for each h ∈ X, i.e.for each h ∈ X there is a polynomial P x,h ∈ Pk(R;Y ) satisfying P x,h(0) = f(x) and f(x+th)−P x,h(t) = o(tk), t → 0.We say that f is directionally T k-smooth on A if it is directionally T k-smooth at every x ∈ A.

Recall that all the polynomials in the above definitions are uniquely determined. Consequently, if f is directionallyT k-smooth at x, then P x,h(t) = P x,th(1) for every h ∈ X and t ∈ R. Further, if f is weakly T k-smooth at x, thenf is directionally T k-smooth at x and for the approximating polynomials from the definitions the following holds:P x,h(t) = P x(th) for h ∈ X, t ∈ R. The converse implication holds under additional assumptions:

Theorem 9. Let X be a Banach space, Y a normed linear space, U ⊂ X open, f : U → Y , a ∈ U , and k ∈ N. Supposethat φ ◦ f is Baire measurable for each φ ∈ Y ∗, f is directionally T k-smooth on U , and further

R(a+ ty, th) = o(tk), t → 0 for each y, h ∈ X, (3)

where R(x, h) = f(x + h) − P x,h(1) and where the polynomials P x,h ∈ Pk(R;Y ) come from the definition of thedirectional T k-smoothness of f at x. Then f is weakly T k-smooth at a.

Proof. Without loss of generality we may assume that a = 0. Set P (h) = P 0,h(1) for each h ∈ X. It is easy to see that itsuffices to show that P ∈ Pk(X;Y ), which will be done using Theorem 2. First we show that φ◦P is Baire measurable

for each φ ∈ Y ∗. So fix φ ∈ Y ∗. For each h ∈ X there are g0(h), . . . , gk(h) ∈ R such that φ◦P 0,h(t) =∑k

j=0 gj(h)tj for

t ∈ R. Note that φ ◦ P (h) = φ ◦ P 0,h(1) =∑k

j=0 gj(h) and so it suffices to show that the functions g0, . . . , gk : X → R

are Baire measurable. Clearly g0(h) = φ ◦ f(0) for each h ∈ X. We proceed by induction, assuming that g0, . . . , gm−1

are Baire measurable for some 1 ≤ m ≤ k. Since for a fixed h ∈ X we have

limt→0

1

tm

(φ ◦ f(th)−

k∑j=0

gj(h)tj

)= φ

(limt→0

f(th)− P 0,h(t)

tm

)= 0,

it follows that gm(h) = limt→01tm

(φ ◦ f(th)−∑m−1

j=0 gj(h)tj). In particular,

gm(h) = limn→∞nm

(φ ◦ f

(h

n

)−

m−1∑j=0

gj(h)

nj

).

Hence on each bounded set gm is a pointwise limit of a sequence of Baire measurable functions and so it is Bairemeasurable.

Now we show that P satisfies the formula (2). Fix x, h ∈ X and define

q(t) =k+1∑j=0

(−1)k+1−j

(k + 1

j

)P (tx+ jth).

Clearly q ∈ Pk(R;Y ). Let r > 0 be such that t(x+ jh) ∈ U for all j ∈ {0, . . . , k + 1} and |t| ≤ r. We have

‖q(t)‖ =

∥∥∥∥∥∥k+1∑j=0

(−1)k+1−j

(k + 1

j

)(P (tx+ jth)− f(tx+ jth) + f(tx+ jth)− P tx,th(j) + P tx,th(j)

)∥∥∥∥∥∥≤

k+1∑j=0

(k + 1

j

)(‖R(0, tx+ jth)‖+ ‖R(tx, jth)‖)for all |t| ≤ r, since

∑k+1j=0 (−1)k+1−j

(k+1j

)P tx,th(j) = 0 by Theorem 2. This estimate combined with the assumption (3)

implies that q(t) = o(tk), t → 0. It follows that q is a zero polynomial (apply the classical fact to φ ◦ q, φ ∈ Y ∗), andin particular q(1) = 0, which is what we wanted to prove.

��As a consequence of the above theorem we obtain that the Converse Taylor theorem holds even if we assume only

directional T k-smoothness:

Theorem 10. Let X be a Banach space, Y a normed linear space, U ⊂ X an open set, f : U → Y , and k ∈ N.Then f ∈ Ck(U ;Y ) if and only if f is a directionally T k-smooth mapping such that φ ◦ f is Baire measurable for eachφ ∈ Y ∗ and

lim(y,h)→(x,0)

h �=0

‖R(y, h)‖‖h‖k = 0 (4)


for every x ∈ U , where R(x, h) = f(x+ h)− P x,h(1) and the polynomials P x,h ∈ Pk(X;Y ) come from the definitionof the directional T k-smoothness of f at x.

Proof. It suffices to notice that the assumption (4) implies (3) in Theorem 9 and it also promotes the weak T k-smooth-ness to T k-smoothness. Hence both implications follow from Theorem 7.

��We end this section by showing that Theorem 9 and Theorem 8 (and the one-dimensional Marchaud’s theorem)

very easily imply Marchaud’s theorem in higher (even infinite) dimensions under the assumption of continuity of f . Inthe next section we give more general versions of the propositions below, which need some additional technical results.

We believe that the following simple proof of Proposition 13 may be interesting also for people working only withfunctions on R

n. We underline that Proposition 13 can be very easily generalised (see Remark 14) to more generalsubsets of Rn than the bounded convex sets (namely the sets with the UCC property); in this way we obtain a resultthat is more general (see Proposition 21) than the currently known multi-dimensional versions of Marchaud’s theorem([JS], which works with LG domains).

Definition 11. Let X, Y be normed linear spaces and U ⊂ X an open set. We say that f : U → Y is Ck,ω-smoothon every open segment in U if for every x ∈ U , h ∈ SX , and (a, b) ⊂ R satisfying x + th ∈ U for all t ∈ (a, b) themapping g : (a, b) → Y , g(t) = f(x+ th), is Ck,ω-smooth.

Proposition 12. Let X be a Banach space, Y a normed linear space, U ⊂ X an open convex bounded set, f ∈ C(U ;Y ),k ∈ N, and ω ∈ M. Suppose that f is Ck,ω-smooth on every open segment in U . Then f is Ck,mω-smooth on U withm = cke

kU , where ck > 0 is a constant depending only on k.

Proof. Clearly f is directionally T k-smooth on U . Let x, x + h ∈ U , h �= 0. Then g(t) = f(x + t h

‖h‖)is Ck,ω-

smooth on (−δ, ‖h‖ + δ) for some δ > 0. Thus Taylor’s theorem (Theorem 5) used on g at 0 with the increment ‖h‖implies that ‖R(x, h)‖ = ‖f(x + h) − P x,h(1)‖ =

∥∥g(‖h‖) − P x,h/‖h‖(‖h‖)∥∥ ≤ 1k!ω(‖h‖)‖h‖k, where the polynomials

P x,h ∈ Pk(R;Y ) come from the definition of the directional T k-smoothness. So (3) is clearly satisfied at each a ∈ Uand thus f is weakly T k-smooth on U by Theorem 9. Consequently, f is UT k-smooth on U with modulus 1

k!ω, so theapplication of Theorem 8 finishes the proof.

��Proposition 13. Let X be a Banach space, U ⊂ X an open convex bounded set, f ∈ C(U), and k ∈ N. Thenf ∈ Ck(U) and

ωdkf (t) ≤ BkekU

∫ t

0

ωk+1(f ; s)

sk+1ds,

provided that the integral on the right-hand side is finite for some t > 0. The constant Bk depends only on k.

Proof. For a fixed x ∈ U and h ∈ SX we define g(t) = f(x+ th) for t ∈ (c, d), where (c, d) is the maximal interval suchthat x + th ∈ U for all t ∈ (c, d). Then it is easy to see (using (1)) that ωk+1(g; s) ≤ ωk+1(f ; s) for all s ∈ [0,+∞).

Therefore ωg(k)(t) ≤ Ak

∫ t

0ωk+1(f ;s)

sk+1 ds by Theorem 6 used on each [a, b] ⊂ (c, d). Hence Proposition 12 implies that

ωdkf (t) ≤ AkckekU

∫ t

0ωk+1(f ;s)

sk+1 ds.��

Remark 14. If U has the UCC property (see Definition 18), then estimating as in (7) we obtain the assertion of theabove proposition with

ωdkf (t) ≤ Ck,U

∫ t

0

ωk+1(f ; s)

sk+1ds,

where the constant Ck,U depends only on k and U .

4. Main results

Remark 15. First we remark that Theorem 9 holds also for incomplete spaces X provided that we additionally assumethat φ ◦ f is continuous for each φ ∈ Y ∗ and

lim(y,t)→(h,0)

t �=0

‖R(a, ty)‖tk

= 0 for each h ∈ X. (5)

Indeed, it suffices to modify the proof of Theorem 9 by showing that in this case φ ◦ P is continuous for each φ ∈ Y ∗

and then use the remark after Theorem 2. It is clear by passing to φ ◦ f and φ ◦ P that it suffices to prove that P iscontinuous if f is continuous in case that Y = R. So, suppose to the contrary that there are ε > 0 and {hn} ⊂ X suchthat hn → h ∈ X and |P (hn) − P (h)| ≥ ε. Put qn(t) = P (thn) − P (th), n ∈ N. Then qn(t) = P 0,hn(t) − P 0,h(t) andso qn ∈ Pk(R;R); recall that a = 0. Let K =

⋃|t|≤r t

({h} ∪ {hn; n ∈ N}) for some r > 0 such that K ⊂ U . Then K

is compact and hence f is uniformly continuous on K. Let ω be a modulus of continuity of f on K. We have

|qn(t)| ≤ |P (thn)− f(thn)|+ |f(thn)− f(th)|+ |f(th)− P (th)| ≤ |R(0, thn)|+ ω(‖thn − th‖) + |R(0, th)|


for |t| ≤ r. Recall the following elementary fact about polynomials: For a given k ∈ N there is M > 0 such that ifQ ∈ Pk(R;R), then |Q(x)| ≤ M‖Q‖|x|k for every x ∈ R, |x| ≥ 1 (it follows easily e.g. from [HJ, Fact 1.42]). By (5)there are 0 < δ ≤ min{r, 1} and n0 ∈ N such that |R(0, thn)| ≤ ε

4M |t|k and |R(0, th)| ≤ ε4M |t|k for all |t| ≤ δ and

n ≥ n0. Hence, applying the above fact to Q(t) = qn(δt), we obtain

ε ≤ |qn(1)| ≤ M

δksup|t|≤δ

|qn(t)| ≤ M

δk(sup|t|≤δ

|R(0, thn)|+ ω(δ‖hn − h‖) + sup|t|≤δ

|R(0, th)|) ≤ ε

2+

M

δkω(δ‖hn − h‖)

for all n ≥ n0. This is a contradiction.

We will need the following easy fact, which is well-known for X = Y = R (see e.g. [DS, p. 215, (5.2)]).

Proposition 16. For each k ∈ N, ω ∈ M, C > 0, R ≥ r > 0 such that ω(R) < +∞, and j ∈ {1, . . . , k} there isMj(k, ω, r, R,C) > 0 such that if X, Y are normed linear spaces, U ⊂ X is an open convex set such that U(a, r) ⊂ U ⊂U(a,R) for some a ∈ U , and f ∈ Ck,ω(U ;Y ) satisfies ‖f(x)‖ ≤ C for x ∈ U(a, r), then ‖djf(x)‖ ≤ Mj(k, ω, r, R,C)for x ∈ U .

Proof. First we claim that for j = k we can take Mk(k, ω, r, R,C) = (4k)kCrk

+ ω(r) + ω(R). For a fixed h ∈ SX andφ ∈ BY ∗ we put g(t) = φ ◦ f(a+ th) for t ∈ (−r, r). Using Facts 3 and 4 it is easy to see that g ∈ Ck,ω

((−r, r)

). Put

s = r2k . It is a well-known fact (see e.g. [Z, Exercise 5.3.14] or [DS, p. 195, (3.34)]) that there is ξ ∈ (0, r) such that

g(k)(ξ) =Δk

sg(0)sk

. It follows from (1) that |g(k)(ξ)| ≤ 2kCsk

and so |g(k)(0)| ≤ |g(k)(ξ)|+ |g(k)(0)− g(k)(ξ)| ≤ 2kCsk

+ ω(r).

Using Facts 3 and 4 again we get∣∣φ(dkf(a)[h])∣∣ = |g(k)(0)| ≤ (4k)kC

rk+ ω(r). By taking the supremum over all h ∈ SX

and φ ∈ BY ∗ we finally obtain ‖dkf(a)‖ ≤ (4k)kCrk

+ω(r). Now for any x ∈ U we have ‖dkf(x)‖ ≤ ‖dkf(a)‖+ ‖dkf(x)−dkf(a)‖ ≤ Mk(k, ω, r, R,C).

Next, fix j ∈ N, C > 0, and R ≥ r > 0. It is sufficient to show the existence of Mj(k, ω, r, R,C) for k ≥j and ω ∈ M with ω(R) < +∞. We have already proved it for k = j. Now assume that k > j and Mj(k −1, ω, r, R,C) exists for all ω ∈ M with ω(R) < +∞. To prove the existence of Mj(k, ω, r, R,C) let ω ∈ M andf ∈ Ck,ω(U ;Y ) satisfying the assumptions be given. We already know that ‖dkf(x)‖ ≤ Mk(k, ω, r, R,C) for x ∈ U .

Hence by Fact 1, ‖Dkf(x)‖ ≤ kk

k! Mk(k, ω, r, R,C) for x ∈ U . It follows that Dk−1f (and consequently also dk−1f) is

Lipschitz with constant kk

k! Mk(k, ω, r, R,C) and so f ∈ Ck−1,ω(U ;Y ), where ω(t) = kk

k! Mk(k, ω, r, R,C) · t. Thus wemay set Mj(k, ω, r, R,C) = Mj(k − 1, ω, r, R,C).

��Let X be a normed linear space. Recall that a set A ⊂ BX∗ is called λ-norming, λ ≥ 1, if supφ∈A φ(x) ≥ 1

λ‖x‖ foreach x ∈ X.

Theorem 17. Let X, Y be normed linear spaces, U ⊂ X open, f : U → Y a locally bounded mapping, k ∈ N, ω ∈ M,and A ⊂ BY ∗ a λ-norming set. If φ ◦ f ∈ Ck,ω(U) for each φ ∈ A, then f ∈ Ck,λω(U ;Y ).

Proof. By passing to the completion of Y we may assume without loss of generality that Y is a Banach space.We prove the theorem by induction on k. More precisely, let X be a normed linear space and U ⊂ X an open set;

for each k ∈ N we prove the following statement: If Y is a Banach space, A ⊂ BY ∗ a λ-norming set, ω ∈ M, andf : U → Y a locally bounded mapping satisfying φ ◦ f ∈ Ck,ω(U) for each φ ∈ A, then f ∈ Ck,λω(U ;Y ).

First assume that k = 1. Fix x ∈ U . To prove that the derivative Df(x) exists we first show that for a given h ∈ Xthe limit L(h) = limt→0

1t

(f(x + th) − f(x)

)exists using the Cauchy criterion. By the Mean value theorem, for any

φ ∈ A, δ > 0 sufficiently small, and s, t ∈ (−δ, δ) \ {0} there are ξ, θ ∈ (−δ, δ) such that∣∣∣∣1t (φ ◦ f(x+ th)− φ ◦ f(x))− 1

s

(φ ◦ f(x+ sh)− φ ◦ f(x))∣∣∣∣ = ∣∣D(φ ◦ f)(x+ ζh)[h]−D(φ ◦ f)(x+ θh)[h]

∣∣≤ ‖h‖ω(2‖h‖δ).

Hence for s, t ∈ (−δ, δ) \ {0}∥∥∥∥1t (f(x+ th)− f(x))− 1

s

(f(x+ sh)− f(x)

)∥∥∥∥ ≤ λ supφ∈A

∣∣∣∣φ(1

t

(f(x+ th)− f(x)

)− 1

s

(f(x+ sh)− f(x)

))∣∣∣∣= λ sup

φ∈A

∣∣∣∣1t (φ ◦ f(x+ th)− φ ◦ f(x))− 1

s

(φ ◦ f(x+ sh)− φ ◦ f(x))∣∣∣∣ ≤ λ‖h‖ω(2‖h‖δ).

(6)

This clearly implies the existence of L(h). From the definition of L(h) clearly φ(L(h)) = D(φ ◦ f)(x)[h] for any φ ∈ A.Since A separates the points of Y , it easily follows that L is linear. Finally, for any h ∈ X sufficiently small we can sett = 1 and pass to the limit as s → 0 in (6) to obtain ‖f(x+ h)− f(x)− L(h)‖ ≤ λω(2‖h‖)‖h‖, which implies that Lis bounded and Df(x) = L. Also, for any x, y ∈ U using Fact 4 we can estimate∥∥Df(x)−Df(y)

∥∥ ≤ suph∈BX

λ supφ∈A

∣∣φ(Df(x)[h]−Df(y)[h])∣∣ = λ sup

φ∈Asup

h∈BX

∣∣D(φ◦f)(x)[h]−D(φ◦f)(y)[h]∣∣ ≤ λω(‖x−y‖).


To continue the induction, assume that the statement holds for k−1 and the assumptions of the statement for k aresatisfied. Fix x ∈ U and let r, C > 0 be such that ‖f(y)‖ ≤ C for y ∈ U(x, r) and ω(r) < +∞. By Proposition 16 for

each φ ∈ A the mapping dk(φ◦f) is bounded by Mk(k, ω, r, r, C) on U(x, r). Hence ‖Dk(φ◦f)(y)‖ ≤ kk

k! Mk(k, ω, r, r, C)

for y ∈ U(x, r) by Fact 1. It follows that Dk−1(φ ◦ f), and consequently also dk−1(φ ◦ f), is Lipschitz with constantkk

k! Mk(k, ω, r, r, C) on U(x, r). Thus the inductive hypothesis (used for the modulus ω(t) = kk

k! Mk(k, ω, r, r, C) ·t) yieldsthat f ∈ Ck−1

(U(x, r);Y

). Let W ⊂ (L(k−1X;Y )

)∗be the set of all functionals for which there exist φ ∈ A and

h1, . . . , hk−1 ∈ BX such that ψ(M) = φ(M(h1, . . . , hk−1)

)for M ∈ L(k−1X;Y ). Then W is clearly a λ-norming set.

We set g = Dk−1f on U(x, r). For any ψ ∈ W determined by φ ∈ A and h1, . . . , hk−1 ∈ BX , and any y ∈ U(x, r) usingFact 4 we obtain (ψ ◦ g)(y) = (

φ ◦Dk−1f(y))[h1, . . . , hk−1] =

(Dk−1(φ ◦ f)(y))[h1, . . . , hk−1] =

(ε ◦Dk−1(φ ◦ f))(y),

where ε ∈ (L(k−1X;R))∗

is the operator of evaluation ε(M) = M(h1, . . . , hk−1). Thus D(ψ ◦ g)(y)[h] = D(ε ◦

Dk−1(φ ◦ f))(y)[h] = ε(D(Dk−1(φ ◦ f))(y)[h]) = Dk(φ ◦ f)(y)[h, h1, . . . , hk−1], again by Fact 4. From Fact 1 it follows

that ψ◦g ∈ C1,mω(U(x, r)), where m = kk

k! . Thus by the first step of the induction applied to the (continuous) mappingg, the λ-norming set W , and the modulus ω = mω we obtain that Dg = Dkf exists on U(x, r).

To finish the proof, for any x, y ∈ U using Fact 4 we can estimate∥∥dkf(x)−dkf(y)∥∥ ≤ sup

h∈BX

λ supφ∈A

∣∣φ(dkf(x)[h]−dkf(y)[h])∣∣ = λ sup

φ∈Asup

h∈BX

∣∣dk(φ◦f)(x)[h]−dk(φ◦f)(y)[h]∣∣ ≤ λω(‖x−y‖).

��

Definition 18. Let N ∈ N and e > 0. We say that an open subset U of a normed linear space has the (N, e)-uniformconvex chain property (or (N, e)-UCC property) if for each x, y ∈ U there is a polygonal path [x0, . . . , xn], n ≤ N ,with x = x0 and y = xn such that ‖xj − xj−1‖ ≤ ‖x− y‖ and the segment [xj−1, xj ] lies in an open convex boundedVj ⊂ U with eVj

≤ e for each j = 1, . . . , n. We say that U has the uniform convex chain property (or UCC property)if it has the (N, e)-UCC property for some N ∈ N and e > 0.

Theorem 19. Let X, Y be normed linear spaces, U ⊂ X an open set with the UCC property, f : U → Y , k ∈ N, andω ∈ M. Suppose that f is Ck,ω-smooth on every open segment in U and that either

(i) f is locally bounded, or(ii) X is complete and φ ◦ f is Baire measurable for each φ ∈ Y ∗.

Then f is Ck,mω-smooth on U for some m > 0. More precisely, if U has the (N, e)-UCC property, then we can setm = ckNek, where ck > 0 is a constant depending only on k.

Proof. First assume that U is additionally convex and bounded. Clearly f is directionally T k-smooth on U . Letx, x + h ∈ U , h �= 0. Then g(t) = f

(x + t h

‖h‖)is Ck,ω-smooth on (−δ, ‖h‖ + δ) for some δ > 0. Thus Taylor’s

theorem (Theorem 5) used on g at 0 with the increment ‖h‖ implies that ‖R(x, h)‖ = ‖f(x + h) − P x,h(1)‖ =∥∥g(‖h‖) − P x,h/‖h‖(‖h‖)∥∥ ≤ 1k!ω(‖h‖)‖h‖k, where the polynomials P x,h ∈ Pk(R;Y ) come from the definition of the

directional T k-smoothness. If (ii) holds, then since (3) is clearly satisfied at each a ∈ U , the mapping f is weaklyT k-smooth on U by Theorem 9.

If (i) holds, then f is continuous. Indeed, for a fixed a ∈ U we find r > 0 such that U(a, r) ⊂ U , f is boundedby C > 0 on U(a, r), and ω(r) < +∞. Then Proposition 16 implies that for each v ∈ SX the Ck,ω-smooth mappingt �→ f(a+ tv) is Lipschitz on (−r, r) with constant M1(k, ω, r, r, C). Further, since (3) and (5) are clearly satisfied ateach a ∈ U , Remark 15 implies that f is weakly T k-smooth on U . Consequently, in both cases f is UT k-smooth on Uwith modulus 1

k!ω, so an application of Theorem 8 yields that f if Ck,mω smooth with m = ckekU , where ck depends

only on k.Now assume that U has the (N, e)-UCC property. Fix any x, y ∈ U and let [x0, . . . , xn], n ≤ N , be the polygonal

path and V1, . . . , Vn the convex sets from the definition of the UCC property. Then using the first part of the proofon each of the sets Vj we obtain

‖dkf(x)− dkf(y)‖ ≤n∑

j=1

‖dkf(xj)− dkf(xj−1)‖ ≤n∑

j=1

ckekVjω(‖xj − xj−1‖) ≤ Ncke

kω(‖x− y‖). (7)

��

Combining the one-dimensional Marchaud’s theorem together with our criterions for Ck,ω-smoothness we obtainthe following general version of Marchaud’s theorem:

Theorem 20. Let X, Y be normed linear spaces, U ⊂ X an open set with the (N, e)-UCC property, f : U → Y , andk ∈ N. Suppose that either

(i) f is locally bounded, or(ii) X is complete, f is bounded on every closed segment in U , and φ ◦ f is Baire measurable for each φ ∈ Y ∗.


Then f ∈ Ck(U ;Y ) and for each t ≥ 0

ωdkf (t) ≤ BkNek∫ t

0

ωk+1(f ; s)

sk+1ds,

provided that the integral on the right-hand side is finite for some t > 0. The constant Bk depends only on k.

Proof. For a fixed x ∈ U and h ∈ SX we define g(t) = f(x + th) for t ∈ (c, d), where (c, d) is the maximal intervalcontaining 0 such that x+ th ∈ U for all t ∈ (c, d). Further, we set gφ = φ ◦ g for each φ ∈ BY ∗ . Then it is easy to see

(using (1)) that ωk+1(gφ; s) ≤ ωk+1(g; s) ≤ ωk+1(f ; s) for all s ∈ [0,+∞). Thus ωg(k)φ

(t) ≤ η(t) = Ak

∫ t

0ωk+1(f ;s)

sk+1 ds by

Theorem 6 used on each [a, b] ⊂ (c, d). It follows that g ∈ Ck,η((c, d)

)by Theorem 17. Hence Theorem 19 implies that

ωdkf (t) ≤ ckNekη(t).��

To put the UCC property into perspective, we recall that the strongest currently known formulation of Marchaud’stheorem is for the bounded LG domains in R

n, which follows from [JS, Theorems 1 and 2]; notice that 0 ≤ t ≤ 1 in [JS,Theorem 1], so the case of unbounded domains is not covered. It is easily seen that each LG domain as defined in [JS,p. 123] satisfies the uniform cone condition of [AF, p. 83]. For our purposes it is sufficient to deal only with boundeddomains, in which case the definition of the uniform cone condition simplifies to the following: An open boundedsubset U of Rn is said to satisfy the uniform cone condition if there exist δ > 0, a finite open covering {Uj}mj=1 of the

set {x ∈ U ; dist(x,Rn \U) ≤ δ}, and a corresponding sequence {Cj}mj=1 of cones all linearly isometric to a fixed conewith vertex at the origin, such that x+ Cj ⊂ U for each x ∈ Uj ∩ U .

Recall that a cone in Rn with vertex at the origin, axis direction v ∈ R

n, ‖v‖ = 1, aperture angle ϕ ∈ (0, π], andheight ρ > 0 is the set C = C(v, ϕ, ρ) = {x ∈ R

n; 〈x, v〉 ≥ ‖x‖ cos ϕ2 , ‖x‖ ≤ ρ}.

The following proposition shows that in Rn our version of Marchaud’s theorem is more general than that of [JS].

Proposition 21. Let U ⊂ Rn be a connected open bounded set that satisfies the uniform cone condition. Then U has

the UCC property. On the other hand, there is a bounded open set in R2 with the UCC property that does not satisfy

the uniform cone condition (and even neither the segment condition nor the weak cone condition, see [AF, pp. 82–84]).

Before proving the proposition we make a couple of simple observations:(a) If U is an open connected subset of a normed linear space X and K ⊂ U is compact, then there is an open

connected set V ⊂ X such that K ⊂ V ⊂ U and dist(V,X \ U) > 0. Indeed, K can be covered by finitely manyballs U(xj , rj), j = 1, . . . , n, such that U(xj , 2rj) ⊂ U . Since U is pathwise connected, there are curves γj ⊂ Uconnecting xj and x1, j = 1, . . . , n. By the compactness of γj there are open connected neighbourhoods Vj of γj suchthat dist(Vj , X \ U) > 0. Thus it suffices to put V =

⋃nj=1

(U(xj , rj) ∪ Vj

).

(b) Let C ⊂ Rn be the cone C = C(v, 2ϕ, ρ). If x, y ∈ R

n satisfy 0 < ‖x−y‖ < ρ sinϕ2+sinϕ , then z = x+ 2

sinϕ‖x−y‖v ∈(x+ IntC)∩ (y+ IntC). Indeed, clearly z ∈ x+ IntC. Further, ‖z− y‖ ≤ ‖z− x‖+ ‖x− y‖ = ‖x− y‖( 2

sinϕ +1)< ρ.

Finally, to estimate the angle γ between v and z − y consider the triangle xyz. Then sin γ = h‖z−x‖ , where h is the

altitude of the triangle xyz from the vertex x, and so sin γ ≤ ‖x−y‖‖z−x‖ = sinϕ

2 < sinϕ.

Proof of Proposition 21. Let U ⊂ Rn be a connected open bounded set satisfying the uniform cone condition. Then

there is a cone C ⊂ Rn with vertex at the origin and δ > 0 such that the set K = {x ∈ U ; dist(x,Rn \ U) ≤ δ} is

covered by finitely many open sets Uj , j = 1, . . . ,m, such that x + Cj ⊂ U for every x ∈ U ∩ Uj , where Cj is a conelinearly isometric to C. Let σ be a Lebesgue number of the covering {Uj ∩K}mj=1 of K (see [E, Theorem 4.3.31]). Thenx+ Cj ⊂ U and y + Cj ⊂ U for some j ∈ {1, . . . ,m} whenever x, y ∈ K ∩ U are such that ‖x− y‖ < σ.

Let C = C(u, 2ϕ, ρ) and let 0 < η ≤ δ be such that C contains a closed ball of radius η. Put W = {x ∈U ; dist(x,Rn \ U) > η}. By (a) above there is an open connected V ⊂ R

n such that W ⊂ V ⊂ U and ζ =

dist(V,Rn \ U) > 0, so ζ ≤ η ≤ δ. Let d = diamC and put ε = min{ζ, σ, ρ sinϕ

2+sinϕ

}. By the compactness the

set V is covered by balls U(xj ,ε2 ), xj ∈ V , j = 1, . . . ,M . We claim that U has the (N, eC)-UCC property, where

N = max{� 2

sinϕ�+ � 2sinϕ + 1�, 2�d

ε �+M + 1}.

Let x, y ∈ U . We distinguish three cases. First assume that ‖x − y‖ < ε and x ∈ V or y ∈ V . If x ∈ V , theny ∈ U(x, ζ) ⊂ U and note that for any ball B we have eB = 2 < eC . Clearly, the polygonal path consisting of thesegment [x, y] lies in U(x, ζ). If y ∈ V , then we proceed analogously.

Next, if ‖x − y‖ < ε and x, y ∈ U \ V , then x, y ∈ K and since ε ≤ σ, there is a cone Cj such that x + Cj ⊂ Uand y + Cj ⊂ U . Let the axis of Cj be given by the vector v ∈ R

n, ‖v‖ = 1. Set z = x + 2sinϕ‖x − y‖v. Then z ∈

(x+IntCj)∩(y+IntCj) by (b) above. Using the compactness of Cj we find t < 0 such that if we set Vx = x+tv+IntCj

and Vy = y + tv + IntCj , then x, z ∈ Vx ⊂ U and y, z ∈ Vy ⊂ U . Thus by partitioning the segments [x, z] and [z, y]we can create a polygonal path between x and y consisting of � 2

sinϕ�+ � 2sinϕ + 1� segments of length at most ‖x− y‖

such that each of the segments lies in Vx or Vy. Clearly eVx = eVy = eC .Finally, we deal with the case ‖x − y‖ ≥ ε. If x ∈ V , then we set wx = x. Otherwise there is a cone Cj such that

x+Cj ⊂ U . By the assumption there is wx ∈ U satisfying B(wx, η) ⊂ x+Cj . It follows that wx ∈ W ⊂ V . By shifting


the cone slightly we obtain an open cone Vx ⊂ U affinely isometric to IntC such that x,wx ∈ Vx. Thus there is a

polygonal path between x and wx that lies in Vx and consists of at most⌈‖wx−x‖

‖x−y‖⌉ ≤ �d

ε � segments of length at most

‖x− y‖. Similarly we construct wy ∈ V and a polygonal path between y and wy with analogous properties.Now it suffices to notice that there is a polygonal path between wx and wy that consists of at most M +1 segments

of length at most ε ≤ ‖x− y‖ such that each segment lies in a ball contained in U . Indeed, recall that V is covered byballs U(xj ,

ε2 ), j = 1, . . . ,M , and each U(xj , ε) ⊂ U . There are p, q ∈ N such that wx ∈ U(xp,

ε2 ) and wy ∈ U(xq,

ε2 ).

Next, define A0 = U(xp,ε2 ) and Al =

⋃{U(xj ,

ε2 ); U(xj ,

ε2 ) ∩Al−1 �= ∅}, l ∈ N. Clearly, Al = AM−1 for l ≥ M . Since

V is connected, xq ∈ AM−1, and hence there is a polygonal path [xp, xk1, . . . , xks

, xq], where s ≤ M − 2 and each ofits segments is of length less than ε.

On the other hand, consider the following set: Let U0 ⊂ R2 be the open triangle with vertices [0, 0], [1, 0], and

[1,−1], let Uj = (21−2j , 22−2j)× (−2−2j , 2−2j), j ∈ N, and U =⋃∞

j=0 Uj .

Then it is easy to see that for each aj = [22−2j , 2−2j ] there is no segment longer than 5·2−2j that lies in U and has aj asan endpoint. Thus the set U does not satisfy the uniform cone condition (consider the open set from the open coveringin the definition that contains the point [0, 0] on the boundary of U). Similarly it can be seen that U neither satisfiesthe segment condition nor the weak cone condition of [AF]. On the other hand, U has the

(3, 2√

2−1

)-UCC property.

Indeed, eU0= 2√

2−1and eUj

= 2√2 < eU0

, j ∈ N. Now assume that x = [x1, x2] ∈ Uj , y = [y1, y2] ∈ Uk, 1 ≤ j < k, and

x2, y2 ≥ 0. Put u = [x1,−a] and v = [y1,−a], where a = min{2−2k−1, 2−2j−x2}. Then ‖x−y‖ ≥ x1−y1 > 2−2j and so‖u−x‖ = x2+a ≤ 2−2j < ‖x−y‖, ‖v−u‖ = x1−y1 ≤ ‖x−y‖, and ‖y−v‖ = y2+a < 2−2k+2−2k−1 < 2−2j < ‖x−y‖.Also, [x, u] ⊂ Uj , [u, v] ⊂ U0, and [v, y] ⊂ Uk. The other possibilities for positions of x, y can be dealt with similarly.��Remark 22. We close the paper with the last remark: if U is an open convex subset of a normed linear space Xthat contains an unbounded cone, i.e. the set u +

⋃t∈(0,+∞) tU(a, r) for some u, a ∈ X and r > 0, then U has the(

1, 2‖a‖+rr

)-UCC property. Indeed, without loss of generality we may assume that u = 0. Now for any x, y ∈ U choose

R > 0 such that x, y ∈ U(0, R) and put V = U(0, R) ∩ U . Since V clearly contains the ball U(

R‖a‖+ra,

R‖a‖+r r

), it

follows that eV ≤ 2RR

‖a‖+rr= 2‖a‖+r

r .

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Department of Mathematical Analysis, Charles University, Sokolovska 83, 186 75 Praha 8, Czech Republic

E-mail address: [email protected]

E-mail address: [email protected]

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Smoothness via directional smoothness and Marchaud's theorem in Banach spaces