ON TWO-SIDED UNIDIRECTIONAL MEAN VALUE INEQUALITY IN A FRÉCHET SMOOTH SPACE Текст научной статьи по специальности «Математика»

URAL MATHEMATICAL JOURNAL, Vol. 9, No. 2, 2023, pp. 132-140

DOI: 10.15826/umj.2023.2.011

ON TWO-SIDED UNIDIRECTIONAL MEAN VALUE INEQUALITY IN A FRECHET SMOOTH SPACE1

Dmitry V. Khlopin

Krasovskii Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, 620108, Russian Federation

khlopin@imm.uran.ru

Abstract: The paper is devoted to a new unidirectional mean value inequality for the Frechet subdifferential of a continuous function. This mean value inequality finds an intermediate point and localizes its value both from above and from below; for this reason, the inequality is called two-sided. The inequality is considered for a continuous function defined on a Frechet smooth space. This class of Banach spaces includes the case of a reflexive space and the case of a separable Asplund space. As some application of these inequalities, we give an upper estimate for the Frechet subdifferential of the upper limit of continuous functions defined on a reflexive space.

Keywords: Smooth Banach space, Frechet subdifferential, Unidirectional mean value inequality, Upper limit of continuous functions.

1. Introduction

Consider the following mean value inequalities.

Proposition 1 [12, Theorems 2.1 and 2.2]. Let a scalar function f be defined and lower semi-continuous on a Frechet smooth Banach space X. Let points u and v in X be given. Then, for arbitrary numbers V < f (v) — f (u) and V > 0, there exist a point z- € [u; v] + VB and a Frechet subgradient Z- € f (z-) such that

V < Z-(V — u) and f (z-) <f (u) + max(0, s) + V. (1.1)

Furthermore, if f is continuous, there are a point z+ € [u; v] + VB and a Frechet subgradient Z+ € <9f (z+) such that

V < Z+ (v — u) and f (z+) > f (V) — max(0, s) — V. (1.2)

Note that inequalities (1.1) and (1.2) are similar. This suggests that, in the case of continuity of f, it is possible to get a common point z+ = z- such that the value f (z) is localized from both above and below. Proving the corresponding two-sided unidirectional mean inequality is the primary goal of this paper.

As part of the historical background, note that the existence of a pair (z-, Z-) satisfying inequalities like (1.1) has been widely studied (see, for example, [13, Subsect. 3.4.8] and [14, Sect. 4.4]). Unlike different variants of Lagrange's mean value theorem for certain classes of Lipschitz continuous functions, they ensure an upper bound of f (v) — f (u) through some subgradient Z. These inequalities apply to any lower semicontinuous function. Furthermore, the corresponding to the

1This study was funded by the RFBR and DFG (project no. 21-51-12007).

Frechet subdifferentials unidirectional mean value inequality is equivalent to the Asplund property of a Banach space [14, 17], and therefore is equivalent to several basic principles of variational analysis [1, 18], for example, to the inspired by [16] and [5] multidirectional mean value inequality [4, Subsection 3.6.1]; for more recent references, see [10] and [8]. However, the multidirectional mean value theorem as well as the previous unidirectional mean value inequality also localizes f (z) on one side only.

The rest of the paper is organized as follows. In Section 3, we will prove the desired two-sided unidirectional mean value inequality for continuous functions. Then, applying this result, in Section 4, we will show an upper estimate for the Frechet subdifferential of the upper limit of continuous functions. But first, we will recall several elementary definitions and notions.

2. Definitions and notation

We will use elementary notions from the set-valued and variational analyses [4, 13, 15]. For a nonempty set X of some real Banach space X, denote by cl X and co X the closure and the convex hull of X. For a point x £ X, the contingent (Bouligand tangent) cone to X at x is the set T(x; X) of all v £ X such that one finds a decreasing to 0 sequence of positive tn and a converging to v sequence of vn £ X such that x + tnvn £ X for all positive integers n. For a point x £ X, we say that Z £ X* is a Frechet normal to X at x if one has x £ X and

Z(Zn - x) hm sup -r-|j- = 0

n—^^O ||zn x||

for all converging to x sequences of zn £ X. Denote by NV(x; X) the set of all Frechet normals to

X at x.

We call a Banach space X Frechet smooth if this space has an equivalent norm that is C1-smooth off the origin. Note that any reflexive Banach space and any separable Asplund space are Frechet smooth [4, Theorem 6.1.6]. It is worth mentioning that each Frechet smooth space has a C 1-smooth Lipschitz function with bounded nonempty support [3, Ex. 4.3.9]. Denote by B and B* the unit closed balls in X and X*, respectively.

Given an extended-real-valued function g : X — R U {-to, to}, define its lower semicontinuous envelope lsc g by the rule:

lsc g(x) △ lim inf inf g(z) for all x £ X.

z€x+xB

Note that this function is lower semicontinuous. In addition, a function g is lower semicontinuous iff its epigraph

epig △ {(x, a) £ X x R | a > g(x)}

is closed. In the case of lower semicontinuous function g, define the Frechet subdifferential of g at

x as

dg(x) △ {Z £ X* | (Z, -1) £ NV(x, g(x); epig)} for a point x £ X with finite g(x); let also dg(x) = 0 if |g(x)| = to.

3. Two-sided mean value inequality

Theorem 1. Let X be a Frechet smooth space. Let a continuous function f : X — R and some closed interval [u; v] in X be given. Then, for a real number s < f (v) — f (u) and positive e, there exist some point V £ [u; v] + eB and Frechet subgradient Z £ Vf (V) such that

s < V(v - u) and |f(V) - f (u)|<|s| + e. (3.1)

Proof. Without loss of generality, we can assume that u = 0 and f (u) = 0. Now, the initial inequality can be written as s < f (v).

Case s < 0. Let s be negative. Choose a positive number e < min(|s|, f (v) -s). Define S = e/4 and s = s + 3e. Since s < 0 = f (0) < |s| and s < f (v), one finds a positive number i < 1 such that

|s| > f(Vv) > -V|s| > —|S| = s (3.2)

because f is continuous on [0; v]. For the same reason, there is a positive k < e such that

|f(z) - f(0)| < e for all z £ [0; v] n kB. (3.3)

We claim that there exist some V £ [0; iv] + kB and V £ f (Z) such that

s < -|s|<Vv and |f (Z)| < |s|+2S < |s|. (3.4)

To this end, consider the continuous map

[0; i] 9 t — h(r) = f (tv) - Tf (iv)/i.

Since h(i) = h(0) = 0 holds, due to the intermediate value theorem, there exists positive V < t that satisfies the equality h(V) = 0 and at least one of the following conditions:

(i) T < k; (ii) h|[0,f] is nonpositive; (iii) h|[0,f] is nonnegative.

Now, the relations 0 < V < t < 1, h(V) = 0, and (3.2) yield the inequality

(3.2) „ „

-|s| < —|s|V < Vf (Vv)/V = f (Vv) - f (0). (3.5)

Let us apply Proposition 1 to this inequality with

u_ = 0, v+= Vv, s = —|s|T, and Kn = K/n

for all positive integers n. This gives some r_, r+ £ [0; V], z_, z+ £ X, Z_ £ df (z_), and Z+ £ df (z+) such that

-|s|V < VZ_v, IIr— v - z_ II < Kn, f (z_) - f (0) < Kn, -|s|V < VZ+v, ||r+v - z+ II < Kn, f (z+) - f (Vv) > -Kn.

Next, taking into account the inequalities V > 0 and f (0) + Kn = Kn < e, we have

-|s| < Z±v, z± = r±v + KnB C [0; v] + kB,

(3.5)

f (z_) <f (0) + Kn < e, and f(z+) > -Kn + f (Vv) > -Kn - |s| > -e - |s|. (3.6)

Now, in the case of V < k (condition (i)) and in the case of nonpositive h|[0;f] (condition (ii)), let us set Vn = z+, Vn = r+, and Zn = Z+ for all positive integers n; and in the case of nonnegative

h|[0;f] (condition (iii)), set zn = z_, Vn = r_, and Zn = Z_ for all positive integers n. Then, in all these cases and for all positive integers n, we have proved the first inequality in (3.4). So, it is required to check only

|f(Vn)| < |s|+2S

for at least one positive integer n.

Note that all Znv lie in the compact set [0; Zv]. Passing to a subsequence, we can assume that this sequence converges. By ||Zn — Znv|| — 0, the both sequences of Zn and Znv has the common limit. The sequences of f (Zn) and f (Znv) are the same by the continuity of f ; in particular, one finds a positive integer N such that

|f (Zn) — f (Znv)| < ê. (3.7)

So, it is required to check only the inequality

|f(Znv)| < |s| + ê.

Now, in the case of nonnegative h|[0;f] (condition (iii)), by the choice of ZN = r_ and ZN = z_, we obtain

0 < h(r_) = f (r_v) — r_f(tv)/Z < f (r_v) + |f (tv)|

and

(3.6) (3.7) A (3.2)

ê > f (Zn ) = f (z_) > f (r_v) — ê >—|f (Zv)|— ê > —|s|— e.

In the case h|[0;f] < 0 (condition (ii)), one has

0 > h(r+) > f(r+v) — |f (iv)|

and

(3.6) (3.7) (3.2)

—ê — |S| < f(ZN)= f (z+) < f(r+v)+ ê <|f (tv)| + ê < |S|+ ê.

Finally, in the case Z < k (condition (i)), (3.3) and (3.7) yield

|f(r_v)| < 2ê < |s| + ê. Inequalities (3.4) have been proved.

Case s > 0. Assume that s is nonnegative. Recall that s < f (v). Choose a positive ê such that s + ê < f (v). Define

This entails

s = s + ê/2.

0 <s <s + e< f (v), and one can choose positive e such that

e + (1 + e)3s < s + e.

Consider the map f : X x R ^ R defined as

/(x, r) △ f (x) — (1 + e)rs for all x € X, r € R. Then, we have /(0,0) = 0,

<9/(x, r) = <9f(x) x {—(1 + e)s}, f(v, 1) = /(v, 1) — f(0,0) = f(v) — (1 + e)s > s — (1 + e)s = — es .

Since —es < 0, we can apply the first case of our theorem to the inequality

—es<7(v, 1) — f (0,0).

Then, there exist some number r £ [0; 1], point

s = (z, r) £ (rv, r) + eB,

and subgradient

C = (Z, -(1+ e)s) £ f(z,r)

that satisfy (3.4); i.e.,

-es < Zv - (1 + e)s and |f (V, V)| < e|s| + S.

Now, the first inequality leads to s < s < Zv by s < s; on the other hand, the second inequality entails

|f(z)| = |/(z,r) + (1 + ^Vsl < es + e + (1 + e)|V|s < es + e + (1 + e)2s < (1 + e)3s + e < s + e

by |V| < |r| + e < 1 + e and the choice of e.

The theorem is proved. □

Remark 1. As [12, Example 2.1] has shown, (1.2) can be violated if f : R — R is only lower semicontinuous. Therefore, the assumption of the continuity of f is essential in this theorem as well.

Remark 2. In the case of Lipschitz continuous function f, for its G-subdifferential, there exists a variant of unidirectional mean value inequality that guaranties the inclusion z £ [u; v] instead of z £ [u; v] + eB (see [9, Theorem 4.70]). However, this is not true for a Frechet subdifferential. Indeed, for the Lipschitz continuous function

R2 9 (x,y) — f(x,y) = -|x|,

its Frechet subdifferential is empty on the interval [(0,0); (0,1)]; in particular, no Frechet subgradient Z satisfies (3.1).

Remark 3. It may mistakenly seem that Theorem 1 does not essentially use the asymmetry of a Frechet subdifferential and can be directly extended to the symmetric case. Indeed, Lebourg's mean value theorem [6, Theorem 2.4] for Clarke subdifferentials, the mean value theorem [2] for MP-subdifferentials, and the symmetric subdifferential mean value theorem [13, Theorem 3.47], [14, Theorem 4.11] give the corresponding gradient Z of f at some V £ [u; v] that satisfies the symmetric bound

|f(v) - f (u)| = lav - u)|. (3.8)

This bound is exactly the limit of bounds

s+ < Z(v - u) + e and - s_ < (-Z)(v - u) + e

as s+ t f(v) - f(u), -s_ t (-f)(u) - (-f)(v), and e 0. Similarly, passing to the limit in |f (z) - f(u)| < |s| + e, we could hope for the eatimate

|f(V) - f (u)|<|av - u)| (3.9)

together with (3.8). However, in the case

f (x) = x(x - 2) and [u, v] = [0,2],

inequalities (3.8) and (3.9) should give f'(V) = Z = 0 and |f(V)| < 0; i.e., 1 = V £ {0,2}. This contradiction negates the hope of adding two-side estimate (3.9) to (3.8).

4. Subdifferentials of the upper limit of continuous functions

Let a family of continuous functions /0 : X ^ R U {-to, to}, 0 € [0; to) be given. Define a function /sup : X ^ R U {-to, to} by the following rule:

/sup(x) = limsup /0 (x) for all x € X. (4.1)

For every positive 5, denote by Z(X) the set of all Z € X* for which there exists a pair (0,x) € [0; to) x X such that Z € d/0(x),

0> 1/5, x € x + 5B, and |/0(x) - (lsc/sup)(x)| <5. (4.2)

The following estimate of the subdifferential of the upper limit function is the enlargement of [11, Lemma 6] on reflexive spaces as well as the refinement of [12, Theorem 6.1(a)] in the case of continuous functions; its proof is similar to that of [12, Theorem 6.1(a)].

Proposition 2. Assume that X is a reflexive space, a family of scalar functions /0, 0 € [0; +to), continuous on X is given, and /sup is defined by (4.1). For all x € X and £ € d lsc /sup(x), for every positive 5, there exist some N € N, a^, a2,..., aN € [0;1], and Zi,Z2,...,Zn € Z(x,lsc/sup(x)) such that a1 + ■ ■ ■ + aN = 1 and

N

£ afcZfc + 5B*. (4.3)

k=1

Proof. The special case: x = 0 is a local minimum of lsc /sup. Assume that x = 0 and £ = 0; furthermore, assume that

(lsclimsup/0)(0) = inf limsup/0(x) = 0

for some positive 50. Then 0 € d/sup(0) = <9 lsc /sup(0).

Note that /sup(z) = infT>0 E(T, z) for all z € X; here E : [0; +to) x X ^ R U {-to, +to} is defined as

E(T, x) = sup /0(x) for all T > 0, x € X.

0>T

Fix a vector v € B and a positive number 5 < min(50,1/3). Define t = 52. Since 0 is a local minimum of lsc /, there exists a point z € tB such that

0 < lsc /sup(z) < /sup(Z) < 54.

Then,

||z + tv|| < 252 < 5 < 5o and /sup(z + tv) > lsc /sup(Z + tv) > 0.

So,

/sup(z + tv) - /sup(z) > -54 = -52t.

Further, we can find positive numbers T > 1/5 and 0 > T such that

52t>E(T,z) - /sup(z) and 52t + /0(z + tv) >E(T,z + tv). (4.4) By definition of E, we also have

0 < E(T, z + tv) - /sup(z + tv) and /(z) < E(T, z). (4.5)

Subtracting the sum of inequalities (4.5) from the sum of inequalities (4.4), we have

252t + f (z + tv) - f (z) > fsup(z + tv) - fsUp(z).

From fsup(z + tv) - fsup(z) > -52t and 5 < 1/3, it follows that

f(z + tv) - /(z) > -5t.

Now, Theorem 1 for f = f with u = z, v = z + tv, s = -5t, and e = 5(5 - t) gives a number r £ [0, t], a point V £ X, and a subgradient Z £ ¿/(T) such that

-5t<tZv, ||T -z|| < ||t -rv|| + r < t + 5(5 -1) < 252, and |f(T) - f(,z)| < 5t + 5(5 -1) = 52. Then, by the choice of z, we obtain

||T|| < p|| + 252 < 352 <5 and |f(T)| < |f(z) - f(u)| + 52 < 252 <5.

So, we show (4.2) for (x,y) = (0,0), (0,x) = (61, T), therefore we obtain Z £ Zs(0,0). Hence, for each v £ B, we have found Z £ Zs(0,0) such that Zv > -5. This entails

-5 < inf sup Zv < inf sup Zv.

veBCez5 (0,0) veB Cecico Z5 (0,0)

The set B is an weak compact subset of X** = X and, together with clcoZs(0,0), is convex. In addition, the map (Z,v) — Zv is continuous and linear in (Z,v) £ X* x X**. Therefore, the nonsymmetrical Minimax Theorem [4, Theorem 3.6.14] ensures

-5 < inf sup Zv = sup inf Zv.

veB CecicoZ5(0,0) CecicoZs(0,0) veB

Since there exists Z £ clco Zs such that 5 > -Zv for all v £ B, we obtain ||Z|| < 5. Therefore, (4.3) holds in the special case. The special case of this lemma is proved.

The general case. Let a point x £ X and a subgradient £ £ o1 lsc fsup(x) be given. Define y = lscfsup(x). Choose a positive number 5 < 1/3.

Since X is a Frechet smooth space, by [7, Theorem 4.6 (i)], there exist a C^smooth function g and a positive number 5i < 52 such that

£ = g'(x), lscfsup(x) = g(x), and lsc fsup(x + tv) - lsc fsup(x) > g(x + tv) - g(x)

if x £ x + 51B. Further, decreasing 51 if necessary, we can also ensure £ £ g'(x) + 52B* and g(x) £ g(x) + 52B for all x £ x + 51B. So,

lsc(fsup - g)(0) < (fsup - g)(x) for all x £ 5iB.

Using the special case for the maps

X 9 x — /(x) = /(x - x) - g(x - x),

and a positive number 52, we find Z £ clcoZ^(x,y) n 52B*. Then, in the account of the fuzzy sum rule, one finds a positive integer N, points x1,...,xN £ X, subgradients Zi £ f!(x1) -g'(x1),... ,Zn £ T/%(xN) - g'(xN), and convex coefficients a £ [0,1] such that

a1 + ... + aN = 1,

xt £ x + 252B, ^ > 5_2, f (xi) - g(xt) - lsc fsup(x) + g(x)| < 252

for all i and

N

^«kZk e *. k=i

Define

Ci = Ci + g'(xi) e <9/01 (xi).

By the choice of a positive number ¿1 and a smooth function g, we obtain

2

/ (X ) - lsc /sup(x)| < |g(Xi ) - g(x)| + 252 < 5,

N N N

Il V akZk -V afcZk < max ||g'(xj) - £|| < 252 <5, and £ eV akZk + 5B*. k=1 k=1 1 J k=1

So, the proposition is proved. □

Remark 4• If X = Rd, by the famous Caratheodory theorem [15, Theorem 2.29], any finite convex sum of a (co)vectors can be represented by some finite convex sum of no more than d + 1 of them. So, we can assume that N < d +1.

Remark 5. If every / is C 1-smooth, we conclude that every <9/7(x) is a singleton; therefore, Zi = /4 (xi) for all i.

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