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Rev. Anal. Num´er. Th´eor. Approx., vol. 33 (2004) no. 2, pp. 141–148



Dedicated to Professor Elena Popoviciu on the occasion of her 80th birthday.

Abstract. In this note we give some characterizations for the differentiability with respect to a bornology of a continuous convex function. The special case of seminorms is treated. A characterization of this type of differentiability in terms of the subgradient of the function is also obtained.

MSC 2000. 58C20, 46A17, 46G05.

Keywords. Convex function, differentiability, bornology, subgradient.


Let E and E1 be Banach spaces, U an open subset in E and xU. A functionf :UE1 is said to be Gˆateaux differentiable atx if there exists a linear continuous mapping denoted df(x) :EE1 such that for eachh inE one has

(1) df(x)(h) = lim




The functionf is said to beFr´echet differentiableatxif there exists a linear continuous mapping denoted f0(x) : EE1 such that for each ε > 0, there existsδ >0 satisfying

(2) kf(x+h)f(x)−f0(x)(h)k ≤εkhk, for each hBE(x, δ).

The two linear mappings df(x), f0(x) are the Gˆateaux and Fr´echet differ- entials and are unique (when they exist).

In the sequel, we shall be interested only by real functions (i.e. E1 =R).

When a real function f is also convex on an open convex set UE, then the limit in (1) exists and is denoted by d+f(x); this directional derivative is generally only sublinear and the Gˆateaux differentiability off atxis equivalent with the linearity of d+f(x), or with the fact that

d+f(x)(h) =−d+f(x)(−h) [=: df(x)(h)], for each hE.

This research was supported in part by CNCSIS under Contract no. 46474/97 code 14.

“Babe¸s-Bolyai” University, Faculty of Mathematics and Computer Science, 1 Kog˘alni- ceanu St., 400084 Cluj-Napoca, Romania, e-mail: [email protected]


It is obvious that any Fr´echet differentiable function is also Gˆateaux dif- ferentiable, and the two differentials coincide. The converse is not true, even for convex functions. For example, the norm of the Banach space`1 is known to be nowhere Fr´echet differentiable, but it is Gˆateaux differentiable at those points (xn)n∈N having only nonzero components.

It is well known that the functionf is Fr´echet differentiable atxif and only if it is Gˆateaux differentiable at x and the limit (1) is uniform with respect to hB[0,1] (=the closed unit ball in E) or, equivalently, with respect to any bounded subset of E. This remark allows a useful generalization of the differentiability.

Letβ be a nonempty family of bounded sets inE whose union is E,which is directed with respect to ⊆ (i.e., for each B1, B2β there exists B3β such that B1, B2B3) and is invariant under scalar multiplication. Such a family is namedbornology in Phelps’ monograph [3].

The function f is said to beβ-differentiable at the point x iff is Gˆateaux differentiable at x and the limit (1) is uniform in hB for each Bβ.

This turns out to be equivalent with the convergence in the uniform struc- ture Fβ(E,R). We shall denote by τβ the topology induced by this uniform structure.

The following interesting special cases of a bornology arise (as pointed out in [3]):

β=G= the family of all finite subsets inE (generating the Gˆateaux differentiability);

β=F = the family of all bounded subsets in E (generating the Fr´echet differentiability);

β = H = the family of all compact subsets in E (generating the Hadamard differentiability);

β=W = the family of all weak compact subsets inE (generating the strong Hadamard differentiability).

One obviously has the inclusions: GβF, GHWF; if f is β2-differentiable and β1β2, then f is also β1-differentiable and the two differentials coincide.

Theorem 1. Let f be a continuous convex function on an open convex subset U in the normed space E and β a bornology on E. Then f is β- differentiable at xU if and only if, for each Bβ, the limit

(3) lim



t(f(x+th) +f(x−th)−2f(x)) = 0, holds uniformly for hB.

Proof. Necessity. Let B be an arbitrary subset in β. Using (1) for B and

−B one obtains the equalities df(x)(h) = limt→0+1

t(f(x+th)f(x)) and df(x)(−h) = limt→0+1t(f(x−th)f(x)), which hold uniformly for hB;

by addition the desired conclusion follows.


Sufficiency. Choose Bβ, ε > 0. Using the continuity of f, one can select a subgradient x∂f(x). The hypothesis guarantees the existence of a positive number δ such that f(x+th) +f(xth)−2f(x) < tε, for each hB and t ∈(0, δ). (B is bounded, so, for sufficiently small δ >0 one has x±thB.)

We have

hx, thi ≤f(x+th)f(x), hx,−thi ≤f(x−th)f(x), and for 0< t < δ,hB one obtains:

0≤f(x+th)f(x)− hx, thi

= f(x+th) +f(xth)−2f(x)+ f(x)−f(x−th)− hx, thi

≤εt+ 0 =εt,

which implies that (1) holds uniformly for hB.

Remark1. a)For the Fr´echet differentiability it is sufficient that the limit (3) holds uniformly on the unit sphere SE; for the Gˆateaux differentiability, the pointwise limit in (3) suffices.

b) The continuity condition imposed on the convex function f cannot be omitted when E is infinite dimensional; in fact it is sufficient to consider a linear discontinuous functional f (cf. [5, p. 251]); in this case, df(x) = f is

not continuous.

Corollary2. Letfn(n∈N)be a sequence of continuous convex functions on an open convex subset in a Banach space E endowed with a bornology β.

If the seriesPn∈

N fn is pointwise convergent having a continuous sumf, and f isβ-differentiable at a pointx0, then each function fn is β-differentiable at x0.

Proof. The statement follows immediately from the preceding theorem, using the relations:

0≤ X

n∈N 1

t(fn(x+th) +fn(x−th)−2fn(x))

= 1t(f(x+th) +f(xth)−2f(x))

(valid forxU,t >0,hBβ provided that x+thU).

The (semi)norms are important special cases of convex functions. The next result represents a simple characterizations for theβ-differentiability of a norm, extending a theorem of Smulian [1].

Theorem 3. Let E be a normed space endowed with a bornology β and x a point on the unit sphere SE of E.


The norm k·k isβ-differentiable atx if and only if the following condition holds: for all sequences xn, ynSE satisfying xn(x) → 1, yn(x) → 1 one has xnyn →0 in Fβ(E,R).

Proof. Necessity. Let Bβ, ε > 0, xn, ynSE satisfy xn(x) → 1, yn(x) → 1. Choosing B0β such that B ∪(−B) ⊆ B0 and applying the preceding theorem, there existsδ >0 such that fort∈(0, δ] one has

kx+thk+kx−thk<2 +εt≤2 +εδ, for each hB0.

The hypothesis implies the existence of a positive integer n0 such that for nn0:

|1−xn(x)|+|1−yn(x)|< εδ.

We have

xn(x+th) +yn(x−th)≤ kx+thk+kx−thk ≤2 +εδ, hence

xn(th)−yn(th)≤1−xn(x) + 1−yn(x) +εδ <2εδ, fornn0. By takingt=δ, one obtainsxn(h)−yn(h)<2ε, fornn0,hB0.

ForhB, we have ±h∈B0, and the last inequality implies

|xn(h)−yn(h)|<2ε, for each nn0, hB.

Sufficiency. Suppose by contradiction that k·kis not β-differentiable atx, and hence there exists ε > 0, Bβ,hnB\{0},tn >0 such that tn → 0, kx+tnhnk+kx−tnhnk ≥2 +εtn.

Choosing xn, ynSE such that xn(x+tnhn) ≥ kx+tnhnk − ktnhnk/n and yn(x−tnhn)≥ kx−tnhnk − ktnhnk/n one obtains



≥ kx+tnhnk −ktnnhnk − ktnhnk

≥1−ktnnhnk −2ktnhnk, hence xn(x)→1.

Similarly, yn(x)→1.


xn(x+tnhn) +yn(x−tnhn)≥2 +εtn−2ktnnhnk, we have

xn(hn)−yn(hn)≥ε−2khnnk2ε, fornn0,

in contradiction withxnyn→0 inFβ(E,R)


Iff is a continuous convex function on an open convex subsetU of a Banach space E, then the subdifferential ∂f is a set-valued operator, having convex, nonempty weak compact values in E.

We shall obtain a characterization of the β-differentiability for f in terms of the subdifferential operator ∂f. Such characterizations are known for the Fr´echet and Gˆateaux differentiability, and are very useful in the analysis of the smoothness off; in the same time such results motivated an intensive research on the set-valued operators.

If (X, τ1 ), (Y, τ2 ) are topological spaces, a set-valued operatorT :X →2Y is said to be τ12 upper semicontinuous (u.s.c.) atxX, if for each subset Wτ2 containing T(x),there exists Vτ1 containing x such that T(V) =

∪{T(v) :vV} ⊆W.

The set dom(T) :={x∈X:T(x)6=∅} is thedomain of T.

We are interested in the case when the operatorT acts betweenE and 2E, where E is a Banach space. Denoting by k·k the norm in E and by k · k its dual norm in E we shall consider the strong topology τk·k (generated by the norm) on E, and the topology τβ of the β-convergence on E, where β is a bornology on E. We remind that τF = τk·k, where F is the Fr´echet bornology (of all bounded subsets), andτGis the weak topology (Gdenoting the Gˆateaux bornology).

Proposition 4. Let E be a normed space, β a bornology on E, x a point in E and T :E → 2E a set-valued operator. Then, the following statements are equivalent:

(i) T is τk·kβ upper semicontinuous at x.

(ii) For each Wτβ, T(x)⊆W, (xn)n∈NE with kxnxk →0, there existsn0 ∈N, such that for nn0, T(xn)⊆W.

(iii) For each Wτβ, T(x) ⊆ W, there exists δ > 0, such that for r ∈ (0, δ], T(B[x, r])⊆W.

If, furthermore, T(x) is a singleton {x0}, the above conditions are also equivalent with

(iv) For each (xn)n∈NE, withkxnxk →0 it follows that

(4) lim

n→∞sup| hxx0, hi |:xT(xn), h∈B = 0, (B ∈β) and, for β=F, (4) may be reformulated as

(5) lim

r→0+diamT(B[x, r]) = 0.

Proof. The proof is standard, similar to Heine’s theorem in general topology.

We shall need the following result (see [3], [2]):

Theorem 5. Let E be a normed space, f a convex continuous function defined on an open convex set DE. Then the subdifferential ∂f :D→2E is a τk·kG upper semicontinuous operator.


Note that in the general case, ∂f will not be τk·kβ upper-semicontinuous for an arbitrary bornology β as the following example shows:

Example 1. Let E = `1 be the Banach space of all summable sequences endowed with the norm kxk=Pn∈

N|x(n)|, and f :E →R, f(x) =kxk. For hE, we have

d+f(x)(h) = lim











= X


(signx(n))h(n) + X



(the permutation of the limit and sum symbols can be legitimated by using the Weiersrass theorem, or the dominated convergence theorem from mea- sure theory applied to the sum as a discrete integral). The function f is G- differentiable at x if and only if d+f(x)(h) =−d+f(x)(−h), (h ∈ E), which means:



|h(n)|= 0, for each hE,i.e., x(n)6= 0,(n∈N).

Choose now x`1, x(n) =αn>0 (n∈N);then f is G-differentiable at x.

Defining xp = (α1, α2, ..., αp,0,0...), i.e. xp(n) =αn fornp and xp(n) = 0 for n > p, we obviously have kxpxk →0.

But d+f(xp)(h) = h(1) +...+h(p) +|h(p+ 1)|+|h(p+ 2)|+..., and by taking xp(h) =h(1) +...+h(p), one obtains:

xpE, xp ≤d+f(xp), hence xp∂f(xp).

On the other hand,kdf(x)−xpk = 1, so ∂f is not τk·kF u.s.c. at x (cf.

(iii), with W =B(df(x),1)).

In this example,f is notF-differentiable atx. This fact will follow from the next theorem which contains also a refinement of the preceding proposition

Theorem6. LetE be a normed space,βa bornology,f a continuous convex function on an open convex set DE, which is β-differentiable at xD.

Then the subdifferential ∂f : D → 2E is an τk·kβ upper semicontinuous operator.

Proof. Suppose by contradiction that∂f is notτk·kβ u.s.c. atx. Applying (iv) from Proposition 4 one obtains that there existxnE, withkxn−xk →0, ε >0,Bβ,hnB,xn∂f(xn), such that

| hxnx0, hni |>2ε, (n∈N), wherex0= df(x).

ChoseB0β such that B∪(−B)⊆B0.


Interchanging if necessary hn with−hn(∈B0), we will have (6) hxnx0, hni>2ε.

From theβ-differentiability off atx, there existsδ >0, such thatB[x, δm]⊆ D, wherem >0 is chosen such thatB0B[0, m], and

f(x+th)f(x)− hx, thi ≤tε, (t∈(0, δ], h∈B0).


(7) f(x+thn)−f(x)− hx, thni ≤tε, (n∈N, t∈(0, δ]).

Using the fact that xn∂f(xn), one obtains hxn, x+δhnxni ≤ f(x+ δhn)−f(xn),hence

(8) hxn, δhni ≤f(x+δhn)−f(x) +hxn, xnxni+f(x)f(xn).

From (6), (7) and (8) we have 2εδ <hxnx0, δhni

=hxn, δhni − hx0, δhni

≤f(x+δhn)−f(x) +hxn, xnxni+f(x)−f(xn)− hx0, δhni

=(f(x+δhn)−f(x)− hx0, δhni) +hxn, xnxni+f(x)−f(xn)


The convex function f being continuous, it is locally Lipschitz, hence the sequencekxnkis bounded (by the Lipschitz constant). Forn→ ∞one obtains

2εδ≤εδ, a contradiction.

Theorem7. LetE be a normed space,βa bornology,f a continuous convex function on an open convex set DE. Then the following statements are equivalent:

(i) f is β-differentiable at xD.

(ii) Each selectionϕ:DE for the subdifferential∂f is τk·kβ contin- uous at xD.

(iii) There exists a selectionϕ:DE for the subdifferential∂fwhich is τk·kβ continuous at xD.

Proof. (i)⇒(ii). According to the previous proposition, ∂f is τk·kβ u.s.c., hence each of its selections will be τk·kβ continuous.

(ii)⇒(iii). This implication is obvious.

(iii)⇒(i). ForyD we have hϕ(x), y−xi ≤f(y)−f(x), because ϕ(x)

∂f(x). Usingϕ(y)∂f(x) one obtains hϕ(y), x−yi ≤f(x)−f(y),hence:

(9) 0≤f(y)f(x)− hϕ(x), y−xi ≤ hϕ(y)ϕ(x), yxi.

ForhE, t >0, replacing in (9)y=x+th, dividing bytand lettingt→0+, one obtains 0≤d+f(x)(h)−ϕ(x)(h)≤0,hence

d+f(x) =ϕ(x)E,


and f is G-differentiable at x.

ForBβ,hB,t >0,y=x+th, we have

0≤ 1t(f(x+th)f(x))−df(x)≤ hϕ(x+th)ϕ(x), hi.

From the τk·kβ continuity of ϕ, the right hand side tends to 0 uniformly forhB (=bounded) as t→0+,and the conclusion follows.


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Received by the editors: April 14, 2004.




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