Journal of Convex Analysis Volume 1 (1994), No.1, 47–60

Differentiability of Convex Functions on a Banach Space with Smooth Bump Function^{1}

Li Yongxin, Shi Shuzhong

Nankai Institute of Mathematics Tianjin, 300071, People’s Repubilic of China

Received 15 November 1993

In this paper, we prove that if a Banach space X admits a Lipschitz β-smooth bump function, then any continuous convex function on a convex open subset Ω of X is β-differentiable in a dense G_{δ} subset of Ω.

48 Y. Li, S. Shi / Differentiability of Convex Functions on a Banach Space

In 1990, Preiss, Phelps and Namioka [15] proved that if a real Banach space X admits a β-smooth norm, then every continuous convex function f on a convex open subset Ω of X is generically β-differentiable; i.e., there exists a dense G_{δ} subset G of Ω such that f is β-differentiable at each point of G. This result with Haydon’s counter example ([9], [10], see also [7], Chap. VII,) finally resolves the “generalized” famous Asplund conjecture (see Asplund [1] and Day [4], but only for Fréchet- and Gâteaux-differentiability), which asks whether the existence of an equivalent β-smooth norm is equivalent to the generic β-differentiablility of any continuous convex function. The first partial answer to the Asplund conjecture was obtained in 1976 by Ekeland– Lebourg [8] who showed that if a Banach space X admits a Fréchet-smooth bump function, then X is an Asplund space, i.e., every continuous convex function on a convex open subset of X is generically Fréchet-differentiable. Since Haydon has constructed a Banach space, which admits a Lipschitz Fréchet-smooth (even C^{1}) bump function, but does not admit any equivalent Gâteaux-smooth norm, the necessary part of Asplund conjecture is false. However, Haydon’s counter example does not negate the following “modified” Asplund conjecture: it is whether the existence of a Lipschitz β-smooth bump function is equivalent to the generic β-differentiablility of any continuous convex function (cf. [7], Problem II.1). In this paper, we will prove the sufficient part of this “modified” Asplund conjecture: if a Banach space X admits a Lipschitz β-smooth bump function, then every continuous convex function on a convex open subset of X is generically β-differentiable. The weaker result, where instead of a dense G_{δ} subset one has only a dense subset, was proved independently by Deville–Godefroy–Zizler [5], [6] and Li–Shi [12]. [5] also announced the main result of this paper for the case of the Gâteaux-differentiability with a sketch of proof. Our contribution is to give a different and general proof. Our method is similar to that in [15] and in [12]. We adapt the construction used in [15] and also invoke a Banach–Mazur game to discuss the generic properties. An essential difference between Preiss–Phelps–Namioka’s proof and ours is that for a maximal monotone mapping T: X → 2^{X^{∗}}, we do not seek a pair (x_{0}, e_{0}) satisfying

1 Research supported by the National Natural Science Foundation of China.

ISSN 0944-6532 / $ 2.50 ◯^{c} Heldermann Verlag Berlin

∀x^{∗} ∈ T (x_{0}),  \frac{⟨x^{∗}, e_{0}⟩}{∥e_{0}∥} = sup_{e∈X\\{0\}} \frac{⟨x^{∗}, e⟩}{∥e∥} = ∥x^{∗}∥^{∗} = const,  (1)

but seek a pair (x_{0}, e_{0}) such that

∀x^{∗} ∈ T (x_{0}),  \frac{⟨x^{∗}, e_{0}⟩}{ρ(e_{0})} = sup_{e∈X} \frac{⟨x^{∗}, e⟩}{ρ(e)} := ρ^{∗}(x^{∗}) = const,  (2)

where ρ^{∗} is a “topologically equivalent” gauge function (positive sublinear function) of the norm on X^{∗} and it is generated by a “β-well function” ρ on X, which is induced by a Lipschitz β-smooth bump function (see (5)). In particular, it makes the process of searching for pairs (x_{0}, e_{0}) easier than that in [15] and a subset of any β-neighbourhood of x_{0}^{∗} ∈ X^{∗} may be characterized by ρ and ρ^{∗} (see Proposition 6). We use also a method by which we proved a generalization of Ekeland’s ε-variational principle and of its Borwein– Preiss smooth variant in [12]. Let us begin with some concepts about β-differentiability. All of them can be found in [14] or [2]. Let X be a real Banach space. A bornology of X, denoted by β, is a family of bounded subsets of X which forms a covering of X, i.e., ⋃_{S∈β} S = X. An (extended real-valued) function f: X → IR ∪ {±∞} is β-superdifferentiable at x ∈ X if f is finite at x and there is an x^{∗} ∈ X^{∗} such that for any S ∈ β and t > 0,

\{ \begin{matrix}f(x + th) − f(x) ≤ ⟨x^{∗}, th⟩ + tε(t, h) \\ with ε(t, h) & → 0 as t → 0 uniformly with respect to h ∈ S. \end{matrix}  (3)

Such an x^{∗} is called a β-superderivative of f at x and all β-superderivatives of f at x form a subset ∂^{β}f(x) of X^{∗}. For convenience, we put ∂^{β}f(x) = ∅ if f(x) = −∞ and say that if f(x) = +∞, then f is β-superdifferentiable at x and ∂^{β}f(x) = X^{∗}. Similarly, we can define β-subdifferentiable, β-subderivative and ∂_{β}f(x). If f is both β-superdifferentiable and β-subdifferentiable at x, then f is β-differentiable at x and ∂^{β}f(x) = ∂_{β}f(x) is necessarily a singleton; its unique element is the β-derivative of f at x, denoted by ∇_{β}f(x). Moreover, if f is β-differentiable at each point of a subset D, we say that f is β-smooth in D. But a β-smooth equivalent norm means that this norm is β-smooth away from the origin. A bornology β of X induces a topological vector structure on X^{∗}, which is generated by the zero-neighbourhood base {D_{S,ε}}, where

∀S ∈ β, ε > 0,  D_{S,ε} = { y^{∗} ∈ X^{∗} | ∀h ∈ S, ⟨y^{∗}, h⟩ < ε }.  (4)

We denote X^{∗} equipped with this topological vector structure by (X^{∗}, τ_{β}), which is a locally convex space. When β is the collection of all bounded subsets of X, τ_{β} becomes the strong topology of X^{∗}, and when β is the collection of all finite subsets of X, τ_{β} is reduced to the w^{∗}-topology of X^{∗}.

Let Ω ⊂ X be an open set. A set-valued mapping T: Ω → 2^{X^{∗}} is said to be β-continuous at x ([15]) if T (x) is a singleton and upper semicontinuous at x from (X, ∥ · ∥) to (X^{∗}, τ_{β}); i.e., for any S ∈ β and ε > 0, there is an open ball B_{δ,x} ⊂ Ω centered at x such that T (B_{δ,x}) ⊂ T (x) + D_{S,ε}. Notice that the notion used here is not the notion of continuity requiring upper and lower semicontinuity. Since T (x) is required to be a singleton, the lower semicontinuity at x is a direct consequence of the upper one. But generally the image of T at x is not required to be a singleton.

Proposition 1. ([14], [15]) A continuous convex function on a convex open subset Ω of X is β-differentiable at x ∈ Ω if and only if its subdifferential mapping ∂f: X → 2^{X^{∗}} is β-continuous at x.

Recall that the subdifferential mapping of a lower semicontinuous convex function on a Banach space X is maximal monotone ([14]); we reduce our problem to showing the generic β-continuity of a maximal monotone mapping on X. The basic assumption of this paper is as follows:

(H) \{ \begin{matrix}There is a Lipschitz β-smooth bump function ν: X → IR_{+}; i.e. \\ ν is nonconstant, Lipschitz, β-smooth and zero outside some ball. \end{matrix}

Proposition 2. If a Banach space X satisfies (H), then one has:

(H′) \{ \begin{matrix}There exists a Lipschitz β-superdifferentiable function \\ μ: X → [0, 1] such that μ(0) = 0 and [∥x∥ ≥ 1] ⟹ [μ(x) = 1]. \end{matrix}

Proof. Set α := sup_{x∈X} ν(x) and R > sup{∥x∥ | ν(x) > 0}. Then, α > 0 and R > 0. Without loss of generality, we can suppose that ν(0) ≥ α/2 and R = 1. Otherwise, we replace ν by ν_{1}(x) = ν(2Rx + x_{0}), where x_{0} satisfies ν(x_{0}) ≥ α/2. Take an increasing function h ∈ C^{∞}(IR_{+}; [0, 1]) such that

h(t) = \{ \begin{matrix}1, \\ 0, \end{matrix}  when t ≥ α/2; when t = 0.

Then the function μ(x) = 1 − h(ν(x)) satisfies (H′).

Actually, we need only a Lipschitz β-superdifferentiable function μ such as in (H′) instead of a Lipschitz β-differentiable bump function (H). It means that we need only the βsubdifferentiability of ν in (H). We say that ρ is a β-well function on X if ρ: X → [1, +∞] is β-superdifferentiable and continuous, and satisfies

ρ(0) < +∞ and [∥x∥ ≥ 1] ⟹ [ρ(x) = +∞],  (5)

where, according to our convention, if ρ(x) = +∞, then ρ is considered to be β-super-differentiable at x; and the continuity of ρ is for the norm topology on X and the usual topology on [1, +∞].

Proposition 3. If (H′) holds, then there exists a β-well function on X.

Proof. Take a real function g: [0, 1) → [1, +∞) which is an increasing C^{∞}-function with g(0) = 1 and lim_{t→1} g(t) = +∞. Set g(1) := +∞. Then, the composite function of g with μ in (H′), ρ_{0} = g ∘ μ: X → [1, +∞] is a β-well function on X. In fact, we need only to show that ρ_{0} is β-superdifferentiable at x ∈ X with μ(x) < 1. Take S ∈ β with M := sup_{h∈S} ∥h∥ < +∞. Since μ is Lipschitz and β-superdifferentiable, and g is differentiable in [0, 1) with g′(μ(x)) ≥ 0, we have that for t > 0,

g(μ(x + th)) − g(μ(x))
= g′(μ(x))(μ(x + th) − μ(x)) + o(|μ(x + th) − μ(x)|)
≤ g′(μ(x))(⟨x^{∗}, th⟩ + tε_{1}(t, h)) + LMtε_{2}(t, h) for any x^{∗} ∈ ∂^{β}μ(x)
= ⟨g′(μ(x))x^{∗}, th⟩ + tε_{0}(t, h),

where L is a Lipschitz constant of μ and ε_{i}(t, h), i = 0, 1, 2, tend to 0 as t → 0 uniformly with respect to h ∈ S. So ρ_{0} is β-superdifferentiable at x.

The β-well function ρ that we will construct to get (2) will be of the form of ρ_{∞} in the following proposition.

Proposition 4. Let ρ_{0} be a β-well function on X, μ be defined as in (H′), μ_{n}(x) = μ(nx)/2^{n}, n = 1, 2, …, and {e_{n}}_{n=1}^{∞} ⊂ X be a sequence. Then

ρ_{n}(x) := ρ_{0}(x) + ∑_{k=1}^{n} μ_{k}(x − e_{k}), n = 1, 2, …,  (6)

and

ρ_{∞}(x) := ρ_{0}(x) + ∑_{n=1}^{∞} μ_{n}(x − e_{n}),  (7)

are all β-well functions on X.

Proof. In fact, μ_{n}: X → [0, 1/2^{n}] satisfies:

\{ \begin{matrix}i) μ_{n}(0) = 0 and [∥x∥ ≥ 1/n] ⟹ [μ_{n}(x) = 1/2^{n}]; \\ ii) μ_{n} is β-superdifferentiable everywhere; \\ iii) μ_{n} is Lipschitz of rank nL/2^{n}, where L is a Lipschitz constant of μ. \end{matrix}  (8)

Hence, from (6), ρ_{n} is obviously a β-well function. On the other hand, ∑_{n=1}^{∞} μ_{n}(x − e_{n}) converges uniformly, and so, is continuous in x. Since μ_{n} is Lipschitz of rank nL/2^{n}, we have that

∑_{n=1}^{∞} μ_{n}(x − e_{n} + th) − ∑_{n=1}^{∞} μ_{n}(x − e_{n})

≤ ∑_{n=1}^{N} μ_{n}(x − e_{n} + th) − ∑_{n=1}^{N} μ_{n}(x − e_{n}) + tL∥h∥ ∑_{n=N+1}^{∞} n/2^{n}.

As any finite sum of β-superdifferentiable functions is also β-superdifferentiable, we can use the Weierstrass M-test to show that ∑_{n=1}^{∞} μ_{n}(x − e_{n}) is β-superdifferentiable too and so is ρ_{∞}.

Now for a β-well function ρ, we define a gauge function ρ^{∗}: X^{∗} → IR on X^{∗} by

∀x^{∗} ∈ X^{∗},  ρ^{∗}(x^{∗}) := sup_{e∈X} \frac{⟨x^{∗}, e⟩}{ρ(e)}.  (9)

Obviously, for two β-well functions ρ_{0} and ρ_{1},

[ρ_{1} ≥ ρ_{0}] ⇒ [ρ_{1}^{∗} ≤ ρ_{0}^{∗}].  (10)

Proposition 5. For a gauge function ρ^{∗} on X^{∗} defined by (9) there exists ε_{0} ∈ (0, 1) such that

∀x^{∗} ∈ X^{∗},  (1 − ε_{0})∥x^{∗}∥^{∗} ≤ ρ^{∗}(x^{∗}) ≤ ∥x^{∗}∥^{∗}.  (11)

In particular, if ρ is symmetric, then ρ^{∗} is an equivalent norm on (X^{∗}, ∥ · ∥^{∗}).

Proof. Since ρ(x) = +∞ whenever ∥x∥ ≥ 1 and ρ ≥ 1, we obtain that

ρ^{∗}(x^{∗}) = sup_{e∈X} \frac{⟨x^{∗}, e⟩}{ρ(e)} = sup_{∥e∥≤1} \frac{⟨x^{∗}, e⟩}{ρ(e)}

≤ sup_{∥e∥≤1}⟨x^{∗}, e⟩ = ∥x^{∗}∥^{∗}.

On the other hand, since ρ is continuous and ρ(0) ∈ [1, +∞), there exists a δ ∈ (0, 2ρ(0)) such that ρ(x) ∈ [1, 2ρ(0)] whenever x ∈ B_{δ} = { x ∈ X | ∥x∥ ≤ δ }. Hence,

ρ^{∗}(x^{∗}) = sup_{e∈X} \frac{⟨x^{∗}, e⟩}{ρ(e)} ≥ sup_{e∈B_{δ}} \frac{⟨x^{∗}, e⟩}{ρ(e)}

≥ \frac{1}{2ρ(0)} sup_{e∈B_{δ}}⟨x^{∗}, e⟩ ≥ \frac{δ}{2ρ(0)}∥x^{∗}∥^{∗},

i.e., (11) holds for 1 − ε_{0} = δ/2ρ(0).

Our main result is based on the following two simple facts.

52 Y. Li, S. Shi / Differentiability of Convex Functions on a Banach Space

Proposition 6. Let ρ be a β-well function on X, e_{0} ∈ X with ρ(e_{0}) finite. If there exists x_{0}^{∗} ∈ X^{∗} such that

c := \frac{⟨x_{0}^{∗}, e_{0}⟩}{ρ(e_{0})} = sup_{e∈X} \frac{⟨x_{0}^{∗}, e⟩}{ρ(e)} = ρ^{∗}(x_{0}^{∗}) > 0,  (12)

then

(i) ρ is β-differentiable at e_{0} and

x_{0}^{∗} = c∇_{β}ρ(e_{0});  (13)

(ii) for any S ∈ β and any ε > 0, there exists δ > 0 such that

D(ρ, e_{0}, x_{0}^{∗}, δ) := \{x^{∗} ∈ X^{∗} | c − δ < \frac{⟨x^{∗}, e_{0}⟩}{ρ(e_{0})} ≤ ρ^{∗}(x^{∗}) < c + δ \}  (14)
⊂ x_{0}^{∗} + D_{S,ε}.  (15)

Proof. (i) By (12), we have

∀e ∈ X, cρ(e) ≥ ⟨x_{0}^{∗}, e⟩.

Joining up with (12), it follows

∀e ∈ X, cρ(e) − cρ(e_{0}) ≥ ⟨x_{0}^{∗}, e − e_{0}⟩.

Hence, x_{0}^{∗} ∈ c∂_{β}ρ(e_{0}). On the other hand, from the definition of β-well function (5), ∂^{β}ρ(e_{0}) is nonempty, and then, ∂^{β}ρ(e_{0}) must be a singleton. Thus, (13) holds. (ii) Suppose that S ∈ β and ε > 0. From (13), we can take a fixed t_{0} > 0 such that

∀h ∈ S,  ρ(e_{0} + t_{0}h) ≤ ρ(e_{0}) + \frac{1}{c}⟨x_{0}^{∗}, t_{0}h⟩ + \frac{εt_{0}}{2c} .  (16)

If x^{∗} ∈ D(ρ, e_{0}, x_{0}^{∗}, δ) with a sufficiently small δ > 0, then from (9), the definition (14) of D(ρ, e_{0}, x_{0}^{∗}, δ) and (16), we have that

\begin{aligned}⟨x^{∗}, e_{0} + t_{0}h⟩ & ≤ ρ^{∗}(x^{∗})ρ(e_{0} + t_{0}h) ≤ (c + δ)ρ(e_{0} + t_{0}h) \\ & ≤ (c + δ)ρ(e_{0}) + \frac{c + δ}{c} ⟨x_{0}^{∗}, t_{0}h⟩ + \frac{c + δ}{2c} εt_{0}, \end{aligned}

and then,

\begin{aligned}⟨x^{∗} − x_{0}^{∗}, t_{0}h⟩ & ≤ (c + δ)ρ(e_{0}) − ⟨x^{∗}, e_{0}⟩ + \frac{δ}{c}⟨x_{0}^{∗}, t_{0}h⟩ + \frac{c + δ}{2c} εt_{0} \\ & ≤ (c + δ)ρ(e_{0}) − (c − δ)ρ(e_{0}) + \frac{δ}{c}⟨x_{0}^{∗}, t_{0}h⟩ + \frac{c + δ}{2c} εt_{0}. \end{aligned}

Hence, for a sufficiently small δ > 0, we obtain that

∀h ∈ S, ⟨x^{∗} − x_{0}^{∗}, h⟩ ≤ \frac{2δ}{t_{0}} ρ(e_{0}) + \frac{δ}{c}⟨x_{0}^{∗}, h⟩ + \frac{c + δ}{2c} ε < ε;

i.e., (15) holds.

To obtain the desired generic properties, we have to introduce a Banach–Mazur game ([11]) on an open subset Ω of X. Let W be a subset of Ω. A Banach–Mazur game with the objective subset W on Ω is a nested open subset (of Ω) sequence U_{1} ⊃ V_{1} ⊃ U_{2} ⊃ V_{2} ⊃ ⋯ ⊃ U_{n} ⊃ V_{n} ⊃ ⋯, where U_{n} and V_{n} are chosen by player A and player B alternatively. We say that B is the winner if ⋂_{n=1}^{∞} V_{n} ⊂ W, where ⋂_{n=1}^{∞} V_{n} is not necessarily nonempty. A strategy for B is a sequence of maps f_{n} such that for any n, f_{n} is defined for every U_{1}, U_{2}, …, U_{n} chosen by A, and f_{n}(U_{1}, U_{2}, …, U_{n}) is an open subset contained in U_{n}. A strategy for B is a winning strategy if for any of A’s choices of {U_{n}} such that U_{1} ⊃ f_{1}(U_{1}) ⊃ U_{2} ⊃ f_{2}(U_{1}, U_{2}) ⊃ ⋯ ⊃ U_{n} ⊃ f_{n}(U_{1}, U_{2}, ⋯, U_{n}) ⊃ ⋯, B is the winner.

Proposition 7. ([11]) (Banach–Mazur Game Theorem) The objective subset W contains a dense G_{δ} subset of Ω if and only if B has a winning strategy.

Finally, we need the following properties of maximal monotone mappings T: X → 2^{X^{∗}}.

Proposition 8. Let T: X → 2^{X^{∗}} be a maximal monotone mapping. If Ω = int { x ∈ X | T (x) ≠ ∅ } is nonempty, then

(i) T is locally bounded on Ω; (ii) T is upper semicontinuous from (X, ∥ · ∥) to (X^{∗}, w^{∗}) on Ω; (iii) let e ∈ X, U be an open subset of Ω and x̄ ∈ U; if for some x̄^{∗} ∈ T (x̄) and some

b ∈ IR, ⟨x̄^{∗}, e⟩ > b, then there exists an open subset V ⊂ U such that

∀x^{∗} ∈ T (V),  ⟨x^{∗}, e⟩ > b.  (17)

Proof. i) and ii) are well-known ([14]). iii) has been posed in [15] in a little weaker form. We show iii) as follows. Take t > 0 such that x̄ + te ∈ U. By the monotonicity of T for any x^{∗} ∈ T (x̄ + te) we have ⟨x^{∗}, e⟩ ≥ ⟨x̄^{∗}, e⟩ > b, i.e., T (x̄ + te) ⊂ { y^{∗} ∈ X^{∗} | ⟨y^{∗}, e⟩ > b }, which means that { y^{∗} ∈ X^{∗} | ⟨y^{∗}, e⟩ > b } is a w^{∗}-open neighborhood of T (x̄ + te). From ii), there exists an open neighbourhood V of x̄ + te, V ⊂ U, such that T (V) ⊂ { y^{∗} ∈ X^{∗} | ⟨y^{∗}, e⟩ > b }.

Now we deal with our main theorem.

Theorem 1. Assume that X admits a Lipschitz β-differentiable bump function, i.e., (H) holds. If T: X → 2^{X^{∗}} is a maximal monotone mapping on X and Ω := int { x ∈ X | T (x) ≠ ∅ } is nonempty, then T is β-continuous in a dense G_{δ} subset of Ω.

Proof. By using Proposition 7, we suppose that player A and player B are playing a Banach–Mazur game on Ω with the objective subset

W = { x ∈ Ω | T is β-continuous at x. }  (18)

It means that given A’s choice of {U_{n}}, we will find B’s choice {V_{n}} such that ⋂_{n=1}^{∞} V_{n} ⊂ W; i.e., if y_{∞} ∈ ⋂_{n=1}^{∞} V_{n}, then T is a singleton and β-continuous at y_{∞}. For proving that T (y_{∞}) is a singleton, we will construct a β-well function of the form ρ_{∞} in Proposition 3 and show that for e_{∞} = lim_{n→∞} e_{n}, we have

∀y_{∞}^{∗} ∈ T (y_{∞}), \frac{⟨y_{∞}^{∗}, e_{∞}⟩}{ρ_{∞}(e_{∞})} = ρ_{∞}^{∗}(y_{∞}^{∗}) = const,

from which, by using Proposition 6 i), it follows y_{∞}^{∗} = const∇_{β}ρ_{∞}(e_{∞}). And then, by using Proposition 6 ii), we will show that for any S ∈ β and any ε > 0, there exists a sufficient large n such that T (V_{n}) ⊂ T (y_{∞}) + D_{S,ε}, which implies the β-continuity of T at y_{∞}. We begin this process. Firstly, we take a β-well function ρ_{0} on X, the existence of which is due to Proposition 2 and 3. ρ_{0}^{∗} is generated as in (9) by ρ_{0} and due to Proposition 5, we may suppose that ε_{0} ∈ (0, 1) satisfies

∀x^{∗} ∈ X^{∗},  (1 − ε_{0})∥x^{∗}∥^{∗} ≤ ρ_{0}^{∗}(x^{∗}) ≤ ∥x^{∗}∥^{∗}.  (19)

For any U_{1} chosen by A, due to Proposition 8 i), B can choose V_{1} ⊂ U_{1} such that T is bounded in V_{1}. Then the playing is going on with U_{2} ⊂ V_{1}. Set

s_{0} = sup_{x^{∗}∈T (U_{2})} ρ_{0}^{∗}(x^{∗}) < +∞.  (20)

Without loss of generality, we suppose that s_{0} > 0; otherwise, the proof is trivial. From (9) and (20), there exist x̄ ∈ U_{2}, x̄^{∗} ∈ T (x̄) and e_{1} ∈ X such that

⟨x̄^{∗}, e_{1}⟩ > ρ_{0}(e_{1})(1 − ε_{0})s_{0}.  (21)

Due to iii) of Proposition 8, B can choose V_{2} ⊂ U_{2} such that

∀x^{∗} ∈ T (V_{2}),  ⟨x^{∗}, e_{1}⟩ > ρ_{0}(e_{1})(1 − ε_{0})s_{0}.  (22)

For any U_{3} chosen by A with U_{3} ⊂ V_{2}, define

D_{1} = { e ∈ X | sup_{x^{∗}∈T (U_{3})}⟨x^{∗}, e⟩ ≥ ρ_{0}(e)(1 − ε_{0})s_{0}}.  (23)

Then (22) implies e_{1} ∈ D_{1}. In addition, because of the boundedness of T (U_{3}), x ↦ sup_{x^{∗}∈T (U_{3})}⟨x^{∗}, x⟩ is continuous, hence, D_{1} is closed.

Denote

ρ_{1}(x) = ρ_{0}(x) + μ_{1}(x − e_{1}),  (24)

where μ_{1} is defined as in Proposition 4. Set

s_{1} = sup_{x^{∗}∈T (U_{3})} ρ_{1}^{∗}(x^{∗}).  (25)

Then from U_{3} ⊂ U_{2} and (10), s_{1} ≤ s_{0} and from (22), (24) and (25), we deduce that

∀x^{∗} ∈ T (U_{3}),  (1 − ε_{0})s_{0} < \frac{⟨x^{∗}, e_{1}⟩}{ρ_{0}(e_{1})} = \frac{⟨x^{∗}, e_{1}⟩}{ρ_{1}(e_{1})} ≤ s_{1} ≤ s_{0}.  (26)

Y. Li, S. Shi / Differentiability of Convex Functions on a Banach Space 55

Take ε_{1} so small that

ε_{1} ∈ (0, (1 − ε_{0})^{2}/2^{2}) and (1 − ε_{0})s_{0} < (1 − ε_{1})s_{1}.  (27)

By the same reasoning, B can choose V_{3} ⊂ U_{3} and e_{2} such that

∀x^{∗} ∈ T (V_{3}),  ⟨x^{∗}, e_{2}⟩ > ρ_{1}(e_{2})(1 − ε_{1})s_{1}.  (28)

For any U_{4} chosen by A with U_{4} ⊂ V_{3}, define

D_{2} = {e ∈ X | sup_{x^{∗}∈T (U_{4})}⟨x^{∗}, e⟩ ≥ ρ_{1}(e)(1 − ε_{1})s_{1}}.  (29)

Then e_{2} ∈ D_{2}, D_{2} ⊂ D_{1} and D_{2} is closed. Now for ρ_{n−1}, s_{n−1}, ε_{n−1}, e_{n}, V_{n+1}, U_{n+2} and D_{n} given as above, n = 1, 2, …, set

ρ_{n}(x) = ρ_{n−1}(x) + μ_{n}(x − e_{n}),  (30)
s_{n} = sup_{x^{∗}∈T (U_{n+2})} ρ_{n}^{∗}(x^{∗}).  (31)

Then

∀x^{∗} ∈ T (U_{n+2}),  (1 − ε_{n−1})s_{n−1} < \frac{⟨x^{∗}, e_{n}⟩}{ρ_{n−1}(e_{n})} = \frac{⟨x^{∗}, e_{n}⟩}{ρ_{n}(e_{n})} ≤ s_{n} ≤ s_{n−1}. (32)

Take ε_{n} so small that

ε_{n} ∈ (0, (1 − ε_{0})^{2}/2^{n+1}) and (1 − ε_{n−1})s_{n−1} < (1 − ε_{n})s_{n}.  (33)

B can choose e_{n+1} and V_{n+2} ⊂ U_{n+2} such that

∀x^{∗} ∈ T (V_{n+2}),  ⟨x^{∗}, e_{n+1}⟩ > ρ_{n}(e_{n+1})(1 − ε_{n})s_{n}.  (34)

For any U_{n+3} chosen by A with U_{n+3} ⊂ V_{n+2}, set

D_{n+1} = {e ∈ X | sup_{x^{∗}∈T (U_{n+3})}⟨x^{∗}, e⟩ ≥ ρ_{n}(e)(1 − ε_{n})s_{n}}.  (35)

Then e_{n+1} ∈ D_{n+1}, D_{n+1} ⊂ D_{n} and D_{n+1} is closed. As the game continues, B obtains sequences {ρ_{n−1}}_{n=1}^{∞}, {s_{n−1}}_{n=1}^{∞}, {ε_{n−1}}_{n=1}^{∞}, {e_{n}}_{n=1}^{∞}, {U_{n}}_{n=1}^{∞}, {V_{n}}_{n=1}^{∞} and {D_{n}}_{n=1}^{∞}. For any x_{n} ∈ D_{n+1}, from (35), there exists x_{n}^{∗} ∈ T (U_{n+3}) satisfying

\frac{⟨x_{n}^{∗}, x_{n}⟩}{ρ_{n−1}(x_{n}) + μ_{n}(x_{n} − e_{n})} ≥ (1 − ε_{n})s_{n} > (1 − ε_{n−1})s_{n−1}

or

ρ_{n−1}(x_{n}) + μ_{n}(x_{n} − e_{n}) < \frac{⟨x_{n}^{∗}, x_{n}⟩}{(1 − ε_{n−1})s_{n−1}} .  (36)

On the other hand, since x_{n}^{∗} ∈ T (U_{n+3}) ⊂ T (U_{n+1}) ⊂ T (U_{2}), by (31), we have that

\frac{⟨x_{n}^{∗}, x_{n}⟩}{ρ_{n−1}(x_{n})} ≤ s_{n−1} or ρ_{n−1}(x_{n}) ≥ \frac{⟨x_{n}^{∗}, x_{n}⟩}{s_{n−1}}.  (37)

Obviously, ∥x_{n}∥ < 1; otherwise ρ_{n−1}(x_{n}) = +∞, which contradicts (36). So, from (19) and (20), it follows

⟨x_{n}^{∗}, x_{n}⟩ ≤ ∥x_{n}^{∗}∥^{∗} ≤ ρ_{0}^{∗}(x_{n}^{∗})/(1 − ε_{0}) ≤ s_{0}/(1 − ε_{0}).  (38)

Therefore, from (36), (37), (33), (27) and (38), it follows that

μ_{n}(x_{n} − e_{n}) ≤ \frac{⟨x_{n}^{∗}, x_{n}⟩}{(1 − ε_{n−1})s_{n−1}} − \frac{⟨x_{n}^{∗}, x_{n}⟩}{s_{n−1}} ≤ \frac{ε_{n−1}⟨x_{n}^{∗}, x_{n}⟩}{(1 − ε_{n−1})s_{n−1}} ≤ \frac{ε_{n−1}}{(1 − ε_{0})^{2}} < \frac{1}{2^{n}} .

Hence, by (8) i), ∥x_{n} − e_{n}∥ < 1/n. Thus, {D_{n}}_{n=1}^{∞} is a nested sequence of closed subsets with diameters decreasing to 0 and hence there exists a unique e_{∞} ∈ X such that

{e_{∞}} = ⋂_{n=1}^{∞} D_{n}.  (39)

Now we show that T is β-continuous at any y_{∞} ∈ ⋂_{n=1}^{∞} V_{n}. In fact, for any y_{∞}^{∗} ∈ T (y_{∞}) ⊂ T (V_{n+2}), from (34), we have that

⟨y_{∞}^{∗}, e_{n+1}⟩ ≥ ρ_{n}(e_{n+1})(1 − ε_{n})s_{n}.  (40)

By (32) and (33), s_{n} converges to some s_{∞} > 0. By using Proposition 4, ρ_{∞}(x) = ρ_{0}(x) + ∑_{k=1}^{∞} μ_{k}(x − e_{k}) is a β-well function on X and ρ_{n} → ρ_{∞} uniformly on {x ∈ X | ∥x∥ < 1}. Joining up with e_{n} → e_{∞} and ε_{n} → 0, from (40), we deduce that

⟨y_{∞}^{∗}, e_{∞}⟩ ≥ ρ_{∞}(e_{∞})s_{∞}.  (41)

On the other hand, for any n,

\frac{⟨y_{∞}^{∗}, e_{∞}⟩}{ρ_{∞}(e_{∞})} ≤ ρ_{∞}^{∗}(y_{∞}^{∗}) ≤ ρ_{n}^{∗}(y_{∞}^{∗}) ≤ s_{n}.

then we also have

\frac{⟨y_{∞}^{∗}, e_{∞}⟩}{ρ_{∞}(e_{∞})} ≤ ρ_{∞}^{∗}(y_{∞}^{∗}) ≤ s_{∞}.  (42)

Thus, (41) and (42) ensure that

ρ_{∞}^{∗}(y_{∞}^{∗}) = \frac{⟨y_{∞}^{∗}, e_{∞}⟩}{ρ_{∞}(e_{∞})} = s_{∞}.  (43)

From Proposition 6 i), y_{∞}^{∗} = s_{∞}∇_{β}ρ_{∞}(e_{∞}), and it follows that T (y_{∞}) is a singleton.

Moreover, for any given δ > 0, we have that for n sufficiently large,

∀y^{∗} ∈ T (V_{n+2}) ⊂ T (U_{n+2}),  ρ_{∞}^{∗}(y^{∗}) ≤ ρ_{n}^{∗}(y^{∗}) ≤ s_{n} < s_{∞} + δ.  (44)

At the same time, from (34), we have

∀y^{∗} ∈ T (V_{n+2}), \frac{⟨y^{∗}, e_{n+1}⟩}{ρ_{n}(e_{n+1})} > (1 − ε_{n})s_{n}  (45)

and noting (by (31)) that ∥e_{∞}∥ < 1 and ρ_{n} ≥ 1,

|\frac{⟨y^{∗}, e_{∞}⟩}{ρ_{∞}(e_{∞})} − \frac{⟨y^{∗}, e_{n+1}⟩}{ρ_{n}(e_{n+1})}|

≤ |⟨y^{∗}, e_{∞}⟩ (\frac{1}{ρ_{∞}(e_{∞})} − \frac{1}{ρ_{n}(e_{n+1})})| + |\frac{1}{ρ_{n}(e_{n+1})}⟨y^{∗}, e_{∞} − e_{n+1}⟩|
≤ ∥y^{∗}∥^{∗} (|\frac{1}{ρ_{∞}(e_{∞})} − \frac{1}{ρ_{n}(e_{n+1})}| + ∥e_{∞} − e_{n+1}∥)
≤ \frac{s_{0}}{1 − ε_{0}} (|\frac{1}{ρ_{∞}(e_{∞})} − \frac{1}{ρ_{n}(e_{n+1})}| + ∥e_{∞} − e_{n+1}∥) .

When n tends to +∞, the last two terms above tend to 0. Hence, joining up with (45), we have also that for n sufficiently large,

\frac{⟨y^{∗}, e_{∞}⟩}{ρ_{∞}(e_{∞})} ≥ s_{∞} − δ.  (46)

Due to Proposition 6 ii), from (44) and (46), we deduce that for any S ∈ β and ε > 0, there exists a sufficiently large n such that

T (V_{n+2}) ⊂ T (y_{∞}) + D_{S,ε},

i.e., T is β-continuous at y_{∞}. Thus, ⋂_{n=1}^{∞} V_{n} ⊂ W.

Corollary. If a Banach space X admits a Lipschitz β-differentiable bump function, i.e., (H) holds, then every continuous convex function on a convex open subset Ω ⊂ X is β-differentiable in a dense G_{δ} subset of Ω.

Remark 1. Recall that a map ϕ: Ω → 2^{X^{∗}}, where Ω may be any Hausdorff space, is said to be a w^{∗}-usco map if ϕ is nonempty compact convex valued and upper semicontinuous for the w^{∗}-topology of X^{∗}. A w^{∗}-usco map ϕ: Ω → 2^{X^{∗}} is said to be minimal if there is no w^{∗}-usco map ψ: Ω → 2^{X^{∗}} such that the graph of ψ is a proper subset of the graph of ϕ ([15]). A maximal monotone mapping on X is an example of minimal w^{∗}-usco map if we take Ω := int { x ∈ X | T (x) ≠ ∅ }. As in [15], we can use the same method to generalize Theorem 1 as follows.

Theorem 2. Assume that X admits a Lipschitz β-differentiable bump function, i.e., (H) holds. If Ω is a Baire space and ϕ: Ω → 2^{X^{∗}} is a minimal w^{∗}-usco map, then ϕ is β-continuous in a dense G_{δ} subset of Ω.

Remark 2. In the proof of Theorem 1, our aim is to show (43); that is, to show that there exists e_{∞} such that for any y_{∞}^{∗} ∈ T (e_{∞}) and any e ∈ X,

\frac{1}{⟨y_{∞}^{∗}, e⟩} (ρ_{0}(e) + ∑_{n=1}^{∞} μ_{n}(e − e_{n})) ≥ \frac{1}{⟨y_{∞}^{∗}, e_{∞}⟩} (ρ_{0}(e_{∞}) + ∑_{n=1}^{∞} μ_{n}(e_{∞} − e_{n}))

It is similar to the generalization of Ekeland’s ε-variational principle in our paper [11], and its proof is also as in [11]. So, combining a Banach–Mazur game, it is possible to prove a more general ε-variational principle. We will discuss this problem in another paper.

Remark 3. The only difference between (H′) and the existence of a β-smooth norm is the convexity. In fact, we have the following proposition.

Proposition 9. Let X be a Banach space. If the following assumption holds:

(H′_{c}) \{ \begin{matrix}There exists a (Lipschitz) β-superdifferentiable function \\ μ: X → [0, 1] such that μ(0) = 0, [∥x∥ ≥ 1] ⟹ [μ(x) = 1] \\ and for some α ∈ (0, 1/2), μ is convex on \{x ∈ X | μ(x) ≤ 2α\}. \end{matrix}

Then there exists a β-smooth equivalent norm on X.

Proof. Without loss of generality, we can assume that μ is symmetric. Otherwise, consider (μ(x) + μ(−x))/2 instead of μ. Define the Minkowski function p: X → IR associated to the open convex subset Ω := { x ∈ X | μ(x) < α }, i.e.,

p(x) = inf{λ > 0 | x ∈ λΩ}.

Since Ω is nonempty, symmetric and bounded, p is an equivalent norm on X. We prove that p is β-smooth away from the origin. Let G := ∂Ω be the boundary of Ω. It is sufficient to show that p is β-smooth at each point of G, or, due to Proposition 1, that ∂p is β-continuous on G.

Let x_{0} ∈ G. Firstly, we notice that μ is convex and continuous on \overline{Ω}. So, μ is subdifferentiable in the sense of convex analysis at x_{0}, and hence is β-differentiable at x_{0}. We show that

⟨∇_{β}μ(x_{0}), x_{0}⟩ > 0.  (47)

Otherwise, μ(0) − μ(x_{0}) ≥ ⟨∇_{β}μ(x_{0}), −x_{0}⟩ ≥ 0, which contradicts the fact that μ(0) = 0 < α = μ(x_{0}). Secondly, we show that ∂p(x_{0}) is a singleton. In fact, it is well-known that for the indicator function δ\overline{Ω}: X → IR ∪ \{+∞\}, defined by

δ\overline{Ω}(x) := \{ \begin{matrix}0, \\ + ∞, \end{matrix}  if x ∈ \overline{Ω}, ; otherwise,

we have that

∂δ\overline{Ω}(x_{0}) = ⋃_{λ≥0} λ∂p(x_{0}) = ⋃_{λ≥0} λ∇_{β}μ(x_{0}),

(see, for instance, [3], pp. 55–56.) and it follows that

∂p(x_{0}) ⊂ ⋃_{λ≥0} λ∇_{β}μ(x_{0}).  (48)

On the other hand, since p is a norm on X, we have that

∀x^{∗} ∈ ∂p(x_{0}),  ⟨x^{∗}, x_{0}⟩ = p(x_{0}) = 1.  (49)

Therefore, from (47), (48) and (49), we deduce that

∂p(x_{0}) = \{ \frac{∇_{β}μ(x_{0})}{⟨∇_{β}μ(x_{0}), x_{0}⟩} \} .

i.e., the Gâteaux derivative of p at x_{0} is

∇p(x_{0}) = \frac{∇_{β}μ(x_{0})}{⟨∇_{β}μ(x_{0}), x_{0}⟩}.  (50)

Finally, we show that ∇p is β-continuous on G. In fact, from (50), it is easy to see that in general,

∀x ≠ 0, ∇p(x) = p(x)\frac{∇_{β}μ(x/p(x))}{⟨∇_{β}μ(x/p(x)), x/p(x)⟩}.

So, for any x_{0} ∈ G and any x, h ∈ X, we have that

⟨∇p(x) − ∇p(x_{0}), h⟩
= ⟨\frac{p(x)∇_{β}μ(x/p(x))}{⟨∇_{β}μ(x/p(x)), x/p(x)⟩} − \frac{∇_{β}μ(x_{0})}{⟨∇_{β}μ(x_{0}), x_{0}⟩}, h⟩

≤ |\frac{p(x) − 1}{⟨∇_{β}μ(x/p(x)), x/p(x)⟩}⟨∇_{β}μ(x/p(x)), h⟩|  (51)

+ |\frac{1}{⟨∇_{β}μ(x/p(x)), x/p(x)⟩} − \frac{1}{⟨∇_{β}μ(x_{0}), x_{0}⟩}| |⟨∇_{β}μ(x/p(x)), h⟩|
+ \frac{1}{⟨∇_{β}μ(x_{0}), x_{0}⟩}⟨∇_{β}μ(x/p(x)) − ∇_{β}μ(x_{0}), h⟩.

Take any S ∈ β. Then, sup_{h∈S} ∥h∥ < +∞. Since ∇_{β}μ is locally bounded and ∥ · ∥- w^{∗}-continuous over {x | μ(x) < 2α} and ⟨∇_{β}μ(x_{0}), x_{0}⟩ > 0, it is easy to deduce that the function x ↦ 1/⟨∇_{β}μ(x/p(x)), x/p(x)⟩ is continuous at x_{0}. Hence, joining up with the continuity of p, when x → x_{0}, the first and second added terms of (51) can be converges to 0 uniformly on h ∈ S. Moreover, due to the β-continuity of ∇_{β}μ over {x | μ(x) < 2α}, the third one also converges to 0. Thus, ∇p is β-continuous on G.

Given a C^{1} bump function, we can apply Leduc’s method to construct a C^{1} positive homogeneous function which is different from an equivalent norm only in the subadditivity ([14], Theorem 3.6). This together with Proposition 9 tells us that the essential obstruction to the transition from a smooth bump function to a smooth norm is exactly the lack of convexity. This is the reason why, in our proof of Theorem 1, we have avoided the use of the subadditivity and the subdifferentiability of norm; in particular, we did not use Cauchy sequences, but used sequences of nested closed subsets to show the convergence of {e_{n}}.

Acknowledgment.The authors are grateful to two anonymous referees for their useful comments and suggestions. In particular, they show us that our main result for the case of the Gâteauxdifferentiability has been announced by [5]. The first author wishes to thank Professor Phelps for his kind letter with helpful comments and a copy of [5]. The second author wishes to thank Professor Godefroy for sending the new book [7] to China.

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