since for (y, r) ∈ ker p (x1 ) — R the inverse image is given as t(y + x1 ) where t can be

calculated from r = p(t (y+x1 )). Since t ’ p(t (y+x1 )) is a di¬eomorphism between

the intervals (0, +∞) ’ (p(0), +∞) this t is uniquely determined. Furthermore, t

depends smoothly on (y, r): Let s ’ (y(s), r(s)) be a smooth curve, then t(s) is

given by the implicit equation p(t (y(s) + x1 )) = r(s), and by the 2-dimensional

implicit function theorem the solution s ’ t(s) is smooth.

(2) By general principles pU is a sublinear mapping, and U = {x : pU (x) < 1} since

U is open. Thus it remains to show that pU is smooth on its open carrier. So let c be

a smooth curve in the carrier. By assumption, there is a di¬eomorphism v, locally

de¬ned on E near an intersection point a of the ray through c(0) with the boundary

‚U = {x : p(x) = 1}, such that ‚U corresponds to a closed linear subspace F ⊆ E.

Since U is convex there is a bounded linear functional » ∈ E with »(a) = 1 and

U ⊆ {x ∈ E : »(x) ¤ 1} by the theorem of Hahn-Banach. Then »(Ta (‚U )) = 0

since any smooth curve in ‚U through a stays inside {x : »(x) ¤ 1}. Furthermore,

b : ‚t |1 v(ta) ∈ F , since otherwise t ’ v ’1 (tb) ∈ ‚U but ‚t |1 »(v ’1 (tb)) = »(a) = 1.

‚ ‚

/

Put f := 1/pU —¦ c : R ’ R. Then f is a solution of the implicit equation (» —¦

dv ’1 (0) —¦ v)(f (t)c(t)) = 0 which has a unique smooth solution by the implicit

function theorem in dimension 2 since

—¦ dv ’1 (0) —¦ v)(sc(t)) = »dv ’1 (0)dv(f (t)c(t))c(t) = 0

‚

‚s |s=f (t) (»

1

for t near 0, since for t = 0 we get »(c(0)) = f (0) . So pU is smooth on its carrier.

13.16. The space c0 (“). For an arbitrary set “ the space c0 (“) is the closure

of all functions on “ with ¬nite support in the Banach space ∞ (“) of globally

bounded functions on “ with the supremum norm. The supremum norm on c0 (“)

is not di¬erentiable on its carrier, see (13.12). Nevertheless, it was shown in [Bonic,

Frampton, 1965] that c0 is C ∞ -regular.

Proposition. Smooth norm on c0 . Due to Kuiper according to [Bonic, Framp-

ton, 1966]. There exists an equivalent norm on c0 (“) which is smooth o¬ 0.

Proof. To prove this let h : R ’ R be an unbounded symmetric smooth convex

function vanishing near 0. Let f : c0 (“) ’ R be given by f (x) := γ∈“ h(xγ ).

Locally on c0 (“) the function f is just a ¬nite sum, hence f is smooth. In fact let

13.16

13.17 13. Di¬erentiability of seminorms 143

h(t) = 0 for |t| ¤ δ. For x ∈ c0 (“) the set F := {γ : |xγ | ≥ δ/2} is ¬nite, and for

y ’ x < δ we have that f (y) = γ∈F h(yγ ).

The set U := {x : f (x) < 1} is open, and bounded: Let h(t) ≥ 1 for |t| ≥ ∆ and

f (x) < 1, then h(xγ ) < 1 and thus |xγ | ¤ ∆ for all γ. The set U is also absolutely

convex: Since h is convex, so is f and hence U . Since h is symmetric, so is f and

hence U .

The boundary ‚U = f ’1 (1) is a splitting submanifold of c0 (“) by the implicit

function theorem on Banach spaces, since df (x)x = 0 for x ∈ ‚U . In fact df (x)(x) =

γ h (xγ )xγ ≥ 0 and at least for one γ we have h(xγ ) > 0 and thus h (xγ ) = 0.

So by (13.14) the Minkowski functional pU is smooth o¬ 0. Obviously, it is an

equivalent norm.

13.17. Proposition. Inheritance properties for di¬erentiable norms.

(1) The product of two spaces with C n -norm has again a C n -norm given by

x1 2 + x2 2 . More generally, the 2 -sum of C n -normable

(x1 , x2 ) :=

Banach spaces is C n -normable.

(2) A subspace of a space with a C n -norm has a C n -norm.

(3) [Godefroy, Pelant, et. al., 1988]. If c0 (“) ’ E ’ F is a short exact sequence

of Banach spaces, and F has a C k -norm, then E has a C k -norm. See also

(14.12.1) and (16.19).

(4) For a compact space K let K be the set of all accumulation points of K.

The operation K ’ K has the following properties:

A⊆B’A ⊆B

(a)

(A ∪ B) = A ∪ B

(b)

(A — B) = (A — B) ∪ (A — B )

(c)

1

{0} ∪ { n : n ∈ N} = {0}

(d)

K = … ” K discrete.

(e)

(5) If K is compact and K (ω) = … then C(K) has an equivalent C ∞ -norm, see

also (16.20).

Proof. (1) and (2) are obvious.

(4) (a) is obvious, since if {x} is open in B and x ∈ A, then it is also open in A

in the trace topology, hence A © (B \ B ) ⊆ A \ A and hence A = A \ (A \ A ) ⊆

(A \ A © (B \ B )) = A © B ⊆ B .

(b) By monotonicity we have ˜⊇™. Conversely let x ∈ A ∪ B , w.l.o.g. x ∈ A ,

suppose x ∈ (A ∪ B) , then {x} is open in A ∪ B and hence {x} = {x} © A would

/

be open in A, i.e. x ∈ A , a contradiction.

/

(c) is obvious, since {(x, y)} is open in A — B ” {x} is open in A and {y} is open

in B.

(d) and (e) are trivial.

13.17

144 Chapter III. Partitions of unity 13.17

For (3) a construction is used similar to that of Kuiper™s smooth norm for c0 . Let

π : E ’ F be the quotient mapping and the quotient norm on F . The dual

sequence 1 (A) ← E — ← F — splits (just de¬ne T : 1 (A) ’ E — by selection of

x— := T (ea ) ∈ E — with x— = 1 and x— |c0 (A) = eva using Hahn Banach). Note

a a a

that for every x ∈ E and µ > 0 the set {± : |x— (x)| ≥ π(x) + µ} is ¬nite. In fact,

±

by de¬nition of the quotient norm π(x) := sup{ x + y : y ∈ c0 (“)} there is a

y ∈ c0 (“) such that x + y ¤ π(x) + µ/2. The set “0 := {± : |y± | ≥ µ/2} is

¬nite. For all other ± we have

|x— (x)| ¤ |x— (x + y)| + |x— (y)| ¤ x— x + y + |y± | <

± ± ± ±

< 1 ( π(x) + µ/2) + µ/2 = π(x) + µ.

Furthermore, we have

x ¤ 2 π(x) + sup{|x— (x)| : ±}.

±

In fact,

x = sup{| x— , x | : x— ¤ 1}

¤ sup{| T (») + y — —¦ π, x | : » ¤ 1, y — ¤ 2}

1

= sup{|x— (x)| : ±} + 2 π(x) ,

±

since x— = T (») + x— ’ T (»), where » := x— |c0 (“) and hence » 1 ¤ x— ¤ 1,

and |T (»)(x)| ¤ » 1 sup{|x— (x)| : ±} ¤ x hence T (») ¤ » 1 , and y — —¦ π =

±

—

x ’ T (»). Let denote a norm on F which is smooth and is larger than the

quotient norm. Analogously to (13.16) we de¬ne

h(x— (x)),

f (x) := h(4 π(x) ) a

a∈A

where h : R ’ [0, 1] is smooth, even, 1 for |t| ¤ 1, 0 for |t| ≥ 2 and concave

on {t : h(t) ≥ 1/2}. Then f is smooth, since if π(x) > 1/2 then the ¬rst factor

vanishes locally, and if π(x) < 1 we have that “0 := {± : |x— (x)| ≥ 1 ’ µ}

±

is ¬nite, where µ := (1 ’ π(x) )/2, for y ’ x < µ also |x± (y) ’ x— (x)| < µ

—

±

—

and hence |x± (y)| < 1 ’ µ + µ = 1 for all ± ∈ “0 . So the product is locally

/

¬nite. The set {x : f (x) > 1 } is open, bounded and absolutely convex and has

2

a smooth boundary {x : f (x) = 1 }. It is symmetric since f is symmetric. It is

2

bounded, since f (x) > 1/2 implies h(4 π(x) ) ≥ 1/2 and h(x— (x)) ≥ 1/2 for all

a

—

a. Thus 4 π(x) ¤ 2 and |xa (x)| ¤ 2 and thus x ¤ 2 · 1/2 + 2 = 3. For the

convexity note that xi ≥ 0, yi ≥ 0, 0 ¤ t ¤ 1, i xi ≥ 1/2, i yi ≥ 1/2 imply

i (txi +(1’t)yi ) ≥ 1/2, since log is concave. Since all factors of f have to be ≥ 1/2

and h is concave on this set, convexity follows. Since one factor of f (x) = ± f± (x)

has to be unequal to 1, the derivative f (x)(x) < 0, since f± (x)(x) ¤ 0 for all ± by

concavity and f± (x)(x) < 0 for all x with f± (x) < 1. So its Minkowski-functional

is an equivalent smooth norm on E.

13.17

13.18 13. Di¬erentiability of seminorms 145

Statement (5) follows from (3). First recall that K is the set of accumulation

points of K, i.e. those points x for which every neighborhood meets K \ {x}, i.e. x

is not open. Thus K \ K is discrete. For successor ordinals ± = β + 1 one de¬nes

K (±) := (K (β) ) and for limit ordinals ± as β<± K (β) . For a compact space K the

equality K (ω) = … implies K (n) = … for some n ∈ ω, since K (n) is closed. Now one

shows this by induction. Let E := {f ∈ C(K) : f |K = 0}. By the Tietze-Urysohn

theorem one has a short exact sequence c0 (K \ K ) ∼ E ’ C(K) ’ C(K ). The

=

equality E = c0 (K \ K0 ) can be seen as follows:

Let f ∈ C(K) with f |K = 0. Suppose there is some µ > 0 such that {x : |f (x)| ≥ µ}

is not ¬nite. Then there is some accumulation point x∞ of this set and hence

|f (x∞ )| ≥ µ but x∞ ∈ K and so f (x∞ ) = 0. Conversely let f ∈ c0 (K \ K ) and

˜ ˜ ˜ ˜

de¬ne f by f |K := 0 and f |K\K = f . Then f is continuous on K \ K , since

˜

K \ K is discrete. For x ∈ K we have that f (x) = 0 and for each µ > 0 the set

˜ ˜

{y : |f (y)| ≥ µ} is ¬nite, hence its complement is a neighborhood of x, and f is

continuous at x. So the result follows by induction.

13.18. Results.

(1) We do not know whether the quotient of a C n -normable space is again C n -

normable. Compare however with [Fitzpatrick, 1980].

(2) The statement (13.17.5) is quite sharp, since by [Haydon, 1990] there is

a compact space K with K (ω) = {∞} but without a Gˆteaux-di¬erentiable

a

norm.

(3) [Talagrand, 1986] proved that for every ordinal number γ, the compact and

scattered space [0, γ] with the order topology is C 1 -normable.

(4) It was shown by [Toru´czyk, 1981] that two Banach spaces are homeomor-

n

phic if and only if their density number is the same. Hence, one can view Ba-

nach spaces as exotic (di¬erentiable or linear) structures on Hilbert spaces.

If two Banach spaces are even C 1 -di¬eomorphic then the di¬erential (at 0)

gives a continuous linear homeomorphism. It was for some time unknown

if also uniformly homeomorphic (or at least Lipschitz homeomorphic) Ba-

nach spaces are already linearly homeomorphic. By [En¬‚o, 1970] a Banach

space which is uniformly homeomorphic to a Hilbert space is linearly home-

omorphic to it. A counter-example to the general statement was given by

[Aharoni, Lindenstrauss, 1978], and another one is due to [Ciesielski, Pol,

1984]: There exists a short exact sequence c0 (“1 ) ’ C(K) ’ c0 (“2 ) where

C(K) cannot be continuously injected into some c0 (“) but is Lipschitz equiv-

alent to c0 (“). For these and similar questions see [Tzafriri, 1980].

(5) A space all of whose closed subspaces are complemented is a Hilbert space,

[Lindenstrauss, Tzafriri, 1971].

(6) [En¬‚o, Lindenstrauss, Pisier, 1975] There exists a Banach space E not iso-

morphic to a Hilbert space and a short exact sequence 2 ’ E ’ 2 .

(7) [Bonic, Reis, 1966]. If the norm of a Banach space and its dual norm are

C 2 then the space is a Hilbert space.

(8) [Deville, Godefroy, Zizler, 1990]. This yields also an example that existence

13.18

146 Chapter III. Partitions of unity 13.19

of smooth norms is not a three-space property, cf. (14.12).

Notes. (2) Note that K \ K is discrete, open and dense in K. So we get for

(n) (n+1)

every n ∈ N by induction a space Kn with Kn = … and Kn = …. In fact

(A — B)(n) = i+j=n A(i) — B (j) . Next consider the 1-point compacti¬cation K∞

of the locally compact space n∈N Kn . Then K∞ = {∞} ∪ n∈N Kn . In fact

every neighborhood of {∞} contains all but ¬nitely many of the Kn , thus we

(n) (i)

have ⊇. The obvious relation is clear. Hence K∞ = {∞} ∪ i≥n Kn . And

(ω) (n)

K∞ = n<ω K∞ = {∞} = …. The space of [Haydon, 1990] is the one-point

±

compacti¬cation of a locally compact space L given as follows: L := ±<ω1 ω1 , i.e.

the space of functions ω1 ’ ω1 , which are de¬ned on some countable ordinal. It is

ordered by restriction, i.e. s t :” dom s ⊆ dom t and t|dom s = s.

(3) The order topology on X := [0, γ] has the sets {x : x < a} and {x : x > a}

as basis. In particular open intervals (a, b) := {x : a < x < b} are open. It is

compact, since every subset has a greatest lower bound. In fact let U on X be a

covering. Consider S := {x ∈ X : [inf X, x) is covered by ¬nitely many U ∈ U}.

Let s∞ := sup S. Note that x ∈ S implies that [inf X, x] is covered by ¬nitely many

sets in U. We have that s∞ ∈ S, since there is an U ∈ U with s∞ ∈ U . Then there

is an x with s∞ ∈ (x, s∞ ] ⊆ U , hence [inf X, x] is covered by ¬nitely many sets in

U since there is an s ∈ S with x < s, so [inf X, s∞ ] = [inf X, x] ∪ (x, s∞ ] is covered

by ¬nitely many sets, i.e. s∞ ∈ S.

The space X is scattered, i.e. X (±) = … for some ordinal ±. For this we have to show

that every closed non-empty subset K ⊆ X has open points. For every subset K

of X there is a minimum min K ∈ K, hence [inf X, min K + 1) © K = {min K} is

open in K.

For γ equal to the ¬rst in¬nite ordinal ω we have [0, γ] = N∞ , the one-point

compacti¬cation of the discrete space N. Thus C([0, γ]) ∼ c0 — R and the result

=

follows in this case from (13.16).

(5) For splitting short exact sequences the result analogous to (13.17.3) is by

(13.17.1) obviously true. By (5) there are non-splitting exact sequences 0 ’ F ’

E ’ E/F ’ 0 for every Banach space which is not Hilbertizable.

(8) By (6) there is a sort exact sequence with hilbertizable ends, but with middle

term E not hilbertizable. So neither the sequence nor the dualized sequence splits.

If E and E would have a C 2 -norm then E would be hilbertizable by (7).

13.19. Proposition. Let E be a Banach space, x = 1. Then the following

statements are equivalent:

(1) The norm is Fr´chet di¬erentiable at x;

e

(2) The following two equivalent conditions hold:

x+h + x’h ’2 x

lim = 0,

h

h’0

x + th + x ’ th ’ 2 x

= 0 uniformly in h ¤ 1;

lim

t

t’0

13.19

13.20 13. Di¬erentiability of seminorms 147

— — — — — —

yn = 1, zn = 1, yn (x) ’ 1, zn (x) ’ 1 ’ yn ’ zn ’ 0.

(3)

Proof. (1)’(2) This is obvious, since for the derivative of the norm at x we have

limh’0 x±h ’ h ’l(±h) = 0 and adding these equations gives (2).

x

x+th ’ x

(2) ’ (1) Since (h) := limt always exists, and since

0 t

x + th + x ’ th ’ 2 x x + th ’ x x + t(’h) ’ x

= +

t t t

≥ l(h) + l(’h) ≥ 0

x±th ’ x

’ (±h) ≥ 0, so the

we have (’h) = (h), thus is linear. Moreover t

limit is uniform for h ¤ 1.

(2) ’ (3) By (2) we have that for µ > 0 there exists a δ such that x+h + x’h ¤

— —

2 + µ h for all h < δ. For yn = 1 and zn = 1 we have

— —

yn (x + h) + zn (x ’ h) ¤ x + h + x ’ h .

— —

Since yn (x) ’ 1 and zn (x) ’ 1 we get for large n that

— — — —

(yn ’ zn )(h) ¤ 2 ’ yn (x) ’ zn (x) + µ h ¤ 2µδ,

— — — —

hence yn ’ zn ¤ 2µ, i.e. zn ’ yn ’ 0.

(3) ’ (2) Otherwise, there exists an µ > 0 and 0 = hn ’ 0, such that

x + hn + x ’ hn ≥ 2 + µ hn .

— —

Now choose yn = 1 and zn = 1 with

1 1

— —

yn (x + hn ) ≥ x + hn ’ hn and zn (x ’ hn ) ≥ x ’ hn ’ hn .

n n

— — — —

Then yn (x) = yn (x + hn ) ’ yn (hn ) ’ 1 and similarly zn (x) ’ 1. Furthermore

2

— —

yn (x + hn ) + zn (x ’ hn ) ≥ 2 + (µ ’ ) hn ,

n

hence

2 2

— — — —

(yn ’ zn )(hn ) ≥ 2 + (µ ’ ) hn ’ (yn + zn )(x) ≥ (µ ’ ) hn ,

n n

— — 2

thus yn ’ zn ≥ µ ’ n , a contradiction.

13.20. Proposition. Fr´chet di¬erentiable norms via locally uniformly

e

rotund duals. [Lovaglia, 1955] If the dual norm of a Banach space E is locally

uniformly rotund on E then the norm is Fr´chet di¬erentiable on E.

e

A norm is called locally uniformly rotund if xn ’ x and x + xn ’ 2 x

implies xn ’ x. This is equivalent to 2( x 2 + xn 2 ) ’ x + xn 2 ’ 0 implies

xn ’ x, since

2

+ 2 xn 2 ) ’ x + xn 2 2 2

’ ( x + xn )2 = ( x ’ xn )2 .

≥2 x

2( x + 2 xn

— — — —

Proof. We use (13.19), so let x = 1, yn = 1, zn = 1, yn (x) ’ 1, zn (x) ’ 1.

Let x— = 1 with x— (x) = 1. Then 2 ≥ x— + yn ≥ (x— + yn )(x) ’ 2. Since

— —

— —

E is locally uniformly rotund we get yn ’ x and similarly zn ’ z, hence

— —

yn ’ zn ’ 0.

13.20

148 Chapter III. Partitions of unity 13.22

13.21. Remarks on locally uniformly rotund spaces. By [Kadec, 1959] and

[Kadec, 1961] every separable Banach space is isomorphic to a locally uniformly

rotund Banach space. By [Day, 1955] the space ∞ (“) is not isomorphic to a locally

uniformly rotund Banach space. Every Banach space admitting a continuous linear

injection into some c0 (“) is locally uniformly rotund renormable, see [Troyanski,

1971]. By (53.21) every WCG-Banach space has such an injection, which is due

to [Amir, Lindenstrauss, 1968]. By [Troyanski, 1968] every Banach space with

unconditional basis (see [Jarchow, 1981, 14.7]) is isomorphic to a locally uniformly

rotund Banach space.

In particular, it follows from these results that every re¬‚exive Banach space has an

equivalent Fr´chet di¬erentiable norm. In particular Lp has a Fr´chet di¬erentiable

e e

norm for 1 < p < ∞ and in fact the p-norm is itself Fr´chet di¬erentiable, see

e

(13.13).

13.22. Proposition. If E is separable then E admits an equivalent norm, whose

dual norm is locally uniform rotund.

Proof. Let E be separable. Then there exists a bounded linear operator T : E ’

such that T — (( 2 ) ) is dense in E (and obviously T — is weak— -continuous):

2

Take a dense subset {x— : i ∈ N} ⊆ E of {x— ∈ E : x— ¤ 1} with x— ¤ 1.

i i

2

De¬ne T : E ’ by

x— (x)

T (x)i := i i .

2

Then for the basic unit vector ei ∈ ( 2 ) we have

x— (x)

—

T (ei )(x) = ei (T (x)) = T (x)i = i i ,

2

i.e. T — (ei ) = 2’i x— .

i

Note that the canonical norm on 2 is locally uniformly rotund. We now claim that

E has a dual locally uniform rotund norm. For x— ∈ E and n ∈ N we de¬ne

1—

x— := inf{ x— ’ T — y — : y — ∈ ( 2 ) } and

2 2 2

+ y