Chapter 4
Convexity and Smoothness
4.1
Strict Convexity, Smoothness, and Gateaux
Di↵erentiablity
Definition 4.1.1. Let X be a Banach space with a norm denoted by k · k.
A map
f : X \ {0} ! X ⇤ \ {0}, f 7! fx
is called a support mapping whenever.
a) f ( x) = fx , for
> 0 and
b) If x 2 SX , then kfx k = 1 and fx (x) = 1 (and thus fx (x) = kxk2 for
all x 2 X).
Often we only define fx for x 2 SX and then assume that fx = kxkfx/kxk ,
for all x 2 X \ {0}.
For x 2 X a support functional of x is an element x⇤ 2 X ⇤ , with kx⇤ k =
kxk and hx⇤ , xh= kxk2 . Thus a support map is a map f(·) : X ! X ⇤ , which
assigns to each x 2 X a support functional of x.
We say that X is smooth at x0 2 SX if there exists a unique fx 2 SX ⇤ ,
for which fx (x) = 1, and we say that X is smooth if it is smooth at each
point of SX .
The Banach space X is said to have Gateaux di↵erentiable norm at x0 2
SX , if for all y 2 SX
kx0 + hyk
h!0
h
⇢0 (x0 , y) = lim
kx0 k
exists, and we say that k · k is Gateaux di↵erentiable if it is Gateaux di↵erentiable norm at each x0 2 SX .
105
106
CHAPTER 4. CONVEXITY AND SMOOTHNESS
Example 4.1.2. For X = Lp [0, 1], 1 < p < 1 the function
f : Lp [0, 1] ! Lq [0, 1],
fx (t) = sign(x(t))
x(t)
kxkp
p/q
1
kxkp = kxkp
p
q
p
|x(t)| q
is a (and the only) support function for Lp [0, 1].
In order establish a relation between Gateaux di↵erentiability and smoothness we observe the following equalities and inequalities for any x 2 X,
y 2 SX , and h > 0:
fx (y)
fx (hy)
=
kxk
hkxk
fx (x) kxk2 + fx (hy)
=
hkxk
fx (x + hy) kxk2
=
hkxk
|fx (x + hy)| kxk2

hkxk
kfx kkx + hyk kxk2

hkxk
kx + hyk kxk
=
h
kx + hyk2 kx + hykkxk
=
hkx + hyk
kx + hyk2 |fx+hy (x)|

hkx + hyk
fx+hy (x + hy) |fx+hy (x)|
=
hkx + hyk
hfx+hy (y) + fx+hy (x) |fx+hy (x)|
=
hkx + hyk
hfx+hy (y)
fx+hy (y)

=
hkx + hyk
kx + hyk
and thus for any x 2 X ,y 2 SX , and h > 0:
(4.1)
fx (y)
|fx (x + hy)|

kxk
hkxk
kxk

kx + hyk
h
kxk

fx+hy (y)
.
kx + hyk
4.1. STRICT CONVEXITY, SMOOTHNESS, AND GATEAUX DIFFERENTIABLITY107
Theorem 4.1.3. Assume X is a Banach space and x0 2 SX The following
statements are equivalent:
a) X is smooth at x0 .
b) Every support mapping f : x 7! fx is norm to w⇤ continuous from SX
to SX ⇤ at the point x0 .
c) There exists a support mapping f(·) : x 7! fx which is norm to w⇤
continuous from SX to SX ⇤ at the point x0 .
d) The norm is Gateaux di↵erentiable at x0 .
In that case
kx0 + hyk
h!0
h
fx (y) = ⇢0 (x0 , y) = lim
kx0 k
for all y 2 SX .
Proof. ¬(b) ) ¬(a). Assume that (xn ) ⇢ SX is a net, which converges
in norm to x0 , but for which fxn does not converge in w⇤ to fx0 , where
f(·) : X ! X ⇤ is the support map. We can assume that there is a w⇤
neighborhood U of fx0 , not containing any of the fxn , and by Alaoglu’s
Theorem 2.3.2 we can assume that fxn has an accumulation point x⇤ 2 U ,
which cannot be equal to fx0 .
As
|x⇤ (x0 )
1|
⇤
fxn (xn )|
⇤
fxn (x0 )| + |fxn (x0
= |x (x0 )
 |x (x0 )
⇤
 |x (x0 )
fxn (x0 )| + kx0
xn )|
xn k !n2M,n!1 0, for some infinite M ⇢ N
it follows that x⇤ (x0 ) = 1, and since kx⇤ k  1 we must have kx⇤ k = 1. Since
x⇤ 6= fx0 , X cannot be smooth at x0 .
(b) ) (c) is clear (since by The Theorem of Hahn Banach there is always
at least one support map).
(c) ) (d) Follows from (4.1), and from applying (4.1) to y instead of y
which gives
kx
hyk kxk
=
hkxk
kx + h( y)k
hkxk
kxk
fx ( y)
= fx (y)
kxk
and
kx
hyk kxk
=
hkxk
kx + h( y)k
hkxk
kxk

fx+h( y) ( y)
fx+h( y) (y)
=
.
kx + h( y)k
kx + h( y)k
108
CHAPTER 4. CONVEXITY AND SMOOTHNESS
(d) ) (a) Let f 2 Sx⇤ be such that f (x0 ) = kx0 k = 1. Since (4.1) is true
for any support function it follows that
f (y) 
kx0 + hyk
h
kx0 k
, for all y 2 SX and h > 0,
and
kx0
hyk
h
kx0 k
kx0 + ( y)k kx0 k

h
for all y 2 SX and h < 0.
=
f ( y) = f (y)
Thus, by assumption (d), ⇢0 (x0 , y) = f (y), which proves the uniqueness of
f 2 SX ⇤ with f (x0 ) = 1.
Definition 4.1.4. A Banach space X with norm k·k is called strictly convex
whenever S(X) contains no non-trivial line segement, i.e. if for all x, y 2 SX ,
x 6= y it follows that kx + yk < 2.
Theorem 4.1.5. If X ⇤ is strictly convex then X is smooth, and if X ⇤ is
smooth the X is strictly convex.
Proof. If X is not smooth then there exists an x0 2 SX , and two functionals
x⇤ 6= y ⇤ in SX ⇤ with x⇤ (x0 ) = y ⇤ (x0 ) = 1 but this means that
kx⇤ + y ⇤ k
(x⇤ + y ⇤ )(x0 ) = 2,
which implies that X ⇤ is not strictly convex. If X is not strictly convex then
there exist x 6= y in SX so that k x + (1
)yk = 1, for all 0   1. So
let x⇤ 2 SX ⇤ such that
⇣x + y ⌘
x⇤
= 1.
2
But this implies that
⇣x + y ⌘ 1
1
1 1
1 = x⇤
= x⇤ (x) + x⇤ (y)  + = 1,
2
2
2
2 2
which implies that x⇤ (x) = x⇤ (y) = 1, which by viewing x and y to be
elements in X ⇤⇤ , implies that X ⇤ is not smooth.
Exercises
4.1. STRICT CONVEXITY, SMOOTHNESS, AND GATEAUX DIFFERENTIABLITY109
1.
Show that `1 admits an equivalent norm ||| · ||| which is strictly convex
and (`1 , ||| · |||) is (isometrically) the dual of c0 with some equivalent
norm.
2.
Assume that T : X ! Y is a linear, bounded, and injective operator
between two Banach spaces and assume that Y is strictly convex. Show
that X admits an equivalent norm for which X is strictly convex.
110
4.2
CHAPTER 4. CONVEXITY AND SMOOTHNESS
Uniform Convexity and Uniform Smoothness
Definition 4.2.1. Let X be a Banach space with norm k · k. We say that
the norm of X is Fréchet di↵erentiable at x0 2 SX if
kx0 + hyk
h!0
h
lim
kx0 k
exists uniformly in y 2 SX .
We say that the norm of X is Fréchet di↵erentiable if the norm of X is
Fréchet di↵erentiable at each x0 2 SX .
Remark. By Theorem 4.1.3 it follows from the Frechét di↵erentiability of
⇤ and
the norm at x0 that there a unique support functional fx0 2 SX
lim
h!0
kx0 + hyk
h
kx0 k
fx0 (y) = 0,
uniformly in y and thus that (put z = hy)
lim
z!0
kx0 + zk
kx0 k
kzk
fx0 (z)
= 0.
In particular, if X has a Fréchet di↵erentiable norm it follows from Theorem 4.1.3 that there is a unique support map x ! fx .
Proposition 4.2.2. Let X be a Banach space with norm k · k. Then the
norm is Fréchet di↵erentiable if and only if the support map is norm-norm
continuous.
Proof. (We assume that K = R) “)” Assume that (xn ) ⇢ SX converges to
x0 and put x⇤n = fxn , n 2 N, and x⇤0 = fx0 . It follows from Theorem 4.1.3
that x⇤n (x0 ) ! 1, for n ! 1. Assume that our claim were not true, and we
can assume that for some " > 0 we have kx⇤n x⇤0 k > 2", and therefore we
can choose vectors zn 2 SX , for each n 2 N so that (x⇤n x⇤0 )(zn ) > 2". But
then
⇣1
⌘
x⇤0 (x0 ) x⇤n (x0 )  x⇤0 (x0 ) x⇤n (x0 )
x⇤n (zn ) x⇤0 (zn )
1
{z
}
" |
>2"
1
= x⇤n (x0 ) x⇤0 (x0 ) + x⇤0 (zn ) x⇤n (zn ) x⇤n (x0 ) x⇤0 (x0 )
"
⇣
⌘
1
= x⇤n x⇤0 x0 + zn x⇤0 (x0 ) x⇤n (x0 )
"
4.2. UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS
Thus if we put
111
⇣
⌘
1
 x⇤n x0 + zn x⇤0 (x0 ) x⇤n (x0 )
"
⇣
⌘
1
⇤
x0 x0 + zn x⇤0 (x0 ) x⇤n (x0 )
"
1 ⇤
 x0 + zn x0 (x0 ) x⇤n (x0 )
"
⇣ 1
⌘
kx0 k x⇤0 zn x⇤0 (x0 ) x⇤n (x0 ) .
"
1 ⇤
x (x0 ) x⇤n (x0 ) ,
" 0
it follows that kyn k ! 0, if n ! 1, and, using the Fréchet di↵erentiability
of the norm that (note that x⇤0 (x0 ) x⇤n (x0 ) /kyn k = ") we deduce that
y n = zn
0<"=
x⇤0 (x0 ) x⇤n (x0 )
kx0 + yn k kx0 k

kyn k
kyn k
x⇤0 (yn )
!n!1 0,
which is a contradiction.
“(” From (4.1) it follows that for x, y 2 SX , and h 2 R
kx + hyk
h
kxk
fx (y)

fx+hy (y)
kx + hyk
 fx+hy (y)
 kfx+hy
fx (y)
fx (y) +
fx k +
fx+hy (y)
kx + hyk
1
1 + |h|
fx+hy (y)
1 kfx+hy k,
which converges uniformly in y to 0 and proves our claim.
Definition 4.2.3. Let X be a Banach space with norm k · k.
We say that the norm is uniformly Fréchet di↵erentiable on SX if
lim
h!0
kx + hyk
h
kxk
fx (y) ,
uniformly in x 2 SX and y 2 SX . In other words if for all " > 0 there is a
> 0 so that for all x, y 2 SX and all h 2 R, 0 < |h| <
kx + hyk
h
kxk
fx (y) < ".
112
CHAPTER 4. CONVEXITY AND SMOOTHNESS
X is uniformly convex if for all " > 0 there is a > 0 so that for all
x, y 2 SX with kx yk " it follows that k(x + y)/2k < 1
.
We call
(
)
kx + yk
: x, y 2 SX , kx yk " , for " 2 [0, 2]
X (") = inf 1
2
the modulus of uniform convexity of X.
X is called uniform smooth if for all " > 0 there exists a
for all x, y 2 SX and all h 2 (0, ]
kx + hyk + kx
> 0 so that
hyk < 2 + "h.
The modulus of uniform smoothness of X is the map ⇢ : [0, 1) ! [0, 1)
(
)
kx + zk kx zk
⇢X (⌧ ) = sup
+
1 : x, z 2 X, kxk = 1, kzk  ⌧ .
2
2
Remark. X is uniformly convex if and only if X (") > 0 for all " > 0. X
is uniformly smooth if and only if lim⌧ !0 ⇢X (⌧ )/⌧ = 0.
Theorem 4.2.4. For a Banach space X the following statements are equivalent.
a) There exists a support map x ! fx which uniformly continuous on SX
with respect to the norms.
b) The norm on X is uniformly Fréchet di↵erentiable on SX .
c) X is uniformly smooth.
d) X ⇤ is uniformly convex.
e) Every support map x ! fx is uniformly continuous on SX with respect
to the norms.
Proof. “(a))(b)” We proceed as in the proof of Proposition 4.2.2. From
(4.1) it follows that for x, y 2 SX , and h 2 R
kx + hyk
h
kxk
fx (y)

fx+hy (y)
kx + hyk
fx (y)
4.2. UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS
 fx+hy (y)
 kfx+hy
fx (y) +
fx k +
fx+hy (y)
kx + hyk
1
1 + |h|
kxk
fx+hy (y)
1 kfx+hy k
which converges by (a) uniformly in x and y, to 0.
“(b))(c)”. Assuming (b) we can choose for " > 0 a
h 2 (0, ) and all x, y 2 SX
kx + hyk
h
113
> 0 so that for all
fx (y) < "/2.
But this implies that for all h 2 (0, ) and all x, y 2 SX we have
kx + hyk + kx
hyk
kx + hyk
=2+h
h
kxk
 2 + "h,
⇣ kx + h( y)k
fx (y) +
h
kxk
fx ( y)
⌘
!
which implies our claim.
“(c))(d)”. Let " > 0. By (c) we can find > 0 such that for all x 2 SX
and z 2 X, with kzk  , we have kx + zk + kx zk  2 + "kzk/4.
Let x⇤ , y ⇤ 2 SX ⇤ with kx⇤ y ⇤ k
". There is a z 2 X, kzk  /2 so
that (x⇤ y ⇤ )(z) " /2. This implies
kx⇤ + y ⇤ k = sup (x⇤ + y ⇤ )(x)
x2SX
= sup x⇤ (x + z) + y ⇤ (x
z)
(x⇤
y ⇤ )(z)
x2SX
 sup kx + zk + kx
zk
" /2
x2SX
 2 + "kzk/4
" /2 < 2
" /4.
“(d))(e)”. Let x 7! fx be a support functional. By (d) we can choose
for " > a so that for all x⇤ , y ⇤ 2 SX ⇤ we have kx⇤ y ⇤ k < " , whenever
kx⇤ + y ⇤ k > 2
.
Assume now that x, y 2 SX with kx yk < . Then
kfx + fy k
1
(fx + fy )(x + y)
2
114
CHAPTER 4. CONVEXITY AND SMOOTHNESS
1
= fx (x) + fy (y) + fx (y
2
2 kx yk 2
,
which implies that kfx
“(e))(a)”. Clear.
1
x) + fy (x
2
y)
fy k < ", which proves our claim.
Theorem 4.2.5. Every uniformly convex and every uniformly smooth Banach space is reflexive.
Proof. Assume that X is uniformly convex, and let x⇤⇤ 2 SX ⇤⇤ . Since
BX is w⇤ -dense in BX ⇤⇤ we can find a net (xi )i2I which converges with
respect to w⇤ to x⇤⇤ . Since for every ⌘ > 0 there is a x⇤ 2 SX ⇤ with
limi2I x⇤ (xi ) = x⇤⇤ (x⇤ ) > 1 ⌘, it follows that limi2I kxi k = 1 and we can
therefore assume that kxi k = 1, i 2 I. We claim that (xi ) is a Cauchy net
with respect to the norm to x⇤⇤ , which would finish our proof.
So let " > 0 and choose so that kx+yk > 2
implies that kx yk < ",
for any x, y 2 SX . Then choose x⇤ 2 SX ⇤ , so that x⇤⇤ (x⇤ ) > 1
/4, and
finally let i0 2 I so that x⇤ ⇤ (xi ) 1
/2, for all i i0 . It follows that
kxi + xj k
x⇤ (xi + xj )
2
whenever i, j
i0 ,
and thus kxi xj k < ", which verifies our claim.
If X is uniformly smooth it follows from Theorem 4.2.4 that X ⇤ is uniformly convex. The first part yields that X ⇤ is reflexive, which implies that
X is reflexive.
Exercises
1. Show that for there is a constant c > 0 so that for all " > 0,
`2 (vp)
(Here
`2
c"2 .
is the modulus of uniform convexity of `2 ).
2. Prove that for every " > 0, C > 1 and any n 2 N there is an N =
(n, ", C) so that the following holds:
If X is an N dimensional space which is C-isomorphic to `N
1 then X
has an n-dimensional subspace Y which is (1 + ") ismorphic to `n1 .
4.2. UNIFORM CONVEXITY AND UNIFORM SMOOTHNESS
115
2
Hint: prove first the following: Assume that ||| · ||| is a norm on `n1 so
that
1
2
kxk1  |||x|||  kxk < for all x 2 `n1 ,
C
then there is a ||| · |||-normalized block sequence (x1 , x2 , . . . xn ) so that :
n
1 X
p
|bi | 
C i=1
3. Show that
convex.
1
n=1
`n1
`2
n
X
i=1
bi x i 
n
X
i=1
|bi |.
does not admit a norm which is uniformly
116
CHAPTER 4. CONVEXITY AND SMOOTHNESS