1.1 Normed space geometry
1.1.3 Support functionals
Let x∈X and f ∈X∗, then we say that f issupport functional of xif ∥f∥= ∥x∥and
f(x) =∥x∥2.
Proposition 1.1.8. Every point in a normed space has a support functional.
Proof. Choose x0 ∈X with∥x0∥= 1. Define the linear functional φ : span{x0} →R
whereφ(ax0) =a for all a ∈R. We note that φ(x)≤ ∥x∥ for all x∈span{x0}, thus
by the Hahn-Banach theorem there exists a linear map f :X→R where f(x)≤ ∥x∥
for all x ∈ X and f(x) = φ(x) for all x ∈ span{x0}. It now follows f is a support
functional of x0.
Choose any x0 ∈ X. If x0 = 0 then the zero map is a support functional of x0.
Supposex0 ̸= 0, then x0/∥x0∥has support functionalf and we note∥x0∥f is a support
functional of x0.
As every point has at least one support functional we shall define for eachx∈X
the set ϕ[x] of support functionals of x.
We say that a non-zero point x is smooth if it has a unique support functional
(i.e. |ϕ[x]| = 1) and define smooth(X) ⊆ X \ {0} to be the set of smooth points of
X. If smooth(X)∪ {0}=X then we say that X is smooth. We define a norm to be strictly convex if ∥tx+ (1−t)y∥<1 for all distinct x, y ∈S1[0] and t∈(0,1).
The dual map of X is the map
ϕ: smooth(X)∪ {0} →X∗, x7→ϕ(x),
where ϕ(0) = 0 and ϕ(x) is the unique support functional of x ∈ smooth(X). It is
immediate that ϕis homogeneous sincef is the support functional ofx if and only if af is the support functional of ax for a̸= 0.
Proposition 1.1.9. LetX be a Euclidean normed space. Then the following holds:
(i) X is strictly convex.
(ii) X is smooth, and for each x∈X,
ϕ(x) :X →R, y 7→ ⟨x, y⟩
is the unique support functional of x.
(iii) The dual map ϕ:X →X∗ is a linear isometric isomorphism. Proof. (i): For any distinct x, y ∈S1[0] and t∈(0,1),
∥tx+ (1−t)y∥2 = t2∥x∥2+ 2t(1−t)⟨x, y⟩+ (1−t)2∥y∥2
< t2+ 2t(1−t) + (1−t)2
= 1,
by the Cauchy-Schwarz inequality, as x̸=y.
(ii): Choose x ∈ X, then by the Cauchy-Schwarz inequality, ϕ(x) supports x.
Suppose that there also exists f ∈ X∗ that supports x. The map z 7→ ⟨z,·⟩ is an
injective (thus surjective) linear map betweenX andX∗, thus there exists some y∈X
such that f =ϕ(y). By the Cauchy-Schwarz inequality,
|ϕ(y)x| ≤ ∥x∥∥y∥=∥x∥2
with equality if and only if y= xor y= −x. As ϕ(y)x=−∥x∥2 ify =−x, theny= x
and f =ϕ(x). As this holds for allx∈X\ {0}, X is smooth.
(iii): As noted in (ii), ϕ is a linear isomorphism. As ϕ(x) is the unique suport
1.1 Normed space geometry 19
The following result shows that if two normed spaces are isometrically isomorphic then they have equivalent support functions in some way.
Proposition 1.1.10. Let X and Y be normed spaces and T : X → Y a isometric
isomorphism. Then the following holds:
(i) If f ∈X∗ is a support functional ofx∈X, then f◦T−1 is a support functional
of T(x).
(ii) x is a smooth point of X if and only ifT(x) is a smooth point of Y.
(iii) X is smooth if and only ifY is smooth.
(iv) X is strictly convex if and only if Y is strictly convex.
Proof. (i): As T is an isometric isomorphism, ∥T(x)∥Y =∥x∥X, thus f◦T−1(T(x)) =
∥T(x)∥2
Y. Choose any y∈Y with∥y∥Y = 1, then as T is an isomorphism there exists
y′ ∈X with T(y′) = y, and ∥y′∥X = 1 also. We now note that
|f◦T−1(y)|=|f(y′)| ≤ ∥x∥X =∥T(x)∥Y,
thus ∥f◦T−1∥Y =∥T(x)∥Y as required.
(ii): Let f, g ∈ X∗ be distinct support functionals of x ∈ X. By (i), f ◦ T−1
and g◦T−1 are support functionals of T(x). We note that f, g can not be linearly
dependent; if f =cg for somec̸= 0,1 then f(x) =c∥x∥X ̸=∥x∥X. As f, g are linearly
independent we may choose any y ∈ kerf such that y /∈ kerg. It now follows that T(y)∈kerf ◦T−1 and T(y)∈/ kerg◦T−1, thus f◦T−1 ̸=g◦T−1. This implies that
if x is a non-smooth point of X then T(x) is a non-smooth point ofY. By symmetry
we note the converse also holds as required. (iii): This follows immediately from (ii).
(iv): Suppose X is strictly convex and choose any two points x, y ∈ Y with
∥x∥Y = ∥y∥Y = 1. As T is an isomorphism there exist x′, y′ ∈ X with T(x′) = x,
T(y′) =y; we note that ∥x′∥X =∥y′∥X = 1 also. It now follows that for anyt ∈(0,1),
∥tx+ (1−t)y∥Y =∥T(tx′+ (1−t)y′)∥Y =∥tx′ + (1−t)y′∥X <1
as X is strictly convex, thus Y is strictly convex as required.
For a d-dimensional normed space X we shall define S ⊂ X to be negligible if
for every ϵ >0 there exists a sequence (xn)n∈N in S and (rn)n∈N in (0,∞) such that
P
n∈Nrnd < ϵand
S ⊂ [
n∈N
Brn(xn).
We note that for two norms ∥ · ∥,∥ · ∥′ of X, if S is a negligible subset of (X,∥ · ∥) it
will also be a negligible subset of (X,∥ · ∥′). The countable union of negligible sets is a
negligible set, and the complement of a negligible set is a dense set (see Proposition B.2.7). If S ⊂Rd, then S is negligible if and only if it has Lebesgue measure zero (see
Theorem B.2.6).
Proposition 1.1.11. For any normed space X the following properties hold:
(i) For x0 ̸= 0, x0 ∈smooth(X) if and only if x7→ ∥x∥is differentiable at x0.
(ii) If x7→ ∥x∥ is differentiable atx0 then it has derivative ∥x10∥ϕ(x0).
(iii) The set smooth(X) is dense in X and smooth(X)c is negligible.
(iv) The map ϕis continuous.
1.1 Normed space geometry 21 Proof. (i) & (ii): By [40, Lemma 1], x 7→ ∥x∥ is differentiable at x0 if and only if
x0 ∈smooth(X) with derivative ∥x10∥ϕ(x0).
(iii): The result follows from (i), [60, Theorem 25.5] and Theorem B.2.6 asx7→ ∥x∥
is convex.
(iv): By [60, Theorem 25.5], the mapx7→ 1
∥x∥ϕ(x) is continuous on smooth(X), thus
ϕ is continuous on smooth(X) also. As ∥ϕ(x)∥=∥x∥ it follows thatϕ is continuous
at 0∈X∗ also as required.
(v): If X is Euclidean then by Proposition 1.1.9 (iii), X is smooth and ϕis linear.
Suppose ϕ is linear. If we define ⟨x, y⟩ := 1
2(ϕ(x)y+ϕ(y)x) for each x, y ∈ X then
⟨·,·⟩ is an inner product on X and ∥x∥2 =⟨x, x⟩, thus X is Euclidean.
The following are some examples of normed spaces and their corresponding dual maps.
Example 1.1.12 (Smooth and strictly convex). For q∈(1,∞), the space ℓd
q is strictly
convex, as the real-valued function a 7→ aq is strictly convex and increasing. Let
sgn : R→ {−1,0,1} be the sign function, i.e.
sgn(a) = 1, if a >0 0, if a= 0 −1, if a <0. For each x := (x(i))d
i=1 ∈ Rd and r > 0 define the vector x(r) ∈ Rd, where for each
i= 1, . . . , d,
Take x∈S1[0], then we obtain the support functional
ϕ(x) :Rd→R, y 7→ 1 ∥x∥qq−2
D
x(q−1), yE.
As we can differentiate ∥ · ∥q at any point in Rd, it follows from Proposition 1.1.11 (i)
that ℓd
q is smooth.
Example 1.1.13 (Neither smooth nor strictly convex). Fix d≥2 and choose a finite
spanning set F ⊂(Rd)∗ such that 0∈/ F and if f ∈F then −f ∈F. We may define a
centrally symmetric (d-dimensional) polytope
P :={x∈Rd:f(x)≤1 for all f ∈F}.
DefineX to be the linear space Rd with norm
∥x∥P := max
f∈F |f(x)|.
By [34, Lemma 3],∥x0∥Pf ∈F is a support functional ofx0 if and only iff(x0) =∥x0∥P.
We refer to these norm spaces aspolyhedral norm spaces.
The set of points x ∈ S1[0] where ∥x∥P = f(x) = g(x) for distinct f, g ∈ F is a
non-empty negligible set as it is exactly the intersection of a finite set of hyperplanes, thusX is not smooth but does have an open set of smooth points. As |F|<∞ there
must exist two pointsx, y ∈S1[0] that obtain their norm for the same linear functional
f ∈F, and so
t∥x∥P + (1−t)∥y∥P = tf(x) + (1−t)f(y)
= f(tx+ (1−t)y)
1.1 Normed space geometry 23
≤ t∥x∥P + (1−t)∥y∥P,
thus X is also not strictly convex.
Example 1.1.14 (Smooth but not strictly convex). LetX be the linear spaceR2 with
norm ∥(x, y)∥= |y|, if |y|>|x| x2+y2 2|x| if |x| ≥ |y|, x̸= 0 0 if x=y= 0
with unit sphere as described in Figure 1.1.
Let z = (x, y) ∈ X and choose any w ∈ X. The norm is differentiable at all
non-zero points (i.e. X is smooth), thus we have ϕ(z)w=∥z∥ ⟨f(z), w⟩, where,
f(z) = (0,1), if |y|>|x| x3−xy2 2|x|3 , y |x| if |x| ≥ |y|, x̸= 0 (0,0), if x=y= 0.
The norm is not strictly convex however, as the points{(t,1) :t∈(−1,1)}all lie in
S1[0].
Example 1.1.15 (Strictly convex but not smooth). Let ∥ · ∥a be a strictly convex and
smooth norm on Rd and∥ · ∥b be a non-smooth norm on Rd. We define X to be the
linear space Rd with the norm ∥x∥:=∥x∥a+∥x∥b.
Choose any non-zero, non-smooth pointx∈(Rd,∥ · ∥
b). If xis a smooth point of
X then by Proposition 1.1.11 (i), both ∥ · ∥and ∥ · ∥a are differentiable at x. However,
Fig. 1.1 The unit ball of the normed space described in Example 1.1.14.
follows that the smooth points of X are exactly the smooth points of (Rd,∥ · ∥b), thus
X is not smooth.
Choose any two points x, y ∈S1[0] and t ∈(0,1), then by the strict convexity of
∥ · ∥a,
∥tx+ (1−t)y∥ = ∥tx+ (1−t)y∥a+∥tx+ (1−t)y∥b
< (t∥x∥a+ (1−t)∥y∥a) + (t∥x∥b+ (1−t)∥y∥b)
= t∥x∥+ (1−t)∥y∥,
thus X is also strictly convex.
The following are some useful properties regarding support functionals.
Proposition 1.1.16. Let X be a normed space andx∈X\ {0}. Then the following
holds:
(i) ϕ[x] is a compact and convex subset of S∥∗x∥[0].
(ii) If dimX = 2 then ϕ[x] = [f, g] for some f, g ∈ X∗, and x ∈ smooth(X) if and
1.1 Normed space geometry 25 Proof. (i): For each f ∈ ϕ[x] we have ∥f∥ = ∥x∥ by definition, thus ϕ[x] ⊂ S∥∗x∥[0].
Given f, g ∈ϕ[x] and t∈[0,1] we note that (tf + (1−t)g)(x) =∥x∥2 and
∥tf + (1−t)g∥ ≤t∥f∥+ (1−t)∥g∥=∥x∥,
thustf+(1−t)g ∈ϕ[x] andϕ[x] is convex. Finally if (fn)n∈Nis a convergent sequence of
support functionals ofxwith limitf then ∥f∥= ∥x∥andf(x) = limn→∞fn(x) =∥x∥2,
thus f ∈ϕ[x]; since this implies ϕ[x] is a closed subset of the compact set S∥∗x∥[0] then
it too is compact.
(ii): If x is smooth then ϕ[x] = {ϕ(x)} = [ϕ(x), ϕ(x)]. Suppose x is not smooth,
then by (i), ϕ[x] is a compact convex subset of the 1-dimensional manifoldS∥∗x∥[0], and
hence is a line segment.
Proposition 1.1.17. Let X be a normed space and x ∈ S1[0]∩smooth(X). Then
the following holds:
(i) The set ϕ(x)−1[{1}]∩S
1[0] is closed and convex.
(ii) If dimX = 2, ϕ(x)−1[{1}]∩S1[0] = [x1, x2].
Proof. (i): Choose y, z ∈ ϕ(x)−1[{1}]∩S
1[0], then ϕ(x)(ty + (1−t)z) = 1 for all
t ∈[0,1]. We further note that
1 =|ϕ(x)(ty+ (1−t)z)| ≤ ∥ty+ (1−t)z∥ ≤1,
thus ty+ (1−t)z ∈S1[0] also andϕ(x)−1[{1}]∩S1[0] is convex. Asϕ(x) is continuous
then ϕ(x)−1[{1}]∩S
1[0] is closed also.
(ii): If dimX = 2 it follows that ϕ(x)−1[{1}] ∩ S1[0] = [x1, x2] as S1[0] is a
Example 1.1.18. LetV be a finite dimensional real vector space with dualV∗. Choose
a non-empty set F ⊂V∗ such that the following holds:
(i) If f ∈F then −f ∈F.
(ii) For each x∈V, there exists M >0 such that f(x)≤M for all f ∈F.
(iii) spanF =V∗.
We may now define the normed space X = (V,∥ · ∥) where for allx∈X,
∥x∥:= sup
f∈F
f(x).
We note that the following holds:
(i) As F is compact it follows that if ∥x∥= 1, there existsf ∈F such thatf(x) = 1.
It is immediate that f is a support functional of x.
(ii) If we define∥ · ∥′ to be the norm generated byF, ∥ · ∥′′ to be the norm generated
by conv(F) and ∥ · ∥′′′ to be the norm generated by ∂conv(F), then
∥ · ∥=∥ · ∥′ =∥ · ∥′′ =∥ · ∥′′′
(iii) By [60, Theorem 14.5], B1∗[0] = conv(F); it follows thatS1∗[0] = ∂conv(F).
(iv) For any x∈S1[0], define Fx ⊂F to be the set
Fx :={f ∈F :f(x) = 1},
1.1 Normed space geometry 27
x1
x2
x3
x4
Fig. 1.2 The unit ball of the normed space described in Example 1.1.19 with the first four elements of the sequence (xn)n∈N of non-smooth points.
Example 1.1.19. Although most normed spaces that are commonly studied have an
open set of smooths points, not all normed spaces do so. We shall now construct a normed plane with a smooth point that does not lie in smooth(X)◦.
For eachi, j ∈ {−1,1} andθ ∈[0,π2], define the linear functional s∗(i, j, θ)∈(R2)∗,
where for anyx∈R2,
s∗(i, j, θ)(x) := D(−1)icos (θ),(−1)jsin (θ)
, xE.
Define the sequence (θn)n∈N where θn := 2πn, and further define
F :={s∗(i, j, θn) :n ∈N, i, j ∈ {−1,1}}.
We note that
with s∗(i,−1,0) =s∗(i,1,0) for i∈ {−1,1}.
Following Example 1.1.18, we define the norm ∥ · ∥ for R2 with F. For any
i, j ∈ {−1,1} and n ∈N, s∗(i, j, θn)((1,0)) = (−1)icos π 2n <1 and s∗(1,1,0)((1,0)) = 1,
thus (1,0) is smooth (and similarly (−1,0) is smooth).
For each n ∈Ndefine
xn:= 2 sinθ n+1 2 sin (θn+1) cos 3θn+1 2 ! ,sin 3θn+1 2 !! ,
and the function fn: [0,π2]→R with
fn(φ) :=s∗(1,1, φ)(xn) = 2 sinθ n+1 2 cos3θ n+1 2 −φ sin (θn+1) .
We now note that
fn(θn) = fn(θn+1) = 1,
thus if ∥xn∥= 1 then s∗(1,1, θn) ands∗(1,1, θn+1) support xn. To see that ∥xn∥= 1
we first note by differentiating that f is strictly increasing on [0,3θn+1
2 ] and strictly
decreasing on [3θn+1
2 ,
π
2], thus for allm ∈N
1.1 Normed space geometry 29
As
s∗(i, j, θm)(xn)≤fn(θm)≤1,
∥xn∥= 1. The sequence (xn)n∈N is a sequence of non-smooth points converging to a
smooth point (1,0), thus smooth(X) is not open.
The following gives a useful characterisation for strictly convex normed spaces.
Proposition 1.1.20. Let X be a normed space, then X is strictly convex if and only
if no support functional supports more than one point in X.
Proof. Suppose there exists f ∈X∗ that supports two distinct points x, y ∈S1[0]. For
all t∈(0,1) we have
1 = tf(x) + (1−t)f(y) =f(tx+ (1−t)y)≤ ∥tx+ (1−t)y∥,
thus X is not strictly convex. Now suppose there exists f′ ∈ X∗ that supports two
distinct points x′, y′ ∈ X, then by definition ∥x′∥ = ∥y′∥ = ∥f′∥. Let c = ∥f1′∥ and
define f := cf′, x:=cx′ and y:=cy′. It follows that f supportsx, y ∈S1[0], thus X∗
is not strictly convex.
Now suppose [x, y]⊂S1[0] for distinct points x, y ∈X. Choose t ∈(0,1) and let
f be a support functional oftx+ (1−t)y. As ∥f∥= 1 then f(x), f(y)≤1. Suppose
f(x)<1, then
1 =f(tx+ (1−t)y) = tf(x) + (1−t)f(y)<1,
thus f(x) = 1. By symmetry, f(y) = 1, thusf supports xand y.
Remark 1.1.21. It follows from Proposition 1.1.20 that if the dual map of X is
We define for S1[0] the(inner) Löwner ellipsoid S of S1[0], the unique convex body
of maximal volume bounded byS1[0] which has a Minkowski functional∥·∥S :X →R≥0
that can be induced by an inner product. It is immediate that ∥x∥S ≥ ∥x∥ for all
x∈X and the Euclidean space (X,∥ · ∥S) has unit sphere S. For more information on
Löwner ellipsoids see [67, Chapter 3.3].
Lemma 1.1.22. Let X be ad-dimensional normed space. Then there exists smooth
points y1, . . . , yd ∈S1[0] so that ϕ(y1), . . . , ϕ(yd) are linearly independent.
Proof. By [3, Lemma 6.1] there existsy1, . . . , yd∈S1[0] that lie on the Löwner ellipsoid
S ofS1[0]. Suppose fi is a support functional foryi with respect to ∥ · ∥ and choose any
x∈S. As S ⊂B1[0] (the unit ball of (X,∥ · ∥)) then |fi(x)| ≤1, thus f is a support
functional for yi with respect to ∥ · ∥S also. As (X,∥ · ∥S) is Euclidean then it follows
that y1, . . . , yd are smooth andϕ(y1), . . . , ϕ(yd) are linearly independent.
Proposition 1.1.23. Suppose dimX =d ≥2. For all
x1, . . . , xn∈S1[0]∩smooth(X)
with n < d there exists y∈S1[0]∩smooth(X) such that
y /∈span{x1, . . . , xn} ϕ(y)∈/span{ϕ(x1), . . . , ϕ(xn)}.
Proof. By Lemma 1.1.22, there exists smooth points y1, . . . , yd ∈ S1[0] so that the
support functionals ϕ(y1), . . . , ϕ(yd) are linearly independent. Define
Z := span{ϕ(x1), . . . , ϕ(xn)},
then dimZ ≤n < d. If ϕ(y1), . . . , ϕ(yd)∈Z then dimZ =d, thus there exists yi ∈/ Z.
1.1 Normed space geometry 31