Isometries between finite-dimensional normed vector spaces James Fennell Supervised by Dr. Stephen Wills

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Isometries between finite-dimensional
normed vector spaces
James Fennell
Supervised by Dr. Stephen Wills
Abstract
An isometry between two n-dimensional normed vector spaces is a
map which preserves both algebraic and metric properties. If an isometry exists the two spaces are termed isometric and are essentially the
same. In this project we first investigate the isometries between the
well known `pn spaces. Motivated by a desire to be more geometrical,
we then develop a rigorous notion of generalized which is an isometric invariant of two dimensional spaces and hence a classification
tool.
Contents
1
Introduction
Isometries and isometrically isomorphic spaces . . . . . . . . . . .
2
Isometries of the p-norm on Rn
Dual space arguments . . . . . . . . . . . . . . . . .
A relationship between operators on dual p-spaces.
Isometries when p and q are not Hölder conjugate .
Existence cases when p and q are conjugate . . . . .
The p-norm classification theorem as a special case
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7
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An isometric invariant: generalized Generalized . . . . . . . . . . . . . .
Curves in normed vector spaces . . . .
Bounds on X . . . . . . . . . . . . . .
X preserved under “dualizing” . . .
Integral representation of arc length .
X in the p-norms . . . . . . . . . . . .
Beyond two dimensions . . . . . . . .
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References
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1
Introduction
1
Introduction
In the theory of normed vector spaces there are a number of types of isomorphism. Even without norms, we have the concept of linear isomorphism:
two vector spaces over the same scalar field are linearly isomorphic if there
exists an invertible linear map (an isomorphism) mapping one to the other.
When we introduce a norm we also introduce a topology, and can talk about
topological equivalence. We say that two norms k k1 and k k2 on a vector space X are topologically equivalent if there exist strictly positive real
numbers ˛ and ˇ such that, for all x 2 X ,
˛kxk1 kxk2 ˇkxk1 :
(1.1)
This condition is equivalent to the two norms giving rise to the same open
sets and hence the same topology.
A fundamental result in functional analysis is that in the finite dimensional case the choice of norm is in a particular sense arbitrary.
Theorem 1.1. Let .X; k kX / and .Y; k kY / be two n-dimensional normed vector
spaces over R. Then X and Y are linearly homeomorphic: there exists a linear isomorphism T W X ! Y such that the norm x 7! kT xkY is topologically equivalent
to k kX .
Proof. The fact that X and Y are linearly isomorphic to each other and to
Rn are basic facts from linear algebra which we assume. From the latter,
it is sufficient to prove the topological equivalence result for the case X D
Y D Rn . We will show that the norm k kX is equivalent to a specific norm
k k1 which is easier to work with, and the general result will follow by a
transitivity argument.
Let fek gnkD1 be the standard basis of Rn . We define for every x 2 Rn ,
kxk1
n
X
D
xk ek kD1
D maxfjxi jg:
i
1
This expression is easily seen to satisfy all of the norm axioms. We can also
see that this norm is equivalent to the usual Euclidean norm:
kxk1 n
X
kD1
!1=2
2
jxk j
n X
kD1
2 !1=2
n1=2 kxk1 :
max jxi j
i
In particular, any set that is compact with respect to the usual Euclidean
norm is compact with respect to k k1 .
1 Introduction
2
For general k kX we have the basic estimate
n
n
X
X
jxk jkek kX
kxkX D x k ek kD1
n
X
X
kD1
kxk1 maxfkei kX g D n maxfkei kX g kxk1 :
kD1
i
i
(1.2)
This leads us to define
ˇX D n maxfkei kX g > 0:
i
It follows from this estimate that the function W .Rn ; k k1 / ! R given by
.x/ D kxkX is continuous. Indeed, if x; y 2 Rn then
j.x/
.y/j D jkxkX
kykX j kx
ykX ˇX kx
yk1 :
The boundary of unit ball with respect to k k1 , @B1 .0; 1/, is closed and
bounded and hence, by the Heine-Borel theorem, compact. The continuous
function thus attains its minimum at some point x0 2 @B1 .0; 1/. As 0 62
@B1 .0; 1/ we cannot have .x0 / D kx0 kX D 0, so that
˛X D .x0 / > 0:
Then for any x 2 Rn we have
x kxk ˛X H) kxkX ˛X kxk1
1 X
(1.3)
Combining (1.2) and (1.3) gives that for any x 2 Rn ,
˛X kxk1 kxkX ˇX kxk1 :
If k kY is any other norm a calculation shows that
˛X
ˇ
kxkY kxkX X kxkY
ˇY
˛Y
and hence the norms k kX and k kY are equivalent.
This result is strong. It asserts that in the finite dimensional case statements concerning limits and continuity are independent of the particular
norm chosen. However, it does not mean that we can completely identify
the two spaces involved. The norm equivalence condition (1.1) only imposes bounds on how much the length of a vector can change, whereas to
fully identify the two spaces we should at least require that the length of
any two specific vectors that we are identifying be the same. This motivates
the definition of an isometry.
3
Isometries and isometrically isomorphic spaces
Isometries and isometrically isomorphic spaces
Definition 1.2. An isometry between two normed vector spaces .X; k kX /
and .Y; k kY / is an injective linear map S W X ! Y such that kxkX D kS xkY
for all x 2 X . If the map is also surjective it is called an isometric isomorphism.
In this case the spaces .X; k kX / and .Y; k kY / are said to be isometrically
isomorphic or isometric, and we can identify them. We write this symbolically
as .X; k kX / Š .Y; k kY /.
There are two basic observations that apply in general.
(a) Every normed vector space is isometrically isomorphic to itself with
one isomorphism given by the identity map.
(b) If .Z; k kZ / is a third normed vector space, and T W Y ! Z is an
isometry, then the compositon map T ı S is an isometry from X to Z.
This is readily seen from
kT .S x/kZ D kS xkY D kxkX :
To say more about isometries we should take account of the dimensions
of X and Y , which greatly determine the existence of isometric isomorphisms. Let n D dim.X / and m D dim.Y /. If m < n then no injective
linear maps X ! Y exist, and hence no isometries. If m D n < 1 then
every injective map is necessarily surjective, and hence every isometry is
automatically an isometric isomorphism. If m > n, or if m D n D 1,
then we can find injective linear maps that are not surjective. In many cases
(though not necessarily) we can then find isometries that are not isometric
isomorphisms.
We now restrict our attention to isometric isomorphisms, which are the
basic item of study of this thesis. For convenience we will denote the set
of isometric isomorphisms from X to Y by I.XI Y /, and will employ the
shorthand I.X / D I.X I X / for the set of all isometric isomorphisms from
X to itself. We can now state a few more observations.
(c) If S 2 I.XI Y / then S 1 exists as S is bijective, and S
This is easily seen from
kS
1
ykX D kS.S
1
1
2 I.Y I X /.
y/kY D kykY :
(d) It follows from (a), (b) and (c) that I.X / forms a group under the associative law of compositon. This fact will not be used much in this
thesis, but we will describe later on how some authors use it to great
effect in classifying isometries.
(e) It also follows from (a), (b) and (c) that the property of being isometrically isomorphic is an equivalence relation: it is reflexive from (a),
symmetric from (b), and transitive from (c).
2 Isometries of the p-norm on Rn
4
Proposition 1.3. If X ¤ Y then either I.XI Y / D ∅ or for some fixed T 2
I.XI Y /,
I.X I Y / D fS T W S 2 I.Y /g :
Proof. If R 2 I.X I Y / then R D RT
1T
D S T for S D RT
1
2 I.Y /.
From now on we will be concerned exclusively with isometric isomorphisms between two finite-dimensional vector spaces of the same dimension. We have seen that in this case all isometries are isometric isomorphisms, and so we will use the term “isometry” exclusively in referring to
these maps, for the sake of brevity.
With the preliminary ideas discussed we now turn towards the two main
components of this thesis. In the first we take the well known p-norms
and attempt to classify the isometries that exist between them. We will see
that the classification theorem, though rather clear on geometric grounds,
is somewhat awkward to prove directly. Motivated by this, the second part
introduces an invariant under isometric isomorphism, generalized , which
serves as a convenient tool for proving that two given norms over R2 are not
isometrically isomorphic. We conclude with some discussion of generalizing the notion to Rn .
2
Isometries of the p-norm on Rn
The most familiar example of a normed vector space is Rn equipped with
one of the “p-norms” defined for 1 p < 1 by
n
X
kxkp D xk ek D
kD1
n
X
!1=p
p
jxk j
(2.1)
kD1
p
and for p D 1 by
kxk1 D
max
k2f1;:::;ng
fjxk jg :
(2.2)
The fact that these expressions define norms is a standard exercise in functional analysis which we omit. (See, for instance, [2, p. 140].) We will,
however, mention the main ingredient in the standard proof, the HölderMinkowski inequality, as we require it later anyway.
Definition 2.1. We say that p; q 2 Œ1; 1 are Hölder conjugates if
1
1
C D1
p
q
for p and q both finite, or if either
p D 1 and q D 1
or
p D 1 and q D 1:
Isometries of the p-norm on Rn
5
Theorem 2.2 (The Hölder-Minkowski inequality). If p and q are Hölder conjugates then for every x; y 2 Rn ,
n
X
jxk yk j kxkp kykq
(2.3)
kD1
with equality if and only if there exists c > 0 such that for all k
jyk j D cjxk jp
jxk j D c
1
if p 2 Œ1; 1/;
if p D 1:
(2.4)
This, again, is a standard result for which we do not provide a proof (see [3,
p. 24]).
The p-norms are special case of a more general set of norms that are
frequently studied in their own right.
Definition 2.3. A norm on Rn is symmetric if
ka1 e1 C : : : C an en k D k ˙ a1 e1 ˙ : : : ˙ an en k
(2.5)
Figure 1 illustrates the unit balls for a number of choices of p. The pnorms are all equivalent by Theorem 1.1, but this is easy to prove directly.
It is clear that, for p < 1,
kxkp
1 n
X
jxk jp kD1
n
X
p
kxkp
1 D nkxk1
kD1
and this yields
1
kxk1 kxkp n p kxk1 :
(2.6)
For any p and q finite we then get
n
1
q
1
kxkq kxkp n p kxkq :
(2.7)
Equation (2.6) explains the motivation for the p D 1 norm by showing that
for fixed x,
lim kxkp D kxk1 :
p!1
Throughout this thesis we will denote Rn as a vector space with norm
p
k kp by `n , which is standard notation.
Given the ubiquity of the p-norms and the desirability for some form
of concreteness, this section of this thesis is devoted to a full classification
p
of the isometries that exist between the `n spaces. For convenience we will
p
q
denote the set of isometries from `n to `n by In .p; q/, and also use the short
hand In .p/ WD In .p; p/. We will always assume n 2, as when n D 1
p
kxkp D jx1 j D kxkq for all p and q, so the `1 spaces are trivially isomorphic.
2 Isometries of the p-norm on Rn
6
pD1
p
D
D
4
4:
p
D
2
p
D
4
1:
p
1
Figure 1: Unit balls for the p-norm for a selection of ps. (Based on similar
figure in [4].)
Before stating the classification theorem we need to recall notations for
two common matrix groups. By O.n/ we shall mean the set of orthogonal
matrices of dimension n; that is, matrices whose rows form an orthonormal
basis for Rn . We denote the group of generalized permutation matrices by
GP .n/. This group consists of matrices A for which there exists a permutation 2 Sn (the group of permutations) and a function ! W f1; :::; ng ! f˙1g
such that Aek D !n e .k/ . Examples for n D 3 are
0
1
@0
0
1
0
0 0
0
A
@
1 0 and 1
0 1
0
1
0
0
1
0
0 A:
1
The classification theorem is as follows.
Theorem 2.4. The set of isometries In .p; q/ is
1. O.n/, if p D q D 2;
2. GP .n/, if p D q ¤ 2;
1 1
3. S ı
W S 2 GP .2/ , if p D 1, q D 1 and n D 2;
1
1
Dual space arguments
4.
1
1
Sı
1
2
7
1
W S 2 GP .2/ , if p D 1, q D 1 and n D 2;
1
5. empty, otherwise.
Let us pause for a moment to reason the components of this theorem
geometrically.
1. The norm for the case p D q D 2 is regular Euclidean distance, with
the unit balls given by circles for n D 2, spheres for n D 3, and so on.
These shapes can be rotated about the origin and reflected in any axis,
operations represented precisely by the group of orthogonal matrices.
2. There is less freedom when p D q ¤ 2. From Figure 1 it is clear that we
cannot rotate the unit ball for p D 1 by an arbitrary angle. However we
can rotate it by =2 radians. This isn’t a rotation as such, but rather a
permutation of the axes. That we can also reflect the unit ball through
the axes explains why the group is that of generalized permutations.
3,4. A special case occurs when n D 2 and p D 1 and q D 1. This case is
again geometrically clear: all we are doing is rotating the p D 1 unit
ball by =4 radians and scaling it up to arrive at the p D 1 ball. The
n D 2, p D 1, q D 1 case is merely the inverse of this operation.
5. For n > 2 the previous case breaks down. In n D 3 the unit balls are
given by an 8-sided octahedron (p D 1) and a 6-sided cube (p D 1).
For other values of p, we would guess visually that it is not possible
to linearly transform any one onto another.
These are merely geometric observations and, of course, do not amount
to a rigorous proof. This we begin now.
Dual space arguments
It is common in functional analysis to make arguments involving dual spaces
and, indeed, the main part of our proof of Theorem 2.4 rests on such an argument. As a result, we will initially be required to determine the nature of
p
the dual spaces of `n .
We first recall the definition of a dual space and its induced norm.
Definition 2.5. The dual space of X , denoted by X , is the vector space of
all linear functionals
WX !R
equipped with the norm
kk D sup fj.x/j W x 2 X; kxkX 1g :
(2.8)
2 Isometries of the p-norm on Rn
8
p
For the `n spaces, the dual spaces are especially nice.
q
p Lemma 2.6. If p is Hölder conjugate to q then `n Š `n .
q
p Proof. Define a map T W `n ! `n by
T ./ D
n
X
.ek /ek :
kD1
We must show that T is an isomorphism (linear, injective and surjective) and also isometric.
Linearity follows immediately from the linearity
p of every element of `n .
p (1) Injective. Let , 2 `n and suppose that T ./ D T . /. Then by
linearity
n
X
..ek /
T .
/D
.ek // ek D 0
kD1
By linear independence we must have .ek / D
and T is injective.
(2) Surjective. Now let
xD
n
X
.ek / for all k, so that D
xk ek 2 `qn
kD1
p
be arbitrary, and define x W `n ! R by
x .a/ D x
n
X
!
D
a k ek
kD1
n
X
ak xk :
kD1
This clearly defines a linear functional with the property
T .x / D
n
X
x .ek /ek D
kD1
n
X
xk ek D x:
kD1
p (3) Isometric. Recall that for any 2 `n we define
˚
kk D sup j.x/j W x 2 `p
n ; kxkp 1 :
(2.9)
We wish to show that kk D kT kq . We may suppose that ¤ 0, for in this
case T D 0 by linearity and both norms are 0. By the Hölder-Minowski
p
inequality (2.3) we get that, for any x 2 `n with kxkp 1,
ˇ
ˇ n
n
ˇ X
ˇX
ˇ
ˇ
xk .ek /ˇ jxk .ek /j kxkp kT ./kq kT ./kq :
j.x/j D ˇ
ˇ
ˇ
kD1
kD1
9
Dual space arguments
Taking the supremum in (2.9) we get kk kT kq .
To show that this inequality is indeed equality we must consider two
cases. First, suppose that q is finite. Define x by
xk D ˛j.ek /jq
2
.ek /
where ˛ 2 R is a positive constant chosen so that kxkp D 1. Then as
q
xk .ek / D ˛j.ek /j 0 H)
n
X
xk .ek / 0
kD1
we have
ˇ n
ˇ
n
n
ˇX
ˇ X
X
ˇ
ˇ
j.x/j D ˇ
xk .ek /ˇ D
xk .ek / D
jxk .ek /j:
ˇ
ˇ
kD1
As jxk j D ˛j.ek /jq
1
j.x/j D
kD1
kD1
the Hölder inequality becomes equality, giving
n
X
jxk .ek /j D kxkp kT ./kq D kT ./kq ;
kD1
as kxkp = 1 by construction.
For the q D 1 case let m an integer such that j.em /j j.ek /j for all k
and define x by
xk D
.ek /=j.ek /j if k D m;
0
otherwise.
Then kxkp D 1 and j.x/j D kT k1 .
In both cases the supremum is attained, so kk D kT kq and T is an
isometric isomorphism.
It is worth mentioning at this point that the proof follows not from the
Hölder-Minkowski inequality per se but also crucially relies on the fact that
for a fixed first vector x the inequality can always be made equality by an
appropriate choice of the second vector y. Let us see how the result might
break down otherwise. One can show that the p-norms are monotonic: if
1 s < r 1 then for every x 2 Rn ,
kxkr kxks
(2.10)
with equality if any only if x D ˛ei for some i (recall Figure 1). We can
then devise a generalized version of the Hölder-Minkowski inequality: if
2 Isometries of the p-norm on Rn
10
r; s 2 Œ1; 1 are such that there exists p and q Hölder conjugate to each
other with p > r and q > s then
ˇ
ˇ n
ˇ
ˇX
ˇ
ˇ
xk yk ˇ kxkp kykq
by Hölder-Minkowski
ˇ
ˇ
ˇ
kD1
kxkr kyks
by (2.10)
The problem is that if x ¤ ˛ei we cannot choose y to make this inequality
equality, and we hence cannot show that the supremum in (2.9) is attained.
In a general sense, this shows that the significance of the Hölder-Minkowski
inequality is as much to do with its cases of equality as with the general
inequality statement itself!
A relationship between operators on dual p-spaces.
We now provide the promised dual-space argument that will underpin most
of our proof of the classification theorem. This part of our work follows the
approach taken in [6], where they assume without proof the next Lemma.
p
First, a definition: let S be any linear operator between `n spaces, and let
2
3
a11 a1n
6
:: 7
A D Œaij nn D 4 :::
: 5
an1 ann
be its matrix with respect to the standard basis. In this representation
Sei D
n
X
aki ek :
kD1
We define the transpose operator S T to be that operator with matrix AT in
the standard basis.
p
q
Lemma 2.7. S W `n ! `n is an isometry if and only if S T W `sn ! `rn is an
isometry, where r is the Hölder conjugate of p and s is the Hölder conjugate of q.
p
q
Proof. Let S W `n ! `n be an isometry. Define
S W `qn ' `sn ! `p
' `rn
n
by S D ı S. By definition
˚
kS k D sup k ı S.x/k W kxkp 1 :
Writing y D S.x/ and using the fact that S 1 is also an isometry we get
˚
kS k D sup k.y/k W kS 1 ykp 1
˚
D sup k.y/k W kykq 1
D kk;
11
Isometries when p and q are not Hölder conjugate
and so S is an isometry. We will now show that S D S T when using
appropriate bases.
p p
Let fe1 ; :::; en g be the standard basis of `n . The dual basis of `n is
f1 ; :::; n g where k .ej / D ıkj . Let A D Œaij nn be the matrix of S with
respect to the standard basis and B D Œbij nn the matrix of S with respect
to the dual basis fk gnkD1 . We have
!
n
X
ŒS .j /.ei / D j ı S.ei / D j
aki ek D aij
kD1
and
ŒS .j /.ei / D
n
X
!
bj k k .ei / D
kD1
n
X
bj k k .ei / D bj i :
kD1
giving aij D bj i and hence S D S T .
The only if condition follows from the fact that S T
T
D S.
Isometries when p and q are not Hölder conjugate
We are ready to begin proving components of the classification theorem.
Lemma 2.8. Suppose that p; q 2 Œ1; 1 are not Hölder conjugate. Then
In .p; q/ GP .n/:
Proof. Take any S 2 In .p; q/. Then by Lemma 2.7, S T 2 In .s; r/ where r is
the Hölder conjugate of p and s is the Hölder conjugate of q. By hypothesis
p and q are not Hölder conjugate. As r is Hölder conjugate to p we must
have r ¤ q.
Take A to be the matrix of S with respect to the standard basis, and B
that of S T . Then A D B T .
As we will be employing the explicit expressions for the p-norms we
need to consider a number of cases. First, suppose q ¤ 1 and p ¤ 1, so
that r ¤ 1. As S is an isometry,
1 D kei kp D kSei kq D
n
X
jaki jq D 1
(2.11)
kD1
for all i . Similarly for B,
1 D kei ks D kS T ei kr D
n
X
jbki jr D
kD1
n
X
jai k jr D 1:
(2.12)
kD1
Summing the two preceding equations over i yields
n X
n
X
i D1 kD1
jaki jq D n D
n X
n
X
i D1 kD1
jai k jr :
(2.13)
2 Isometries of the p-norm on Rn
12
Now if r < q then jai k jr jai k jq with equality if and only if ai k D 0
or ai k D ˙1, with a similar conclusion if q < r. Then in either double
sum of (2.13) we must have precisely n terms with absolute value 1 and all
remaining terms equal to 0. By (2.11) we get that every row of A contains
precisely one non-zero term. Similarly, (2.12) says that every column of A
contains precisely one non-zero term. The matrix A is thus a generalized
permutation matrix.
Next, suppose q D 1. As r ¤ q, r is finite. The analogue of (2.11) states
that for all k
maxfjaki jg D 1 H) aj i D ˙1
(2.14)
k
for some j . There are n choices of k and so n elements such that jaij j D 1.
Equation (2.12) and the right-hand side of equation (2.13) still apply as r is
finite. The latter shows that all elements of A are 0 besides the n elements
with absolute value 1. Equation (2.12) shows that every column contains
exactly one non-zero element while (2.14) shows that every row contains
exactly one non-zero element. A is again a generalized permutation matrix.
The final case is p D 1. This gives r D 1 and hence q ¤ r finite.
The proof of this case is identical to the last with the roles of r and q interchanged.
Corollary 2.9. In .p/ D GP .n/ when p ¤ 2.
Proof. If p ¤ 2 then p is not Hölder conjugate to itself, and Lemma 2.8
applies. To conclude we have to show that GP .n/ In .p/.
Take any S 2 GP .n/. There exists a permutation and numbers ˛k D
˙1 such that Sek D ˛k e .k/ for all k. Then for any x,
kxkpp
D
n
X
kD1
p
jxk j D
n
X
j˛k x .k/ jp D kS xkpp
kD1
so that S is an isometry of the p-norm. For p D 1,
kxk1 D maxfjxk jg D maxfj˛k x .k/ jg D kS xk1 :
k
k
Corollary 2.10. If p and q are distinct and not Hölder conjugate then In .p; q/ D
∅ when n 2.
Proof. We already know by Lemma 2.8 that if S 2 In .p; q/, then S 2 GP .n/.
Now ke1 C e2 kp D 21=p but
kS.e1 C e2 /kq D k˛1 e .1/ C ˛2 e .2/ kq D 21=q
as j˛1 j D j˛2 j D 1. Hence S is not an isometry if p ¤ q.
13
Existence cases when p and q are conjugate
Existence cases when p and q are conjugate
Theorem 2.4 identifies three special cases (2, 3 and 4) where isometries exist
when p and q are Hölder conjugate, and hence not covered by Lemma 2.7.
Proposition 2.11. In .2/ D O.n/.
Proof. Suppose S 2 O.n/ and that A is its associated matrix in the standard
basis. Denote the i th column of A by ai D Sei , so that .ai /j D aj i . As A
is an orthogonal matrix, hai ; aj i D ıij , where h; i denotes the usual inner
product on Rn . Then
* n
+
n
X
X
2
kS.x/k2 D hS x; S xi D
xk ak ;
xj aj
j D1
kD1
D
n
X
n
X
xk xj ıkj D
kD1 j D1
n
X
jxk j2 D kxk22
kD1
so that S is an isometry.
Conversely, suppose that S 2 In .2/. We have 1 D kSei k22 D hSei ; Sei i.
Then, if i ¤ j ,
˝
˛
2 Sei ; Sej D kSei C Sej k22 kSei k22 kSej k22
D kS.ei C ej /k22
D kei C ej k22
kSei k22
kei k22
kSej k22
kej k22 D 0:
˝
˛
So Sei ; Sej D ıij , and S is orthogonal.
Proposition 2.12.
1 1
I2 .1; 1/ D S ı
W S 2 GP .2/
1
1
1
1 1
I2 .1; 1/ D
Sı
W S 2 GP .2/
1
1
2
(2.15)
(2.16)
Proof. It is only necessary to prove (2.15) as (2.16) follows from inverting.
Let
1 1
AD
;
1
1
and let x D x1 e1 C x2 e2 2 `12 be arbitrary. We have kxk1 D jx1 j C jx2 j. Then
Ax D .x1 C x2 /e1 C .x1
H) kAxk1 D maxfjx1 C x2 j; jx1
x2 /e2
x2 jg jx1 j C jx2 j D kxk1 :
If x1 and x2 have the same sign then jx1 C x2 j D jx1 j C jx2 j; otherwise,
jx1 x2 j D jx1 j C jx2 j. In either case kAxk1 D kxk1 and A 2 I2 .1; 1/.
2 Isometries of the p-norm on Rn
14
By Proposition 1.3 we have
1
I2 .1; 1/ D S ı
1
1
W S 2 I2 .1/ :
1
But by Proposition 2.9, I2 .1/ D GP .2/.
Recalling that X is self-dual if it is isometrically isomorphic to its dual
space X , the above incidentally proves the existence of a finite-dimensional
self-dual Banach space which is not a Hilbert space. This shows that the
converse of the Riesz representation theorem — that every Hilbert space is
self-dual — is not in general true.
The preceding Proposition was easy to prove as the expressions for k k1
and k k1 are particularly nice. We can exploit this again to prove the n > 2
analogue.
Proposition 2.13. If n > 2 then In .1; 1/ D ∅.
Proof. Suppose S W `1n ! `1
n is an isometry and that A is its matrix representation in the standard basis. As kSei k1 D 1 we have jaij j 1 for all i
and j . Now applying S to elements in `1n of the form
vD
n
X
˛j ej
j D1
for ˛j D ˙1, we get that there for each such v there exists i with
ˇ
ˇ
ˇ
ˇ
ˇX
ˇ
ˇX
ˇ
ˇ n
ˇ
ˇ n
ˇ
ˇ
ˇ
ˇ
aij ˛j ˇ D n because kS vk1 D max ˇ
aij ˛j ˇˇ D kvk1 D n
ˇ
i ˇ
ˇj D1
ˇ
ˇ
j D1
As j˛j aij j 1, we must have j˛j aij j D jaij j D 1 and
n
X
aij ˛j D ˙n
j D1
However there are only n choices of i whereas there are 2n 1 pairwise linearly independent choices of v. Hence if n > 2 the system is overdetermined, and we reach a contradiction.
The p-norm classification theorem as a special case
So far we have proven all of Theorem 2.4 except the cases when p and q are
Hölder conjugate, both finite, and p ¤ 2. Our attempts to prove these final
cases in an elementary way were unsuccessful.
15
The p-norm classification theorem as a special case
The approaches in the literature to this problem are more abstract. A
group theoretic approach is taken in [5]. They initially focus on the isometry group I.X / of a given space X, and then relate isometries from X to
another space Y back to the group I.X /. In this way they obtain a number of conditions on the form of isometries between X and Y . One of their
theorems is especially applicable to our final case here. We describe it informally, as the formal statement takes some space to conextualize.
Let k k be a symmetric norm (recall Definition 2.3) on Rn and write
X D .Rn ; k k/. Then X is isometrically isomorphic to X if and only if one
of the following conditions apply.
Pn
2 1=2 for some > 0; that is, k k is a scaled version
(a) kxk D i D1 jxi j
of standard Euclidean distance.
(b) n D 4, and the isometry group I.X / contains both GP .4/ and additional elements that are specified. These additional elements are not
in GP .4/.
(c) n D 2, and the isometry group I.X / consists of reflections through
the standard basis and rotations by =.2k/ for some k 2 N. In this
case, there exists an isometry from X to X that is a rotation by =.4k/
composed with a possible scaling.
p
q
As p and q are Hölder conjugate, letting X D `n gives X D `n . It
is clear that case (a) does not apply as p ¤ 2. We have seen that I.X / D
I4 .p/ does not contain elements other than GP .4/, so case (b) does not apply
either. For case (c), we observe that we must have k D 1 as I.X / D I2 .p/
does not contains rotations of =4 if p ¤ 2. We should then be able to find
an isometry T that is a rotation of =4 radians composed with a scaling; it’s
associated matrix in the standard basis would be of the form
for some . From
1 D ke1 kp D kT e1 kq D ke1
e2 kq D 21=q
21=p D ke1 C e2 kp D kT .e1 C e2 /kq D k2e1 kq D 2
we see that if p and q are finite and p ¤ 2 then no such exists.
p
q
Hence X D `n and X D `n are not isometric if p and q are both finite
and p ¤ 2.
This concludes our proof of Theorem 2.4.
3 An isometric invariant: generalized 16
3
An isometric invariant: generalized In the previous section we found that the classification theorem describp
ing isometries between the `n spaces, though easy to reason on geometric
grounds, was relatively difficult to prove. In this section we take a different
approach to classification, seeking a relatively easy to compute isometric
invariant that we can use to demonstrate that two spaces are not isometric.
Our work will have the advantage of being mostly independent of the form
of the particular norms in question, but the disadvantage of being restricted
(until the very end) to two dimensional spaces.
Generalized The number D 3:1415 : : : is traditionally defined as the ratio of the circumference of a circle to its diameter. This definition is readily generalized
to two dimensional real normed vector spaces. If BX denotes the unit ball
of the normed vector space X we define
X D
circumference of BX
length of @BX
D
:
diameter of BX
diameter of BX
(3.1)
In order for this to make sense we have to describe what we mean by “length”
and “diameter”.
In general, the diameter of a set S is
diam.S / D supfkx
yk W x; y 2 S g:
If S D BX and x; y 2 BX the triangle equality gives
kx
yk kxk C kyk 2:
Now choosing some x with kxk D 1 we have
kx
. x/k D k2xk D 2
and hence that diam.BX / D 2, irrespective of the norm. Our definition of
X then simplifies to
X D
1
length of @BX :
2
(3.2)
We next develop an appropriate notion of the length of a curve that we
can use to give (3.2) a strictly defined meaning. The length of a curve will
be invariant under isometric isomorphism in the following sense: if C X
is a curve in X and T W X ! Y is an isometry then D D T .C/ is a curve in Y
with
length of D D length of C:
17
Curves in normed vector spaces
However, in general, given C and a hypothesized isometry T W X ! Y it
is impossible write down an explicit formula for D and hence calculate its
length in the norm of Y . The advantage of focusing on @BX is that we know
what its image under T will be: namely, @BY ! Hence X will be an isometric
invariant: if X and Y are isometric we will have
X D Y :
Curves in normed vector spaces
We temporarily relax the dimensionality constraint, for maximum generality. Let X be an n-dimensional real normed vector space with norm k k.
Definition 3.1. A curve C in X is the image of a continuous function ˛ W
Œ0; 1 ! X that is injective on Œ0; 1/ and .0; 1. The curve is said to be closed
if ˛.0/ D ˛.1/.
The function ˛ will not in general be unique: different functions from
Œ0; 1 to X can have the same image and so give rise to the same curve. In
differential geometry such functions would be referred to as different parameterizations of the curve, and we will employ this terminology here.
Definition 3.2. Recall that a partition of Œ0; 1 is a finite sequence ftk gm
kD0
with t0 D 0, tm D 1 and tk < tkC1 for all k. Fix a parameterization ˛ of C.
The length of C, denoted by L.C/, is
( m
)
X
L.C/ D sup
k˛.tk / ˛.tk 1 /kX W ftk gm
kD0 a partition of Œ0; 1 : (3.3)
kD1
The idea behind the definition is to approximate the length of the curve
by polygonal paths whose lengths can be easily computed in terms of the
norm.
x5
x1
x0 x4
x2
k˛.t5 /
˛.t4 /kX
xk D ˛.tk /
x3
To show that (3.3) is well defined — that is, independent of the parameterization ˛ — we make an appeal to an inverse function theorem in pointset topology (see [9, p. 88]).
3 An isometric invariant: generalized 18
Theorem 3.3. Suppose that f W X ! Y is an injective continuous function from
a compact space X to a Hausdorff space Y . Then f 1 W f .X / ! X is continuous.
Now suppose that ˛ and ˇ are two parameterizations of the same curve
is a partition of Œ0; 1. If C is not closed then ˇ is injective
C, and that ftk gm
kD0
on all of Œ0; 1 and we can apply Theorem 3.3 to ˇ to construct a continuous
bijection
ˇ 1 ı ˛ W Œ0; 1 ! Œ0; 1
which, by the familiar result in one-variable real analysis, must be either
strictly increasing or decreasing. If C is closed we can assume ˛.0/ D ˇ.0/.
We can then apply Theorem 3.3 to the functions
ˇ
ˇ
ˇ ˇŒ0;3=4 W Œ0; 3=4 ! X and ˇ ˇŒ1=4;1 W Œ1=4; 1 ! X
to construct a continuous bijection ˇ 1 ı ˛ on all of Œ0; 1 which must be
strictly increasing. In all instances the sequence
˚
n
sk D .˛ 1 ı ˇ/.tk / kD1
is a partition of Œ0; 1. We have
m
X
kD1
k˛.tk /
˛.tk
1 /kX
D
m
X
kˇ.sk /
ˇ.sk
1 /kX
kD1
and hence the definition (3.3) of L.C/ is independent of the parameterization chosen.
We will state formally our earlier remark relating curves and isometries.
Proposition 3.4. If C is a curve in X with parameterization ˛ W Œ0; 1 ! X and
T W X ! Y is an isometry then D D T .C/ is a curve in Y with parameterization
T ı ˛ W Œ0; 1 ! Y . Moreover, L.C/ D L.D/.
Proof. This follows immediately from Definitions 3.1 and 3.2 and the fact
that as an isometry between finite-dimensional spaces T is automatically
continuous and injective.
Bounds on X
Now that we have a notion of length and hence a precise definition of X ,
we might wonder in what range X can take values. Can we find, for any
a > 0, a norm on R2 such that X D a? It turns out that X is restrictive in
this sense. First, it’s easy to get a lower bound: take some unit vector v and
“approximate” @BX with the polygonal path joining the points fv; v; vg.
This gives
1
X .k.v/ . v/kX C k. v/ .v/kX / D 2:
2
For an upper bound we restrict our attention to a particularly nice set of
norms.
19
Bounds on X
Definition 3.5. A norm on a two-dimensional real vector space X is reflective
if there exists a basis fe1 ; e2 g of X such that for all a; b 2 R
kae1 C be2 kX D k
ae1 C be2 kX D kae1
be2 kX :
The set of norms with this reflective property contains the symmetric
norms of Definition 2:3, but includes many more besides.
Theorem 3.6. Suppose that the norm on X is reflective. Then X 4.
There is a very geometric reason why
the theorem holds in this specific case. The
key part is the convexity of the unit ball in
any normed vector space. As we traverse
the boundary of the unit ball the projection of our path onto one of the axes moves
monotonically towards or away from the
origin. Then the length of the projection,
which we can use to bound the length of
the curve, is just the length of the unit vector in the direction of the axis.
x4
b3
x3
x2
b2
x1
b1
a3
a2
a1 x 0
Proof. Let fe1 ; e2 g be the basis about which the norm is reflective. We can
write every x 2 X as x D ae1 C be1 . Let C be the segment of @BX lying
in the upper right quadrant; that is, @BX in the region a; b 0. Let ˛ be a
parameterization of C with ˛.0/ D e1 and ˛.1/ D e2 . We will show that the
length of C is less than or equal to 2, and summing over the quadrants will
yield the bound.
Let ftk gnkD0 be a partition of Œ0; 1. For all k write
˛.tk / D xk D ak e1 C bk e2 :
By the reflective condition we have
kak e1
bk e2 kX D kak e1 C bk e2 kX D 1;
and combined with the convexity of the unit ball this gives
1
1 ..ak e1 bk e2 / C .ak e1 C bk e2 //
D kak e1 kX D jak j:
2
X
A similar argument shows that jbk j 1.
We now show inductively that for all k < n, ak akC1 and bk bkC1 .
For the k D 0 case, we have a0 D 1 and b0 D 0. The bounds above show
us that a1 1 D a0 and b1 0 D b0 , as required.
Now suppose that for some m, am am 1 and bm bm 1 . We refer to
the next figure.
3 An isometric invariant: generalized 20
e2
xmC1 in this
region
L
B
bm
xm
A
bm
xm
1
1
e1
am am
1
Let us denote the line joining the origin and the point xm by L. If xmC1
lay below the line L then when going from xmC1 D ˛.tmC1 / to e2 D ˛.1/
the curve would have to cross the line L again. As k˛.t /kX D 1 always, the
curve would have to pass through xm , contradicting injectivity. Hence xmC1
lies above the line L.
The point xmC1 cannot lie in the region A. For it it did we could extend
a line from the origin through xmC1 that would intersect the line joining
xm and e2 . The intersection point would be further from the origin than
xmC1 , and so would have norm strictly greater than one. However by the
convexity of the unit ball every point on the line joining xm and e2 must
have norm less than or equal to 1, a contradiction.
The point xmC1 cannot lie in the region B either. If it did we could draw
a line from xmC1 to xm 1 that would intersect L at some point further from
the origin to xm . Similar to the A case, the intersection point would simultaneously have to have norm greater than 1 (being further from the origin
than xm 2 @BX ) and less than or equal to 1 (being on a line joining two
points on the unit circle).
Hence xmC1 cannot be in either region A or B. We then have bmC1 bm
and amC1 am . (Here we implicitly use the fact that bm 1.)
We now calculate a bound on the length of the polygonal path:
n
X
k˛.tk /
˛.tk
1 /kX D
kD1
n
X
k.ak e1 C bk e2 /
.ak
k.ak
C k.bk
bk
.bk
1/
1 e1
bk
1 e2 /kX
kD1
D
n
X
kD1
n
X
.ak
ak
1
1 /e1 kX
ak / C
kD1
D .a0
n
X
bk
1 /e2 kX
kD1
an / C .bn
b0 / D 2:
This is true for every polygonal path, so L.C/ 2. There are four quadrants,
so X 4.
21
X preserved under “dualizing”
a6
a5
b1
a1
b2
by translation invariance
L.ai / D L.bi / D 1
H)
b6
b3
X D
a4
b5
b4
1
2
P6
i D1 L.ai /
D3
a2
a3
Figure 2: X D 3 if the unit ball is a regular hexagon.
The main reason our proof cannot be adapted to remove the reflection
constraint on the norm is that we look at each quadrant in isolation. For two
general axes we would expect some interplay between quadrants so that the
length of @BX in some quadrant would be larger than 2 but that the sum of
the lengths over any two adjacent quadrants would still be less than 4. The
reflective constraint precludes such interaction.
In the general case the X 4 bound still holds, and is one component
of Golab’s theorem which describes the optimal bounds on X (see [8]).
Theorem 3.7 (Golab’s Theorem). 3 X 4 with X D 3 if and only if BX is
a regular hexagon and X D 4 if and only if BX is a parallelogram.
The proof that X 4 in general involves constructing a set of axes such
that the argument in our proof can be applied. Showing that X D 3 if the
unit ball is a regular hexagon is easy; see Figure 2.
X preserved under “dualizing”
In this section we investigate the relationship between X and X . As in
the last section, the proof of our main theorem will require a restriction. We
will denote the unit ball in X by B, and the unit ball in X by B .
Definition 3.8. A normed vector space is strictly convex if the boundary of
the unit ball contains no line segment.
Refering to the illustrations of the p-norms in Figure 1 (page 6), we
p
would be led to hypothesize that `n is strictly convex if and only if p 2
.1; 1/, and this is indeed the case.
3 An isometric invariant: generalized 22
Theorem 3.9. If X and X are both strictly convex then X D X .
The basic idea of the proof is simple. Let ˛ W Œ0; 1 ! X be some parameterization of @B and consider an equally spaced partition ftk D k=ngnkD0
of Œ0; 1. This gives rise to a polygonal path P in B with vertices fk D
˛.tk /gnkD0 . We have
length.P/ D
n
X
kk
k
1 k:
kD1
Now we constuct a polygonal path Q in B with vertices fxk gnkD1 in @B chosen so that
.k k 1 /.xk / D kk k 1 k:
(3.4)
This gives
length.P/ D
n
X
.k
kD1
n
X1
k
1 /.xk / D
n
X1
k .xk
kD0
n
X1
kk kkxk
xkC1 k D
kD0
kxk
xkC1 /
xkC1 k D length.Q/:
kD0
Hence for every polygonal path P approximating @B we can find a longer
polygonal path Q approximating @B. It follows that X X . The reflexive
property of finite dimensional Banach spaces, .X / D X, then gives X D
X .
The problem with this argument is that we haven’t shown that Q is actually an admissible approximation of @B. For instance, Q could intersect
itself making the approximation meaningless.
x1
x0
x2
x3
To ensure the proof is correct we need to prove the existence of a parameterization ˇ of @B such that ˇ.tk / D xk . The injectivity of ˇ would then
prevent a situation like above. An immediate problem is our choice of xk
in (3.5). The next result says that in order for this proof approach to work
smoothly we really need strict convexity, and this explains the restriction in
our theorem.
Lemma 3.10. A two-dimensional real normed vector space X is strictly convex if
and only if for every 2 @B there is a unique point x 2 @B where attains its
norm in the sense that .x/ D kk D 1.
23
X preserved under “dualizing”
Proof. Suppose that there exists 2 @B such that .x1 / D 1 D .x2 / for
distinct points x1 ; x2 2 @B. For any 2 Œ0; 1 we have by the convexity of B,
kx1 C .1
/x2 k 1:
Then
1 D . / C .1
/
D .x1 C .1
/x2 /
kkkx1 C .1
D kx1 C .1
/x2 k
/x2 k 1:
Hence x1 C .1 /x2 2 @B for all 2 Œ0; 1, so X is not strictly convex.
Now suppose that X is not strictly convex. Then there exist distinct
x1 ; x2 2 @B such that
kx1 C .1 /x2 k D 1
for all 2 Œ0; 1.
Set y1 D x1
so that
x2 . Choose y2 2 X to be perpendicular to y1 and scaled
x1 D ˛1 y1 C y2 and x2 D ˛2 y1 C y2 :
Here perpendicularity is with respect to the usual Euclidean inner product.
Now define by
.a1 y1 C a2 y2 / D a2 :
We then have .x1 / D .x2 / D 1. Furthermore, by the convexity of B, we
have that
a1 y1 C a2 y2 2 B H) ja2 j 1
so that kk D 1. This norm is attained at the distinct points x1 and x2 .
Now in the case of a strictly convex normed vector space we can define
the “attainment map” A W @B ! @B such that A is the unique element of
@B where attains its norm kk D 1.
Lemma 3.11. Suppose X is strictly convex so that the attainment map is welldefined. X is strictly convex if and only if the attainment map A is injective.
Proof. Suppose A is injective and let 1 ; 2 2 @B be arbitrary. Write x1 D
A1 ¤ A2 D x2 and set
kx1 x2 k
rD
:
2
Consider the two closed subsets of @B defined for i D 1; 2 by
Si D @BnB.xi ; r/:
3 An isometric invariant: generalized 24
By the compactness of S1 and the continuity of 1 , 1 jS1 attains its upper
bound at some point y1 2 S1 . As the unique attainment point is not in S1
we have
1 .y1 / D 1 1 < 1:
Defining y2 similarly gives
2 .y2 / D 1
2 < 1:
Now for any 2 Œ0; 1 we have
.1 C .1
/2 /.x/ D 1 .x/ C .1
/2 .x/:
If x 2 S1 we get
.1 C .1
/2 /.x/ .1
1 / C .1
/.1/ D 1
1 ;
while if x 2 S2 we get
.1 C .1
/2 /.x/ .1/ C .1
/.1
2 / D 1
A similar argument provides lower bounds on .1 C .1
k1 C .1
/2 k maxf1
1 ; 1
.1
.1
/2 :
/2 /.x/ to give
/2 g < 1
and X is strictly convex.
Now suppose that A is not injective: that there exists distinct points
1 ; 2 2 B such that A1 D x D A2 for some x 2 @B. Then by the
convexity of B we have for all 2 Œ0; 1
k1 C .1
/2 k 1:
On the other hand,
.1 C .1
so that 1 C .1
/2 /.x/ D C .1
/ D 1
/2 2 @B for all 2 Œ0; 1, and B is not strictly convex.
From now on we will assume, as in the statement of Theorem 3.9, that
both X and X are strictly convex and hence that the preceding two lemmas
apply.
Lemma 3.12. The chord map
C˛ .t / D
is continuous and injective.
˛.t /
k˛.t /
˛.t
˛.t
1=n/
1=n/k
25
X preserved under “dualizing”
Proof. Continuity is a direct consequence of ˛ being continuous.
Suppose C˛ .t / D C˛ .s/. This means that the two vectors
˛.t /
˛.t
1=n/
and ˛.s/
˛.s
1=n/
point in the same direction. (This is a stronger requirement than the vectors
merely being parallel.) If s ¤ t then the four distinct points
˛.t /; ˛.t
1=n/;
˛.s/ and ˛.s
1=n/
cannot be collinear, as then B would contain the line segment connecting
them and would not be strictly convex.
If the points are not collinear the picture will be like so:
˛.s/
˛.t /
˛.t
˛.s
1=n/
1=n/
The injectivity of ˛ is breached: as ˛ traverses the unit boundary it will
pass through ˛.t / after ˛.s/ if and only if it passes through ˛.t 1=n/ after
˛.s 1=n/. Hence t D s, and C˛ is injective.
Lemma 3.13. The attainment map is continuous and injective.
Proof. It is injective by Lemma 3.11.
For continuity choose 2 @B and fix > 0. Define the compact set
S D @BnB.x ; /
where x D A. As S is compact jS attains its maximum at some point
x0 . As attains its norm at the unique point x ¤ x0 , we have .x0 / < 1.
Define ı D 1 .x0 / > 0.
Now suppose 2 @B satisfies k
k < ı. Then
1
.x / D
.x /
.x / < ı D 1
Thus .x / > .x0 / and x 62 S . This means
kA
and so A is continuous at .
A k D kx
x k<
.x0 /:
3 An isometric invariant: generalized 26
We finally have sufficient machinary to prove Theorem 3.9.
Proof of Theorem 3.9. Recall that we had a parameterization ˛ of @B , an equally
spaced partition ftk D k=ngnkD0 of Œ0; 1, and a polygonal path P with vertices
f˛.tk / D k gnkD0 . (That we can approximate the length arbitarily well by
equally-spaced partitions is the content of Lemma 3.15 below.) We defined
xk 2 @B so that
.k k 1 /.xk / D kk k 1 k
(3.5)
which is equivalent to defining xk so that
k
k k 1
C˛ .tk /.xk / D
.xk / D k
kk k 1 k
k
k
k
1
1
D 1:
k
(3.6)
We had the estimate
n
X
kk
1k k
n
X
kxk
xkC1 k:
kD1
kD1
Now define ˇ W Œ0; 1 ! @B by
ˇ.t / D A.C˛ .t //:
This is a continuous and injective function as the composition of continuous
and injective functions. It image is contained in the image of A, @B. We have
ˇ.tk / D xk
and hence
n
X
kxk
xkC1 k [email protected]/;
kD1
as required.
Integral representation of arc length
So far we have proven two general facts about X : that when the norm is
sufficiently nice it takes values in Œ2; 4 and is equal to X . However, our
initial motivation to study X was rather more concrete: we wished to develop a quantity that is invariant under isometric isomorphism and hence
a tool to show that two given spaces are not isometric. The main problem
with actually implementing this idea in specific cases is that the definition
of the length of a path given in (3.3) is not easy to compute.
In differential geometry there is an easily evaluated integral “arc-length”
formula for the length of a path. One might wonder if such a formula exists
when we equip Rn with an arbitrary norm rather the standard Euclidean
norm. We are pleased to report that this is indeed the case.
27
Integral representation of arc length
Theorem 3.14. Let X D .Rn ; k k/ be a normed vector space, and C a curve with
parameterization ˛ W Œ0; 1 ! X. Suppose that ˛ is differentiable on .0; 1/, that ˛
possesses left and right derivatives at 0 and 1 respectively, and that ˛ 0 is continuous
on Œ0; 1. Then
Z 1
L.C/ D
k˛ 0 .t /kdt:
(3.7)
0
The hypotheses of this theorem can be weakened to those of the ordinary arc-length theorem, but this entails developing more integration theory which we avoid owing to space considerations.
We will approximate the length of the curve by sums of the form
m X
˛ j
Sm D
m
j D1
˛
j
1 m
(3.8)
and show that these converge to the integral in (3.7). First we must show
that the actual length L.C/ can be approximated arbitrarily well by these
kinds of sums, which arise when we restrict our attention to partitions of
Œ0; 1 into parts of equal length.
Lemma 3.15. lim Sm D L.C/.
m!1
Proof. Let ftk gN
be any partition of Œ0; 1 and fix > 0. As ˛ is continuous
kD0
and hence uniformly continuous on Œ0; 1 there exists ı > 0 such that
k˛.t /
˛.s/k <
whenever kt
2N
sk < ı:
Choose m > 1=ı. Then we can find a sequence flk gN
N such that
kD0
˛.tk / ˛ lk < m 2N
for all k. In particular, l0 D 0 and lN D m. Then
lk
n
n X
k˛.tk / ˛.tk 1 /k D ˛.tk /
˛
˛
˛.tk 1 /
m
m
nDlk 1
lk C ˛ lk 1
˛.t
/
˛
˛.t
/
k
k
1
m m
lk
X
n
n 1 C
˛
˛ m
m nDlk
C
N
1 C1
lk
X
nDlk
1 C1
˛ n
m
˛
n
1 :
m 3 An isometric invariant: generalized 28
Now summing over k,
N
X
k˛.tk /
˛.tk
m X
˛ n
1 /k C
m
˛
n
nD1
kD1
1 D C Sm :
m The choice of was arbitrary so
N
X
k˛.tk /
˛.tk
1 /k
lim inf Sm :
m!1
kD1
The choice of partition was arbitrary, so
L.C/ lim inf Sm :
m!1
But of course Sm L.C/ for each m 2 N, so
lim sup Sm L.C/ lim inf Sm ;
m!1
m!1
and the result is proven.
Proof of Theorem 3.14. Recall that we partition Œ0; 1 into parts of equal length.
The mean value theorem, which the component functions of ˛ satisfy by
the hypotheses of this theorem, says that for every j 2 f1; :::; mg and every
k 2 f1; :::; ng we can find
j 1 j
tjk 2
;
m m
such that
1 0 k
˛ t D ˛k
m k j
j
m
˛k
j
1
:
m
(3.9)
Now it would not be correct to claim that
j 1
1 0 k
j
˛ tj D ˛
˛
:
m
m
m
as, in general, tjk will vary with k. An approach involving bounds is necessary.
Fix > 0. By the reverse triangle inequality we have
jkyk
kxkj < whenever kx
yk < :
Now because k k is equivalent to the infinity norm k k1 , there is a positive
constant A such that
kx
yk1 D maxfjxi
i
yi jg Akx
yk:
29
Integral representation of arc length
Letting ı D =A > 0 gives
kxkj < whenever kx
jkyk
yk1 < ı:
(3.10)
Now for each k, ˛k0 is continuous on Œ0; 1, hence uniformly continuous.
We can then choose k so that
ˇ 0
ˇ
ˇ˛ .t / ˛ 0 .s/ˇ < ı whenever jt sj < k :
(3.11)
k
k
Let D minfk g.
Set m > 1= and for every 1 j m choose some
j 1 j
xj 2
;
:
m m
Then
ˇ
1
ˇ
tjk ˇ <
< k ;
(3.12)
m
where the numbers tjk are selected as earlier.
Now letting vj be the vector with k th component ˛k0 tjk we can write
(3.8) as
m
X
1
kvj k:
Sm D
m
ˇ
ˇ 0
ˇ˛k .xj /
ˇ
ˇ
ˇ
ˇ
˛k0 tjk ˇ < ı as ˇxj
j D1
The condition of (3.12) holds, so (3.10) holds too. Hence
ˇ
ˇ
ˇkvj k k˛ 0 .xj /kˇ < and
C
m
m
X
X
1 0
1 0
k˛ .xj /k Sm k˛ .xj /k C :
m
m
j D1
j D1
R1
But the sum here is a just a Riemann sum for the integral 0 k˛ 0 .t /kdt. This
integral exists as k˛ 0 .t /k is a continuous function from Œ0; 1 to R, and hence
is integrable. As is arbitrary, taking limits we must have
Z 1
L.C/ D lim Sm D
k˛ 0 .t /kdt:
m!1
0
All of the authors we consulted [4, 6] actually take (3.7) as the definition
of the length of a curve. While this was probably motivated by a desire
for brevity, it is an unsatisfactory approach for two reasons. First, it is not
geometrically obvious that the integral formula genuinely gives the length
of the curve, whereas it is completely obvious that our definition (3.3) does.
Secondly, we simply observe that our presentation of the arc length formula
has come after our proofs that X is bounded and X D X . Taking an
elementary approach gives us a more general machinery for investigating
the lengths of curves in the abstract.
3 An isometric invariant: generalized 30
X in the p-norms
Now that we are in a position to compute values of X we will apply our
p
results to the classification of the isometries between `2 . We will denote by
p the value of X in the p-norm. By the previous theorem this is given by
Z 1
k˛ 0 .t /kp dt
0
p
where ˛ 0 .t / is a parametrization of the boundary of the unit disk in `n .
Let us find such a parametrization. In order to use the integral arc length
formula the derivative of the curve must be bounded. We will parameterize
the section of the boundary that includes points .x; y/ satisfying 0 x 2 1=p and y > 0.
y
˛.t /
Sy
m
ab
m
ou
et
ric
ty
D
x
.1
1
t p/ p
t
Starting with x p C y p D 1 we find
1=p ˛.t / D t; 1 t p
I t 2 Œ0; 2
parametrizes this section. Then
˛ 0 .t / D 1;
tp
1
1=p
k˛ 0 .t /kp D 1 C
tp
1 tp
t
p
Z
0
2
1=p
"
1C
tp
1 tp

#
1 1=p
:
Hence
1
p D 8
2
1=p
1 p 1
and
"
x
1
p
2
p
#
1 1=p
dt:
31
X in the p-norms
We will simplify this expression through an integral substitution. Let u D
t p , or equivalently u1=p D t. This gives du D pt p 1 dt, or
1 1=p 1
u
p
p
du D dt:
Then
u p 1 1=p 1 1 p
1C
u1=p
du
1 u
p
0
Z 1=2 u p 1 1=p 1=p
1C
u1 p
du
1 u
0
1=p
Z 1=2 1
1
C
du:
up 1
.1 u/p 1
0
Z
p D 4
D
4
p
D
4
p
1=2
This integral is symmetric about the interval Œ0; 1, so we can write this as
p D
2
p
Z
0
1
u1
p
C .1
u/1
p 1=p
du:
(3.13)
This formula proves incredibly difficult to work with directly. However
we can apply our general results from before. Theorem 3.9 tells us that if
p
q
if p and q are Hölder conjugate and both finite so that `2 and `2 are both
strictly convex then
p D q :
(3.14)
For the p D 1 and q D 1 case we can calculate p manually as the boundary
of the unit disk is a just parallelogram in both cases. This calculation gives
1 D 1 D 4, and (3.14) still holds. (An elementary proof of (3.14) is given
in [4].)
p
We also know that the `n spaces are reflective about the standard basis.
This implies that
p 4:
Finally, of course,
2 D D 3:14159 : : :
For this final year project, we did not analyze the integral representation
in any depth. In fact, we initially attempted to prove (3.14) directly for this
specific case, but found it easier to prove the general Theorem 3.9! However,
it was easy to use Mathematica to calculate (3.13) as a function of p, and the
plot is shown in Figure 3. The only difficulty was the integrand becoming
numerically unstable for large p; this problem was dealt with by applying
(3.14) in the form
p D p.1 p/ 1
to avoid having to evaluate the integral (3.13) for p > 2.
3 An isometric invariant: generalized 32
p
4
3
p
1 4=3
2
3
4
5
6
7
8
9
10
Figure 3: Plot of p , including an illustration that p D q when p and q
are Hölder conjugate.
Figure 3 displays clear behaviour. The function monotonically decreases
on Œ1; 2/, attains its global minimum at p D 2, and then monotonically increases on .2; 1/. We found a proof of the latter assertions in the literature;
see [1]. Assuming monotonicity, and recalling that X is an isometric invariant, these results combined are sufficient to prove
p
q
Corollary 3.16. If p and q are not equal and not Hölder conjugate then `2 and `2
are not isometric.
Beyond two dimensions
The primary weakness of X is that it can only be applied in two dimensional spaces. Of course, it is not hard to dream up invariants in n dimensions based on X ; for instance, we could define, for n > 2,
X D minfY W Y a two-dimensional subspace of Xg:
This quantity is called the “girth” in the literature. The n-dimensional version of Theorem 3.9, that X D X , is true and was proven in [7]. The
proof uses topological properties of the set of all n-dimensional normed
vector spaces equipped with the Banach-Mazur distance. In this topology
the “girth” is a continuous function.
According to the same paper, a higher-dimensional analogue of Theorem 3.7 is still an open problem.
Conjecture 3.17. If n > 2 then X D 3:1415 : : : with equality if and only if
X is Euclidean.
References
33
References
[1] C. L. Adler and James Tanton. is the minimum value for pi. The College
Mathematics Journal, 31(2):102–106, March 2000.
[2] Marek Capiński and Ekkhard Kopp. Measure, Integral and Probability.
Springer –Verlag London, second edition, 2004.
[3] G. Hardy, J.E. Littlewood, and G. Pólya. Inequalities. Cambridge University Press, second edition, 1952.
[4] Joseph B. Keller and Ravi Vakil. p , the value of in `p . The American
Mathematical Monthly, 116(10):931–935, December 2009.
[5] Chi-Kwong Li. Isometries, and isometry groups. The American Mathematical Monthly, 107(4):334–340, April 2000.
[6] Chi-Kwong Li and Wasin So. Isometries of the `p –norm. The American
Mathematical Monthly, 101(5):452–453, May 1994.
[7] Juan Carlos Alvarez Paiva. Dual spheres have the same girth. American
Journal of Mathematics, 128(2):361–371, 2006.
[8] Juan Carlos Alvarez Paiva and Anthony Thomson. On the perimeter
and area of the unit disk. The American Mathematical Monthly, 112(2):141–
154, February 2005.
[9] W.A. Sutherland. Introduction to Metric and Topological Spaces. Oxford
University Press, first edition, 1975.
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