1. UNIVERSIDAD DE CÁDIZ
FACULTAD DE CIENCIAS
OPEN PROBLEMS ON TOPOLOGICAL VECTOR
SPACES WITH APPLICATIONS TO INVERSE
PROBLEMS IN BIOENGINEERING
Justin Robert Hill
2. Open problems on topological vector spaces with
applications to inverse problems in bioengineering
Directores: Dr. Francisco Javier García Pacheco and Dr. Clemente Cobos Sánchez
Firma Doctorando
Firma del Director Firma del Director
Cádiz, October 2016
9. CHAPTER
1
Abstract
An abstract of this dissertation in both English and Spanish will be presented first thing.
1.1 Abstract
A totally anti-proximinal subset of a vector space is a non-empty proper subset which does not
have a nearest point whatever is the norm that the vector space is endowed with. A Hausdorff
locally convex topological vector space is said to have the (weak) anti-proximinal property
if every totally anti-proximinal (absolutely) convex subset is not rare. Ricceri’s Conjecture,
posed by Prof. Biaggio Ricceri, establishes the existence of a non-complete normed space
satisfying the anti-proximinal property. In this dissertation we approach Ricceri’s Conjecture
in the positive by proving that a Hausdorff locally convex topological vector space enjoys
the weak anti-proximinal property if and only if it is barreled. As a consequence, we show
the existence of non-complete normed spaces satisfying the weak anti-proximinal property.
v
10. 1. ABSTRACT
We also introduce a new class of convex sets called quasi-absolutely convex and show that a
Hausdorff locally convex topological vector space satisfies the weak anti-proximinal property
if and only if every totally anti-proximinal quasi-absolutely convex subset is not rare. This
provides another partial positive solution to Ricceri’s Conjecture with many applications to
the theory of partial differential equations. We also study the intrinsic structure of totally
anti-proximinal convex subsets proving, among other things, that the absolutely convex hull
of a linearly bounded totally anti-proximinal convex set must be finitely open. As a con-
sequence of this, a new characterization of barrelledness in terms of comparison of norms
is provided. Another of our advances consists of showing that a totally anti-proximinal ab-
solutely convex subset of a vector space is linearly open. We also prove that if every totally
anti-proximinal convex subset of a vector space is linearly open then Ricceri’s Conjecture
holds true. We also demonstrate that the concept of total anti-proximinality does not make
sense in the scope of pseudo-normed spaces. Falling a bit out of Ricceri’s Conjecture, we also
study some geometric properties related to the set ΠX := {(x, x∗
) ∈ SX × SX∗ : x∗
(x) = 1}
obtaining two characterizations of Hilbert spaces in the category of Banach spaces. We also
compute the distance of a generic element (h, k) ∈ H ⊕2 H to ΠH for H a Hilbert space. As
an application of our mathematical results, an inverse boundary element method and effi-
cient optimisation techniques were combined to produce a versatile framework to design
truly optimal TMS coils. The presented approach can be seen as an improvement of the
work introduced by Cobos Sanchez et al. where the optimality of the resulting coil solutions
was not guaranteed. This new numerical framework has been efficiently applied to produce
TMS coils with arbitrary geometry, allowing the inclusion of new coil features in the design
process, such as optimised maximum current density or reduced temperature. Even the
structural head properties have been considered to produce more realistic TMS stimulators.
Several examples of TMS coils were designed and simulated to demonstrate the validity of
the proposed approach.
vi
11. 1.2 Resumen
1.2 Resumen
Un subconjunto totalmente anti-proximinal de un espacio vectorial es un subconjunto propio
no vacío que no tiene un punto más cercano, cualquiera que sea la norma con la que esté
dotado el espacio vectorial. Se dice que un espacio vectorial topológico localmente convexo
y de Hausdorff tiene la propiedad anti-proximinal (débil) si cada subconjunto totalmente
anti-proximinal (absolutamente) convexo es no raro. La Conjetura de Ricceri, planteada
por el profesor Biaggio Ricceri, establece la existencia de un espacio normado no com-
pleto que satisface la propiedad anti-proximinal. En esta tesis doctoral nos acercamos a
la Conjetura de Ricceri positivamente demostrando que un espacio vectorial topológico de
Hausdorff y localmente convexo goza de la propiedad anti-proximal débil si y sólo si es
tonelado. Como consecuencia, mostramos la existencia de espacios normados no completos
que satisfacen la propiedad anti-proximal débil. También introducimos una nueva clase de
conjuntos convexos llamados cuasi absolutamente convexos y demostramos que un espacio
vectorial topológico localmente convexo de Hausdorff satisface la propiedad anti-proximal
débil si y sólo si cada subconjunto casi absolutamente convexo totalmente anti-proximinal
es no raro. Esto último proporciona otra solución positiva parcial a la Conjetura de Ricceri
con muchas aplicaciones a la teoría de ecuaciones diferenciales parciales. También estu-
diamos la estructura intrínseca de los subconjuntos convexos totalmente anti-proximinales
demostrando, entre otras cosas, que la envoltura absolutamente convexa de un conjunto
convexo totalmente anti-proximinal linealmente acotado debe ser finitamente abierta. Como
consecuencia de esto, se proporciona una nueva caracterización del concepto de tonelación
en términos de comparación de normas. Otro de nuestros avances consiste en demostrar que
un subconjunto totalmente anti-proximal absolutamente convexo de un espacio vectorial es
linealmente abierto. También probamos que si cada subconjunto convexo totalmente anti-
proximinal de un espacio vectorial es linealmente abierto entonces la Conjetura de Ricceri
es verdadera. También demostramos que el concepto de anti-proximinalidad total no tiene
vii
12. 1. ABSTRACT
sentido en el ámbito de los espacios pseudo-normados. Saliéndonos un poco de la Con-
jetura de Ricceri, estudiamos también algunas propiedades geométricas relacionadas con
el conjunto ΠX := {(x, x∗
) ∈ SX × SX∗ : x∗
(x) = 1} obteniendo dos caracterizaciones de los
espacios de Hilbert en la categoría de espacios de Banach. También calculamos la distan-
cia de un elemento genérico (h, k) ∈ H ⊕2 H a ΠH para H un espacio de Hilbert. Como
aplicación de nuestros resultados matemáticos, combinamos un método de elementos de
contorno inverso y técnicas de optimización eficientes para producir un marco versátil para
diseñar bobinas TMS verdaderamente óptimas. El enfoque presentado puede ser visto como
una mejora del trabajo introducido por Cobos Sánchez et al. donde la optimalidad de las
soluciones de las bobinas resultantes no estaba garantizada. Este nuevo marco numérico
ha sido aplicado eficientemente para producir bobinas TMS con geometría arbitraria, per-
mitiendo la inclusión de nuevas características de la bobina en el proceso de diseño, tales
como la densidad de corriente máxima optimizada o la temperatura reducida. Incluso las
propiedades estructurales de la cabeza se han considerado para producir estimuladores TMS
más realistas. Se diseñaron y simularon varios ejemplos de bobinas TMS para demostrar la
validez del enfoque propuesto.
viii
13. CHAPTER
2
Introduction
Modern Mathematics is defined as a Former First-Order Language. It finds its origins in Pro-
positional Logic. When the symbols ∈, ⊆, ∃, and ∀ are added to the syntax of Propositional
Logic, then the Zermelo-Fraenkel Axioms, together with the Axiom of Choice or weakenings
of it, give birth to what is called Modern Mathematics. The reader may recall that the four
previous symbols are not independent as, for instance, ⊆ can be derived from ∈ as follows:
A ⊆ B ↔ ∀x (x ∈ A → x ∈ B).
Four major areas compose Modern Mathematics: Algebra, Topology, Analysis, and Geometry.
These four areas are integrated by Category Theory.
Definition 2.0.1. A category is a pair of classes = (ob( ),hom( )) consisting of:
• A class of objects, denoted by ob( ). The elements of ob( ) are sets but ob( ) is not a
set itself, it is a class.
ix
14. 2. INTRODUCTION
• A class of morphisms, hom( ). The elements of hom( ) are non-empty sets denoted by
hom (A, B) for all A, B ∈ ob( ), that is,
hom( ) := {hom (A, B) : A, B ∈ ob( )}.
The elements of hom (A, B) are not necessarily maps or relations from A to B. Again,
hom( ) is not a set but a class.
• For any three objects A, B, C ∈ ob( ), a binary operation exists
hom (A, B) × hom (B, C) → hom (A, C)
(f , g) → g ◦ f ,
called composition of morphisms, verifying the following two properties:
1. Associativity: For all A, B, C, D ∈ ob( ), all f ∈ hom (A, B), all g ∈ hom (B, C),
and all h ∈ hom (C, D), (h ◦ g) ◦ f = h ◦ (g ◦ f ).
2. Identity: For every A ∈ ob(C) there exists an element IA ∈ hom (A,A), called the
identity morphism for A, such that for all B, C ∈ ob( ), all f ∈ hom (A, B), and
all g ∈ hom (C,A), f ◦ IA = f and IA ◦ g = g.
Among others, one observation the reader may quickly notice is that hom (A,A) endowed
with the composition becomes a monoid. What really integrates the four areas previously
mentioned into a single and unique mathematical view is the concept of functor, which
will not be defined nor treated in this manuscript. Nonetheless, general theorems about
categories can be proved that apply to all four areas of Modern Mathematics providing a
general and abstract view of the tight connections between those areas. This way we have
• global concepts relative to morphisms such as monomorphisms, epimorphisms, bi-
morphisms, isomorphisms, injections, projections, etc.;
x
15. 2.1 Scope of this work
• global concepts relative to objects such as sub-objects, quotient objects, initial objects,
final objects, etc.;
• and global concepts relative to objects and morphisms such as products, co-products,
universal properties, etc.;
2.1 Scope of this work
The scope of this joint work can be summarized into two main objectives:
1. Proving Ricceri’s Conjecture true or finding a counter-example.
2. Designing truly optimal TMS coils.
The birth of this work begins at a meeting between one of the advisors of the PhD candidate
with Italian Mathematician Biaggio Ricceri at the Department of Mathematics and Computer
Sciences of the University of Catania. Ricceri stated a conjecture (see (28)) on the topological
structure of certain subsets of normed spaces (see (28)). As a result of working on that
conjecture, four papers came out (18; 19; 20; 21). The last three of those four papers are
a joint work between the PhD candidate and one the advisors. Those four papers contain
original results which are not directly framed in the scope of Ricceri’s Conjecture, but are
crucial towards accomplishing our approaches to the conjecture. Therefore, we have placed
those results in a preliminary chapter in this manuscript. However, we would like the reader
to beware about the originality of the results in the Preliminary Chapter.
After maintaining conversations with some members of the Department of Electronic Engin-
eering of the College of Engineering of the University of Cadiz, the PhD candidate together
with his two advisors realized that some of the results on the Ricceri’s Conjecture could
actually be applied to optimize the norm of certain matrices that represent physical vec-
xi
16. 2. INTRODUCTION
tor magnitudes such as the electrical field or the magnetic field. The resolution of those
optimization problems afforded the possibility of designing certain Transcranial Magnetic
Stimulation Coils.
Transcranial Magnetic Stimulation (TMS) is a non-invasive technique to stimulate the brain,
which is applied to studies of cortical effective connectivity, presurgical mapping, psychi-
atric and medical conditions, such as major depressive disorder, schizophrenia, bipolar de-
pression, post-traumatic, stress disorder and obsessive-compulsive disorder, amongst others
(31).
In TMS, strong current pulses driven through a coil are used to induce an electric field stimu-
lating neurons in the cortex. The efficiency of the stimulation is determined by coil geometry,
orientation, stimulus intensity, depth of the targeted tissue and some other factors, such as,
stimulus waveform and duration.
The TMS stimulator most commonly employed is the so called round or figure-of-eight or
butterfly coil, but since the invention of TMS numerous coil geometries have been proposed
to improved the performance and spatial characteristics of the electromagnetic stimulation
(13).
The problem in TMS coil design is to find optimal positions for the multiple windings of
coils (or equivalently the current density) so as to produce fields with the desired spatial
characteristics and properties (24) (high focality, field penetration depth, low inductance,
low heat dissipation, etc.).
Similar problems to TMS coil design can be found in engineering, in which it is also required
to determine a quasi-static spatial distribution of electric currents flowing on a conductive
surface subjected to electromagnetic constraints. Some of these problems have been suc-
cessfully solved by modelling the current under search in terms of the stream function using
xii
17. 2.1 Scope of this work
a boundary element method (BEM). A relevant application can be found in magnetic res-
onance imaging (MRI), where gradient coils have been efficiently designed following this
technique (27),(11).
Recently, Cobos Sanchez et al. (8) have used this numerical strategy to formulate a new
TMS coil design method, in which a stream-function based current model is incorporated
into an inverse boundary element method (IBEM). In that work, the desired current distri-
bution is eventually obtained by solving an optimization problem, where a cost function or
functional formed with a weighted linear combination of all the objectives, is minimized by
using classical techniques, such as simple partial derivation.
The computational approach in Cobos Sanchez et al. (8) has demonstrated a remarkable
flexibility for the inclusion of new coil features in the design process, such as the minimiz-
ation of the magnetic stored-energy, minimization of power dissipation or minimization of
the undesired electric field induced in non target regions of the cortex.
Although this stream function IBEM has proved to design efficient TMS stimulators, unfor-
tunately it was not known how optimal these coil solutions were. Especially since the asso-
ciated optimisation problem has a maximisation part, which has to be rigorously tackled so
as to produce the most effective stimulation of the desired cortex regions.
On the other hand, applications of TMS for diagnostic and therapeutic purposes are con-
stantly growing, being often restricted by technical limitations. The versatility of stream
function IBEM therefore opens up the possibilities of overcoming some of these restrictions
with the design of a new generation of TMS stimulators with improved performance and
novel properties, such as reduced mechanical stress, minimum coil heating, optimized max-
imum current density amongst others.
Nonetheless, most of these new performance features increase the mathematical complexity
xiii
21. CHAPTER
3
Preliminary results
3.1 Geometric preliminary results
3.1.1 The metric projection and suns
Let X be a pseudo-metric space and consider A and B to be non-empty subsets of X.
• The distance from A to B is defined by d (A, B) := inf d (A× B).
• The diameter of A is defined as d (A) := sup d (A× A) and the radius of A is defined as
r (A) := d (A)/2.
• A is said to be bounded provided that d (A) < ∞.
Definition 3.1.1. Let X be a pseudo-metric space and consider x, y ∈ X. The segment of
1
22. 3. PRELIMINARY RESULTS
extremes x and y is defined as
[x, y] := {z ∈ X : d (x, y) = d (x,z) + d (z, y)}.
A subset C of X is said to be convex provided that [x, y] ⊆ C for all x, y ∈ C.
Several properties verified by segments and convex sets follow:
• x, y ∈ [x, y] for all x, y ∈ X.
• [x, y] = [y, x] for all x, y ∈ X.
• [x, y] = [x,z] ∪ [z, y] for all z ∈ [x, y].
As well, the non-empty intersection of any family of convex sets is also convex and thus the
convex hull of a set is defined as the intersection of all convex sets containing it. Finally,
other segments and related sets usually defined are
◦ (x, y) := [x, y] {x, y}.
◦ (x, y] := [x, y] {x}.
◦ The semi-straight line or ray joining two different points x and y of extreme x is
usually defined as the set [x, y] ∪ {z ∈ X : y ∈ (x,z)}.
◦ The straight line joining two different points x and y is usually defined as the set
{z ∈ X : x ∈ (z, y)} ∪ [x, y] ∪ {z ∈ X : y ∈ (x,z)}.
A set-valued function on psuedo-metric spaces that will become useful is:
Definition 3.1.2. Let X be a pseudo-metric space and consider a subset A of X. The metric
2
23. 3.1 Geometric preliminary results
projection of A is defined as
PA : X → (A)
x → PA (x) := {a ∈ A : d (x,A) = d (x, a)}.
We refer the reader to (14, Chapter 12) for a better perspective on metric projections in inner
product spaces.
Proposition 3.1.3. Let X be a pseudo-metric space and consider a non-empty proper subset A
of X. Let x ∈ X A and a ∈ PA (x). Then:
1. (a, x) ∩ A = ∅.
2. a ∈ PA (y) for all y ∈ (a, x).
If X is a metric space, y ∈ X, a ∈ PA (y), and a = y, then y /∈ A (which means that condition 1
is implied by condition 2 in metric spaces).
Proof. Before showing 1 and 2 we will assume that X is a metric space, y ∈ X, a ∈ PA (y),
and a = y. If y ∈ A, then 0 = d (y,A) = d (y, a), which implies that y = a in virtue of the
fact that d is a metric. Now we will show 1 and 2 under the original assumption that d is a
pseudo-metric.
1. If y ∈ (a, x) ∩ A, then we fall in the contradiction that
d (x,A) = d (x, a) = d (x, y) + d (y, a) > d (x, y) ≥ d (x,A).
2. If y ∈ (a, x) and there exists b ∈ A with d (y, b) < d (y, a), then we fail in the contra-
3
24. 3. PRELIMINARY RESULTS
diction that
d (x,A) = d (x, a)
= d (x, y) + d (y, a)
> d (x, y) + d (y, b)
≥ d (x, b)
≥ d (x,A).
An immediate use of metric projections is in the concept of a sun.
Definition 3.1.4. Let X be a pseudo-metric space. A non-empty proper subset A of X is said to
be:
• an α-sun provided that for every x ∈ X A there exists a ray starting from x such that
d (x,A) = d (x,z) + d (z,A) for all z ∈ .
• a sun provided that for every x ∈ X A there exists a ∈ PA (x) such that
{y ∈ X : x ∈ (a, y)} ∩ A = ∅
and a ∈ PA (y) for all y ∈ X with x ∈ (a, y).
• a strict sun provided that for every x ∈ X A and every a ∈ PA (x) we have that
{y ∈ X : x ∈ (a, y)} ∩ A = ∅
and a ∈ PA (y) for all y ∈ X with x ∈ (a, y).
4
25. 3.1 Geometric preliminary results
It is well known that every strict sun is a sun, and every sun is an α-sun. Examples of strict
suns are the Thales convex sets.
Definition 3.1.5. Let X be a metric space. A convex subset C of X is said to be Thales provided
that for all a = b ∈ C, all y ∈ X {a}, and all x ∈ (a, y) there exists c ∈ (a, b) verifying that
d (y, b)
d (y, a)
=
d (x, c)
d (x, a)
.
The reason why we call these convex sets Thales is because the classical Thales Theorem is
verified.
Proposition 3.1.6. Let X be a metric space and consider a non-empty proper subset A of X. If
A is a Thales convex set, then A is a sun.
Proof. Fix arbitrary elements x ∈ X A and a ∈ PA (x). Now fix another arbitrary element
y ∈ X with x ∈ (a, y).
• Assume that y ∈ A. In this case we have that x ∈ (a, y) ⊂ A because A is convex, which
contradicts the fact that x ∈ X A.
• Let b ∈ A {a}. By hypothesis there exists c ∈ (a, b) verifying that
d (y, b) =
d (y, a)
d (x, a)
d (x, c).
Now simply observe that
d (y, b) =
d (y, a)
d (x, a)
d (x, c) ≥
d (y, a)
d (x, a)
d (x,A) =
d (y, a)
d (x, a)
d (x, a) = d (y, a).
5
26. 3. PRELIMINARY RESULTS
3.1.2 Convexity, balancedness, and absorbance
Given a real vector space and x, y ∈ X, the segment of extremes x and y is usually defined
as [x, y] := {t x + (1 − t) y : t ∈ [0,1]}. Rays and straight lines are defined as follows:
◦ The semi-straight line or ray joining two different points x and y of extreme x is
defined as the set {t y + (1 − t) x : t > 0}.
◦ The straight line joining two different points x and y is defined as the set
{t y + (1 − t) x : t ∈ }.
In the last section of the preface we will show that, in fact, the linear concept and the metric
concept of segment and convexity agree on normed spaces.
Definition 3.1.7. Let X be a real or complex vector space. A non-empty subset A of X is said to
be
• convex provided that [a, b] ⊆ A for a, b ∈ A,
• balanced provided that B (0,1)A ⊆ A, and
• absorbing provided that for all x ∈ X there exists δx > 0 such that B (0,δx )A ⊆ A.
A set that is convex and balanced at the same time is usually called absolutely convex. The
next proposition can be found in [(16),Lemma 2.4].
Proposition 3.1.8. Let X be a real or complex vector space. If M is a convex and balanced
subset of X, then M is absorbing if and only if span(M) = X.
Proof. It is pretty obvious that every absorbing set is a generator system. Conversely, assume
that M is a generator system. Let x ∈ X {0} and consider λ1,...,λn ∈ and m1,..., mn ∈ M
6
27. 3.1 Geometric preliminary results
such that x = λ1m1 +···+λnmn. Because x = 0, we have that |λ1|+···+|λn| > 0, and thus
we can consider
δx :=
1
|λ1| + ··· + |λn|
.
Now, take any α ∈ with |α| ≤ δx . We have that αx = (αλ1) m1 + ··· + (αλn) mn and
|αλ1| + ··· + |αλn| ≤ 1, therefore, since M is absolutely convex, we have that αx ∈ M.
The reader may easily check that the non-empty intersection of any family of convex or
balanced sets is convex or balanced, respectively. And the intersection of any finite family of
absorbing sets is absorbing.
Given a real or complex vector space X and a non-empty subset A of X, then:
• The balanced hull of A is defined as the intersection of all balanced subsets of X con-
taining A and denoted by bl(A). Furthermore,
bl(A) = {λa : λ ∈ B , a ∈ A}.
• The convex hull of A is defined as the intersection of all convex subsets of X containing
A and denoted by co(A). Furthermore,
co(A) =
n
i=1
tiai : ti ∈ [0,1], ai ∈ A,
n
i=1
ti = 1 .
• The absolutely convex hull of A is defined as the intersection of all convex and balanced
subsets of X containing A and denoted by aco(A). Furthermore,
aco(A) =
n
i=1
tiai : ti ∈ , ai ∈ A,
n
i=1
|ti| ≤ 1 .
7
28. 3. PRELIMINARY RESULTS
The convex hull of a balanced set is absolutely convex but the balanced hull of a convex set
may not be convex. In notation form, co(bl(A)) = aco(A) and bl(co(A)) ⊆ aco(A).
3.1.3 Quasi-absolute convexity
The usual concepts of balancedness and absolute convexity will be non-trivially generalized
to fit our purposes.
Definition 3.1.9. Given a real or complex vector space X and a subset A of X, we will say that
A
• almost contains 0 provided that there is a ∈ A such that a − A ⊆ bl(A);
• is quasi-balanced provided that there are a ∈ Aand δ ∈ U (0,1){0} such that a−|δ|A ⊆
A;
• is quasi-absolutely convex provided that it is convex, quasi-balanced, and 0 ∈ A.
The reader may easily find examples of sets almost containing 0 but not containing 0, of
quasi-balanced sets which are not balanced, and of quasi-absolutely convex sets which are
not absolutely convex.
Remark 3.1.10. Let X be a real or complex vector space and consider a subset A of X.
1. If 0 ∈ A, then A almost contains 0.
2. If A is balanced, then it is quasi-balanced.
3. If A is quasi-balanced, then it almost contains 0.
4. If A is convex and 0 ∈ A, then [0,1]A ⊆ A and A+ A = 2A.
5. If A is absolutely convex, then A is quasi-absolutely convex.
8
29. 3.1 Geometric preliminary results
3.1.4 Convexity, segments, and suns
As promised earlier, we will first prove that the vector and metric definitions for convexity
and segment are equivalent on normed spaces.
Theorem 3.1.11. Let X be a normed space, A ⊂ X, and x, y ∈ X. Then:
1. [x, y] = {t y + (1 − t) x : t ∈ [0,1]};
2. A is vector convex if and only if it is metric convex.
Proof. 1. First suppose z ∈ {t y + (1 − t) x : t ∈ [0,1]}. Then
x − z + z − y = x − (t y + (1 − t) x) + t y + (1 − t) x − y
= t x − t y + (1 − t) x − (1 − t) y
= t x − y + (1 − t) x − y
= x − y
Second, suppose z ∈ [x, y].
2. Trivial.
This fact is the linchpin that allows us to prove many theorems on normed spaces from either
a metric or vector perspective. For instance:
Theorem 3.1.12. Let X be a finite dimensional normed space. If A is a bounded convex subset
of X containing 0, then A is quasi-absolutely convex.
Proof. Let Y := span(A). In accordance to (16, Theorem 2.1) we deduce that intY (A) = ∅.
9
30. 3. PRELIMINARY RESULTS
Let a ∈ A and ,τ > 0 such that BY (a, ) ⊆ A ⊆ BY (a,τ). Finally, it suffices to notice that
a −
τ +
A ⊆ a −
τ +
BY (a,τ) = a −
τ +
(a + τBY )
=
τ
τ +
a −
τ
τ +
BY =
τ
τ +
(a − BY )
=
τ
τ +
BY (a, ) ⊆
τ
τ +
A ⊆ A
if we take into consideration Remark 3.1.10(1).
As well:
Theorem 3.1.13. Let X be a normed space and A a non-empty proper convex subset of X. Then
A is Euclidean and thus a sun.
Proof. Consider a = b ∈ A, y ∈ X {a}, and x ∈ (a, y). There exists t > 1 such that
y = t x + (1 − t) a. Observe that (t x + (1 − t) a) − a = t x − a and hence
t =
(t x + (1 − t) a) − a
x − a
.
We have that t−1
t a + 1
t b ∈ A since A is convex. Finally
(t x + (1 − t) a) − b = t x −
t − 1
t
a +
1
t
b .
3.1.5 Linear boundedness
Boundedness is a concept proper of the pre-ordered spaces that can be extended to pseudo-
metric spaces and vector spaces.
10
31. 3.2 Topological preliminary results
Definition 3.1.14. Let X be a real vector space. A subset A of X is said to be linearly bounded
provided that A does not contain rays or straight lines
The reader may quickly notice that a set is linearly bounded if and only if every segment of it
is contained in a maximal segment. As a consequence, linearly bounded sets do not contain
non-trivial vector subspaces. The converse to this last assertion does not hold even under
the hypothesis of balancedness.
Example 3.1.15. The set
(x, y) ∈ 2
: x < 0, y ∈ (−1,0) ∪ (x, y) ∈ 2
: x > 0, y ∈ (0,1) ∪ {(0,0)}
is balanced, does not contain non-trivial vector subspaces of 2
, and is not linearly bounded.
Proposition 3.1.16. Let X be a real vector space. If A is an absolutely convex subset of X, then
A is linearly bounded if and only if A contains no non-trivial vector subspaces of X.
Proof. Assume that A is absolutely convex and contains no non-trvial vector subspaces of X.
Suppose to the contrary that A is not linearly bounded and consider a = b ∈ A such that
{a + t (b − a) : t ∈ [0,∞)} ⊆ A. It is not difficult to see that (b − a) ⊆ A.
3.2 Topological preliminary results
3.2.1 Diagonals
For a topological space X the diagonal of X × X is denoted by
DX := {(x, y) ∈ X × X : x = y}.
11
32. 3. PRELIMINARY RESULTS
In case X is a topological vector space, then the anti-diagonal is defined as
D−
X := {(x, y) ∈ X × X : x = −y}.
Lemma 3.2.1. Let X be a topological vector space.
1. For every (x, y) ∈ X × X we have
(x, y) =
x + y
2
,
x + y
2
+
x − y
2
,
y − x
2
.
2. DX and D−
X are topologically complemented in X × X and both isomorphic to X.
Proof.
1. Immediate.
2. It suffices to notice that the linear projection
P : X × X → DX
(x, y) → P (x, y) =
x+y
2 ,
x+y
2
is continuous and (I − P)(x, y) =
x−y
2 ,
y−x
2 for all (x, y) ∈ X × X.
3.2.2 The finest locally convex vector topology
Theorem 3.2.2 (The finest locally convex vector topology). Let X be a real or complex vector
space. There exists the finest locally convex vector topology τX on X, that is, if ν is a locally
convex vector topology on X, then ν ⊆ τX .
12
33. 3.2 Topological preliminary results
An explicit proof of the previous theorem will not be presented here. Instead, we will sketch
it through a series of definitions and remarks.
Definition 3.2.3. Let X be a real or complex vector space and consider a non-empty subset A
of X.
• We say that x ∈ X is an internal point of A when for every y ∈ X, there exists δy > 0
such that x + λy ∈ A for all λ ∈ 0,δy .
• The set of internal points of A is called the linear interior of A and is denoted by inter(A).
• A is said to be linearly open provided that A = inter(A).
The linearly open sets are precisely the open the sets of the finest locally convex vector
topology.
Remark 3.2.4. Let X be a real or complex vector space. Then
τX := {A ∈ (X) ∅ : A = inter(A)} ∪ {∅}.
3.2.3 Rareness and quasi-absolute convexity
Theorem 3.2.5. Let X be a real or complex topological vector space and consider a quasi-
absolutely convex subset A of X. If the absolutely convex hull of A is not rare, then A is not rare
either.
Proof. By hypothesis we may consider a ∈ A and ∈ (0,1) such that a − A ⊆ A. We will
follow several steps:
• In the first place, we will prove that a + ∈ co(A∪ −A) ⊆ A + A. Indeed, let b, c ∈ A
13
34. 3. PRELIMINARY RESULTS
and t ∈ [0,1]. Notice that
a + (t b + (1 − t)(−c)) = ( t) b + (a + (1 − t)(−c))
∈ A+ A
in virtue of Remark 3.1.10(1).
• In the second and last place, observe that A + A is not rare in virtue of the previous
point, and A+ A = 2A according to Remark 3.1.10(1), therefore 2A is non-rare and so
is A.
Lemma 3.2.6. Let X be a Hausdorff locally convex topological vector space. Let A be an ab-
sorbing subset of X. If X is a Baire space, then A is not rare.
Proof. Since A is absorbing we have that X = n∈ nA. By hypothesis, there exists n ∈ so
that nA is non-rare, and so is A.
3.2.4 Comparison of norms and barrelledness
Let X be a vector space and consider two norms |·| and · on X. It is well known that the
following four assertions are equivalent:
• There exists K > 0 such that |·| ≤ K · .
• The topology induced by |·| is contained in the topology induced by · .
• The unit ball of · , B · , is bounded in (X,|·|).
• B|·| has non-empty interior in (X, · ).
14
35. 3.2 Topological preliminary results
In particular, any of the conditions above implies that B|·| is closed in (X, · ). We will show
now that this last assertion is equivalent to all four points above only when X is barrelled.
Let X be a vector space. Let A be a non-empty subset of X. Note that if A is absorbing, then
the Minkowski functional on A, φA, is well defined. Recall that
φA (x) := inf{λ > 0 : x ∈ λA}
for all x ∈ X. If, in addition, A is absolutely convex, then φA is a semi-norm on X which
verifies that
UφA
⊆ A ⊆ BφA
.
Finally, if, on top of everything else, A is linearly bounded, then φA is a norm on X.
Lemma 3.2.7. Let X be a topological vector space. Let A be a barrel of X and denote by |·| the
semi-norm on X given by the Minkowski functional of A. Then A = B|·|.
Proof. Observe that U|·| ⊆ A ⊆ B|·|. Let x ∈ B|·| and consider any u ∈ U|·|. It is well known
that [u, x) ⊂ U|·|, therefore 1
n u + 1 − 1
n x n∈
is a sequence in A which converges to x in
the original vector topology of X. Since A is closed in that vector topology, we deduce that
x ∈ A.
With these results in mind, we can prove the following then:
Theorem 3.2.8. Let X be a normed space with norm · . The following conditions are equi-
valent:
1. X is barrelled.
2. If |·| is a norm on X whose unit ball is closed in the topology induced by · , then there
exists K > 0 such that |·| ≤ K · .
15
36. 3. PRELIMINARY RESULTS
Proof.
(1)⇒(2) Assume that X is barrelled and let |·| be a norm on X whose unit ball, B|·|, is closed
in the topology induced by · . Finally, notice that B|·| is a barrel of X and thus it has
non-empty interior.
(2)⇒(1) Let A be any barrel of X. Notice that we may assume without any loss of generality
that A is bounded since we can intersect it with the unit ball of X. So, let us suppose
that A is bounded. Denote by |·| the norm on X given by the Minkowski functional of
A. Since A is closed in (X, · ), by Lemma 3.2.7 we have that A = B|·| and hence B|·| is
closed in (X · ). By hypothesis, there exists K > 0 such that |·| ≤ K · , which means
that A is a neighborhood of 0 in (X · ).
16
37. CHAPTER
4
Total anti-proximinality
4.1 Anti-proximinality
4.1.1 Anti-proximinality in pseudo-metric spaces
Definition 4.1.1. Let E be a pseudo-metric space. A non-empty proper subset A of E is said to
be anti-proximinal provided that for all e ∈ E A, the distance from e to A, d (e,A), is never
attained at any a ∈ A. In other words, PA (E A) = {∅}.
Pathological phenomena always occur when dealing with awkward pseudo-metrics as there
are such spaces free of anti-proximinal subsets.
Example 4.1.2. Let X be a set with more than one point.
1. If X is endowed with the null pseudo-metric, that is, d(x, y) = 0 for all x, y ∈ X, then
no non-empty proper subset of X is anti-proximinal. Indeed, if A is a non-empty proper
17
38. 4. TOTAL ANTI-PROXIMINALITY
subset of X and x ∈ X A, then it suffices to consider any a ∈ A to deduce that d (x,A) =
0 = d (x, a).
2. If X is endowed with the discrete metric, that is, d(x, y) = δx y for all x, y ∈ X, then
no non-empty proper subset of X is anti-proximinal. Indeed, if A is a non-empty proper
subset of X and x ∈ X A, then it suffices to consider any a ∈ A to deduce that d (x,A) =
1 = d (x, a).
In order to assure the existence of anti-proximinal sets we need to jump to relatively good
metric spaces.
Proposition 4.1.3. Let X be a metric space. If A is a proper dense subset of X, then A is anti-
proximinal in X.
Proof. Let x ∈ X A and assume that there exists a ∈ A such that d (x, a) = d (x,A). Observe
that d (x,A) = 0 as A is dense in X, thus d(x, a) = 0 which means that x = a ∈ A since d is a
metric. This contradicts the fact that x /∈ A.
Proposition 4.1.4. Let X be a pseudo-metric space. Let A be an anti-proximinal subset of X.
Then:
1. If Y is another pseudo-metric space and f : X → Y is a surjective k-isometry, then f (A)
is anti-proximinal in Y .
2. If X is a metric space and B is a dense subset of A, then B is also anti-proximinal in X.
Proof.
1. Let y ∈ Y f (A) and assume we can find a ∈ A such that d (y, f (a)) = d (y, f (A)).
Since f is surjective there exists x ∈ X such that f (x) = y. Now observe that x /∈ A
and
d (x, a) = kd (y, f (a)) = kd (y, f (A)) = d (x,A),
18
39. 4.1 Anti-proximinality
which means that A is not anti-proximinal in X.
2. Let x ∈ X B and suppose there is b ∈ B such that d (x, b) = d (x, B). We will distin-
guish between two cases:
• Assume that x ∈ A. Then d (x, b) = d (x, B) = 0, which means the contradiction
that x = b ∈ A.
• Assume that x /∈ A. Then d (x, b) = d (x, B) ≤ d (x,A) ≤ d (x, b), which means
the contradiction that A is not anti-proximinal.
Proposition 4.1.5. Let X be a pseudo-metric space. Let {Ai}i∈I be a family of anti-proximinal
subsets of X. If i∈I Ai = X, then i∈I Ai is anti-proximinal.
Proof. Assume to the contrary that there are x ∈ X i∈I Ai and c ∈ i∈I Ai such that
d x, i∈I Ai = d (x, c). There exists j ∈ I such that c ∈ Aj. Observe now that x /∈ Aj and
d (x, c) = d x,
i∈I
Ai ≤ d x,Aj ≤ d (x, c),
which means that Aj is not anti-proximinal.
4.1.2 Anti-proximinality in semi-normed spaces
Proposition 4.1.6. Let E be a semi-normed space. If A is an anti-proximinal subset of E, then
A+ e and λA are both anti-proximinal for every e ∈ E and every λ = 0.
Proof. Fix arbitrary elements e ∈ E and λ = 0. Observe that the maps x → x +e and x → λx
are an isometry and a |λ|-isometry, respectively. Therefore, in accordance to Proposition
19
40. 4. TOTAL ANTI-PROXIMINALITY
4.1.4 we deduce that A+ e and λA are both anti-proximinal.
In Hilbert spaces, anti-proximinal sets verify interesting properties.
Theorem 4.1.7. Let H be a Hilbert space. If A is an anti-proximinal subset of H, then A is
fundamental, that is, span(A) = H.
Proof. Suppose to the contrary that span(A) = H. Take any a ∈ A and any h ∈ span(A)⊥
.
Notice that h + a /∈ A and
(h + a) − a = d (h + a,span(A)) ≤ d (h + a,A) ≤ (h + a) − a ,
which means that A is not anti-proximinal.
Trivial examples of anti-proximinal subsets of semi-normed spaces are the open subsets.
4.1.3 Anti-proximinal convex sets
As expected, the convex hull of an anti-proximinal set is anti-proximinal. As of today, it
is unknown whether there exists an anti-proximinal set whose balanced hull is not anti-
proximinal.
Theorem 4.1.8. Let X be a normed space. Let A be an anti-proximinal subset of X. Then:
1. co(A) is also anti-proximinal.
2. If A almost contains 0, then bl(A) and aco(A) are both anti-proximinal.
Proof.
20
41. 4.1 Anti-proximinality
1. Let x /∈ co(A) and suppose that there exist t1,..., tn ∈ [0,1] and a1,..., an ∈ A such
that t1 = 0, t1 + ··· + tn = 1, and
d (x,co(A)) = x −
n
i=1
tiai .
Notice that
1
t1
x −
n
n=2
tiai /∈ A
and it is not difficult to check that
d
1
t1
x −
n
n=2
tiai ,A =
1
t1
x −
n
n=2
tiai − a1 ,
which means that A is not anti-proximinal.
2. First off, note that aco(A) = co(bl(A)), thus in virtue of 1 of this theorem it only
suffices to show that bl(A) is anti-proximinal. Let x /∈ bl(A) and suppose that there
exists γ ∈ B and a ∈ A such that d (x,bl(A)) = x − γa . We will distinguish between
two cases:
• γ = 0. In this case we have that x = d (x,bl(A)) ≤ x − λb for all b ∈ A
and all λ ∈ B . By hypothesis we can find a0 ∈ A with a0 − A ⊆ bl(A). We will
show that y /∈ A and y − a0 = d (y,A), where y := x + a0. Assume first that
there exists b ∈ A such that x + a0 = b. Then x = −(a0 − b) ∈ −bl(A) = bl(A),
which is not possible. Now consider any b ∈ A. By hypothesis, a0 − b ∈ bl(A),
therefore y − b = x + (a0 − b) ≥ x = y − a0 . This shows that A is not
anti-proximinal.
21
42. 4. TOTAL ANTI-PROXIMINALITY
• γ = 0. Observe that
1
γ
x /∈ A and
d
1
γ
x,A =
1
γ
x − a ,
which means that A is not anti-proximinal.
4.2 Total anti-proximinality
As we already remarked in Example 4.1.2, if a set with more than one point is endowed
with the null pseudo-metric or the discrete metric, then no non-empty proper subset of X
is anti-proximinal. Therefore it really makes no sense to define total anti-proximinality for
pseudo-metric spaces or metric spaces.
4.2.1 Total anti-proximinality in semi-normed spaces
Let X be a semi-normed space. According to (2, Theorem 2.1) we have the following:
• The set V := {x ∈ X : x = 0} is a closed vector subspace of X.
• For every v ∈ V and every x ∈ X we have that v + x = x .
• If A ⊆ x + V for some x ∈ X, then d (y,A) = y − x for all y ∈ X.
• If A ⊆ i∈I (xi + V) for some family {xi : i ∈ I} ⊆ X, then d (y,A) = infi∈I y − xi for
all y ∈ X, and thus d (y,A) is always attained provided that {xi : i ∈ I} is compact.
Lemma 4.2.1. Let X be a semi-normed space and consider the set V = {x ∈ X : x = 0}. Let
A be a non-empty proper subset of X. If there exists x ∈ X A such that (x − A) ∩ V = ∅, then
22
43. 4.2 Total anti-proximinality
A is not anti-proximinal for the pseudo-metric given by the semi-norm.
Proof. Simply notice that
0 ≤ d (x,A) ≤ x − a = 0,
where a ∈ A is so that x − a ∈ V.
Remark 4.2.2. Let X be a semi-normed space. If x ∈ X BX , then
d (x,BX ) = x − 1 = x −
x
x
.
Indeed,
d (x,BX ) ≤ x −
x
x
= x − 1,
and if y ∈ BX , then
x − 1 ≤ x − y ≤ | x − y | ≤ x − y .
Lemma 4.2.3. Let X be a vector space. Let A be a totally anti-proximinal subset of X. Let ·
be a semi-norm on X. If there exists e ∈ X A such that d · (e,A) > 0, then d · (e,A) is not
attained.
Proof. Assume the existence of a ∈ A so that d · (e,A) = e − a . Consider any norm |·| on
X such that |e − a| ≤ e − a . Define a new norm on X given by · := max{ · ,|·|}. Notice
that
e − a ≥ d · (e,A) ≥ d · (e,A) = e − a = e − a ,
which contradicts the fact that A is totally anti-proximinal.
Now with Lemma 4.2.3 and Lemma 4.2.1 in mind, we can prove the following:
23
44. 4. TOTAL ANTI-PROXIMINALITY
Proposition 4.2.4. Let X be a vector space. Let A be a totally anti-proximinal subset of X.
Let · be a semi-norm on X which is not a norm and consider V = {x ∈ X : x = 0}. The
following conditions are equivalent:
1. A is anti-proximinal for the pseudo-metric given by the semi-norm · .
2. For every x ∈ X A we have that (x − A) ∩ V = ∅.
Proof.
1 ⇒2 Let x ∈ X A such that (x − A) ∩ V = ∅. In virtue of Lemma 4.2.1 we deduce that A is
not anti-proximinal for the pseudo-metric given by the semi-norm · .
2 ⇒1 Let x ∈ X A. All we need to show is that d · (x,A) is not attained. Suppose to
the contrary then that d · (x,A) is attained. Bearing in mind Lemma 4.2.3, we may
assume that d · (x,A) = 0. Then there exists a ∈ A such that 0 = d · (x,A) = x − a .
This means that x − a ∈ V and hence (x − A) ∩ V = ∅.
And with Proposition 4.2.4 nailed down, we can demonstrate then that, in fact, total anti-
proximinality can not be defined for semi-normed spaces.
Corollary 4.2.5. Let X be a vector space. No non-empty proper subset of X is anti-proximinal
for every semi-norm defined on X.
Proof. Suppose the existence of a non-empty proper subset A of X which is anti-proximinal
for every semi-norm defined on X. Since every norm is a semi-norm, in particular we have
that A is totally anti-proximinal. We will construct a semi-norm on X and find an element
x ∈ X A such that (x − A) ∩ V = ∅, which will constitute a contradiction in virtue of
24
45. 4.2 Total anti-proximinality
Proposition 4.2.4. By hypothesis there exist x ∈ X A and a ∈ A. At this stage it only suffices
to consider any semi-norm on X whose set V of null-norm vectors contains (x − a).
4.2.2 Total anti-proximinality in normed spaces
However, in normed spaces, total anti-proximinality not only makes sense but plays a fun-
damental role in the geometry of those spaces.
Definition 4.2.6. A subset A of a vector space E is said to be totally anti-proximinal when it is
anti-proximinal for every norm on E.
Theorem 4.2.7. Let E be a vector space. Let A be a non-empty proper subset of E. If A is totally
anti-proximinal in E, then A is a generator system of E, that is, span(A) = E.
Proof. Denote P := span(A) and suppose P = E. Let Q be an algebraical complement for
P in E, that is, P ⊕ Q = E. Let · P and · Q be any norms on P and Q respectively, and
consider the following norm on E given by
p + q := p 2
P + q 2
Q
where p ∈ P and q ∈ Q. It is not difficult to check that d (p + q, P) = q Q for all p ∈ P and
q ∈ Q. Note then that if a ∈ A and q ∈ Q, then a + q /∈ A and
q Q ≥ d (a + q,A) ≥ d (a + q, P) = q Q ,
which means that A is not anti-proximinal.
By bearing in mind Proposition 4.1.5 and Theorem 4.1.8 we have the following remark.
Remark 4.2.8. Let X be a vector space.
25
46. 4. TOTAL ANTI-PROXIMINALITY
1. If {Ai}i∈I is a family of totally anti-proximinal subsets of X such that i∈I Ai = X, then
i∈I AiX is also totally anti-proximinal.
2. If A is a totally anti-proximinal subset of X, then co(A) is also totally anti-proximinal.
3. If A is a totally anti-proximinal subset of X which almost contains 0, then both bl(A) and
aco(A) are totally anti-proximinal.
Remark 4.2.9. Let X be a vector space. We will denote by to the following fields: if X is
real and + i if X is complex. It is not difficult to check that if B is a Hamel basis for X, then
Y := l1 b1 + ··· + lp bp : lj ∈ , bj ∈ B,1 ≤ j ≤ p, p ∈
is dense in X no matter what the vector topology X is endowed with. Notice that Y is a -vector
space.
Proposition 4.2.10. Let X be a vector space. Let A be a non-empty linearly open subset of X.
Then A ∩ Y is totally anti-proximinal but neither linearly open nor convex, where Y is the set
considered in Remark 4.2.9.
Proof. Firstly, notice that the density of Y in X endowed with the finest locally convex vector
topology implies that A∩Y = ∅. In fact, A∩Y is dense in A. Let x ∈ X (A∩ Y ) and a ∈ A∩Y .
Since A is linearly open, there exists q ∈ (0,1) ∩ such that qx + (1 − q) a ∈ A. Consider
p ∈ , λ1,...,λp ∈ , and b1,..., bp ∈ B such that x = λ1 b1 + ··· + λp bp, where B is a
Hamel basis for X. Now let · be any norm on X. Again because A is linearly open we can
find l1,..., lp ∈ such that
|li − λi| <
x − a
p bi
for 1 ≤ i ≤ p
and
q l1 b1 + ··· + lp bp + (1 − q) a ∈ A∩ Y.
26
47. 4.2 Total anti-proximinality
We have the following:
x − q l1 b1 + ··· + lp bp + (1 − q) a
≤ x − (qx + (1 − q) a)
+ (qx + (1 − q) a) − q l1 b1 + ··· + lp bp + (1 − q) a
= (1 − q) x − a + q |λ1 − l1| b1 + ··· + q λp − lp bp
< (1 − q) x − a + q x − a
= x − a .
As a consequence, d · (x,A∩ Y ) is never attained and thus Ais anti-proximinal in X endowed
with the norm · .
The reader may notice that the main ideas of the proof of Proposition 4.2.10 can be taken
advantage of to show the following more general result, which in fact is a direct consequence
of Proposition 4.1.4(2).
Proposition 4.2.11. Let X be a vector space. If A is a totally anti-proximinal subset of X and
B is a subset of A which is dense in A for any norm on X, then B is also totally anti-proximinal
in X.
This proposition will be of much use throughout the rest of the manuscript. As well, it serves
to show counter-examples to many assertions on the properties of totally anti-proximinal
sets.
Example 4.2.12.
• The intersection of totally anti-proximinal sets is not always totally anti-proximinal.
Indeed, the sets
(−1,1) ∩ and (−1,1) ∩ ({0} ∪ )
27
48. 4. TOTAL ANTI-PROXIMINALITY
are both totally anti-proximinal subsets of , however their intersection is {0} which is
not totally anti-proximinal.
• Total anti-proximinality is not hereditary to vector subspaces. Indeed, let X := 2
and consider F := × {0} and
A := (x, y) ∈ × ( {0}) : x2
+ y2
≤ 1 ∪ {(0,0)}.
It is not difficult to check that A is a totally anti-proximinal non-convex subset of X.
However, A∩ F = {(0,0)} is not a generator system of F.
• A non-linearly open totally anti-proximinal set which has internal points. Indeed,
let X be the real line and A := (0,1) ∪ ( ∩ (0,2)). Then A is a totally anti-proximinal
subset of X such that inter(A) = ∅ but A is not open.
Another proposition that will come in extremely handy throughout the rest of the manuscript
comes from (18):
Proposition 4.2.13. Let X be a vector space. Let A be a non-empty subset of X.
1. If A = inter(A), then A is totally anti-proximinal.
2. Conversely, if A is totally anti-proximinal, absolutely convex and contains no half-line of
X, then A = inter(A).
Proof.
1. Let · be any norm on X and consider x ∈ X A. Take any a ∈ A and consider
the straight line passing through a. By hypothesis, there is t ∈ (0,1) such that t x +
(1 − t) a ∈ A. Observe now that
x − a > (1 − t) x − a = x − (t x + (1 − t) a) ≥ d · (x,A).
28
49. 4.2 Total anti-proximinality
2. Assume the existence of a ∈ A inter(A). Since span(A) = X (Theorem 4.2.7), by
Lemma 3.1.8 we have that A is absorbing in X, therefore the Minkowski functional of
A in X defines a norm on X which we will denote by · . Simply observe now that
a = 1, since otherwise a ∈ inter(A). Indeed, observe that
U · = inter(A) ⊂ A ⊂ B · .
By hypothesis, d · (2a,A) is never attained. However,
d · (2a,A) = d · 2a,B · = 2a − 1 = 1 = 2a − a ,
which is a contradiction.
4.2.3 Totally anti-proximinal convex sets
Lemma 4.2.14. Let E be a vector space. If A is totally anti-proximinal subset of A contained in
the closed unit ball of a semi-norm on X, then A is actually contained in the open unit ball of
that semi-norm.
Proof. Let · be any semi-norm on X whose closed unit ball B · contains A. Suppose to the
contrary that there exists a ∈ A∩ S · . By applying Remark 4.2.2 we have that
1 = 2a − a ≥ d (2a,A) ≥ d 2a,B · = 2a − 1 = 1,
which contradicts Lemma 4.2.3.
We remind the reader that a subset of a topological vector space is said to be finitely open
29
50. 4. TOTAL ANTI-PROXIMINALITY
provided that its intersection with every finite dimensional subspace is open in the Euclidean
topology. According to [(20), Theorem 3.2], a set is finitely open if and only if it is linearly
open.
Theorem 4.2.15. Let X be a vector space. Suppose that A is a totally anti-proximinal convex
subset of X. Then the absolutely convex hull of A coincides with the open unit ball of the semi-
norm that it generates and hence it is finitely open.
Proof. It is well known that in this case and since A is convex, the absolutely convex hull of
A is given by co(A∪ −A). Denote by · the semi-norm generated by the absolutely convex
hull of A. Since
A,U · ⊆ co(A∪ −A) ⊆ B · ,
by applying Lemma 4.2.14 we deduce that A ⊆ U · , which automatically implies in virtue of
the triangular inequality that co(A∪ −A) = U · .
Corollary 4.2.16. Let X be a vector space and A a non-empty proper subset of X. If A is totally
anti-proximinal and absolutely convex, then A is finitely open.
Hence we’ve removed the hypothesis of linear boundedness from Theorem 4.2.13(2).
Though we have met the goal of this section, we can go a little bit farther. It will not get us
the complete removal of absolute convexity from Theorem 4.2.13(2), but it will get us close.
Corollary 4.2.17. Let X be a vector space. Let A be a totally anti-proximinal subset of X. If
f : X → is non-zero and linear, then sup f (A) is never attained.
Proof. Suppose to the contrary that a ∈ A is so that f (a) = sup f (A). Take any x ∈ X A
such that f (x) > f (a). Consider the semi-norm on X given by · := |f (·)|. Since 0 <
|f (x − a)| ≤ |f (x − b)| for all b ∈ A, we deduce that d (x,A) = x − a , which contradicts
Lemma 4.2.3.
30
51. 4.2 Total anti-proximinality
Scholium 4.2.18. Let X be a vector space. If A is a totally anti-proximinal convex subset of X
such that inter(A) = ∅, then A = inter(A), that is, A is linearly open.
Proof. Suppose to the contrary the existence of an element a ∈ A inter(A). Assume X
endowed with the finest locally convex vector topology. In this situation, int(A) = inter(A) =
∅. According to the Hahn-Banach Separation Theorem, there exists f ∈ X∗
{0} such that
f (a) ≥ sup f (int(A)) = sup f (A), which indeed implies that f (a) = sup f (A). This fact
contradicts Corollary 4.2.17.
At the very end of the next chapter it is shown that the previous scholium does not hold
true if we remove the hypothesis of convexity. In fact, we will explain why the hypothesis of
absolute convexity cannot be removed from Theorem 4.2.13(2).
31
52.
53. CHAPTER
5
Ricceri’s Conjecture
5.1 Anti-proximinal properties
In this section we find the original anti-proximinal property established by Ricceri (see (28))
and a couple of weakenings (see (19; 20)) that are helpful to understand how to approach
Ricceri’s Conjecture. In concrete terms, we have that the anti-proximinal property implies
the quasi anti-proximinal porperty which is equivalent to the weak anti-proximinal property.
5.1.1 The weak and the quasi anti-proximinal properties
Definition 5.1.1. A Hausdorff locally convex topological vector space is said to enjoy
• the weak anti-proximinal property if every totally anti-proximinal absolutely convex sub-
set is not rare;
33
54. 5. RICCERI’S CONJECTURE
• the quasi anti-proximinal property if every totally anti-proximinal quasi-absolutely con-
vex subset is not rare.
Theorem 5.1.2. Let X be a Hausdorff locally convex topological vector space. The following
conditions are equivalent:
1. X satisfies the weak anti-proximinal property.
2. X satisfies the quasi anti-proximinal property.
3. X is barrelled.
Proof.
1 ⇒ 2 If every totally anti-proximinal quasi-absolutely convex subset of X is not rare, then X
satisfies the weak anti-proximinal property since Remark 3.1.10(2) assures that every
absolutely convex subset of X is quasi-absolutely convex.
1 ⇒ 2 Suppose to the contrary that X is not barrelled. By hypothesis X has a barrel M with
empty interior. Consider A := inter(M), which is an absolutely convex set since M is
so. In accordance to Theorem 4.2.13(1) we deduce that A is totally anti-proximinal.
However, int(cl(A)) = int(M) = ∅.
Conversely, assume that X has the weak anti-proximinal property and consider any
totally anti-proximinal quasi-absolutely convex subset A of X. We may assume without
any loss of generality that 0 ∈ A in virtue of (18, Remark 3.1(2)). By applying (18,
Remark 3.7(3)) we have that the absolutely convex hull of A is totally anti-proximinal,
therefore it will be not rare by hypothesis. Finally, Theorem 3.2.5 allows us to deduce
that A is not rare either.
3 ⇒ 1 Let A be a totally anti-proximinal absolutely convex subset of E. Notice that A is a
generator system of E in virtue of Theorem 4.2.7. Next, A is absorbing in view of
34
55. 5.1 Anti-proximinal properties
Lemma 3.1.8, therefore its closure is a barrel of E. By hypothesis, cl(A) has non-empty
interior.
Now, with the following in mind from (26; 30),
Theorem 5.1.3. (Saxon and Wilanski)
Let X be an infinite dimensional Banach space. The following conditions are equivalent:
1. X admits an infinite dimensional separable quotient.
2. There exists a non-barrelled dense subspace Y of X.
we can prove the following:
Corollary 5.1.4. Let X be an infinite dimensional Banach space admitting an infinite dimen-
sional separable quotient. Then X satisfies the weak anti-proximinal property but admits a
proper dense subspace not enjoying it.
Proof. In accordance to Theorem 5.1.3, X has a non-barrelled dense subspace Y . Now The-
orem 5.1.2 assures that X has the weak anti-proximinal property and that Y does not.
5.1.2 Spaces without the weak anti-proximinal property
This section is devoted to explicitly construct proper dense subspaces without the weak anti-
proximinal property in infinite dimensional separable Banach spaces. First, we need the
following three results from (17):
Lemma 5.1.5 (Garcia-Pacheco, (17)). Let X be an infinite dimensional separable Banach
35
56. 5. RICCERI’S CONJECTURE
space. Let en, e∗
n n∈
⊂ SX × X∗
for X be a Markushevich basis for X. The linear operator
1 → X
(tn)n∈ →
∞
n=1
tnen
maps ω∗
-closed, bounded subsets of 1 to sequentially ω-closed subsets of X. As a consequence,
the set
∞
n=1
tnen : (tn)n∈ ∈ B 1
is closed in X, and therefore it has empty interior in X if an only if it has empty interior in its
linear span.
Lemma 5.1.6 (Garcia-Pacheco, (17)). Let X be an infinite dimensional separable Banach
space. Let en, e∗
n n∈
⊂ SX × X∗
for X be a Markushevich basis for X. Then the following
statements are equivalent:
1. The basis (en)n∈ is a Schauder basis equivalent to the 1-basis.
2. The operator
1 → X
(tn)n∈ →
∞
n=1
tnen
is an isomorphism.
3. The set
∞
n=1
tnen : (tn)n∈ ∈ B 1
has non-empty interior.
Lemma 5.1.7 (Garcia-Pacheco, (17)). Let X be an infinite dimensional separable Banach
space. There exists a normalized Markushevich basis for X which is not a Schauder basis equi-
36
57. 5.1 Anti-proximinal properties
valent to the 1-basis.
Then, continuing our look at Markushevich bases, we find:
Remark 5.1.8. Let X be an infinite dimensional separable Banach space and consider a Markushev-
ich basis en, e∗
n n∈
⊂ SX × X∗
for X. Every point of the absolutely convex set
∞
n=1
tnen : (tn)n∈ ∈ U 1
is internal in
∞
n=1
tnen : (tn)n∈ ∈ 1 .
Indeed, it is a direct consequence of the fact that U 1
is open in 1.
As well, we will need the following technical lemma:
Lemma 5.1.9. Let X be a topological space. Let Z be a subset of X. If M is a subset of Z which
is closed in X, then
intZ (M) = intcl(Z) (M).
Proof. Let x ∈ intZ (M) and consider an open set U in X such that x ∈ U ∩ Z ⊆ M. First,
we will show that U ∩ cl(Z) ⊆ cl(U ∩ Z). Let y ∈ U ∩ cl(Z) and consider any open set
V containing y. Since y ∈ cl(Z) and U ∩ V is an open neighborhood of y we have that
V ∩ (U ∩ Z) = (U ∩ V) ∩ Z = ∅. As a consequence, y ∈ cl(U ∩ Z) and hence U ∩ cl(Z) ⊆
cl(U ∩ Z). Therefore, x ∈ U ∩ Z ⊆ U ∩ cl(Z) ⊆ cl(U ∩ Z) ⊆ cl(M) = M and hence x ∈
intcl(Z) (M). Conversely, let x ∈ intcl(Z) (M) and consider an open set U in X such that x ∈
U ∩ cl(Z) ⊆ M. Notice that x ∈ Z since M ⊆ Z. Therefore, x ∈ U ∩ Z ⊆ U ∩ cl(Z) ⊆ M and
hence x ∈ intZ (M).
Now we are in the right position to state and prove the main result of this section. Since
Lemma 5.1.7 tells us that every infinite dimensional separable Banach space has a Schauder
37
58. 5. RICCERI’S CONJECTURE
basis which is not equivalent to the 1-basis, we can state the result as follows:
Theorem 5.1.10. Let X be an infinite dimensional separable Banach space and consider a
Markushevich basis en, e∗
n n∈
⊂ SX × X∗
for X which is not equivalent to the 1-basis. Then
∞
n=1
tnen : (tn)n∈ ∈ 1
is a dense subspace of X which does not satisfy the weak anti-proximinal property.
Proof. Consider the absolutely convex set
A :=
∞
n=1
tnen : (tn)n∈ ∈ U 1
.
Notice that every point of A is internal in the dense subspace
Y :=
∞
n=1
tnen : (tn)n∈ ∈ 1
in virtue of Remark 5.1.8. As a consequence, A is totally anti-proximinal in Y if we bear in
mind Theorem 4.2.13(1). According to Lemma 5.1.5 the set
B :=
∞
n=1
tnen : (tn)n∈ ∈ B 1
is closed in X. This fact, with the collaboration of Lemma 5.1.9, brings up two consequences:
• The closure of A in X is B. Indeed, it suffices to realize that the closure of A in Y is B
and that B is closed in X.
• The interior of B in X coincides with the interior of B in Y . Indeed, it is enough to take
a look at Lemma 5.1.9.
On the other hand, in accordance with Lemma 5.1.6, the fact that en, e∗
n n∈
is not equivalent
38
59. 5.2 Ricceri’s Conjecture
to the 1-basis implies that B has empty interior in X and so does B in Y . In other words,
A is a totally anti-proximinal absolutely convex subset of Y which is also rare. This implies
that Y does not enjoy the weak anti-proximinal property. And it is clear that Y is a dense
subspace of X.
5.1.3 The anti-proximinal property
Definition 5.1.11 (Ricceri, (28)). A Hausdorff locally convex topological vector space is said
to have the anti-proximinal property if every totally anti-proximinal convex subset is not rare.
Theorem 5.1.12. Every non-zero finite dimensional Hausdorff topological vector space enjoys
the anti-proximinal property.
Proof. Let X be any non-zero finite dimensional Hausdorff topological vector space X and
consider A to be any totally anti-proximinal convex subset of X. We may assume without
any loss of generality that 0 ∈ A. By Theorem 4.2.7 we deduce that span(A) = X. Finally, in
view of Theorem (16, Theorem 2.1) we have that int(A) = ∅.
5.2 Ricceri’s Conjecture
The most famous conjecture ever stated by Ricceri states the following:
Conjecture 5.2.1 (Ricceri’s Conjecture, (28)). There exists a non-complete normed space en-
joying the anti-proximinal property.
5.2.1 Weak (positive) approach
Theorem 5.2.2. There exists a non-complete real normed space enjoying the weak anti-proximinal
property.
39
60. 5. RICCERI’S CONJECTURE
Proof. Let X be an infinite dimensional Banach space X and consider a non-continous lin-
ear functional f : X → . It is well known that ker(f ) is not closed and hence not com-
plete either. Now X is complete, therefore it is barrelled. Since ker(f ) is of countable
co-dimension in X (see (29; 32)) we deduce that ker(f ) is also barrelled and thus it enjoys
the weak anti-proximinal property (see Theorem 5.1.2).
5.2.2 Quasi (positive) approach
The reader may notice that the previous corollary constitutes a partial positive solution to
Ricceri’s Conjecture in the following sense:
Theorem 5.2.3. If every totally anti-proximinal convex set containing 0 is quasi-absolutely
convex, then Ricceri’s Conjecture holds true.
Proof. Indeed, if every totally anti-proximinal convex set containing 0 is quasi-absolutely
convex, then the weak anti-proximinal property and the anti-proximinal property are equi-
valent, and according to (18, Theorem 1.3) there exists a non-complete normed space en-
joying the anti-proximinal property.
5.2.3 Intern (positive) approach
In the whole of this subsection we will assume that every totally anti-proximinal convex set
has an absorbing translate.
Theorem 5.2.4. Let X be a Hausdorff locally convex topological vector space. If X is a Baire
space, then X satisfies the anti-proximinal property.
Proof. Let A be a totally anti-proximinal convex subset of X. By our assumption we deduce
the existence of a translate of A which is absorbing. In virtue of Lemma 3.2.6 we have that
40
61. 5.2 Ricceri’s Conjecture
that translate of A is non-rare, and so is A.
Theorem 5.2.4 together with several classic results (and our assumption) will give us the
key to prove Ricceri’s Conjecture true.
Example 5.2.5 (Positive Solution to Ricceri’s Conjecture). There exists a non-complete normed
space satisfying the anti-proximinal property. Indeed, it suffices to consider any non-complete
normed space which is a Baire space and apply Theorem 5.2.4. For an example of a non-complete
normed space which is Baire take a look at (3, Chapter 3) where it is observed that if E is a
separable, infinite-dimensional Banach space, then E contains a dense subspace M of countably
infinite co-dimension which is a Baire space.
41
62.
63. CHAPTER
6
Geometric characterizations of
Hilbert spaces
6.1 The set ΠX
In (4) the authors formally introduce the set ΠX := {(x, x∗
) ∈ SX × SX∗ : x∗
(x) = 1} for X
a normed space and they use it to define a modulus of the Bishop-Phelp-Bollobás property
for functionals. However, the set ΠX appears implicitly in other indices or moduli such as
the numerical index of a Banach space, since the numerical range of a continuous linear
operator T ∈ (X) can be rewritten as V(T) := {x∗
(T(x)) : (x, x∗
) ∈ ΠX }. We refer the
reader to (22) for an excellent survey paper on the numerical index of a Banach space.
It is well known that if H is a Hilbert space, then its duality mapping JH is a surjective linear
isometry, and so we can identify H with H∗
via its dual map. After this identification, ΠH
turns out to be the intersection of SH × SH with the diagonal of H × H.
43
64. 6. GEOMETRIC CHARACTERIZATIONS OF HILBERT SPACES
6.1.1 Extremal structure of ΠX
Given a normed space X we will define the set EX := (ext(BX ) × SX∗ ) ∪ (SX × ext(BX∗ )).
Theorem 6.1.1. Let X be a normed space. The following conditions are equivalent:
1. ΠX ⊆ EX .
2. SX = ext(BX ) ∪ smo(BX ).
Proof.
1. ⇒ 2. Let x ∈ SX ext(BX ). If x /∈ smo(BX ), then there are x∗
= y∗
∈ SX∗ such that x∗
(x) =
y∗
(y) = 1. Notice that x,
x∗
+y∗
2 ∈ ΠX but neither x nor
x∗
+y∗
2 are extreme points of
their respective balls.
2. ⇒ 1. Let (x, x∗
) ∈ ΠX . Assume that x /∈ ext(BX ). By hypothesis x ∈ smo(BX ). Now if
y∗
,z∗
∈ SX∗ and x∗
=
y∗
+z∗
2 , then y∗
(x) = z∗
(x) = 1 which means that y∗
= x∗
by
the smothness of x.
We recall the reader that an exposed face is the set of all vectors of norm 1 at which a given
functional of norm 1 attains its norm. An edge is a maximal segment of the unit sphere
which is an exposed face.
Corollary 6.1.2. Let X be a normed space.
1. If ΠX ⊆ EX , then every edge of BX is a maximal face of BX .
2. If X is real and 2-dimensional, then ΠX ⊆ EX .
Proof.
44
65. 6.1 The set ΠX
1. Let [x, y] ⊂ SX be an edge of BX and consider u∗
∈ SX∗ such that [x, y] = (u∗
)−1
(1) ∩
BX . Suppose to the contrary that [x, y] is not a maximal face of BX , so then it must be
contained in a maximal face C. According to the Hahn-Banach Separation Theorem,
maximal faces are exposed faces, so there exists v∗
∈ SX∗ such that C = (v∗
)−1
(1)∩BX .
Note that u∗
= v∗
since [x, y] C. Finally,
x+y
2 ∈ SX but
x+y
2 /∈ ext(BX ) ∪ smo(BX ).
2. If x ∈ SX ext(BX ), then x belongs to the interior of a segment entirely contained in
the unit sphere. Since X is real and has dimension 2, there is only one hyperplane
supporting BX on that segment, and hence x ∈ smo(BX ).
The next example shows the existence of Banach spaces which can never be equivalently
renormed to achieve that ΠX ⊆ EX . For this we will need a bit of background.
Let ω1 denote the first uncountable ordinal. The space of all bounded real-valued functions
on [0,ω1] will be denoted by ∞ (0,ω1), which becomes a Banach space endowed with the
sup norm. The subspace of ∞ (0,ω1) composed of those functions with countable support
is denoted by m0.
Theorem 6.1.3. No equivalent norm on m0 makes Πm0
⊆ Em0
.
Proof. We will divide the proof in two steps:
1. Πm0
Em0
when m0 is endowed with the sup norm. Indeed, note that in this case, m0
endowed with the sup norm isometrically contains 3
∞. Now observe that Theorem
6.1.1 shows that the condition ΠX ⊆ EX is an hereditary property. Finally, it is sufficient
to realize that Π 3
∞
E 3
∞
in virtue of Corollary 6.1.2(1).
2. Assume that m0 is endowed with any equivalent norm. In accordance to (15, Theorem
7.12), m0 endowed with any (non-necessarily equivalent) norm has a subspace which
45
66. 6. GEOMETRIC CHARACTERIZATIONS OF HILBERT SPACES
is linearly isometric to m0 endowed with the sup norm. Again, the hereditariness of
the condition ΠX ⊆ EX together with 1. concludes the proof.
6.1.2 The distance to ΠH
Our final aim is at finding the distance of a generic element (h, k) ∈ H ⊕2 H to ΠH for H
a Hilbert space. In order to accomplish this we will make use of the following couple of
lemmas. However, we will first study this issue in a more general situation.
Proposition 6.1.4. Let X be a normed space and consider ΠX in X ⊕2 X∗
. Let x ∈ SX and
y∗
∈ SX∗ .
1. d ((x, y∗
),ΠX ) ≤ d y∗
, x−1
(1) ∩ BX∗ .
2. If y is norm-attaining, then d ((x, y∗
),ΠX ) ≤ d x,(y∗
)−1
(1) ∩ BX .
3. |y∗
(x) − 1| ≤ 2d ((x, y∗
),ΠX ).
Proof.
1. Let x∗
∈ x−1
(1)∩BX∗ , then (x, x∗
) ∈ ΠX and so d ((x, y∗
),ΠX ) ≤ (x, y∗
)−(x, x∗
) 2 =
y∗
− x∗
, which means that d ((x, y∗
),ΠX ) ≤ d y∗
, x−1
(1) ∩ BX∗ .
2. It follows a similar proof as in 1.
46
67. 6.1 The set ΠX
3. Let (z,z∗
) ∈ ΠX . Note that
|y∗
(x) − 1| = |y∗
(x) − z∗
(z)|
≤ |y∗
(x) − z∗
(x)| + |z∗
(x) − z∗
(z)|
≤ y∗
− z∗
+ x − z
≤ 2 (x, y∗
) − (z,z∗
) 2
which implies that |y∗
(x) − 1| ≤ 2d ((x, y∗
),ΠX ).
Corollary 6.1.5. Let X be a normed space and consider ΠX in X ⊕2 X∗
. If x ∈ SX and y∗
∈ SX∗
is norm-attaining, then
|y∗
(x) − 1|
2
≤ d ((x, y∗
),ΠX ) ≤ min d y∗
, x−1
(1) ∩ BX∗ , d x,(y∗
)−1
(1) ∩ BX .
It is time now to take care of computing the distance of a generic element (h, k) ∈ H ⊕2 H to
ΠH.
Lemma 6.1.6. Let X be a normed space. If x ∈ X {0}, then d (x,SX ) = x − x
x = | x − 1|.
Proof. Indeed, d (x,SX ) ≤ x − x
x = | x − 1| and if y ∈ SX , then
x −
x
x
= | x − 1| = | x − y | ≤ x − y . (6.1.1)
Lemma 6.1.7. Let X be a normed space and assume that X = M ⊕p N with 1 ≤ p ≤ ∞. Fix
arbitrary elements m ∈ M and n ∈ N.
1. d (m + n, M) = n .
47
68. 6. GEOMETRIC CHARACTERIZATIONS OF HILBERT SPACES
2. d (m + n,SM ) =
p
n p
+ | m − 1|p
if p < ∞,
max{ n ,| m − 1|} if p = ∞.
Proof.
1. Indeed, d (m + n, M) ≤ m + n − m = n and if m ∈ M then
n ≤ m − m
p
+ n p
1
p
= m + n − m p
for p < ∞,
n ≤ max m − m , n = m + n − m p
for p = ∞.
2. Indeed, we may assume that m = 0 and attending to Equation (6.1.1) we have that
d (m + n,SM ) ≤ m + n −
m
m p
=
p
n p
+ | m − 1|p
if p < ∞,
max{ n ,| m − 1|} if p = ∞,
and if m ∈ SM then
p
n p
+ | m − 1|p
≤
p
n p
+ m − m p
= m + n − m p
for p < ∞,
max{ n ,| m − 1|} ≤ max n , m − m = m + n − m p
for p = ∞.
The reader may notice that Lemma 6.1.7(1) still holds if M and N are simply 1-complemented
in X.
Theorem 6.1.8. Let H be a Hilbert space and consider H ⊕2 H. For every h, k ∈ H we have that
d ((h, k),DH) =
h−k
2
d (h, k),SDH
=
h−k 2
2 +
h+k
2
− 1
2
1
2
d (h, k), 2SDH
=
h−k 2
2 +
h+k
2
− 2
2
1
2
48
69. 6.1 The set ΠX
Proof. First off, notice that H ⊕2 H = DH ⊕2 D−
H in virtue of Theorem 6.2.1. By applying
Lemma 6.1.7(1) we deduce that
d ((h, k),DH) =
h − k
2
,
k − h
2 2
=
h − k
2
.
In accordance with 2. of Lemma 6.1.7 we have that
d (h, k),SDH
=
h − k 2
2
+
h + k
2
− 1
2
1
2
.
Finally,
d (h, k), 2SDH
= d 2
1
2
(h, k) , 2SDH
= 2d
h
2
,
k
2
,SDH
= 2
h − k 2
4
+
h + k
2
− 1
2
1
2
=
h − k 2
2
+
h + k
2
− 2
2
1
2
As we mentioned at the beginning of this section, ΠH = 2SDH
, so we immediately deduce
the following final corollary.
Corollary 6.1.9. Let H be a Hilbert space and consider H ⊕2 H. If h, k ∈ H, then
d ((h, k),ΠH) =
h − k 2
2
+
h + k
2
− 2
2
1
2
.
49
70. 6. GEOMETRIC CHARACTERIZATIONS OF HILBERT SPACES
6.2 Geometric characterizations of Hilbert spaces
6.2.1 Using diagonals
Theorem 6.2.1. Let H be a Hilbert space and consider H ⊕2 H. Then (DH)⊥
= D−
H .
Proof. Let h, k ∈ H. By the Parallelogram Law we have that
(h, k) 2
2 = h 2
+ k 2
=
h + k 2
2
+
h − k 2
2
=
h + k
2
2
+
h + k
2
2
+
h − k
2
2
+
h − k
2
2
=
h + k
2
,
h + k
2
2
2
+
h − k
2
,
k − h
2
2
2
.
Corollary 6.2.2. Let X be a Banach space. If DX and D−
X are L2
-complemented in X ⊕2 X, that
is, X ⊕2 X = DX ⊕2 D−
X , then X is a Hilbert space.
Proof. Indeed, it suffices to look at the proof of Theorem 6.2.1 to realize that, under these
assumptions, X verifies the Parallelogram Law and thus it is a Hilbert space.
6.2.2 Using ΠX
If H denotes a Hilbert space, then it is clear that
ΠH = (SH × SH) ∩ DH = 2SDH
provided that H × H is endowed with the · 2-norm.
50
71. 6.2 Geometric characterizations of Hilbert spaces
Theorem 6.2.3. Let X be a Banach space. If there exists a vector subspace V of X ⊕2 X∗
such
that ΠX = 2SV , then X is a Hilbert space and V = DX .
Proof. We will divide the proof in two steps:
1. First off, we will show that X is smooth. Suppose to the contrary that X is not, then
we can find (x, x∗
),(x, y∗
) ∈ ΠX such that x∗
= y∗
. Then
(0, x∗
− y∗
) = (x, x∗
) − (x, y∗
) ∈ ΠX − ΠX ⊆ V.
Thus
2
(0, x∗
− y∗
)
x∗ − y∗
∈ 2SV = ΠX ,
which is impossible.
2. According to (1, Theorem 3.2) it is sufficient to show that JX (x + y) = JX (x)+JY (y)
for all x, y ∈ SX . So fix arbitray elements x, y ∈ SX . We may assume that x and y are
linearly independent. Note that
(x + y,JX (x) + JX (y)) = (x,JX (x)) + (y,JX (y)) ∈ ΠX + ΠX ⊆ V.
Therefore
2
(x + y,JX (x) + JX (y))
x + y 2
+ JX (x) + JX (y) 2
∈ 2SV = ΠX .
So there exists z ∈ SX such that
2
(x + y,JX (x) + JX (y))
x + y 2
+ JX (x) + JX (y) 2
= (z,JX (z)).
This implies that
z =
x + y
x + y
51
72. 6. GEOMETRIC CHARACTERIZATIONS OF HILBERT SPACES
and
JX
x + y
x + y
= 2
JX (x) + JX (y)
x + y 2
+ JX (x) + JX (y) 2
. (6.2.1)
Taking norms and solving for JX (x) + JX (y) we obtain that
JX (x) + JX (y) = x + y .
Going to back to Equation (6.2.1), we deduce that
JX (x + y) = JX (x) + JY (y).
52
73. CHAPTER
7
Applications to transcranial
magnetic stimulation
In this chapter an inverse boundary element method and efficient optimisation techniques
are combined to produce a versatile framework to design truly optimal TMS coils. The
presented approach can be seen as an improvement of the work introduced by Cobos Sanc-
hez et al. (8) where the optimality of the resulting coil solutions was not guaranteed. In
fact, (8) is improved and extended to produce a computational optimisation framework for
designing true optimal TMS coils of arbitrary shape. The presented technique is based on the
combination of general optimisation techniques with a stream function IBEM, which permits
the modelling of most of the TMS coil performance features as convex objectives. To illus-
trate the versatility of this computational framework, novel requirements and constraints are
prototyped here, such as minimum mechanical stress, minimum coil heating or maximum
current density.
53
74. 7. APPLICATIONS TO TRANSCRANIAL MAGNETIC STIMULATION
This new numerical framework has been efficiently applied to produce TMS coils with ar-
bitrary geometry, allowing the inclusion of new coil features in the design process, such
as optimised maximum current density or reduced temperature. Even the structural head
properties have been considered to produce more realistic TMS stimulators.
The structure of this chapter is as follows. Firstly we review some of the most relevant
requirements and parameters that described the performance of a TMS coil. Secondly an
outline of the stream function IBEM is presented, which allows to formulate the TMS coil
design as an optimisation problem.
7.1 TMS coil requirements and performance
In the following, the properties that assess the efficiency of a TMS coil are listed.
7.1.1 Stored magnetic energy
Minimum stored magnetic energy (or equivalently minimum inductance) is an important
requirement in TMS coil design, as it enables the most rapid switching possible of the TMS
fields.
7.1.2 Power dissipation
Power requirements often limit the performance of the TMS coil. An ideal TMS stimulator
should also have a low power dissipation (or equivalently low resistance) in order to reduce
the unwanted Joule heating.
54
75. 7.1 TMS coil requirements and performance
7.1.3 Coil Heating
Similar to the power dissipation, the conduction of high current pulse through resistive coils
leads to considerable heating that can damage the coil. Solutions such as cooling systems are
required in cases intended for prolonged high-speed stimulation, adding significant weight
and bulk.
7.1.4 Induced electric field
An ideal TMS coil should produce a strong stimulation in a prescribed region, and minimum
electric field in the rest of non target regions.
More precisely, the spatial characteristics of the TMS electromagnetic stimulation can be
described with the following parameters.
7.1.5 Penetration
Or depth, d1/2, is the radial distance from the cortical surface to the deepest point where the
electric field strength is half of its maximum value on the surface.
7.1.6 Focality
There are several definitions of focality in the literature (24); in general, more focal stimu-
lation means a smaller stimulation area with the maximum field. Here, we have employed
the focality defined through the effective surface area (13)
S1/2 =
V1/2
d1/2
(7.1.1)
55
76. 7. APPLICATIONS TO TRANSCRANIAL MAGNETIC STIMULATION
where V1/2 is the volume inside the brain where the stimulus is over 50% of the maximum.
This metric takes into account that it is harder to have a focal stimulus deeper in the head.
7.2 Numerical Model
7.2.1 The current density
A model of the current under search can be achieved by using a constant boundary element
method (BEM), that allows the current distribution to be defined in terms of the nodal values
of the stream function and elements of the local geometry (see (10)).
So let us assume that the surface, S ⊆ 3
, on which we want to find the optimal current,
is divided into T triangular flat elements with N nodes, which are lying at each vertex of
the element. If we consider the barycenters of the mesh triangles as RT = {r1,...,rT }, the
current density at each element can be written as
J : RT × N
→ 3
(r,ψ) → J(r,ψ) ≈
N
n=1 ψn n
(r),
(7.2.1)
where ψ = (ψ1,ψ2,...,ψN )T
is the vector containing the nodal values of the stream function
and n
: RT → 3
are functions related to the curl of the shape functions (10) known as
current elements. In the following, ψ ∈ N
is going to be the optimization variable.
If we denote by jn
x , jn
y, jn
z to the Cartesian components of n
, then
J(r,ψ) ≈
N
n=1
ψn n
(r) =
N
n=1
ψn n
x (r),
N
n=1
ψn n
y(r),
N
n=1
ψn n
z (r)
56
77. 7.2 Numerical Model
and the absolute current density is j(ψ) = (j(r1,ψ),..., j(rT ,ψ))T
where
j(r,ψ) :=
N
n=1
ψn n
x (r)
2
+
N
n=1
ψn n
y(r)
2
+
N
n=1
ψn n
z (r)
2
.
7.2.2 The magnetic field
The use of this current model allows the discrete formulation of all the magnitudes involved
in the problem, for instance the magnetic field at a given point is given by
B(r,ψ) ≈
N
n=1
ψnbn
(r), r ∈ 3
(7.2.2)
where bn
(r) = (bn
x (r), bn
y(r), bn
z (r)) is the magnetic induction vector produced a unit stream
function at the nth-node (10).
By applying the current model, Eq. (7.2.1), matrix equations that transform ψ to the various
coil properties and objectives can be then constructed.
The magnetic field at a series of H points, r = {r1,r2,...,rH}
bxi
(r ,ψ) = Bxi
(r )ψ, bxi
∈ H
, Bxi
∈ H×N
, xi = x, y,z. (7.2.3)
The coefficient Bxi
(h, n) = bn
xi
(rh), is the xi−component of the magnetic induction produced
by the current element associated to the nth
-node in the prescribed hth
-point in r .
7.2.3 The stored energy in the coil
W(ψ) = ψT
Lψ, L ∈ N×N
, (7.2.4)
57
78. 7. APPLICATIONS TO TRANSCRANIAL MAGNETIC STIMULATION
where L is the inductance matrix, which is a full symmetric matrix, and since the amount of
stored magnetic energy is always a positive
ψT
Lψ > 0, ∀ψ ∈ N
, ψ = 0 (7.2.5)
then L is positive definite.
7.2.4 The resistive power dissipation of the coil
P(ψ) = ψT
Rψ, R ∈ N×N
. (7.2.6)
where R is the resistance matrix, which is symmetric and positive-definite. Moreover, the
power dissipation can be related to the current at the surface as R ∝ JT
x Jx + JT
y Jy + JT
z Jz.
7.2.5 The electric field
The electric field induced in a series of M points inside of the conducting system (11), r =
{r1,r2,...,rM }
exi
(r ,ψ) = Exi
(r )ψ, exi
∈ M
, Exi
∈ L×N
, xi = x, y,z. (7.2.7)
7.2.6 The temperature
The temperature above ambient of the coil surface (9),
t(rRT
,ψ) = Λ (ψ), t ∈ T
, Λ ∈ T×T
, (ψ) ∈ T
. (7.2.8)
58
79. 7.3 Problem formulation
The matrix Λ is defined completely by the geometry of the mesh (9), and is a vector
containing the constant value of the the Joule heating coefficient at every mesh element
(l,ψ) =
ρr
kew2
J2
(rl,ψ), rl ∈ RT. (7.2.9)
where w is the thickness, ke the effective thermal conductivity and ρr the resistivity of the
conducting surface (9).
7.3 Problem formulation
Cobos Sanchez et al. (8) employed the discretized current model presented in section 7.2
to pose the TMS coil design as an optimization, in which a cost function of ψ (that con-
tains terms to control the electric and magnetic fields induced, stored magnetic energy and
power dissipation) is minimized by using classical techniques, such as simple partial de-
rivation and subsequent matrix inversion of the consequential system of linear equations.
In the following, and for sake of comparison, this type of approach used in will be noted
as partial derivation optimisation (PDO); where it is also worth stressing that PDO cannot
handle neither linear nor quadratic requirements, such as an optimised maximum current
density (27) or optimised maximum temperature, highlighting the need of more versatile
optimization techniques.
Moreover, a key issue when designing a TMS coil is to maximise the electric field induced in
the desired cortex region. In Cobos Sanchez et al. work (8), in order to produced maximal
stimulation, the stored magnetic energy and/or power dissipation (which can be both seen
as smoothing norms of the solution), are minimized for an acceptable level of strength of
the electromagnetic fields in the target volume.
Although this scheme has proved to produce TMS coils with efficient performance, it is not
59
80. 7. APPLICATIONS TO TRANSCRANIAL MAGNETIC STIMULATION
clear how optimal these coil solutions were, and especially whether the induced electromag-
netic fields were truly maximal in the target region.
Therefore, in order to accurately handle new coil requirements and to guarantee optimality
of the resulting solutions, we have to resort to a more mathematically rigorous approach
of the optimisation problem. In this work, it is shown that by using the suggested physical
model (Section 7.2) and after suitable mathematical derivations, the TMS coil design prob-
lem can be stated in the form of a convex optimisation. More precisely, all the most relevant
design problems can be written as
min f0(ψ)
fi(ψ) ≤ bi, 1 ≤ i ≤ m
(7.3.1)
or as
max f0(ψ)
fi(ψ) ≤ bi, 1 ≤ i ≤ m
(7.3.2)
where fi : n
→ are convex functions for i = 0,1,··· , m.
Equations (7.3.1) and (7.3.2) represent a quite convenient formulation of the TMS coil
design problem, as they can be straightforwardly tackled by using one of the several op-
timisation packages available for the solution of convex problems.
In this work, two main optimization schemes have been used to solve problems in Eqs.
(7.3.1) and (7.3.2)
• Singular vector analysis.
• CVX (12) a modelling system for convex optimization problems.
Moreover, it is also worth stressing, that the solution , ψ, of problems described by Eqs.
(7.3.1) and (7.3.2) is the optimal value of the stream function at the conducting surface; the
60
81. 7.3 Problem formulation
final wire arrangement that approximates the continuous current distribution is produced
by contouring ψ (24).
In the following, we present a set of relevant TMS coil design example cases, which have
been chosen to demonstrate the efficiency of the optimisation framework to produce truly
optimal solutions, and to illustrate its versatility to prototype many different performance
requirements and constraints.
For sake of briefness, the coil stimulation is defined by controlling the electric field. Nonethe-
less, the corresponding formulation in terms of the induced magnetic field can be analogously
obtained by interchanging E-field and B-field.
Moreover, due to the similar nature of L and R (symmetric and positive-definite matrices),
results involving the minimum magnetic stored energy TMS coil condition, can be straight-
forwardly exported for the case of designing TMS coils with minimum power dissipation.
7.3.1 Minimum stored magnetic energy
In TMS, the brain responds maximally when the induced current is perpendicular to the
sulcus (24), it is then worth considering the design of TMS stimulator capable of inducing a
maximum electric field in a given optimal direction.
Firstly we study the problem of designing a TMS coil with minimum stored energy (induct-
ance) that maximizes one component in a given of the electric field produce in a target region
formed from a distribution of H points.
max Eψ 2
minψT
Lψ
(7.3.3)
61
82. 7. APPLICATIONS TO TRANSCRANIAL MAGNETIC STIMULATION
where M ∈ with N > M, L ∈ N×N
and E ∈ M×N
, which can be Ex , Ey, Ez or the E-field
matrix in any other given direction.
Equation (7.3.3) can be transformed into a more suitable form by taking into account that
L = CT
C is the Cholesky decomposition of the inductance matrix L (which is symmetric and
positive-definite) and considering a new optimisation variable given by ˜ψ = Cψ. We have
then the following equivalences:
max Eψ 2
minψT
Lψ
⇔
max EC−1 ˜ψ 2
min ˜ψ 2
⇔
max EC−1 ˜ψ 2
˜ψ 2 = 1
⇔
min ˜ψ 2
EC−1 ˜ψ 2 = EC−1
2
.
A solution of problem defined by Eq. (7.3.3) is simply a supporting vector of EC−1
for the
· 2-norm, which is nothing else but a singular vector associated to the largest singular
value of EC−1
, that is, any normalized eigenvector associated to the largest eigenvalue of
the symmetric matrix (EC−1
)T
EC−1
.
7.3.2 Full field maximization
In this section, we investigate the design of a minimum stored energy TMS coil capable
of inducing an electric field with maximum magnitude in a target region formed from a
distribution of H points; this particular problem can be formulated as
max Ex ψ 2 + Eyψ 2
+ Ezψ 2
minψT
Lψ
(7.3.4)
62
83. 7.3 Problem formulation
where M ∈ with N > M, Ex , Ey, Ez ∈ H×N
and L ∈ N×N
.
By using again the Cholesky decomposition L = CT
C of the inductance matrix L and that
˜ψ = Cψ, we have that Problem (7.3.4) is equivalent to
max EzC−1 ˜ψ 2
2 + Ey C−1 ˜ψ 2
2 + Ex C−1 ˜ψ 2
2
˜ψ 2 = 1
(7.3.5)
Solving the above problem simply consists of finding the generalized supporting vectors of
the matrices EzC−1
, Ey C−1
and Ex C−1
. In other words, it suffices to find the normalized
eigenvectors associated to the largest eigenvalue of the symmetric matrix (EzC−1
)T
EzC−1
+
(Ey C−1
)T
Ey C−1
+ (Ex C−1
)T
Ex C−1
. The mathematical foundations on which this last fact
relies are shown in Appendix A.1.1.
It is worth recalling that the magnetic induction (and the electric field for for regions with
uniform conductivity and no electric charge) is divergence-free field , so here we would like
to make the reader notice that the problem of designing a TMS coil that maximises Bx , By
and Bz fields is not equivalent to one that maximises two components. A full proof of this
statement can be found in Appendix A.2.1.
7.3.3 Reduction of the undesired stimulation
Precise spatial localization of stimulation sites is one of the keys of an efficient TMS pro-
cedure, especially to prevent perturbation of non-target cortex regions. In order to achieve
this, we can study the design of a TMS coil capable of producing a maximum E-field in a
prescribed cortex volume, while maintaining the stimulation in other of non-target regions
of interest below a given threshold.
The problem of designing a TMS coil with minimum stored energy which produces a max-
imum electric field in a first target region of M points and minimizes the magnetic field in a
63
