Banach Spaces and Linear Operators Part IMUIC Seminar Banach Spaces and Linear Operators Part I...

Banach Spaces and Linear OperatorsPart I

MUIC Seminar

September 21, 2016

MUIC Seminar () Banach Spaces and Linear Operators Part I September 21, 2016 1 / 23

Banach Spaces

Definition (Banach Space)

Let (X , ‖ · ‖) be a vector space (possibly infinite–dimensional) and‖ ‖ : X → R+

The space X is said to be a normed space if‖x‖ = 0⇔ x = 0,‖λx‖ = |λ|‖x‖ for λ ∈ R(C),‖x + y‖ ≤ ‖x‖+ ‖y‖.

The function ‖ ‖ is called a norm. (X , ‖ ‖) is a metric space underthe metric d(x , y) = ‖x − y‖, x , y ∈ X .The function ‖ · ‖ : X → R determines a metric on X as follows

d(x , y) := ‖x − y‖ for x , y ∈ X .

It is easy to see that so defined function d is, indeed, a (translationinvariant) metric on X , i.e. (X ,d) is a metric (and, therefore,topological ) space .X is said to be a Banach Space if it is complete in the topology ofthe metric d .


STEFAN BANACH [1892–1945]


Let x ∈ X and ε > 0. The set B(x , ε) = {y ∈ X : ‖x − y‖ ≤ ε} is calleda ball (centered about x and of radius ε). The set of all ballsB(x , ε), ε > 0 is a neighbourhood basis of the point x . Due to thetranslation invariance of the metric d associated with the norm ‖ · ‖, itis easy to see that B(x , ε) = x + B(0, ε), and so the neighbourhoodbasis of the origin (i.e. vector 0) determines the neighbourhood basisof every point x ∈ X .Additionally, since the topology of X is determined by the metric, wecan select a countable set of ε’s, e.g. εn = n−1,n = 1,2, . . . todetermine the neighbourhood basis of the origin.


Given the properties (2) and (3) of the norm , we can easily see that ifx , y ∈ B(0, ε) and 0 ≤ λ ≤ 1 then

‖λx + (1− λ)y‖ ≤ λ‖x‖+ (1− λ)‖y‖ ≤ λε+ (1− λ)ε = ε ,

i.e. λx + (1− λy) ∈ B(0, ε) or, otherwise, B(0, ε) is a convex set.

This can be expressed as

TheoremA normed space (X , ‖ · ‖) is a locally convex topological space underits natural metrizable topology derived from the norm. This topology isgenerated by the base of neighbourhoods of the originBn = {x ∈ X : ‖x‖ ≤ n−1}.


The concept of a Banach Space is one of the most important conceptsof modern Functional Analysis. It originated from the XIX centurydevelopments in the Mathematical Analysis and it was given itsmodern form in the 1920’s, mainly through the work of Stefan Banachand his collaborators in Lwow, Poland (Hugo Steinhaus, StanisławMazur, Władysław Orlicz, and Juliusz Schauder, to name a few). Thespaces were called Banach Spaces in honour of Stefan Banach but, ofcourse, he himself did not use this term. The name Banach Spacescame in general use after 1945 when Banach already died. The theoryof what is now known as Banach Space Theory was given its solidform in the 1932 with the appearance of the book by Banach himself,Theorie des operations lineaires . Despite the passage of 80 years ofintensive research and tremendous progress in all areas discussed inthe book, this monograph remains extermely useful. In fact, manyquestions raised in the book were solved only very recently or remainunanswered.


Banach Spaces are complete normed spaces. The concept ofcompleteness is one of the most useful concepts of Analysis.Whenever we have an infinite sequence, we can only inspect a finitenumber of terms. If the space is complete, we are assured that whathappens with the very last terms of the sequence is similar to thebehaviour of the initial terms, i.e. the limit behaves similarly to the restof the sequence (provided that the terms show some tendency towardsconvergence, as expressed by the Cauchy Condition . Thus does notmean, of course, that the incomplete normed spaces are uninteresting.In fact, as we shall see later, some of them are quite useful.


Classical Banach Spaces

Rn : the space of all (real) n–sequences x = (x1, x2, . . . , xn) withthe Euclidean norm ‖x‖ = (

∑ni=1 |xi |2)

12 .

It is known from elementary real analysis that this space is aBanach space.P(0,1) – the space of all polynomials on the interval [0,1]. Thenorm in the space is the so–called supremum norm :

‖x‖∞ = sup{‖x(t)‖ : t ∈ [0,1]}

The space (P(0,1), ‖ · ‖∞) is not complete, i.e. it is not a Banachspace. In fact, the sequence of polynomials

xn(t) = 1 + t +t2

2!+

t3

3!+ · · ·+ tn

n!,n = 1,2, . . .

satisfies the Cauchy condition but xn(t)→ et as n→∞ and et isnot a polynomial.


c0 – the space of all sequences of real numbers which convergeto 0. In other words, x ∈ c0 if x = (x1, x2, x3, . . . ) andxn → 0 (n→∞). This space, equipped with the sup norm :‖x‖∞ = supn∈N|xn|, is a Banach space.c – the space of all convergent sequences with the sup norm (asin the case of c0 above) is a Banach space.`∞ – the space of all bounded sequences. This space is also aBanach space under the sup norm.


`p ,1 ≤ p <∞ – the spaces of all p–summable sequences. Herethe norm is

‖x‖p =

( ∞∑n=1

|xn|p) 1

p

,

and (`p, ‖x‖p) is a Banach space for every 1 ≤ p <∞.

In order to prove that `p’s are vector spaces and normed spaces onehas to use the Minkowski and Holder inequalities. We shall prove theHolder inequality separately. Also note that the space `2 is an infinitedimensional version of the Euclidean n–space of Example 1. This is animportant space, it is called a Hilbert Space and it was perhaps thefirst example of a vector space thoroughly investigated as an object inFunctional Analysis. It also is the ”best” Banach space, meaning that ithas the most regular properties of all spaces.


C(0,1) – the space of all continuous functions on the interval[0,1], with the sup norm

‖x‖∞ = sup{|x(t) : t ∈ [0,1]} ,

is a Banach space.The spaces Lp(0,1), 1 ≤ p <∞ of p–power Lebesgue integrablefunctions on [0,1] with the norm

‖x‖ =

(∫ 1

0|x(t)|pdt

) 1p

.

Here we identify functions functions that differ only on a set ofzero measure.L∞(0,1) – the space of all (essentially) bounded functions on[0,1] with the norm

‖x‖ = esssup{|x(t)| : t ∈ [0,1]} .

L∞(0,1) is a Banach space.MUIC Seminar () Banach Spaces and Linear Operators Part I September 21, 2016 11 / 23

Bounded Linear Operators I

Definition

Let X and Y be two normed spaces. A function T : X → Y is called alinear operator if

T (λx + µy) = λT (x) + µT (y) ,

for all x , y ∈ X and λ, µ ∈ R.

A linear operator is continuous if xn → x ⇒ T (xn)→ T (x) for everyconvergent sequence {xn} in X .Clearly, due to linear nature of X ,Y and T , a linear operator iscontinuous if and only if it is continuous at 0, i.e. T (xn)→ 0 wheneverxn → 0. Here xn → 0 means that ‖xn‖ → 0 and T (xn)→ 0 means that‖T (xn)‖ → 0.


Bounded Linear Operators II

A linear operator T : X → Y is bounded if there exists a constantM > 0 such that ‖T (x)‖ ≤ M‖x‖ for all x ∈ X . This implies severalthings. First that T is bounded on a unit ball of X , i.e. the setB(0,1) = {x ∈ X : ‖x‖ ≤ 1}. That’s why the operator is calledbounded.Secondly, it follows immediately that a bounded linear operator iscontinuous. It is less obvious but equally easy to show that a linearoperator which is continuous is also bounded.


TheoremThe Following Are Equivalent for a linear operator T between twonormed spaces.

T is continuous ,T is continuous at a point x ∈ X ,T is continuous at 0 ,T is bounded .

The set of all bounded operators between two normed spaces istypically denoted by L (X ,Y ).Let T ∈ L (X ,Y ). Then, since T is bounded, there exists a constantM > 0 such that ‖T (x)‖ ≤ M‖x‖ for all x ∈ X . Set‖T‖ := inf{M ≥ 0 : M‖x‖ ≥ ‖T (x)‖}. (The infimum exists; the set ofM ’s is bounded below by 0). The number ‖T‖ associated with T is anoperator norm (or simply a norm ) of an operator T .


Therefore (L (X ,Y ), ‖ · ‖) is a normed space. If Y is a Banach space itis easy to show that (L (X ,Y ), ‖ · ‖) is a Banach space.


Bases in Banach Spaces I

Definition

Let X be a Banach space. A sequence {en} of elements of X is calleda Schauder basis (or, simply, a basis if every element of X can beuniquely represented as a series

∑∞n=1 anen, where

an ∈ R,n = 1,2, . . . .

Not every Banach space has a basis. It is easy to see from thedefinition that if a space X has a basis it must be separable. It is sobecause lin{xn} = X and {xn} is, of course, countable.But most separable Banach spaces have bases. In fact, the spacesc0, `p , where 1 ≤ p <∞ have the simplest basis : the set of all unitvectors {δi,j}, i , j = 1,2, . . . . A more complicated bases also exist in


Bases in Banach Spaces II

Lp(0,1),1 ≤ p <∞ : the Haar system . The space C(0,1) also has abasis. The spaces `∞ and L∞(0,1)) are nonseparable, and as such,do not have a basis.It was for a long time an open problem whether every separableBanach space has a basis. It was only in 1973 that Per Enflo showedthat there exists a separable Banach space without a basis. Today weknow more spaces of this kind, some of them quite natural.


Enflo receiving a goose from Mazur


It was already known to Banach that every Banach space (separableor not) has a weaker property.

Definition

A Banach space X has a basic sequence if there exists a sequence{en} of elements of X such that {en} is a basis for its closed linearspan.

Theorem (Banach)Every infinite – dimensional Banach space contains a basic sequence.


Definition

A linear operator T : X → Y is called a compact operator, if the imageof the unit ball BX of X is a precompact set in Y .

Definition

A sequence {xn}n=∞n=1 is called a unconditional basic sequence if it is abasic sequence for all permutations π(N) of the index set {1,2,3, . . . }.

ProblemDoes every Banach Space have an unconditional basic sequence ?

Theorem (Gowers–Maurey, 1993)

There exists a Banach Space Xgm without an unconditional basicsequence.


TIMOTHY GOWERS – Fields Medallist, 1994


Nuclear Spaces

ALEXANDRE GROTHENDIECK (1928 – 2014) – Fields Medallist,1966


Nuclear Spaces, II


Banach Spaces and Linear Operators Part IMUIC Seminar Banach Spaces and Linear Operators Part I...

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