introduction to hilbert spaces

Introduction to Hilbert Spaces

Dr. Md. Asaduzzaman Professor Department of Mathematics University of Rajshahi . Email: md_asaduzzaman@hotmail.com

Prerequisite concept : Fields, Vector spaces, Subspaces,

Linearly dependent and independent sets of vectors, Basis

and dimension of vector spaces, Linear maps, Linear

functionals, Metric spaces, Interior points, Limit points,

Closure of a subset of a metric space, Open and Closed

subsets of metric spaces, Bounded sets, Compact sets,

Dense sets, Convergent and Cauchy sequences,

Continuous mappings, Homeomorphisms.

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Let be a vector space over the field (where is the real or complex field). A function : is saidto be an inner product on if it satisfies the following conditions: x,y,z and ,(i) x,x0 and x,x=0 iff x=0.(ii) (conjugate symmetry).(iii) (linear in the first argument).

Inner Product Spaces:

xy,yx, zy,βzx,αzβy,αx

A vector space over the real or complex field is called a linear space and a linear space with an inner product is called an inner product space or a pre-Hilbert space.

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How to define inner product in a finite dimensionallinear space:

Let be an n-dimensional linear space over the complex field. Let S={e1, e2, …, en} be a basis of and let A=(aij) be a positive definite Hermitian matrix of order n. Define

nji,1,ae,e ijji

and extend this function to linearly in the first argument and conjugate linearly in the second argument then will define an inner product in .

: SS ℂ by

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ThusIf x,y then scalars i,i (1 i n) such that

.eyandeαx1

ii1

ii

n

i

n

i

where denotes the column vector [1, 2 ,…, n]t and * is the conjugate transpose of the column vector [1, 2,…, n]t .

.Aaβαyx, *

1 1ijji

n

i

n

j

Then

In particular, when we consider real field then positive definite Hermitian matrix A will be replaced by positive definite symmetric matrix A and * will be replaced by t, the row vector [1, 2,…, n] .Thus, in this case,

.Aaβαyx, t

1 1ijji

n

i

n

j5

Remark: The inner product depends on(i) the choice of basis {ei},(ii) the arrangement of the basis,(iii)the choice of the matrix A.Example: Let =ℝn, ei be an i-th column vector of the identity matrix In , (the standard basis of ) and let A be the diagonal matrix with positive diagonal elements d1, d2, … ,dn. i.e., .

1

ndO

OdA Then .yxdAxyyx,

n

1iiii

t

In particular, if A=In then we have the usual inner product .yxxyyx,

n

1iii

t

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Example: Let =ℝ2, .

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and1

121

ee

Then for

.2111

ALet

.21and

21

21

21

21

21

yyyy

xxxx

the coordinate vectors of x and y with respect to the basis e1, e2 are respectively

X,inyandx2

1

2

1

yy

xx

).5(41Ayx, 22211211

t yxyxyxyx

Hence

defines an inner product in X.7

Normed Linear Spaces

Definition: Let X be a linear space over the field K. Then a function ║║: X→ is called aℝ norm in X if it satisfies the following properties:

(i) ║x║≥0;(ii) ║x║=0 iff x=0; (iii) ║x║=||║x║;

(iv) ║x+y║≤║x║+║y║.

A linear space with a norm is called a normed linear space.

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Metric Induced from Norms

For any normed linear space we have the following theorem:

Theorem: Suppose X is a normed linear space . Then for any x, y, zX , we have ║x-y║≤ ║x-z║+ ║y-z║.

Thus in a normed linear space X if we defined(x,y)= x-y then d satisfies all conditions of a metric. This metric is ║ ║called metric induced by norm.

It is clear that different norms on the same linear space induce different metrics.

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An inner product on a linear space defines a norm on given by x║ ║2= xx.

An inner product defines a norm and a norm defines a metric on a linear space .

A norm . on a linear space ║ ║ defines a metric on given by d(x,y)= x-y .║ ║An inner product space is a metric space under the metric induced by its inner product.

But not all norm on a linear space can be obtained from an inner product.Also not every metric on a linear space can be obtained form a norm.

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Hence the metric induced from a norm on a non trivial linear space can not be bounded. Hence a bounded metric can not be obtained from a norm.

The metric d induced from a norm . on a normed linear ║ ║space satisfies (i) d(x,y)=d(x-y,0), (ii) d(x,0)=||d(x,0).

The metric d defined on X by

y.xwhen0,yxwhen,1

),( yxd

is bounded.

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Also, given any metric d on X, the metrics d1 and d2 defined by

)}yx,(,1min{)yx,()y,x(1

)y,x()y,x( 21 ddandddd

are examples of bounded metrics.

Hence d1 and d2 can not be obtained from any norm on X.

If in an inner product space every Cauchy sequence converges then it is called a Hilbert space. Every finite dimensional inner product space is a Hilbert space. It can be shown that there are inner product spaces which are not Hilbert spaces.

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A norm induced by an inner product satisfies theparallelogram equality .y2x2yxyx 2222

Geometrical interpretation: If x and y denote two adjacent sides of a parallelogram then x+y and x-y represent two diagonal vectors and norm measures their lengths.

x-y

x

yx+y

y

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However, a norm on a linear space may not satisfy

parallelogram equality. If a norm satisfies parallelogram

equality then it must be induced by a norm.

Now the question arises, if a norm satisfies parallelogram

identity then how can we determine the corresponding

inner product ? The following theorem answers this

question.

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Theorem: A norm . on a linear space X is induced by an ║ ║inner product on it if and only if it satisfies the parallelogram identity .y2x2yxyx 2222

If it so, the inner product is given by the polarization identity

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The following examples show that all norms are not induced by inner product and all inner product spaces are not Hilbert spaces.

(1) The linear space equipped with the norm given by

2,1, ppp

,x/1

1i

p

i

p

px

pxx ),,(x 21

is not an inner product space and hence not a Hilbert space.

For, if x=(-1,-1,0,0,0,…..) and y=(-1,1,0,0,0,….) then pyx,

.2yxand2yx,2y,2x /1/1 pp

pp

pp

Hence .y2x2yxyx 2222pppp

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(2) The linear space equipped with the norm given by ],[ baC],[,)(sup

],[baCxtxx

bat

is not an inner product space and hence not a Hilbert space.

a b

1

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(3) The linear space equipped with the norm given by ],[ baC

],[,)(2/12

baCxdttxxb

a

is an inner product space since the norm is induced by the inner product .)()()(),(

b

adttytxtytx

But this inner product space is not complete and hence it is not a Hilbert space.

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We now give some examples of infinite dimensional Hilbert spaces

(1) The linear space equipped with the induced norm given by is a Hilbert space.

],[2 baL,)(,

2/122/1

2

b

adttxxxx ],[2 baLx

(2) The linear space of all square sumable sequences equipped with the induced norm given by is a Hilbert space.

,2,

2/1

1

2

2

iixx

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Theorem(Cauchy-Schwarz inequality): For any two vectors x and y in an inner product space

.yxyx,

and the equality holds iff x and y are linearly dependent.

Remark: From the Schwarz inequality we see that

,1yxyx,

holds for all non zero vectors x and y.

This relation motivates us to define angle between two nonzero vectors x and y as .

yxyx,

cos: 1,

yx

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However, the angle between two nonzero vectors is not invariant under all inner product. For example, any two distinct standard basis vectors ei and ej in ℝn are orthogonal with respect to the usual inner product but with respect to other inner product they may not orthogonal. Although, if two vectors are parallel with respect to one inner product, they are parallel with respect to every inner product.

If two vectors x and y are such that x,y0 then we say that the vectors x and y are orthogonal.

x,y=0 iff x=y for some scalar , and x,y=/2 iff xy0.

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Theorem: Every nonempty closed convex set contains a unique element of smallest norm. Proof: Let , then a sequence such that . Since is convex therefore , and . Hence .By parallelogram equality

as and

A subset of a vector space is said to convex iff for all , we have .

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Hence is a Cauchy sequence in . Since is closed , is convergent in .Hence there exists an such that . Since the map is continuous ,Hence . Let such that . Then since . By parallelogram equality

.Hence If is a closed convex subset of the Hilbert space and is an arbitrary point of , there is a unique point of closest to .

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DefinitionIf and in such that , then we say that is orthogonal to . We define

and for any subset of

TheoremLet be a subset of .Then is a closed subset of .Proof Since ,. Let .Then and for all Hence for any scalar and , for all .Hence .Thus is a subspace of .

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Next , since for a fixed , the map defined by is continuous and is closed subset of ,ker.Hence is a closed subset of . Since and each is closed .Therefore is closed.Theorem (Projection Theorem)If is closed subspace of then .ProofLet .Then the set is closed and convex.Hence contains a unique element of smallest norm. Let this element be . Then and for all and for scalar .

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Put and .We now show that .From (1) we have for all and for all scalar

.This inequality holds for all and for all scalar .Thus for any non zero ,if we put in the inequality. Then we see that .Hence for all .

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Thus . Hence Since implies that and.Hence .Thus . Hence .

Definition is called the orthogonal complement of .Corollary : If is a closed subspace of , then Fsubset of , then

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Theorem Let be a closed subspace of . Then there exist a unique pair of maps and such that(1) .These mappings have the following further properties:

(2) for all and for all.(3) (4) if .(5) and are linear mappings.

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Proof Since every can be expressed uniquely as ,where and .Define and .Now ..Let , then , Then and are well defined maps and satisfies (1) and (2). 29

To prove (3)Let Then , for a unique and a unique . Hence

To prove (4)Let and . Then , for a unique and a unique . Hence

since and ..Hence for all . The equality holds if . Hence .

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To prove (5) Let and .Then by (1)

Hence

Since the left hand side is in and right hand is in , both are .Hence

and31

Finally, to prove uniqueness of and Let there are maps such that .Then for

where , and

Since and and We have and .Hence we have and .

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Definition and are respectively called the Projections of on and .From the above theorem , we see that among the points of has the smallest distance from . We call as the distance of from .

If , then a projection mapping.

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Remark The projection theorem shows that every closed linear subspace of has at least one complementary closed linear subspace . One may note that in some Banach spaces a closed may fail to have complementary closed linear subspaces; for instance, the closed subspace of the Banach space is not complemented in . To each closed subspace of ,there is associated a projection of with rage space of and null space of .

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Projection theorem provides a characterization of closed subspaces of a Hilbert space in terms of orthogonality.A subspace of Hilbert space is closed in if and only if

Projection theorem provides a characterization of sets in Hilbert spaces whose span is dense in Let be a non-empty subset of a Hilbert space .Then, span is dense in if and only if .

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Theorem If is a closed subspace of , if is the space spanned by and , then is closed. Corollary: Every finite dimensional subspace of is closed.Proof We know that every finite subset of a metric space is closed .Hence is a closed subspace of .Let be finite dimensional subspace of and let be a basis of .Let be the subspace of generated by and, by the above theorem, is closed. Let be the subspace of generated by, by the above theorem, is closed.Proceeding in this we can show that is a closed subspace of. Since generated by

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TheoremFor any fixed , the mapping is a continuous linear functional on .ProofClearly the map is a linear functional on. To prove continuity, let given. Choose and choose any if . Then for any ,

when .Hence the map is a continuous linear functional on .

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Representations of Continuous Linear Functionals( Riesz representation theorem)Let be a continuous linear functional on the Hilbert space then there is a unique such that for all .ProofIffor all we take .Otherwise, define.

Then clearly is a subspace of .Sinceis continuous and is a closed subset is closed. Since for some , there exists an such that . Thus does not consists of alone.Hence there exists a with Put . Since we have .

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Thus . Hence

Put .Then we have for all .To prove the uniqueness of Suppose there are such that .Then for all . In particular

Hence . 39

Definition A set of vectors in is said to be an orthonormal set if

DefinitionLet be a nonempty subset of . We define to be the set of all finite linear combinations of vectors from , which is a subspace of .Clearly contains and it is the smallest subspace of that contains . We call it the subspace of generated by .

TheoremLet be an orthonormal set of vectors from . Let . Then for any (1) .(2)

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ProofLet , then there exists scalars such that .Taking inner product on both sides with , we have .Thus for each , .Hence . Again, by

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Theorem An orthonormal set of vectors in a Hilbert space is linearly independent.ProofLet be an orthonormal set in . Assume is finite .Writing A= , consider the equation where are scalars. Taking inner product both sides with , ()we have

.An arbitrary set (finite or infinite) is said to be linearly independent if every non-empty finite subset of is linearly independent. Hence it follows that our assertion is also valid for the case when is infinite. 42

Advantage of orthonormal sets over arbitrary linearly independent sets

(1) The determination of the unknown coefficients is simple.(2) The determination of the norm of a vector by inner product.(3) If we wish to add a term to in order to obtain, in that situation, we need to determine only one more coefficient since the other coefficients remain unchanged.

.

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DefinitionLet be an orthonormal set . Let and consider the function : defined by .Then the numbers are call the Fourier coefficients of with respect to the orthonormal set .

TheoremIf is an orthonormal set in and . Then

is the projection of on the subspace generated by .

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Proof Let be the subspace of generated by . Then is a closed subspace of . Hence . Since , there exist a unique and a unique such that .

Since , .Now , for each

Hence

is the projection of on the subspace generated by .45

TheoremLet be a finite orthonormal set in an Hilbert space .Then for any in , we have (a) (Bessel Inequality)(b) Proof We have

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(b)

The inequality (a) can be given “ The sum of the squares of the components of a vector in various perpendicular direction does not exceed the square of the length of the vector itself ”The relation (b) implies “ If we subtract from a vector its components in several perpendicular directions, then the resultant has no component left in any of these directions”. Hence, the resultant vector is perpendicular to each of these perpendicular directions. Bessel Inequality for is essentially the Cauchy-Schwarz Inequality. 47

Theorem ( Bessel’s Inequality): If is an orthonormal set in , then for every.Proof Let be any finite subset of .Then is a finite orthonormal set in .We have

……. (1)Taking supremum on both side over all finite subsets of , we have 48

Keeping in view the usefulness and convenience of orthonormal sequences over the linearly independent sequences , one is

interested to generate orthonormal sequences from the linearly independent sequences .

This is done by a constructive procedure, known as the Grahm (1883) – Schmidt (1907) process.

Let be a (finite or countable infinite) linearly independent set of vectors in inner product space . The problem is to convert this set into an orthonormal set such that , for each .

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Step 1. Normalize , which is necessarily nonzero , so as to obtain as

Step 2. Write , so that

.Clearly ,since is linearly independent. Also, , since

We can take by normalizing ; namelyi.e.,

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Now , since and it is clear that is an orthonormal set in . Further , is the linear combination of , and is a linear combination of and , we have .Step 3.Similarly, the vector is given by

is non-zero, and .Therefore,

Obviously, is an orthonormal set in and it can easily verified that 51

Step 4.Proceeding in the same way, construct , and so on. Step n. The vector

is non zero and.Thus we take

Hence, this process leads to an orthonormal set with the required property.

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NoteThe sum which is subtract from in order to obtain is , in fact, the projection of on .In otherwards, in each step we subtract from its components in directions of previously orthonormalized vectors giving for which is then multiplied by , so we get a vector of norm 1.

If n were the smallest subscript for which ,the then relation

Shows that would be a linear combinations of , contradicting the assumptions that { is linearly independent.

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⟨ 𝒙𝟐 ,𝒖𝟏 ⟩𝒖𝟏𝑢1

− ⟨𝒙𝟐 ,𝒖𝟏 ⟩𝒖𝟏

𝒖𝟐 𝑥 2𝒗𝟐

Illustrating step 2

𝒙𝟏

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Definition Let be an orthonormal set in an Hilbert space .Then is said to be an orthonormal basis of if it is a maximal orthonormal set in .

TheoremLet be an orthonormal set in an Hilbert space . Then is an orthonormal basis if and only if .

TheoremEvery Hilbert spaces has an orthonormal basis.Any two orthonormal bases have the same cardinality.

This fact may be used to classify all possible Hilbert spaces.

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TheoremLet be an arbitrary set, and let be a Hilbert space with an orthonormal basis having the same cardinality as . Then there is an isometric isomorphism ( a one-to-one, linear, norm-preserving map) between and .Proof Let . If , then gives , where . The map of into is therefore norm-preserving; since it is also linear, it must be one-to-one. To show that the map is onto, consider any collection of complex numbers , with .Say except for and let .Since the are orthonormal, it follows that for all , so that maps onto .

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We may also characterize Hilbert spaces that are separable, that is , have a countable dense set.

Theorem A Hilbert space is separable if and only if it has a countable orthonormal basis. If the orthonormal basis has n elements, is isometrically isomorphic to I is infinite, is isometricallyisomorphic to .

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