# FUNCTIONAL ANALYSIS BOOK PDF

Purpose of the book. Functional analysis plays an increasing role in the applied sciences as well as in mathematics itself. Consequently, it becomes more and. As is usual practise in functional analysis, we shall frequently blur the . (Older books call TF the weak topology generated by F: the adjective. Functional Analysis, Banach space, Hilbert space,. Measure .. unbounded operators in Banach spaces following the book by Kato [23] (e.g.. Sections ,

Author: | JOSH FLAMAND |

Language: | English, Arabic, Hindi |

Country: | Libya |

Genre: | Lifestyle |

Pages: | 785 |

Published (Last): | 17.05.2016 |

ISBN: | 426-6-16270-103-2 |

ePub File Size: | 21.55 MB |

PDF File Size: | 13.75 MB |

Distribution: | Free* [*Register to download] |

Downloads: | 27210 |

Uploaded by: | CARLETTA |

PDF Drive is your search engine for PDF files. As of today we have 78,, eBooks for you to download for free. No annoying ads, no download limits, enjoy . PDF | Notes from a course taught by Palle Jorgensen in the fall semester of The course covered central themes in functional analysis and operator theory, with an emphasis on topics of special relevance Theorem of Rudin's book. These notes evolved from the introductory functional analysis course The first section of the book introduces the basic properties of topological vector spaces.

We now show conversely that if a family of subsets satisfies the conditions of Theorem 16, it forms a neighborhood base for a vector topology. This will give a convenient means of constructing vector topologies. Let X be a vector space. Condition iii implies that the intersection of open sets is open so r is a topology on X, and condition iv implies that each element of W is open.

Condition ii implies that addition is continuous. Since U is balanced, tV c U]. The uniqueness of r is clear. A similar result is given by Theorem As in Theorem 17 this defines a vector topology r on X and 2l is a neighborhood base at 0 for elements of 2l are not T.

We give several examples of TVS. Example Let -wLZj be the weakest topology on X such that each f E 5r is continuous. Thus, w 3 is a vector topology on X. Show that c and c0 are linearly homeomorphic. Describe X x Y ' in terms of X' and Y'. X - Y linear. X -4 Y linear.

If T is bounded on some neighborhood of 0, show T is continuous. Define L: Show L is continuous, linear and compute JIL Complete Example Show there exists a linear map T: X - Y which is not continuous. Use a Hamel basis. Let qP: It is routine to check that the neighborhoods of 0 defined above satisfy the conditions of 1.

We show the quotient topology is semi-normable in this case. II' is a norm if and only if M is closed. II[x]II' iv: II II' by 2. Follows from iv and ii. Follows from Proposition 3. If A c X is bounded, [A] is bounded by Proposition 3 iii.

Show the induced map Chapter 6 63 Exercise 3. Show the induced map T: If X is a NLS, show that these spaces are isometric.

M1 is called the annihilator of M. First we show that a Hausdorff vector topology on a finite dimensional TVS is unique. Theorem 1. Let i be a vector Hausdorff topology on Fn. Let p also denote the topology induced by p. Denote the vector with a 1 in the kth coordinate and 0 elsewhere by ek.

First, we show r c p. Let V be a r -neighborhood of 0. Then B is p-compact and, hence, -r -compact since neighborhoods of 0.

VE Y property with respect to B. Therefore, 3 Vl, On a finite dimensional Hausdorff TVS every linear functional is continuous. Theorem 1 and Exercise 1. Corollary 3. Corollary 4. Remark 5. The converse is false. The proof utilizes a result of F. Lemma 6 Riesz's Lemma. Example 7. In general, x0 cannot be chosen to be distance 1 from X0 although this is the case when X0 is finite dimensional.

Let X be a NLS and suppose the unit ball x: Then X is finite dimensional. Suppose that X is not finite dimensional. Then X1 c X and X1 is closed by Corollary 3. But, then B is not sequentially compact. Corollary 9. Let X be a NLS. It is actually the case that i and iv are equivalent for general Chapter 7 69 TVS. A locally compact Hausdorff TVS is finite dimensional.

Let K be a compact neighborhood of 0. Since K contains a closed, balanced neighborhood of 0, we may assume that K is balanced. Let be qP: If Fn is normed by p, show any linear functional on Fn is continuous with respect to p.

Show that any infinite dimensional B-space has uncountable algebraic dimension.

Apply Corollary 3 to the subspace spanned by x0. X -4 Y is continuous. Give an example of 2 norms which are not equivalent. For an interesting discussion of the history of the Hahn-Banach Theorem, see [Ho]. A function p: A semi-norm is obviously sublinear but not conversely. The Hahn-Banach Theorem guarantees that any linear functional defined on a subspace of a vector space which is dominated by a sublinear functional can be extended to a linear functional defined on the entire vector space and the extension is still dominated by the sublinear 73 The Hahn-Banach Theorem 74 functional.

Theorem 2 Hahn-Banach; real case. Let X be a real vector space and p: Let M be a linear subspace of X. Let X be the class of all linear extensions g of f such that g x dx E. Note ' 0 since f E X. X has a maximal element F. Let M1 be the linear subspace spanned by. F and x1. If we can show that it is possible to choose z such that F1 y ' and contradict the maximality of F.

To obtain a complex form of the Hahn-Banach Theorem, we need the following interesting observation which shows how to write a complex linear functional in terms of its real part. Lemma 3 Bohnenblust-Sobczyk. Then for C. X -4R is IR-linear. Conversely, if f: The Hahn-Banach Theorem 76 Proof: The converse is easily checked. Theorem 4 Hahn-Banach; complex case.

Let X be a vector space and p: X -4R a semi-norm. Let M be a linear subspace of X and f: M -4 IF a linear functional. Then f x dx E M so Theorem 2 implies 3 a linear extension F: By the first part, 3 a real linear functional f 1: X - IR which extends. Then F is C-linear and extends f by Lemma 3.

Give an example of a sublinear functional which is not a seminorm. Chapter 8 77 8. But, certainly, jjy' jjS jjx' Ij. The Hahn-Banach Theorem 78 Proof: Let M0 be the linear subspace spanned by M and x0.

Now apply Theorem 1 to f. We establish a dual result; we show that the norm of an element x E X can be found by computing the sup of x over the unit sphere in X'. Corollary 5. Ilx' jj and the sup is attained. If X' E X', lix' 11 S 11x1j. Corollary 6. Later we will show that S carries a natural topology under which it is a compact Hausdorff space and the space B S can be replaced by C S. We next consider some separability results for a NLS and its dual.

The Hahn-Banach Theorem 80 be dense in X'. For each k choose xk e X Proof: Since xk is dense in X', 3 a subsequence x'nk such that x'nk -4 x' and since llx' - x' II? Ilx' k Ilxnkll 0. The converse is false; consider lI and its dual l Exer. Even though a separable space needn't have a separable dual, it does have a countability property that can sometimes be substituted for separability.

Definition 9. By Corollary 5, x' E X': However, some spaces have much smaller norming sets. In C S , the set of Dirac measures, St: Chapter 8 81 Proposition Let X be a separable NLS. Then X' contains a countable norming set for X. Let xk be dense in X. The Canonical Map and Reflexivity: Let X" be the dual of X' with the dual norm and assume that X" carries its dual norm from X. X" is called the second dual or bidual of X.

Thus, n the map JX: A NLS is said to be reflexive if the canonical imbedding J is onto. Note from 5. Also note that for a B-space X to be reflexive, X and X" must be linearly isometric under the canonical imbedding J; R. James has given an example of a non-reflexive B-space X such that X and X" are linearly isometric [J1].

Also, co and c are not reflexive. Here we are following the common procedure of identifying X and JX under the linear isometry J. We now establish several properties of reflexive spaces. A B-space X is reflexive if and only if X' is reflexive. Suppose X is reflexive and let x' ' ' e X' ' '.

Suppose X' is reflexive. A closed linear subspace M of a reflexive space X is reflexive. Let m" E M".

Another interesting property of reflexive spaces is given by Theorem Let X be a reflexive B-space. Then every continuous linear functional x' E X' attains its maximum on the closed unit ball of X.

It is an interesting result of R. James that the converse of Theorem 17 holds [J2]. Then h E Cl: Let yk be Cauchy in Y. IIx'II and the sup is attained compare Corollary 5. Show the dual of a separable reflexive space is separable.

## Notes on Functional Analysis

If X is reflexive and X' contains a countable set which separates the points of X, show X' is separable. Chapter 8 85 Exercise 5. If X is reflexive, show X' has no proper closed subspaces which separates the points of X.

Show there is no norm on L0[0, 1] Example 5. Give an example of a countable norming set for ll. Show that the quotient of a reflexive space by a closed linear subspace is reflexive. The Hahn-Banach Theorem 86 8. I go -4 l " is the shift operator which shifts the coordinates of a sequence one place to the left.

A Banach limit is a continuous linear functional L on l" which satisfies i L x? Any Banach limit agrees with L on convergent sequences.

More generally, we have Proposition 1. Choose N such that ikf xk 5 XN 0 V k. Similarly, L x S sup xk. We now show that the Hahn-Banach Theorem can be used to show Chapter 8 87 the existence of Banach limits.

## Navigation Bar

Banach limits exist. Then is a sublinear functional with p x 5 IIxil.. Banach limits have applications in the theory of finitely additive set functions. See Exer. See also [Jor] for further such applications. Banach also used the Hahn-Banach Theorem to show the existence of a translation invariant, positive, finitely additive set function which is defined on all subsets of IR and assigns unit measure to [0, 1].

Recall that no such countably additive set function exists. Show that there exists a ,' which is not countably additive. That is, in probability terms, when do the moments of a probability distribution determine the distribution? Since the dual of C[0, 1] can be identified with normalized functions of bounded variation Example 5. Given a NLS X, a subset xa: We use the Hahn-Banach Theorem to give a complete answer to this question.

**You might also like:**

*FUNCTIONAL REACTIVE PROGRAMMING ON IOS EPUB*

Helly solved a moment problem of F. Riesz by developing a form of the Hahn-Banach Theorem in C[0, 1]. For an interesting discussion of Helly's paper, which also contains an early version of the Uniform Boundedness Principle, see [Ho].

These early versions of the UBP were established by Hahn and Banach by employing "sliding hump" arguments. In this section we use a theorem about infinite matrices to establish a very general version of the UBP which requires no completeness assumptions on the domain space and then give several applications to illustrate its utility. We begin by establishing the matrix theorem, called the Basic Matrix Theorem, due to Antosik and Mikusinski; the Basic Matrix Theorem can be viewed as an abstract "sliding hump" result.

First, a simple lemma. As one of its We now establish the Basic Matrix Theorem. For this reason the result is sometimes referred to as a Diagonal Theorem. The result for metric linear spaces is due to Antosik and Mikusinski see [AS] for the metric case and many applications. Theorem 2 Basic Matrix Theorem. Choose k0 such that iii - xiJ.

Jk U0 and x1. A matrix which satisfies conditions I and II is called a X-matrix. T E 91 is bounded in Y V x e X] is uniformly bounded on some family of subsets of the domain space [ 5 is uniformly bounded on a family ,A of subsets of X if [Tx: The Chapter 9 95 classical UBP for NLS asserts that any pointwise bounded family of continuous linear operators on a B-space X is uniformly bounded on the family of all bounded subsets of X see Corollaries 6 and 7 below.

If the domain space X does not satisfy some type of completeness condition, then as the following example illustrates the family of bounded subsets of the domain space is in general too large to draw such a conclusion. Example 3. Consider the sequence kek in l 1, the dual of c00 Exer. Whereas the family of all bounded subsets of the domain space is too large a family to draw the conclusion that a pointwise bounded family of continuous linear operators is uniformly bounded on each member of this family, we show that the families of X convergent sequences and X bounded sets do allow such a conclusion to be drawn.

Note that w 3 is weaker than w L X, Y which is weaker than the original topology of X.

## An introduction to functional analysis

Theorem 4 General UBP. Suppose that 9 is pointwise bounded on X, i. T e 5,jEIN i is not absorbed by U. In particular, tiT xi e U for large i which contradicts the construction above. For ii , let Tx: Hence J -a 0 is bounded 4.

Recall 4. Note that the families of bounded sets in i and ii depend upon the family of operators In order to obtain a family of bounded subsets, ,4, which has the property that my pointwise bounded family of continuous linear operators is uniformly bounded on the members of A, we can take the topology w L X, Y which is stronger than w ,9. If X is an.

A-space, then any pointwise bounded family 51c L X, Y is uniformly bounded on bounded subsets of X. Without some form of completeness assumption, such as the. A-space assumption in Corollary 6, the conclusion in Corollary 6 is false see Example 3. For semi-NLS, Corollary 6 takes the following form. Corollary 7.

## About this book

T E 5 is bounded and ii 5 is equicontinuous. Mazur and Orlicz [MO] generalized conclusion ii of Corollary 7 to complete quasi-normed spaces. We can also obtain this form of the UBP from Corollary 6. Corollary 8. Let X be a quasi-normed X-space.

**Also read:**

*KATIES HOPE PDF*

If Sc L X, Y is pointwise bounded, then 51 is equicontinuous. Chapter 9 99 Proof: In general, any equicontinuous family of continuous linear operators is uniformly bounded on bounded sets Exer. However, it is not generally the case that a pointwise bounded family of continuous linear operators or even a family of operators which is uniformly bounded on bounded sets is equicontinuous as the following example shows.

Example 9. Let l2 have the topology w 12 , where we consider each element of l 2 to be a member of the dual of l 2 Example 5. As we will see later 12, w l2 is an. This result gives sufficient conditions for the pointwise limit of a sequence of continuous linear operators to be continuous. We first use the Basic Matrix Theorem to establish a general form of this result. In particular, we have Corollary 11 Banach-Steinhaus. The classical form of the Banach-Steinhaus Theorem assumes that X is a complete quasi-normed space.

Since any such space is a X-space, Corollary 11 gives a generalization of the classical Banach-Steinhaus Theorem to s-spaces. Finally, we employ the UBP to obtain two interesting results which will be employed later. First, we establish a simple and useful test for determining the boundedness of a subset in a NLS.

Let X be a semi-NLS. Then B c X is norm bounded if and only if is bounded d x' E X'. The family Jb: We will establish a more general form of this result later Finally, we give an application of the UBP to sequence spaces see also Exercises 4 and 5. For each n define In: Sargent Spaces: This method of proof was refined by W.

We present her results here. A QNLS of second category such a space is usually called a Baire space is obviously a Sargent space, but, remarkably, there are Sargent spaces which are first category. See [S] for examples of such spaces. For Sargent spaces we have the following UBP. Then 9 is equicontinuous.

## Functional Analysis Books

Hence, some Ek must be somewhere dense, i. It ij Exercise 2.

Establish the analogue of Proposition 13 with the space co replaced by c and l p for 1 Exercise 5. Show Corollary 11 is false if the condition on X is dropped.

We first establish the analogue of 5. A bilinear map B: X x Y -4Z is jointly continuous with respect to the product topology if and only if B is continuous at 0, 0. A map f: Even for bilinear maps separate continuity does not imply joint continuity. Define B: Theorem 3. The result now follows from 9. As an immediate consequence of Theorem 3, we obtain a classic of Mazur and Orlicz on joint continuity.

Corollary 4 Mazur-Orlicz. Chapter 9 If B: XxY-Z is separately continuous and bilinear, then B is continuous. Corollary 4 gives a generalization of their result to X-spaces.

Using Corollary 4 we can weaken axiom v in Definition 2. If we assume that scalar multiplication is only separately continuous with respect to the metric topology induced by the , then it follows that scalar multiplication is actually function continuous since the scalar field is complete. Bourbaki introduced a concept, called hypocontinuity, for bilinear maps which is intermediate between separate continuity and continuity. Let B: X x Y -4 Z be bilinear and separately continuous, and let Y be a family of bounded subsets of Y.

If X is metrizable, the same statement holds with sequences replacing nets. If ,fix 2y is the family of all bounded subsets of X Y and B is both ,fix hypocontinuous. However, we show that any separately continuous bilinear map is hypocontinuous with respect to the family of X bounded sets.

Theorem 6. X x Y -4Z bilinear, separately continuous. Pick tk - co such that tkxk -a 0 2. The conclusion of Theorem 6 is also valid for the family of subsets of Y which are 96 bounded with respect to the original topology of Y.

Any continuous bilinear map is hypocontinuous Exer.

Examples of such bilinear mappings are given in We consider results analogous to Theorem 3 and Corollary 4 for families of bilinear mappings. Let 51 be a family of separately continuous bilinear mappings from X x Y into Z.

The family 5 is 1cft righx equicontinuous if d y E Y V x e X , the family 9! The Chapter 9 sequence in Example 9 below shows that a family can be left equicontinuous but not right equicontinuous. S The family is equicontinuous if it is equicontinuous as a family of mappings from X x Y into Z.

As in Proposition 1, 51 is equicontinuous if and only if S is equicontinuous at 0, 0 Exer. Analogous to Theorem 3 we have Theorem 7.

Let 9 be left equicontinuous. Bi c For any increasing of rr. We have the immediate corollary analogous to Corollary 4. From Corollary 8 we obtain an analogue of Corollary 9.

Corollary If 9 is a family of separately continuous bilinear maps which is pointwise bounded on X X Y, then 5 is equicontinuous. For y E Y the family B -, y: That is, 3 is left equicontinuous. The result follows from Corollary 8. Chapter 9 Corollary 10 gives a generalization of Theorem The proofs in these references use the theory of uniform spaces while the proof above uses only basic properties of convergent sequences and the Basic Matrix Theorem.

Show any continuous bilinear map is hypocontinuous.

Show a bilinear map B: Show a separately continuous bilinear map B: Show that a family of bilinear maps is equicontinuous if and only if it is equicontinuous at 0, 0. Show that if 9 is left equicontinuous, then 9 is pointwise bounded on X x Y. Thus, the converse of Corollary 10 holds. Show that if 9 is equicontinuous, then 5 is uniformly bounded on sets of the form C x D, where C is bounded in X and D is bounded in Y.

Compare with Exer. The theorem has been extended to the case of bounded finitely additive measures and we consider this more general version. We begin by establishing a remarkable lemma of Drewnowski [Dr] which shows that in some sense a bounded finitely additive set function is "not too far" from being countably additive.

Let E be a a-algebra of subsets of a set S. Recall that ba s is the B-space of all bounded, finitely additive set functions p: If p E ba s and E is a pairwise disjoint sequence from E, then n n E-: J Chapter 9 Proof: J Proof: Lemma 3.

Let f be an algebra of subsets of set S, and let pi: Then [pi A: A is bounded if and only if [pi A: Suppose from 4. Now either sup lpi HnEl I: This establishes the sufficiency; the necessity is J clear.

We now establish the Nikodym Boundedness Theorem for bounded, finitely additive set functions. Theorem 4. By Lemma 3 it suffices to show that Ni E. Let c E be the space of Z-simple, real-valued functions equipped with the sup-norm 2. The dual of 5. Also note that the sequence CE above is not norm - ,convergent so J 9. If the conclusion is written in this form, the theorem has the flavor of the classical UBP for B-spaces 9. For later use, we give a statement of this special case. We can now obtain an analogue of Proposition 9.

For each n define Rn: For example, the a -algebra assumption has been relaxed and the range of the measures has been replaced by TVS's.

Let ,4 be the family of all subsets of IN which are either finite or have finite complements. Let n: X - UZ by otherwise. Thus, the o-algebra assumption in Theorem 4 cannot be replaced with E being an algebra. E -4X be finitely additive and have bounded range d n e W. One of the problems of Fourier analysis is to determine the convergence of the Fourier series of a function f. We show that the UBP can be used to show the existence of a continuous periodic function whose Fourier series diverges at 0.

For this, the following lemma is useful. Let g: Define G: We claim that 0 j2, I Dn s I ds is unbounded. Of course, the same argument applies to any point in [0, 27r].

We can use a construction of Banach and Steinhaus, called Chapter 9 condensation of singularities, to show the existence of a continuous, periodic function whose Fourier series diverges at a sequence of points in [0, 2ir]. Theorem 2 Banach, Steinhaus. Example Let -wLZj be the weakest topology on X such that each f E 5r is continuous. Thus, w 3 is a vector topology on X. As a special case we have, Example Let b be a collection of vector topologies on the vector space X. The sup topology, V 1, or sup c, on X generated by D is the weakest topology on X which is stronger than each topology in D.

Thus, sup b is a vector topology on X. We do not need this general result and refer the reader to [H1] 2. See, however, 8. J Exercise 1.

Show that bal S can fail to be closed even when S is closed. Exercise 3. Exercise 4.Show that if 9 is equicontinuous, then 5 is uniformly bounded on sets of the form C x D, where C is bounded in X and D is bounded in Y. Suppose that X and Y are in duality.

Let 2l be a family of subsets of X satisfying: J Chapter 9 Proof: Then h E Cl: For the convenience of the reader not familiar with these spaces their basic properties are described in the appendix.