FUNCTIONAL ANALYSIS PDF
The present manuscript was written for my course Functional Analysis given at the University of Vienna in winter and It was adapted and extended. 1, Linear spaces, metric spaces, normed spaces (PDF). 2, Linear maps between 5, Lebesgue integrable functions form a linear space (PDF). 6, Null functions. FUNCTIONAL ANALYSIS. PIOTR HAJ LASZ. 1. Banach and Hilbert spaces. In what follows K will denote R of C. Definition. A normed space is a pair (X.
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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. Sep 30, A thorough understanding of the Oxford third-year b4 analysis course . As is usual practise in functional analysis, we shall frequently blur the. Jun 8, These are notes for the lecture course “Functional Analysis I” held by the .. functional analysis is the study of Banach spaces and bounded.
Mathematically, a subgroup is represented by H G. In analogy to set theory, the improper subgroups of a group are feg and G. The identities and inverses of the group and subgroup should necessarily coincide proof? Example 38 Let n 2 Z: Then nZ is a subgroup of Z; the group of integers under addition. First, we have to show that nZ is closed under addition. Therefore, nZ is closed under addition. Next, the identity element of Z is 0.
Finally, suppose nx 2 Z. Thus, nZ is closed under taking inverses. Therefore, nZ is a subgroup of Z In this case, the binary operation of G is carried over. Associateive law, therefore, trivially follows. The identity for a subgroup H is the same as that for the group G proof? One way to check whether we have a bona…de subgroup is by checking that it satis…es the axioms for a group in its own right.
Another way is to use the following: We should …rst prove that H is non-empty. If it is a subgroup, then it will at least contain feg. Since H is a subgroup, if b 2 H, then b 1 2 H. Conversely, suppose that for any a; b 2 H, ab 1 2 H. First we prove associativity. For a; b; c 2 H, we know that a; b; c 2 G, so associativity is trivially proved for any three elements in H.
Since a; b 2 H implies ab 1 2 H, we can reverse the roles of the elements and can conclude that ba 1 2 H. Lastly, 1 a 2 H and b 1 2 H implies a b 1 2 H or that ab 2 H, establishing that H is closed under the binary operation of G. We only prove the left cancellation law. The second proof is similar. In simple cases of the integers, rationals, reals and complex numbers, this is very clear and usually applied without the need for justi…cation as a part of school training.
Some situations can be constructed in which there is no clear answer. The distinction between left and right cancellation is important in non- commutative algebra. If inverses exists, then clearly, the cancellation property holds. In fact, every group is therefore a semi-group set with binary operation and asso- ciative law in which the cancellation law holds.
In the …nite case, the existence of inverses and the cancellation law coincide but in the in…nite case, this is not so.
(PDF) Functional Analysis Notes | Abdullah Naeem Malik - cittadelmonte.info
Can you come up with an example? De…nition 43 Suppose H is a subgroup of a group G. It will be shown below that H partitions G into right cosets. It also partitions G into left cosets, and in general these partitions are distinct.
De…nition 44 Let G be a group, H be a subgroup of G. Notice that a b mod H is de…ned for the subgroup relation. The right cosets form a partition of G, i. Suppose H is a subgroup of G. H itself is a right coset. Proceeding in this way we can get all distinct right cosets of H in G. No right coset of H in G is empty since e 2 H.
Any two right cosets of H in G are either disjoint or identical. The union of all right cosets of H in G equal to G.
Therefore the union of all right cosets of H in G gives us a partition of G. Lemma 46 Let Ha, Hb be any two right cosets. Then, f: Thus, in the …nite case, any two right cosets of the same subgroup H have the same number of elements.
In the commutative case, the left and right cosets agree. This will be denoted by N E G. If N is a proper subgroup, then N C G. For any group G, G and e are normal subgroups. Theorem 48 If H is a subgroup of a group G, then the following are equivalent. Every right coset is a left coset, i. Corollary 49 Subgroups of an Abelian group are normal Proof. Example 50 Let G be a group of all 2 2 non-singular matrices. Let S be the set of nonsingular matrices of the form a 0 0 a Notice that a cannot be zero, for otherwise the determinant would be zero and the inverse would not exist.
First we need to …nd out that S is a subgroup. Let 2 S. This implies that S is a subgroup of G: Let 2 S; Let be an 0 a z w arbitrary nonsingular 2 2 matrix. Suppose N is a normal subgroup of G, and C and D are cosets. We wish to de…ne a coset E which is the product of C and D. If c 2 C and d 2 D, de…ne E to be the coset containing cd, i. The coset E does not depend upon the choice of c and d since there are other elements belonging to the respective equivalence class. Once multiplication is well de…ned, the group axioms are immediate.
In the left side of the equality of 1. It follows that they honor identities and inverses. De…nition 53 If G; and G0 ; 0 are groups, a function f: Let a function f: H be a homomorphism. If f is also a one-one correspondence, then f is called an isomorphism.
A group isomorphism is, therefore, a function between two groups that sets up a one-to-one correspondence between the elements of the groups in a way that respects the given group operations.
If there exists an isomorphism between two groups, then the groups are called isomorphic. From the standpoint of group theory, isomorphic groups have the same properties and need not be distinguished. Lattices are isomorphic when orders between elements are preserved. Graphs are isomorphic when they both have a similar structure, with vertices and nodes being the same. Metric spaces are isomorphic when distance is preserved.
Topo- logical spaces are isomorphic when arbitrarily small distances are preserved be- tween the two topological spaces. In short, two mathematical structures are isomorphic or the same if their underlying structures are similar, with a disre- gard to nature of the elements.
The de…nition introduced here is only to make the reader familiar with the rigour of this central concept. Group isomorphism is hardly used afterwards. We shall make use of the concept of isomorphism for metric spaces, vector spaces, norm spaces and inner product spaces.
However, since any vector space is also a group in its own right, it is for the bene…t of the reader that the following be read carefully enough. Example 54 The constant map f: Example 55 If H is a subgroup of G, the inclusion i: G is a homomor- phism.
Example 56 The function f: Example 57 The function h: To prove that this is an isomorphism, we should check tha f: Then G is isomorphic to the additive group of integers: Let us de…ne ': First we show that ' is a homomorphism. Now we show that ' is a surjection. Thus, 8x 2 G: Thus ' is surjective. Thus is surjective. Now we show that ' is an injection. Suppose G and G0 are groups and f: G is a homomorphism. Then, the following properties can be proved and are left to the reader as an exercise.
The composition of homomorphisms is a homomorphism, i. G" is a homomorphism, then h f: G00 is a homomorphism. G0 is a bijection, then the function f 1: G is an isomorphism. Orders of elements are also preserved. Suppose H is a normal subgroup of G. Suppose K is a group.
R and: Then, R is called a ring if 1. Lemma 62 Let R be a ring. Then 1. Thus, a "ring" may not nec- essarily have an identity.
If R does not have the identity element, the jocular term "rng" will be used. Remark 65 Henceforth, the. Similarly, for the right zero divisor. In commutative algebra, this distinction is blurred. Lemma 69 Let R be a ring and let: Then, a is a zero divisor if and only if is not injective. That is, the cancellation laws hold, implying the existence of the inverse of a, a 1. Then, a is a not a zero divisor if and only if is injective. Invertible elements will be called units. Note that it is strictly not true that if an element is not a zero divisor, then it is necessarily a unit.
In fact, it will be seen later, a nonzero divisor has an inverse but only in a larger ring. Then, b 1 and a 1 exist. Since b 1 and a 1 are members of the larger ring, therefore the product b 1 a 1 exists. Associativity follows from the structure of the ring. The multiplicative identity is the inverse of itself and hence trivially a unit. If there is a smallest positive integer n such that n: If there is no such positive integer, then R is said to be of characteristic 2 zero.
Lemma 73 A nilpotent element a of a nonzero ring is always a two-sided zero divisor. Example 75 Zc for any c 2 Z P is not an integral domain.
An element of a ring that is not a zero divisor is called regular, or a non- zero-divisor. In such a case, we will simply refer to the resulting structure as an ideal. In the commutative case, of course this distinction is blurred. Therefore, to prove a subset I of a ring R is an ideal, it is necessary to prove that I is nonempty, closed under subtraction and closed under multiplication by all the elements of R.
This naturally also includes the elements of I. Trivially, f0g and R are ideals. In the case of integers, 2Z is an ideal because addition and subtraction of even numbers preserves evenness, and multiplying an even number by any other integer results in another even number.
Similarly, the set of all integers divisible by a …xed integer n is an ideal denoted nZ. The set of all n n matrices whose last row is zero forms a right ideal in the ring of all n n matrices.
It is not a left ideal. The set of all n n matrices whose last column is zero forms a left ideal but not a right ideal. De…nition 82 Let R be a ring, I be an ideal of R.
An introduction to functional analysis
This relation is an equivalence relation. From hereon, we will assume that R is abelian. We now show that this is well-de…ned. Since I is a commutative subgroup under addition, it is a normal subgroup hence the need to justify the plus operation can be foregone. Another way to see this is as follows: Using the de…nition of cosets: If a c; b d 2 I, what must be true about I to assure that ab cd 2 I?
We need to have the following element always end up in I: Because b d and a c can be any elements of I and either one may be 0 , and a, d can be any elements of R, the property required to assure that this element is in I, and hence that this multiplication of cosets is well-de…ned, is that, for all s in I and r in R, sr and rs are also in I.
In summary, we have De…nition 87 Let F be a set. That is, inverse for the additive identity does not exist. This de…nition should not be surprising, given what we just covered. This de…nition needs to be remembered from here on. The same rules as for the …eld are applied to each jth-tuple for 1 j n: This is one way of obtaining for ourselves a vector space, as we will see in the next chapter.
Example 88 Rational Numbers: Distribution law of multiplication over addition also holds for the set of rational numbers i. The reader is invited to show that these are indeed …elds. Example 90 Fp forms a …eld for any prime p. We can go on to show how polynomials are constructed but that divert us from our main focus.
Let us assume that the reader is familiar with polynomials generally. A high school mathematical book will tell you that the complex numbers are algebraiclly closed whereas the reals are not. The complex numbers are, in some sense, stronger than the reals. However, there is a great deal more of properties the reals enjoy which the complex numbers do not.
We are allowed to "add inequalities": The Archimedean Property guarantees there is such an n. The least number principle for integers or the well ordering principle guarantees there is a least such n: Examples of Archimedean ordered …elds include the reals R and the rationals Q.
The intuition needed to be able to say that we can have a bracket function will come in handy when we consider the modulus operation jxj on a real number. In passing, we mention the following: By de…nition, an in…nitesimal number is not a real number but belongs to an extension of R, R, called the hyperreal numbers. De…nition 94 Let F be a …eld. A subset K that is itself a …eld under the operations of F is called a sub…eld of F.
The …eld F is called an extension …eld of K. Field extensions are the main object of study in …eld theory. The general idea is to start with a base …eld and construct in some manner a larger …eld that contains p the base p …eld and satis…es additional properties.
Such an extension is called a quadratic extension. The elements of F are the "vectors" and the elements of K are the "scalars", with vector addition and scalar multiplication obtained from the corresponding …eld operations.
The dimension of this vector space is called the degree of the extension and is denoted by [F: In the above example, the dimension of the extension is 2.
It is common to construct an extension …eld of a given …eld K as a quotient ring of the polynomial ring K[X] in order to "create" a root for a given polynomial f X. If a number is not algebraic, it is called transcendental.
This is isomorphic to the Complex Numbers. The right picture to carry is that there exists n such that they have a nontrivial linear combination that equals zero. If there exists no such n, then the set of powers of looks like the set of powers of x, and then we can see why F would be isomorphic to F [X]. Two elements , of a …eld K, which is an extension …eld of a …eld F, are called conjugate over F if they are both algebraic over F and have the same minimal polynomial.
This is a sub- ring. Thus, 2 S. S between two rings R and S such that 1. Addition is preserved: The zero element is mapped to zero: Multiplication is preserved: A ring homomorphism for unit rings i. In the commutative case, this is unnecessary. A …eld homomor- phism is a function: K such that: Prove that X , the set of permutations on any set X, is a group.
Decide whether or not the following is a …eld. If it is, state a proof. Prove that M A; G is an Abelian group. In a group G; and for any a; b; c 2 G, prove that following a the identity e and the inverse a 1 for any element a are both unique. Show that the intersection of countably …nite groups is a group but that the union of countably …nite groups is not necessarily a group.
Prove that the existence of left inverse and left identity, as has been de…ned for the de…nition of a group, is implied by the existence of a right inverse and right identity and conversely.
Let G; be a group. The idea of a three-dimensional vector can be viewed as a member of R3. This is called the Euclidean space. The addition, subtraction, scalar and cross multiplication of two vectors is generally well-known and so is the algebra of matrices a recollection is added in the appendix.
However, a mathematical treatment such as the one at our disposal is far more general and has in store some great surprises. Vectors are not vectors anymore in the usual sense yet sound very familiar. Their algebra, too, looks downright distinct but a closer analysis reveals surprising commonalities.
In mathematics, the general idea is look for a common structure, formulate some rules such structures obey and then see where else the structure lies. Of course, intuition may play a role in the converse. Anyway, vector spaces are exactly such spaces. It is to be remarked that mathematically, a space is any set endowed with a particular structure. Hence, vector spaces are just that — a set of vectors with an additional structure. If you want to prove that a particular collection of elements, say integers, forms a group under a certain binary operation, you prove that each and every element of the set satis…es the axioms.
Similarly, a vector space satis…es certain axioms and to prove that a collection is a vector space, one follows the same route. The axioms for a vector space concern with addition and scalar multiplication of vectors, which can be intuitively understood. A scalar is an object that is a part of a …eld F.
We say that the vectors are scaled over the …eld F. More rigorously, De…nition 98 Let V be a non-empty set and F be a …eld. V and: If is any scalar and u is any object in V , then u is in V , as implied by the scalar multiplication function. Also, note that the bold dot " " has been dropped in favour of the juxtaposition.
The …rst …ve axioms for the vector space may be compressed to say that V is an Abelian group under vector addition. One can imagine that such concepts can readily be interpreted physically to be the usual convention of "arrows" in physics.
However, the above axiomatisation is a vast generalisation of other spaces, as well. Functions and even sequences can be interpreted as vectors in the above sense, as can be seen from the examples. Exercise 99 Prove the following: A vector over here is a "ray" from the origin on the x-axis.
We say that R is a vector space over itself. In general, any …eld is a vector space over itself. Rn Rn! Rn such that: Therefore, Rn is an abelian group. Axiom 6 is satis…ed by de…nition. For the 8th Axiom, we follow a similar series of steps: Notice how everything reduces to the pre-established axioms and the fact that the base is a …eld and that the tuples form from the same …eld.
This painstaking mode of argument was only meant to serve as a motivation for how a vector space ought to be proved: To prove that Rn is a vector space over the …eld Q, we can similarly resort to the fact that each tuple, that is, every real number when added or multiplied by a rational number will yield another real number. Equivalently, we can take n-tuple of the Complex plane C and generate for ourselves a vector space.
We will prove that the sum of continuous numbers is continuous. Hence the de…nitions for the addition and scalar multiplication are jus- ti…ed. It is now rather routine to prove that the space of continuous functions is a vector space by showing that under the de…ned "vector" addition and scalar multiplication, the remaining 8 axioms hold. In particular, one should prove that the zero function is continuous; any additive inverse of a continuous function is also continuous. We will pause over here to recall a fact from real analysis: However, if the function under consideration is uniformly continuous, then this will not depend upon the domain so that the lenghts of each pre-image of open sets is the same.
Thus, continuity of a function is a local property since it depends upon x whereas uniform continuity of a function is a global property which does not depend upon x. Needless to say, if a function is uniformly continuous, then it is continuous but the converse is not true in geneal.
Hopefully, you have just proved that this collection of functions forms a vector space. In other words, functions can be viewed as vectors in a sense — a single point in space. Such a space is also called the function space. Wherever possible, vectors will be made bold. However, this will not be strictly followed, especially when it comes to functions and sequences, since this may be a cause of confusion with some pre-established notations. This should in no way mean that functions are not vectors.
Example The set of polynomials P [a; b] of order at most n over the set [a; b] is also a vector space. Try to prove this yourself. Before moving on to another vey important example, let us pause to consider the following: For instance, in the …rst example, we see that R is a vector space over …eld R and Rn , too, is a vector space over R. Generally, if we have vector spaces X1 ; X2 ; De…ne x1 ; x2 ; This is natural because for each i-th tuple, the addition makes sense because the underlying vector space Xi is closed under addition and mulitplication.
It is now a routine matter to verify that the resulting space is indeed a vector space. Such an X is a vector space and is called the Cartesian product see Set Theory in preliminaries for details. Usually, when the context is clear, the word Cartesian is dropped and just the words "product of spaces" is used. However, this is not standard since there are other ways in which a product for a vector space may be de…ned.
Xn Before we move on to consider our next example, a de…nition is in order: In reality, it actually measures the distance of a number from the zero point. In other words, the magnitude of the one-dimensional ray formed on the real line is measured by this de…nition.
The above de…nition was meant to serve as a gentle introduction what the modulus actually, norm is about. But enough with gentle introductions! We start with jsj s jsj and jtj t jtj for s; t 2 R. Rn consists of n-tuples. Now, a sequence may converge or diverge.
This idea is made more formal in metric which you can read ahead for yourself for a refresher. You have probably passed a Calculus, Real Analysis and Topology course so I will just go ahead with myself: We can collect for ourself bounded sequences, convergent sequences or even convergent series.
Now we can move ahead with the real axioms! This should not to be confused with the Hilbert space, which will be studied in the coming chapters. A remark, however, has to be in order; this was the …rst example of a Hilbert space pre- sented in history. From this, one can get a sense that Hilbert spaces are not just about vectors in the usual sense and that their development had rather broad applications in mind.
In order to avoid confusion, we will refer to the space in the example as the sequence space. This is natural since the the bound will depend on the sequence x but has to be valid for all i.
In fact, we might even have a divergent series if this constant depended on each i, giving us an unbounded sequence. This topic can be put aside until normed spaces are de…ned.
Instead of real numbers i , we can also have complex numbers as the base elements of the sequences. Note that this should not be confused with the Lebesgue Space Lp , which deals with functions with a p-norm. The lp space is a special case of the Lp space, which is not covered in MTH However, because of the knowledge of measure that is required, such spaces are studied in MTH Lp spaces are de…ned using natural generalisations of p-norms for …nite-dimensional vector spaces.
In a sense, p and q are inverses or conjugates of each other. These are called conjugate exponents. Theorem For convergent why? Then, we can think of ab as an area of a rectangle with sides a and b. Multiplying these two and placing in the equality 1: Exercise Show that the geometric mean of two positive numbers does not exceed their arithmetic mean. Trivial for zero vectors. Speci…cally, for p 1 1! If we think of the real number line as one-dimensional arrows eminating from zero, we can interpret the mod- ulus as a function that tells how far the real number is from zero.
Of course there is no reason to have a magnitude function the modulus for real numbers only. We can move ahead and generalise it for other vectors as well strictly, elements of a vector space. A norm on N is a real-valued function k: Then, jus- tify to yourself that they apply to three dimensional vectors as well.
Moving on, as can be sensed, this is a generalisation of the concept of "measure" of a vector or its length. A linear vector space together with a norm de…ned on it is called a normed space, denoted by N; k: Since the use of the extra brackets is tedious, we will often shorten it to just the name of the set. There is absolutely no reason why we should restrict ourselves to two …elds but is only because it makes matter a little easier to handle.
We also have the added advantage of completeness in both and that of order in the former. We could have very well made use of the …eld of rational numbers, which may prove to be easy but because of the absence of completeness, we will not be covering this more general case. The imagination dries when the quaternions or even the …eld of functions are used as scalars. The …rst part of N1 is the positivity property or positive de…niteness, N2 is called the Absolute Homogenity Axiom whereas condition N3 is called the Triangle Inequality or the Subadditive property.
The second part of N1 ensures that any two given vectors are two separate points. The …rst intuitive example of the magnitude of two vectors will serve as a motivation for the de…nition. Example Let R2 be the Euclidean space.
Prove that this is indeed a normed space. This is also known as the uniform norm. Note that the norm de…ned as above, the Euclidean norm, is not the only norm on Rn. For instance, Example We can de…ne k: Exercise The norm of lp was hinted at in the example for vector spaces; for any sequence x 2 lp , we have 1!
The proof that this de…nition satis…es the axioms for a norm, making it a bona …de normed space, follows a similar pattern as that for Rn. The proof of N3 is the proof of the Minkowski inequality.
N3 follows by the use of the triangle inequality of real numbers. The concept of a norm is important because length of a vector, distance be- tween two vectors and hence the idea of convergence can be de…ned accordingly.
Needless to say, the idea of convergence is essential in almost all areas of mathe- matics. Also, based on the de…nition of the norm and the vectors, convergence, magnitude and distance can then be decided for, as we shall see. Set Topology 1. A metric de…ned on X is a function d: A metric space is a pair X; d where X is a set and d is a metric on X. The UBP and Equicontinuity The Transpose of a Linear Operator Subdivisions of the Spectrum and Examples The Spectrum of a Compact Operator Banach Algebras Notation Appendix: Definition 1.
X, r is a topological vector space TVS is r is a topology on X such that: Let X, r be a TVS. U E 2 is a open neighborhood base at x0.
The first part of i follows from Definition 1 and the fact that joint continuity implies separate continuity. Definition 3. Remark 4. Definition 5. I t 1 5 1, Chapter 1 5 Remark 6. Note that a balanced set is symmetric and contains 0. Definition 7. Proposition 8.
Let X be a TVS. If S is balanced, then S is balanced. Theorem 9. That is, 3 a base at 0 consisting of balanced sets. Since 0. Definition and Basic Properties 6 Corollary Let X, ti be a TVS. The following are equivalent: Suppose iii holds and x, y E X with x y. There exists a neighborhood of 0, U, such that x - y e U.
Proposition Let U be a neighborhood of 0. Theorem If X is a TVS, then there is a base at 0 which consists of closed, balanced sets. Every neighborhood of 0, U, contains a closed neighborhood of 0, V, which contains a balanced neighborhood of 0, W. Definition A subset S c X is absorbing if it absorbs all singletons.
Notes on Functional Analysis
Every neighborhood of 0 in a TVS is absorbing. Let U be the family of all balanced, open neighborhoods of Definition and Basic Properties 8 0. This is a neighborhood base by Theorem 9 and i holds by Proposition 15 while ii and iii hold by Proposition 2. 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. To show that scalar multiplication is continuous, first observe that: The uniqueness of r is clear.
A similar result is given by Theorem Let 2l be a family of subsets of X satisfying: 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. In general, the open [assumption iv in Theorem 17 guarantees this].
We give several examples of TVS. Example Let X be a vector space and 51 a family of linear maps f: 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.
Chapter 1 11 Completeness: 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. If Z is a TVS, show a map g: Z -4X is continuous with respect to w g if and only if fog: Z -4Yf is continuous Yf E 9. Exercise 4. If Z is a TVS, show that a map f: Z - X is continuous with respect to sup D if and only if it is continuous with respect to 'c d r E D.
Exercise 5. Exercise 6. Exercise 7. Show that any compact subset of a TVS is complete. Exercise 8. A quasi-norm on X is a function: We will show in 9.
Example 2. Let p be a quasi-norm on X. Let be a sequence of quasi-norms on pn X. If I I is a quasi-norm on X, then d x, y semi- metric on X which is a metric if and only if x-yI I defines a is total.
If the quasi- norm is understood, we often say that X is a quasi-normed space. Note that a quasi-normed space is a TVS under the metric topology induced by the quasi- norm the continuity of addition follows from the triangle inequality.
We always assume that a quasi-normed space carries this metric topology. Proposition 4. Let X, p be a quasi-normed space. X - U is continuous. A sequence xj in a TVS X is said to be Mackey convergent to 0 locally null if 3 a sequence of positive scalars Co 1 such that t 1 -4 0. We now show that in a quasi-normed space every sequence which converges to 0 is Mackey convergent to 0. Proposition 5. Let X, 1 I be a quasi-normed space and 1 - 0 in X.
Quasi-normed and Normed Linear Spaces 16 Then 3 an increasing sequence of non-negative integers kj -9 w such that k. For nk 5 j For an example of a sequence which converges to 0 but is not Mackey convergent to 0 see Exercise Definition 6. A semi-normed linear space semi-NLS is an ordered pair X, 11 where 11 similarly. Proposition 7.
Then Iitkxk- txII ii follows from the triangle inequality. From i it follows that a semi-norm norm is a quasi-norm so iii follows from Proposition 4.
We consider completeness for a QNLS. For such spaces the notions of completeness and sequential completeness coincide. Suppose that X is sequentially complete and let x5 be a Cauchy net in X. A quasi-normed space X, I I is complete if and only if every absolutely convergent series in X is convergent.
For each k choose nk such that n? It will be seen later that the notion of X convergence can often be used as an effective substitute for completeness. We can give examples of some of the classical quasi-normed and NLS which we will often use later in examples. We begin with some of the classical sequence spaces. Let s be the space of all real or complex valued sequences. Addition and scalar multiplication are defined coordinatewise. I I is just coordinatewise convergence Example 3 , s is a complete quasi-normed space.
We assume that c is equipped with the sup-norm. We show that c is complete by showing that it is a closed subset of the complete space l Chapter 2 21 Corollary Since Ixn - xII. We assume that c0 is equipped with the sup-norm, and just as in Corollary 13, co is a B-space. We assume that c00 is equipped with the sup-norm.
Let the sequence with a If coordinates. The sequence tj also furnishes an example of a sequence which converges to 0 but which is not X convergent. Let m0 be the subspace of consisting of all sequences with finite range. B S is a B-space Exer. Let S be a compact Hausdorff space. C S is the subspace of B S consisting of all continuous functions. We assume that C S is equipped with the sup-norm; C S is a B-space under the sup-norm since convergence in the sup-norm is just uniform convergence on S.
Note I I is not a semi-norm. This is Riesz's Theorem, [Ro], p. Let a, b E R, a b [a, b] be the space of all b Riemann integrable functions defined on [a, b]. Each l and if I In is a norm on X D , I is the Frechet quasi-norm of Example 3 induced by 1 Id' then M D is a complete quasi-normed space since convergence in is just uniform convergence on compact subsets of D. Note that the topology of BB D does not depend upon the particular sequence Kn chosen.
Let Ck[a, b] be the subspace of C[a, b] which consists of all functions which have at least k continuous derivatives. Then Ck[a, b] is a B-space under Example Let K c " be compact.
Let 11 IL ,k [DeS], p. IR' -4F which have continuous partial derivatives of all orders with support contained in K. For f e.
It is not clear that OK contains anything other than the zero function. We give an example of a non-zero function in. Define cp: Qin - [R defined by f xl, Let S2 c Rn be open. S - IR which have continuous partial derivatives of all orders in Q. A sequence fk converges to 0 if and only if the sequence Dafk converges uniformly on compact subsets of SZ V multi-index a.
Thus, S S2 is complete [DeS], p. Note that the topology of X S2 does not depend upon the particular sequence K. Rn and Cn are B-spaces under any of the norms Ix. Let S 0 and let s be a o algebra of subsets of S. Any bounded finitely additive set function v: Chapter 2 27 Example Let B S, E be the space of all bounded E-measurable functions f: S -4R.
Under the sup-norm, UflI. CE , where t. Finite Products: Let X and Y be quasi-normed spaces. Each of these quasi-norms induces the product topology on X x Y so any of them can be used on the product. A further important example of NLS are given by the inner product and Hilbert spaces. For the convenience of the reader not familiar with these spaces their basic properties are described in the appendix.
Give an example of a function satisfying i - iv but not v. If xk is a Cauchy sequence in a metric space and has a Quasi-normed and Normed Linear Spaces 28 subsequence xn which converges to x, show xk -4 X. If F,xk is a convergent series in a TVS, show xk Show c00 is dense in co with respect to Show c00 is dense in IP with respect to p for 1 Show t p is complete for 0 subspace. Note that p is no a norm for 0 Exercise 7. Show B S is complete under Exercise 9. Show 11 Exercise Give an explicit example of an absolutely convergent series in c00 m0 which is not convergent.
Give an example of a convergent series in I P 1 which is not absolutely convergent. If the sequence xk is Mackey convergent to 0, show mat it converges to 0. Show that COO' c, c0, e p 1 are separable. Show is not separable.
Show B S, E of Example 31 is complete. This simple necessary condition turns out to be sufficient for the topology of a TVS to be semi-metrizable even by a quasi-norm. To establish this result, we require two lemmas. Lemma 1. Let X be a vector space and q a non-negative function n n defined on X. Lemma 2. Then for n n xl, These inequalities and the hypothesis give the result.
Theorem 3 Kakutani. The topology of a first countable TVS X is given by a quasi-norm.
We claim that xn - 0 in X if and only if q xn Suppose xn and let m be a positive integer. Thus, xn E Urn eventually and xn -9 0. We may assume that not all x, y, z E 0. Thus, We now have established the conditions of Lemma 2 and we define q x 5 x by the formula in Lemma 1. We claim that q x. A TVS X is called asemi-metric linear space metric linear space if its topology is semi-metrizable metrizable.
From Theorem 3 we see Chapter 3 35 that a space is a semi-metric linear space if and only if it is first countable if and only if its topology is generated by a quasi-norm.
A complete metric linear space or equivalently a complete quasi-normed space is often referred to as an F-space in honor of M. Whereas the topology of a TVS is generated by a single quasi-norm if and only if the topology is semi-metrizable, Burzyk and P.
We use the notion of boundedness as originally introduced by Banach [MO]. We have the following criteria for boundedness. Theorem 2. Clearly iii iv. Suppose vi holds. Pick a balanced neighborhood V of 0 such that V c U 1. Suppose vii holds. Let U be a balanced neighborhood of 0. Hence, tkxk -1 0. Remark 3. Condition iii was introduced by von Neumann and is often used for the definition of boundedness [vN]. Chapter 4 39 Corollary 4.
In particular, no non-trivial subspace of a Hausdorff TVS is bounded. Every neighborhood of 0 contains 0 so 0 is bounded. So L is not bounded. In a quasi-normed space a bounded subset is metrically bounded, but the converse does not hold in general. Let B be bounded. But nS 0, 1 c S 0, n so B c S 0, n is metrically bounded. Note the space L0 z of Example 2. For semi-NLS, we do have Proposition 6. A subset B of a semi-NLS is bounded if and only if it is metrically bounded.
The definition is due to P. Antosik [A]. Clearly a X bounded set is bounded, but the following example shows that the converse is false.
Example 8. Clearly B is bounded with respect to , but is not X bounded as noted in Example 2. As will be seen later, the notion of X boundedness will be very useful in formulating general results which usually require some sort of completeness conditions. Chapter 4 41 Example 9. Define a norm on m0 see 00 Example 2.
However, no subseries of converge to an element of m0 with respect to II II II II k ek can since convergence in clearly implies coordinatewise convergence. A TVS in which bounded sets are X bounded is called an. However, we show in Proposition 10 below that any quasi-normed. If X is a quasi-normed. Show that a finite set is bounded. Exercise 2. Show a compact subset of a TVS is bounded. Show that a subset of a bounded set is bounded. Show that a finite union or vector sum of bounded sets is bounded.
Show that a convergent sequence is bounded. If X and Y are vector spaces and T: Proposition 1. E -4 F linear. The following are equivalent. Assume ii holds and let x e E. Let V be a neighborhood of Tx. A linear map between TVS is said to be bounded if it carries bounded sets to bounded sets.
Proposition 3. E -4F sequentially continuous. Then T is bounded. Let B c E be bounded. Then bounded by 4. As we will see later E -4 F is bounded, then T is continuous. Assume iv holds. L X, Y denotes the space of all linear, continuous mappings called operators from X into Y. L X, Y is a vector space under the pointwise operations of addition and scalar multiplication. We have the following easily checked properties of the operator norm.
Concerning completeness, we have Proposition 8. Suppose that Tk is Cauchy with respect to the operator norm. Then T: X -i Y is clearly linear. Remark 9. We will establish the converse of Proposition 8 later 8. A linear map from X into the scalar field is called linear functional. The dual of X, denoted by X', is the space of all continuous linear functionals on X, i.
IIxII s 1 ; this norm is called the dual norm on X'. We give isometric descriptions of the duals of some of the classical function spaces. It is assumed that the reader is familiar with some of the classic dual spaces from real analysis; references to such duals are given.
Let f E c6. In particular, if u is counting measure on IN, we have Example We now show that the dual of B S, s is ba s Example 2.
For this we require the integration of bounded E-measurable functions with respect to bounded finitely additive measures belonging to n ba E. The inequality 2 still holds if p belongs to B S, E. There is a partition E: We show U is onto. Then ,u is clearly finitely additive, and if E.: This requires the development of a general integration theory with respect to finitely additive measures which we will not give. The reader can consult [DS] or [HS].
The dual of s is c If Y E c00, y induces a continuous 00 linear functional fy: We claim that the dual of LO[0, 1] is exactly 0.
LO[0, 1]'. This quasi-norm has a trivial dual space exactly like L0[0, 1] above. See [Ki] p. Let 5c L X, Y and consider the conditions: The first implications are clear. For semi-NLS, we have the following norm criterion for equicontinuity.
T E 91 is bounded. We give one further result on equicontinuity for future use. X - Y is linear and continuous.
Let V be a neighborhood of 0 in Y. T is clearly linear and TU c W c V since. W is closed so T is continuous. Define f: Show f E B S ' and compute 11f Describe the dual of c00 and m0 under 11 Compute the dual norm of in terms of the dual norms of X and Y. Show X and L F, X are linearly isometric. 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: Unitary operators.
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