FUNCTIONAL ANALYSIS-NORMED SPACE

Transcription

1 MAT641- MSC Mathematics, MNIT Jaipur FUNCTIONAL ANALYSIS-NORMED SPACE DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR 1. Normed space Norm generalizes the concept of length in an arbitrary vector space. Norm on a linear space X defined over field K is a function. : X R which satisfies the following axioms: For x, y X x 0 and x = 0 iff x = 0. ax = a x for a K. x y = y x x + y x + y, Triangle inequality. A linear space with norm defined on it is called Normed Space. Exercise 1.1. Every normed space is a metric space with metric defined as x y = d(x, y). But converse is not necessarily true. Example 1. Set S of all bounded or unbounded sequence of real or complex numbers with metric d defined as 1 ξ i η i d(x, y) = 2 i 1 + ξ i η i where x = (ξ i ), y = (η i ) S, is a metric space. But (S,. ) is not a normed space for the norm defined as 1 ξ i x = 2 i 1 + ξ i since (Prove it) Example 2. Examples of normed spaces: αx α x. a. Real line R is a normed space with norm x = x, x R. b. Euclidean space R 2 is a normed space with norm x = (ξ 1 ) 2 + (ξ 2 ) 2 x = (ξ 1, ξ 2 ) R 2. c. Euclidean space R n and Unitary Space C n are normed spaces with norm x = ξ ξ n 2, x = (ξ 1,..., ξ n ) R n or C n. d. B[a, b] set of all bounded functions defined on [a, b] with norm defined as x = sup t [a,b] x(t), x B[a, b] is a normed space. Date: February 24, ragarwal.maths@mnit.ac.in Web: drrituagarwal.wordpress.com

2 2 DR. RITU AGARWAL e. Sequence space l { } l = x = (ξ i ) s.t. sup ξ i < i with norm defined as x = sup i ξ i, x l. f. Sequence space l p l p = { x = (ξ i ) s.t. ( ) 1/p, with norm defined as x l p = ξ i p x l p, p 1. } ξ i p < g. Function space C[a, b]: This is the set of all real valued functions x, y,... which are functions of independent real variable t and are defined and continuous on a given closed interval [a, b]. The norms on this set can be defined as: x 1 = max t [a,b] x(t) x 2 = b x(t) dt represents the area between the curve bounded by the ordinates x = a and a x = b and x-axis. (C[a, b], x 1 ) is a Banach space while (C[a, b], x 2 ) is not. (Give counter example). Lemma 1.2. A metric induced by a norm on a normed space (X,. ) satisfies (i) Homogeneity: d(x + z, y + z) = d(x, y), (ii) Translation invariance: d(αx, αy) = α d(x, y), x, y, x X and α is a scalar. (Prove it). In Example 1, (ii) is not satisfied. Theorem 1.3. Norm is a continuous function i.e. x x is a continuous mapping of (X,. ) into R. Proof: A function T is continuous if d(t x, T x 0 ) < ɛ for d(x, x 0 ) < δ. Choose δ = ɛ. Clearly, if d(x, x 0 ) = x x 0 < δ, then d( x, x 0 ) = x x 0 < x x 0 < δ Therefore. is a continuous function. Definition 1.4 (Banach Space). A complete normed linear space is called Banach space. Definition 1.5 (Subspace of a normed space). A subspace Y of a normed space (X,. ) is a subspace of X considered as a vector space with the norm obtained by restricting the norm on X to the subset Y. This norm on Y is said to be induced by the norm on X as. Y =. X/Y Remark 1.6. If (X,. ) is a Banach space, (Y,. ), Y X need not necessarily a Banach space. (Give example) Example 3. In l, let Y be a subset of all spaces with only finitely many non-zero terms. Show that Y is a subspace of l but bot a closed subspace. Proof: Here Y = {y = (η i ) where η i = 0, i > n} l but Y is not closed in l. Consider the subsequence y n = (1, 12,..., 1n ),... Then y = lim n y n = ( 1, 1 2,...) l but y / Y as it contains infinitely many terms. Hence Y is not complete.

3 MAT641-NORMED SPACE 3 Theorem 1.7. A subspace M of a Banach space (X,d) is complete iff M is closed in X. Proof: Let Y is a complete normed space. To show that Y is closed, i.e. Ȳ = Y. Since Ȳ Y, it is sufficient to show that Ȳ Y. Let y Ȳ. Then yiny or y ia a limit point of y. if y Y, there is nothing to prove. So let y is a limit point of Y. So a sequence (y n ) in Y which converges to y. Hence the sequence (y n ) is a Cauchy sequence. Since Y is complete, (y n ) is convergent in Y and hence the point of convergence i.e. limit point y Y. Conversely, let Y is closed and (y n ) be an arbitrary Cauchy sequence in Y. Then y n y X, since X is complete. Also y is a limit point of a sequence in Y which is closed. Therefore, y Y and hence Y is complete i.e. Y is a Banach space. 2. Convergence in normed space A sequence (x n ) in a normed space (X,. ) is convergent if X contains an x such that lim x n x = 0 n That is d(x n, x) = d(x n x, 0) = x n x 0 as n. We write x n. x. Definition 2.1 (Cauchy sequence). A sequence (x n ) in a normed space (X,. ) is Cauchy if for every ɛ > 0, there is a n 0 such that x n x m < ɛ for m, n n 0. Definition 2.2 (Convergence of infinite series). Let (x n ) be a sequence in normed space (X,. ). An infinite series is said to be convergent in the normed space, if the sequence of partial sums converge i.e. s n s 0 as n where s n is the sequence of partial sums defined as s n converges in (X,. ) and having the sum s. = n x k. We say that the series x k Definition 2.3 (Absolute convergence). If the series x k converges, we say that x k converges absolutely. Remark 2.4. In case X = R or C, then an absolute convergence of series implies the convergence of original series. i.e. x k < = x k <. Converse is not necessarily true in general means a convergent series is not necessarily absolutely convergent. E.g. ( 1)n converges but not absolutely. n Example 4. (1) In an arbitrary normed space (X,. ), absolute convergence not necessarily implies convergence. If in a normed space X, absolute convergence of any series implies convergence of that series, show that X is complete. (2) Show that in a Banach space, an absolutely convergent series is convergent. Again, a subspace of Banach space need not be closed for convergence. Example 5. X = l, set of all bounded sequences of real or complex numbers. Let Y be the set of all sequences with only finitely many non-zero terms, i.e. But Y is not closed in l as and y / Y. Y = {y = (y i ) where y i = 0, i > n} l lim y n = y = (1, 12, 13,..., 1n ),... l n Absolute convergence need not imply convergence in space. E.g.

4 4 DR. RITU AGARWAL Example 6. Suppose Y be the subspace of l consisting of all sequences with only finitely many terms non-zero. Consider the sequence (y n ) Y defined as y n = (η (n) j ), η (n) j = Then y 1 = 1, y 2 = 1/4,..., y n = 1/n 2 and y k = is convergent series but 1 n 2, j = n 0, j n 1 k 2 y k = (1, 1/4, 1/9,..., 1/k 2,...) / Y. Theorem 2.5. Prove that a normed space is a Banach space (i.e., complete) if and only if every absolutely convergent series is convergent. 3. Schauder Basis A series x k is said to be convergent if the series (s n ) of partial sums s n = n x k converges. If a normed space X contains a sequence (e n ) with the property that for every x X, there is a unique sequence of scalars (α n ) such that x (α 1 e α n e n ) 0 (3.1) as n, then (e n ) is called a Scahuder basis for X. The series x = α ke k is called expansion of x with respect to (e n ). Theorem 3.1. Show that if a normed space has a Schauder basis, it is separable. Proof: Let X be a real normed space and M = {x X : α 1 e 1 + α 2 e α n e n }, α i, i = 1, 2,..., n are rational numbers and {e 1,..., e n } is the Schauder basis. Since the set of rational numbers is countable, M is a countable set. It remained to show that M is dense in X. Suppose z X then a sequence of scalars β i such that z (β 1 e 1 + β 2 e β n e n ) 0 as n If β i i = 1, 2,..., n are rational, β 1 e 1 + β 2 e β n e n M. If some β i are not rational, because of denseness of rational numbers, β i can be approximated by a rational number α i so that β 1 e 1 + β 2 e β n e n α 1 e 1 + α 2 e α n e n M so that z (α 1 e 1 + α 2 e α n e n ) 0 as n Hence M is dense in X i.e. X is separable. In case, X is complex normed space, M can be defined as M = {x X : α 1 e 1 + α 2 e α n e n }, α k = a k + ib k, a k, b k, k = 1, 2,..., n are rational numbers. Proceeding as above, it can be shown that M is dense in X. The harder question is whether a separable Banach space necessarily has a Schauder basis. This question was raised originally by Banach. For a long time all known examples of Banach spaces were found to have such a basis. In 1973, Enflo constructed a separable Banach space that does not have a Schauder basis. Exercise 3.2. Does every separable Banach space has a Schauder basis? See P. Enflo (1973) for the answer. Example 7. Following are the examples of spaces about their nature of separability R is separable.

5 MAT641-NORMED SPACE 5 C is separable. l is not separable. (Does not have a Schauder basis) l p is separable. (Have a Schauder basis) Remark 3.3. The p spaces are generic normed spaces. The cases of p and p = are inherently different for the above mentioned reason. 4. Finite dimensional spaces Spanning Set: If S is a finite subset of V, then the span of S is the set of all linear combinations of the elements of S. It is denoted by Span(S). Definition 4.1 (Basis). A set of vectors in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set. In more general terms, a basis is a linearly independent spanning set. Every vector space has a Hamel basis, which is a purely algebraic construct. the dimension of a vector space V is the cardinality (i.e. the number of vectors) of a basis of V over its base field. For every vector space there exists a basis,[b] and all bases of a vector space have equal cardinality;[c] as a result, the dimension of a vector space is uniquely defined. We say V is finite-dimensional if the dimension of V is finite, and infinite-dimensional if its dimension is infinite. The dimension of the vector space V over the field F can be written as dim(v). Theorem 4.2 (Lemma). Let {x 1,..., x n } be a linearly independent set of vectors in a normed space X (of any dimension). Then there is a number c > 0 such that for every choice of scalars α 1,..., α n, we have α 1 x α n x n c( α α n ) (4.1) Proof: Let S = α α n. If S = 0, all α i = 0 and equality holds for both sides of (4.1). Consider S 0, then from (4.1), we get or (α 1 x α n x n )/S c β 1 x β n x n c, where α i S = β i and We are required to prove the existence of a number c > 0 such that (4.2) holds. β i = 1 (4.2) Suppose (4.2) doesn t hold. Then the length of the vector (norm) β 1 x β n x n, n β i = 1 is not greater than zero. Therefore, there exists a sequence (y m ) of vectors y m = β (m) 1 x β (m) n x n, β (m) i = 1 such that y m 0 as m. Then for fix j, β (m) j 1. The sequence (β (m) j ) is a bounded sequence of scalars and by Bolzano- Weierstrass theorem, every bounded sequence has a convergent subsequence. Hence suppose that (β (m) j ) j = 1, 2,..., n has a convergent subsequence (γ (m) j ) that converges to β j as m. Therefore, y m = β (m) 1 x β n (m) x n has a subsequence y 1,m = γ (m) 1 x β n (m) x n where

6 6 DR. RITU AGARWAL γ (m) 1 β 1 as m. y 1,m has a subsequence y 2,m = γ (m) 1 x 1 + γ2 m x β n (m) x n where γ (m) 2 β 2 as m. Proceeding similarly, y n,m has a subsequence y n,m = γ (m) 1 x 1 + γ2 m x γ n (m) x n, n γ(m) i = 1 where γ n (m) β n as m. Hence y n,m y = n β ix i, n β i = 1 as m. Since all β i s are not zero as {x i } is set of LI vectors. Therefore y 0 and since norm is a continuous function y n,m y 0 which is contradiction to our assumption, according to which y = 0 and hence y = 0. Therefore, there exists c > 0 such that (4.2) is satisfied. Theorem 4.3. Every finite dimensional subspace Y of a normed space X is complete. In particular, every finite dimensional normed space is complete. Proof: Let dim Y = n and {e 1, e 2,..., en} be the basis for Y. Consider aa arbitrary Cauchy sequence (y m ) in Y. Then by definition N(ɛ), ɛ > 0 such that for m, r N Also, y m Y, therefore Thus, y m y r < ɛ y m = α (m) 1 e α (m) e n, α i are scalars. (α (m) 1 α (r) 1 )e (α (m) n By application of 4.2, there exists c > 0 such that ɛ > For fixed j (α (m) j α (m) j n α (r) j )e j c α n (r) )e n < ɛ α (m) j α (r) j < ɛ c, m, r N α (r) j (α (m) j ) is a Cauchy sequence of scalars in R (or C) which is complete. Hence (α (m) j ) is a convergent sequence and converges to α j as m. Therefore y m = α (m) 1 e α n (m) e n y = α 1 e α n e n as m. Clearly y Y. It remained to show that y m y in norm. y m y = (α (m) j α j )e j K α (m) j α j where K = max 1 j n e j. Since α (m) j α j, as m, y m y in norm. Hence Y is complete. Note: For an infinite dimensional subspace, the result may not hold true. For example, set of polynomials in C[a, b]. Theorem 4.4. Every finite dimensional subspace Y of a normed space X is closed. Definition 4.5 (Equivalent norms). Two norms. α and. β on a vector space V are called equivalent if there exist positive real numbers C and D such that for all x in V C x α x β D x α. Equivalent norms on X defines the same topology on X. Theorem 4.6 (Exercise). Any norm. 0 is a continuous function on V under the topology induced by the norm. 1. Theorem 4.7. On a finite dimensional vector space X, any norm. 0 is equivalent to any other norm. 1.

7 MAT641-NORMED SPACE 7 Proof: Let dim X = n and {e 1, e 2,..., en} be the basis for X. Any x X, therefore Thus, x = α (m) 1 e α (m) e n, α i are scalars. x = α 1 e α n e n = By application of 4.2, there exists c > 0 such that x c α j = For another norm. 0, x 0 = by triangle inequality. Let us denote K = max 1 j n e j 0, then x 0 K n α j e j 0 α j e j α j x c α j e j 0 α j K x = β x (4.3) c for β = K/c > 0. Interchanging the role of x 0 and x and proceeding in similar way, we get for some α > 0. Combining (4.3) and (4.4), we get the desired result. Exercise 4.8. Axioms of equivalence relations hold on equivalent norms. x α x 0 (4.4) Exercise 4.9. Prove that the norms. 2,. defined over n-dimensional space R n are equivalent by a direct proof. Exercise Prove that the norms. 1,. 2 defined over n-dimensional space R n satisfy 1 n Compactness and finite dimension Definition 5.1 (Sequentially Compact). A metric space X is said to be compact if every sequence in X has a convergent subsequence. Theorem 5.2. A compact subset M of a metric space is closed and bounded. Proof: For every x M, there is a sequence (x n ) such that (x n ) x. Since M is compact, x M. Hence M is closed. M is bounded for if M is unbounded, it has an unbounded sequence (y n ) such that d(b, y n ) > n for some fixed b. This sequence could not have a convergent subsequence since a subsequence must be bounded. Converse is not necessarily true. Exercise 5.3. Find the counter example. [Hint: Point set in l 2 ] Theorem 5.4. In a finite dimensional normed space X, any subset M of X is compact if and only if M is closed and bounded.

8 8 DR. RITU AGARWAL Proof: Compactness of M implies M is closed and bounded. Conversely let M is closed and bounded. Since the space X is finite dimensional, let dim X = n and {e 1, e 2,..., e n } be the basis for X. Consider a sequence (x m ) in M. Then x m = α m 1 e 1 + α m 2 e α m n e n Since M is bounded, so is (x m ) i.e. x m k for some positive real number k. k x m = αj m e j c αj m, c > 0 Hence, the sequence of number α m j is bounded for fixed j. By Bolzano-Weierstrass theorem, α m j has a point of accumulation α j. (x m ) has a subsequence (z m ) which converges to z = n α je j. Since M is closed, z M. Thus, an arbitrary sequence in M has a subsequence which converges in M. Hence M is compact. 6. Riesz s Lemma Riesz s lemma (after Frigyes Riesz) is a lemma in functional analysis. It specifies (often easy to check) conditions that guarantee that a subspace in a normed linear space is dense. The lemma may also be called the Riesz lemma or Riesz inequality. It can be seen as a substitute for orthogonality when one is not in an inner product space. Before stating the result, we fix some notation. Let X be a normed linear space with norm. and x be an element of X. Let Y be a closed subspace in X. The distance between an element x and set Y is defined by d(x, Y ) = inf y Y x y. Theorem 6.1 (Riesz s Lemma). Let Y and Z be subspaces of a normed space X (of any dimension) and suppose Y is closed and a proper subset of Z. Then for every real number θ in the interval (0, 1), there is a z Z such that z = 1 and z y θ for all y Y. Proof: Let v Z \ Y and let a = inf y Y v y. As Y is closed and v / Y, we have a > 0. Now let θ (0, 1) and note that there exists a y 0 Y such that a v y 0 a θ (as θ < 1, we have a θ > a). Let, z = v y 0 v y 0. Obviously, z Z and z = 1. Let y be any element of Y. We have the following. z y = 1 (v y v y 0 0) y = 1 v y v y 0 0 ( v y 0 )y a a = θ v y 0 a/θ Theorem 6.2. If a normed space X has the property that the closed unit ball M = {x : x 1} is compact then X is finite dimensional. In a finite dimensional space, closed unit ball is compact. This theorem is converse of it and is ensured by Riesz s Lemma. Proof: Let M is compact but dim X =. Choose x 1 X such that x 1 = 1. This x 1 generates a one dimensional space which is closed and a proper subspace X 1 of X which is infinite dimensional. By Riesz s Lemma, there is an x 2 X of norm 1 such that x 2 x 1 θ = 1 2 The elements x 1, x 12 generate a two dimensional proper closed subspace X 2 of X. Again, by Riesz s Lemma, there is an x 3 X of norm 1 such that x 3 x 1 2, x X 2

9 MAT641-NORMED SPACE 9 Proceeding inductively in a similar manner, we obtain a sequence (x n ) of elements in M such that x m x n 1 2, which cannot be a convergent sequence. Hence, M is not compact which is a contradiction. This implies, our assumption is false i.e. dim X <. m n Exercise 6.3. Let T : X Y be a continuous map from one metric space to other. Then the image of a compact subset M of X under T is compact. References [1] Kreyszig, Introductory Functional Analysis With Applications, Wiley India Edition, [2] Walter Rudin, Functional Analysis, Tata Mc-Graw Hill, New Delhi. [3] Balmohan Vishnu Limaye, Functional analysis, New Age International, [4] Aldric Loughman Brown anda. Page, Elements of functional analysis, Van Nostrand-Reinhold, 1970 [5] J.B. Conway, A Course in Functional Analysis, 2nd Edition, Springer, Berlin, Contents 1. Normed space 1 2. Convergence in normed space 3 3. Schauder Basis 4 4. Finite dimensional spaces 5 5. Compactness and finite dimension 7 6. Riesz s Lemma 8 References 9