Tài liệu Functional problems anhle-full-www.mathvn.com

Functional Analysis Problems with Solutions ANH QUANG LE, Ph.D. September 14, 2013 www.MATHVN.com - Anh Quang Le, Ph.D Contents Contents 1 1 Normed and Inner Product Spaces 3 2 Banach Spaces 15 3 Hilbert Spaces 27 3.1 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 3.2 Weak convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 4 Linear Operators - Linear Functionals 45 4.1 Linear bounded operators . . . . . . . . . . . . . . . . . . . . . . . . 45 4.2 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 5 Fundamental Theorems 73 6 Linear Operators on Hilbert Spaces 87 7 Compact Operators 99 8 Bounded Operators on Banach Spaces and Their Spectra 113 9 Compact Operators and Their Spectra 131 10 Bounded Self Adjoint Operators and Their Spectra 143 1 www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 2 CONTENTS Notations: • B(X, Y ): the space of all bounded (continuous) linear operators from X to Y . • Image (T ) ≡ Ran(T ): the image of a mapping T : X → Y . w • xn − → x: xn converges weakly to x. • X ∗ : the space of all bounded (continuous) linear functionals on X. • F or K: the scalar field, which is R or C. • Re, Im: the real and imaginary parts of a complex number. www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D Chapter 1 Normed and Inner Product Spaces Problem 1. Prove that any ball in a normed space X is convex. Solution. Let B(x0 ; r) be any ball of radius r > 0 centered at x0 ∈ X, and x, y ∈ B(x0 ; r). Then kx − x0 k < r and ky − x0 k < r. For every a ∈ [0, 1] we have kax + (1 − a)y − x0 k = k(x − x0 )a + (1 − a)(y − x0 )k ≤ akx − x0 k + (1 − a)ky − x0 k < ar + (1 − a)r = r. So ax + (1 − a)y ∈ B(x0 ; r). ¥. Problem 2. Consider the linear space C[0, 1] equipped with the norm Z 1 kf k1 = |f (x)|dx. 0 Prove that there is no inner product on C[0, 1] agreed with this norm. 3 www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 4 CHAPTER 1. NORMED AND INNER PRODUCT SPACES Solution. We show that the norm k.k1 does not satisfy the parallelogram law. Let f (x) = 1 and g(x) = 2x. Then Z Z 1 kf k1 = 0 while Z 1 kf − gk1 = 0 1 1.dx = 1, kgk1 = |2x|dx = 1, 0 1 |1 − 2x|dx = , kf + gk1 = 2 Thus, kf − gk21 + kf + gk21 = Z 1 |1 + 2x|dx = 2. 0 17 6= 2(kf k21 + kgk21 ) = 4. 4 ¥ Problem 3. Consider the linear space C[0, 1] equipped with the norm kf k = max |f (t)|. t∈[0,1] Prove that there is no inner product on C[0, 1] agreed with this norm. Solution. We show that the parallelogram law with respect to the given norm does not hold for two elements in C[0, 1]. Let f (t) = t, g(t) = 1 − t, t ∈ [0, 1]. Then f, g ∈ C[0, 1] and kf k = max t = 1, kgk = max (1 − t) = 1, t∈[0,1] t∈[0,1] and kf + gk = max 1 = 1, and kf − gk = max | − 1 + 2t| = 1. t∈[0,1] t∈[0,1] Thus, kf − gk21 + kf + gk21 = 2 6= 2(kf k21 + kgk21 ) = 4. ¥ Problem 4. Prove that: If the unit sphere of a normed space X contains a line segment [x, y] where x, y ∈ X and x 6= y , then x and y are linearly independent and kx + yk = kxk + kyk . www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 5 Solution. Suppose that the unit sphere contains a line segment [x, y] where x, y ∈ X and x 6= y. Then kax + (1 − a)yk = 1 for any a ∈ [0, 1]. Choose a = 1/2 then we get k 12 (x + y)k = 1, that is kx + yk = 2. Since x and y belong to the unit sphere, we have kxk = kyk = 1. Hence kx + yk = kxk + kyk. Let us show that x, y are linearly independent. Assume y = βx for some β ∈ C. We have 1 = kax + (1 − a)βxk = |a + (1 − a)β|. For a = 0 we get |β| = 1 and for a = 1/2 we get |1 + β| = 2. These imply that β = 1, and so x = y, which is a contradiction. ¥ Problem 5. Prove that two any norms in a finite dimensional space X are equivalent. Solution. Since equivalence of norms is an equivalence relation, it suffices to show that an arbitrary norm k.k on X is equivalent to the Euclidian norm P k.k2 . Let {e1 , ..., en } be a basis for X. Every x ∈ X can be written uniquely as x = nk=1 ck ek . Therefore, !1/2 !1/2 à n à n n X X X kek k2 ≤ Akxk2 , |ck |2 |ek |kek k ≤ kxk ≤ k=1 k=1 k=1 P 1/2 where A = ( nk=1 |ek |2 ) is a non-zero constant. This shows that the map x 7→ kxk is continuous w.r.t. the Euclidian norm. Now consider S = {x : kxk2 = 1}. This is just the unit sphere in (X, k.k2 ), which is compact. The map S → R defined by x 7→ kxk is continuous, so it attains a minimum m and a maximum M on S. Note that m > 0 because S 6= ∅. Thus, for all x ∈ S, we have m ≤ kxk ≤ M. Now, for x ∈ X, x 6= 0, x kxk2 ∈ S, so m≤ kxk ≤ M. kxk2 www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 6 CHAPTER 1. NORMED AND INNER PRODUCT SPACES That is mkxk2 ≤ kxk ≤ M kxk2 . Hence, the two norms are equivalent. ¥ Problem 6. Let X be a normed space. (a) Find all subspaces of X which are contained in some ball B(a; r) of X. (b) Find all subspaces of X which contain some ball B(x0 ; ρ) of X. Solution. (a) Let Y be a subspace of X which is contained in some ball B(a; r) of X. Note first that the ball B(a; r) must contain the vector zero of X (and so of Y ); otherwise, the question is impossible. For any number A > 0 and any x ∈ Y , we have Ax ∈ Y since Y is a linear space. By hypothesis Y ⊂ B(a; r), so we have Ax ∈ B(a; r). This implies that kAxk < r + kak. Finally kxk < r + kak . A A > 0 being arbitrary, it follows that kxk = 0, so x = 0. Thus, there is only one subspace of X, namely, Y = {0}, which is contained in some ball B(a; r) of X. (b) Let Z be a subspace of X which contain some ball B(x0 ; ρ) of X. Take any x ∈ B(0; ρ). Then x + x0 ∈ B(x0 ; ρ) and so x + x0 ∈ Z since Z ⊃ B(x0 ; ρ). Now, since x0 ∈ Z, x + x0 ∈ Z and Z is a linear space, we must have x ∈ Z. Hence B(0; ρ) ⊂ Z. ρx Now for any nonzero x ∈ X, we have 2kxk ∈ B(0; ρ) ⊂ Z. Hence x ∈ Z. We can conclude that Z = X. In other words, the only subspace of X which contains some ball B(x0 ; ρ) of X is X itself. ¥ Problem 7. Prove that any finite dimensional normed space : (a) is complete (a Banach space), (b) is reflexive. Solution. Let X be a finite dimensional normed space. Suppose dim X = d. www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 7 (a) By Problem 5, it suffices to consider the Euclidian norm in X. Let {e1 , ..., ed } be a basis for X. For x ∈ X there exist numbers c1 , ..., cd such that à d !1/2 d X X x= ck ek and kxk = |ck |2 . k=1 k=1 P (n) Let (x(n) ) be a Cauchy sequence in X. If for each n, x(n) = dk=1 ak ek then !1/2 à d X (n) (m) → 0 as n, m → ∞. kx(n) − x(m) k = |ak − ak |2 k=1 Hence, for every k = 1, ..., d, (n) (m) |ak − ak | → 0 as n, m → ∞. (n) Therefore, each sequence of numbers (ak ) is a Cauchy sequence, so (n) (0) ak → ak Let a = Pd k=1 as n → ∞ for every k = 1, 2, ..., d. (0) ak ek then x(n) → a ∈ X. (b) Let f ∈ X ] where X ] is the space of all linear functionals on X. We have ! à d d d X X X ck α k , ck f (ek ) = ck ek = f (x) = f k=1 k=1 k=1 where αk = f (ek ). Let us define fk ∈ X ] by the relation fk (x) = ck , k = 1, ..., d. For any x ∈ X and f ∈ X ] , we get f (x) = d X fk (x)αk , i.e., f = Hence, X ] ≤ d. Pdim d Let α f = 0. Then, for any x ∈ X, Pd k=1 k k x = k=1 ᾱk ek , we obtain fk (x) = ᾱk , and k=1 αk fk (x) = αk fk . k=1 k=1 d X d X d X Pd k=1 αk fk (x) = 0, and by taking |αk |2 = 0. k=1 Hence, αk = 0 for all k = 1, ..., d and thus, dim X ] = d. For the space X ∗ we have X ∗ ⊂ X ] , so dim X ∗ = n ≤ d and dim(X ∗ )] = n. From the relation X ⊂ (X ∗ )∗ ⊂ (X ∗ )] we conclude that d ≤ n. Thus, n = d, and so X = (X ∗ )∗ . ¥ www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 8 CHAPTER 1. NORMED AND INNER PRODUCT SPACES Problem 8. ( Reed-Simon II.4) (a) Prove that the inner product in a normed space X can be recovered from the polarization identity: i 1h hx, yi = (kx + yk2 − kx − yk2 ) − i(kx + iyk2 − kx − iyk2 ) . 4 (b) Prove that a normed space is an inner product space if and only if the norm satisfies the parallelogram law: kx + yk2 + kx − yk2 = 2(kxk2 + kyk2 ). Solution. (a) For the real field case, the polarization identity is 1 hx, yi = (kx + yk2 − kx − yk2 ). (∗) 4 We use the symmetry of the inner product and compute the right hand side of (∗): ¤ 1 1£ (kx + yk2 − kx − yk2 ) = hx + y, x + yi − hx − y, x − yi 4 4 ¤ 1£ = hx, yi + hy, xi 2 = hx, yi. For the complex field case, we again expand the right hand side, using the relation we just established: i 1h 2 2 2 2 (kx + yk − kx − yk ) − i(kx + iyk − kx − iyk ) 4 ¤ i£ ¤ 1£ = hx, yi + hy, xi − hx, iyi + hiy, xi 2 2 1 1 i2 i2 = hx, yi + hy, xi − hx, yi + hy, xi 2 2 2 2 = hx, yi. (b) If the norm comes from an inner product, then we have kx + yk2 + kx − yk2 = hx + y, x + yi + hx − y, x − yi = 2hx, xi + 2hy, yi + hx, yi + hy, xi − hx, yi − hy, xi = 2(kxk2 + kyk2 ). www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 9 Now suppose that the norm satisfies the parallelogram law. Assume the field is C, and define the inner product via the polarization identity from part (a). If x, y.z ∈ X, we write x+y =x+ y+z y−z y+z y−z + , x+z =x+ − , 2 2 2 2 and we have hx, yi + hx, zi = + = − = − = − = ¢ 1¡ kx + yk2 + kx − yk2 − kx − yk2 − kx − zk2 4 ¢ i¡ kx + iyk2 + kx + izk2 − kx − iyk2 − kx − izk2 4à ° ° °2 ° ° °2 ° °! ° ° y + z °2 ° ° ° y − z °2 y + z y + z 1 ° ° ° °x + ° +° ° ° ° ° 2 ° − °x − 2 ° − ° 2 ° 2 ° 2 ° ð °2 ° ° ° °2 ° °! ° ° y + z °2 ° ° ° y − z °2 i ° y + z y + z °x + i ° + °i ° ° ° ° ° ° 2 ° − °x − i 2 ° − °i 2 ° 2 ° 2 ° ð °2 ° ° ° °2 ! ° ° ° ° ° y + z °2 ° y − z °2 ° 1 ° y + z y + z °x + ° +° ° −° ° − °x − ° ° 2 ° ° 2 ° ° 2 ° 2 ° 2 ° ð °2 ° ° ° °2 ! ° ° ° ° ° y + z °2 ° y − z °2 ° i ° y + z y + z °x + i ° + °i ° − °i ° − °x − i ° ° 2 ° ° 2 ° ° 2 ° 2 ° 2 ° ¢ 1¡ kx + y + zk2 + kxk2 − kx − (y + z)k2 − kxk2 4 ¢ i¡ kx + i(y + z)k2 + kxk2 − kx − i(y + z)k2 − kxk2 4 hx, y + zi. This holds for all x, y, z ∈ X, so, in particular, hx, nyi = nhx, yi for n ∈ N. And it also satisfies hx, ryi = rhx, yi for r ∈ Q. Moreover, again by the polarization identity, we have ¢ i¡ ¢ 1¡ kx + iyk2 − kx − iyk2 − kx − yk2 − kx + yk2 4 4 = ihx, yi. hx, iyi = Combining these results we have hx, αyi = αhx, yi for α ∈ Q + iQ. www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 10 CHAPTER 1. NORMED AND INNER PRODUCT SPACES Now, if α ∈ C, by the density of Q + iQ in C, there exists a sequence (αn ) in Q + iQ converging to α. It follows that hx, αyi = αhx, yi for α ∈ C. Thus the h., .i is linear. Since ki(x − iy)k = kx − iyk, we have hy, xi = hx, yi, and ¢ 1 i¡ hx, xi = (k2xk2 ) − |1 + i|kxk2 − |1 − i|2 kxk2 = kxk2 . 4 4 So this shows that the norm is induced by h., .i and that it is also positive definite, and thus it is an inner product. ¥ Problem 9. (Least square approximation. Reed-Simon II.5) Let X be an inner product space and let {x1 , ..., xN } be an orthonormal set. Prove that ° ° N ° ° X ° ° cn xn ° °x − ° ° n=1 is minimized by choosing cn = hxn , xi. Solution. For every x ∈ X, we write x= N X hxn , xixn + z, for some z ∈ X. n=1 We observe that for all n = 1, ..., N , hxn , zi = hxn , xi − N X hxn , xihxn , xk i k=1 = hxn , xi − hxn , xi = 0. Therefore z⊥xn . Then due to (∗) we can write x− N X n=1 c n xn = N X ¡ |n=1 ¢ hxn , xi − cn xn +z. {z zN www.MATHVN.com } (∗) www.MATHVN.com - Anh Quang Le, Ph.D 11 Since z⊥zN , we have ° °2 N ° ° X ° ° x − c x = kzN k2 + kzk2 ° n n° ° ° n=1 N X ¯ ¯ ¯hxn , xi − cn ¯2 + kzk2 , = n=1 which attains its minimum if cn = hxn , xi for all n = 1, ..., N. ¥ Review: Quotient normed space. • Let X be a vector space, and let M be a subspace of X. We define an equivalence relation on X by x ∼ y if and only if x − y ∈ M. For x ∈ X, let [x] = x + M denote the equivalence class of x and X/M the set of all equivalence classes. On X/M we define operations: [x] + [y] = [x + y] α[x] = [αx], α ∈ C. Then X/M is a vector space. If the subspace M is closed, then we can define a norm on X/M by k[x]k = inf kyk = inf kx + mk = inf kx − zk = d(x, M ). y∈[x] m∈M m∈M What a ball in X/M looks like? B([x0 ]; r) := {[x] : k[x] − [x0 ]k < r} = {x + M : kx − x0 + M k < r}. • Suppose that M is closed in X. The canonical map (the natural projection) is defined by π : X → X/M, π(x) = [x] = x + M. It can be shown that kπ(x)k ≤ kxk, ∀x ∈ X, so π is continuous. Problem 10. Let X be a normed space and M a closed subspace of X. Let π : X → X/M be the canonical map. Show that the topology induced by the standard norm on X/M is the usual quotient topology, i.e. that O ⊂ X/M is open in X/M if and only if π −1 (O) is open in X. www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 12 CHAPTER 1. NORMED AND INNER PRODUCT SPACES Solution. • If O is open in X/M , then π −1 (O) is open in X since π is continuous. • Now suppose that O ⊂ X/M and that π −1 (O) is open in X. We show that O is open in X/M . Consider an open ball B(0; r), r > 0 in X. Let x ∈ B(0; r). Then kxk < r, and so k[x]k ≤ kxk < r. On the other hand, if k[x]k < r, then there is an y ∈ M such that kx + yk < r. Hence x + y ∈ B(0; r), and so ¡ ¢ [x] = π(x + y) ∈ π B(0; r) . If [x0 ] ∈ O, then x0 ∈ π −1 (O). Since π −1 (O) is open in X, there is an r > 0 such that B(x0 ; r) ⊂ π −1 (O). This implies that ¡ ¢ ¡ ¢ O = ππ −1 (O) ⊃ π B(x0 ; r) = π x0 + B(0; r) = {x + M : kx − x0 + M k < r}. The last set is the open ball of radius r > 0 centered at [x0 ] ∈ O. Thus O is open in X/M. ¥ Problem 11. Let X = C[0, 1], M = {f ∈ C[0, 1] : f (0) = 0}. Show that X/M = C. Solution. Given [f ] ∈ X/M , let ϕ([f ]) = f (0). Then the map ϕ : X/M → C is well-defined. Indeed, if [f ] = [g], then (f − g)(0) = 0 so f (0) = g(0). It is clearly linear. If f ∈ X, then g = f − f (0) ∈ M , and so f − g = f (0) is constant, which tells us that k[f ]k = |f (0)| = |ϕ([f ])|, so ϕ is an isometry (and thus injective and continuous). Finally, constants are in X = C[0, 1], so ϕ is surjective and thus an isometric isomorphism. ¥ Problem 12. P If 0 < p < 1 then `p is a vector space but kxkp = ( n |xn |p )1/p is not a norm for `p . www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 13 Solution. Recall that if x = (x1 , x2 , ...), y = (y1 , y2 , ...) ∈ `p and α ∈ C then x + y = (x1 + y1 , x2 + y2 , ...) and αx = (αx1 , αx2 , ...). It is clear that αx ∈ `p . We show that x + y ∈ `p . For t ≥ 0 it not hard to see that (1 + t)p ≤ 1 + tp , 0 < p < 1. This implies that (a + b)p ≤ ap + bp , 0 < p < 1 and a, b ≥ 0. Therefore, kx + ykpp ≤ kxkpp + kykpp . Since both kxkpp and kykpp are bounded, kx + ykpp is bounded. Hence x + y ∈ `p . To show k.kp is not a norm for `p , let us take an example: If x = (1, 0, ...) and y = (0, 1, 0, ...) then kxkp = kykp = 1 but kx + ykp = 21/p > 2 since 1/p > 1. Therefore, kx + ykp > kxkp + kykp . ¥ Problem 13. Suppose that X is a linear space with inner product h., .i. If xn → x and yn → y as n → ∞, prove that hxn , yn i → hx, yi as n → ∞. Solution. Using the Cauchy-Schwarz and triangle inequalities, we have |hxn , yn i − hx, yi| ≤ ≤ ≤ ≤ |hxn − x, yn i| + |hx, yn − yi| kxn − xkkyn k + kxkkyn − yk kxn − xk(kyn − yk + kyk) + kxkkyn − yk kxn − xkkyn − yk + kxn − xkkyk + kxkkyn − yk. Since kxn − xk → 0 and kyn − yk → 0 as n → ∞, we see that hxn , yn i → hx, yi as n → ∞. www.MATHVN.com ¥ www.MATHVN.com - Anh Quang Le, Ph.D 14 CHAPTER 1. NORMED AND INNER PRODUCT SPACES Problem 14. Prove that if M is a closed subspace and N is a finite dimensional subspace of a normed space X, then M + N := {m + n : m ∈ M, n ∈ N } is closed. Solution. Assume dim N = 1, say N = Span{x}. The case x ∈ M is trivial. Suppose x ∈ / M. Consider the sequence zk := αk x + mk , where mk ∈ M, αk ∈ C, and suppose zk ∈ M + N → y as k → ∞. We want to show y ∈ M + N . The sequence (αk ) is bounded; otherwise, there exists a subsequence (αk0 ) such that 0 < |αk0 | → ∞ as k 0 → ∞. Then zk 0 mk 0 and → 0 as k 0 → ∞, αk0 αk0 so x must be 0, which is in M . This is a contradiction. Consequently, (αk ) is bounded and therefore it has a subsequence (αk0 ) which is converging to some α ∈ C. Thus mk0 = zk0 − αk0 x → y − αx as k 0 → ∞. Hence, y − αx is in M since M is closed. Thus y ∈ M + N . The solution now follows by induction. ¥ www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D Chapter 2 Banach Spaces Problem 15. Let if the series P∞X be a normed space. Prove that X is a Banach space if andPonly ∞ n=1 an converges, where (an ) is any sequence in X satisfying n=1 kan k < ∞. Solution. P∞ Suppose that Pn X is complete. Let (an ) be a sequence in X such that n=1 kan k < ∞. Let Sn = i=1 ai be the partial sum. Then for m > n, ° m ° m °X ° X ° ° kSm − Sn k = ° ai ° ≤ kai k. ° ° i=n+1 i=n+1 P Pm By hypothesis, the series ∞ n=1 kan k converges, so i=n+1 kai k → 0 as n → ∞. Therefore, (Sn ) is P a Cauchy sequence in the Banach space X. Thus, (Sn ) converges, that is, the series ∞ n=1 P an converges. P∞ Conversely, suppose ∞ n=1 an converges in X whenever n=1 kan k < ∞. We show that X is complete. Let (yn ) be a Cauchy sequence in X. Then 1 whenever m > n1 , 2 1 − ym k < 2 whenever m > n2 > n1 . 2 ∃n1 ∈ N : kyn1 − ym k < ∃n2 ∈ N : kyn2 Continuing in this way, we see that there is a sequence (nk ) strictly increasing such that 1 kynk − ym k < k whenever m > nk . 2 15 www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 16 CHAPTER 2. BANACH SPACES In particular, we have kynk+1 − ynk k < 1 for all k ∈ N. 2k Set xk = ynk+1 − ynk . Then n X kxk k = k=1 n X kynk+1 − ynk k < k=1 n X 1 . k 2 k=1 P∞ It k=1 kxk k < ∞. By hypothesis, there is an x ∈ X such that Pmfollows that k=1 xk → x as m → ∞. But we have m X k=1 xk = m X (ynk+1 − ynk ) k=1 = ynm+1 − yn1 . Hence ynm → x + yn1 in X as m → ∞. Thus, the sequence (yn ) has a convergent subsequence and so must itself converges. ¥ Problem 16. Let X be a Banach space. Prove that the closed unit ball B(0; 1) ⊂ X is compact if and only if X is finite dimensional. Solution. • Suppose dim X = n. Then X is isomorphic to Rn (with the standard topology). The result then follows from the Heine-Borel theorem. • Suppose that X is not finite dimensional. We want to show that B(0; 1) is not compact. To do this, we construct a sequence in B(0; 1) which have no convergent subsequence. We will use the following fact usually known as Riesz’s Lemma: (See the proof below) Let M be a closed subspace of a Banach space X. Given any r ∈ (0, 1), there exists an x ∈ X such that kxk = 1 and d(x, M ) ≥ r. Pick x1 ∈ X such that kx1 k = 1. Let S1 := Span {x1 }. Then S1 is closed. According to Riesz’s Lemma, there exists x2 ∈ X such that 1 kx2 k = 1 and d(x2 , S1 ) ≥ . 2 www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 17 Now consider the subspace S2 generated by {x1 , x2 }. Since X is infinite dimensional, S2 is a proper closed subspace of X, and we can apply the Riesz’s Lemma to find an x3 ∈ X such that 1 kx3 k = 1 and d(x3 , S2 ) ≥ . 2 If we continue to proceed this way, we will have a sequence (xn ) and a sequence of closed subspaces (Sn ) such that for all n ∈ N 1 kxn k = 1 and d(xn+1 , Sn ) ≥ . 2 It is clear that the sequence (xn ) is in B(0; 1), and for m > n we have 1 kxn − xm k ≥ d(xm , Sn ) ≥ . 2 Therefore, no subsequence of (xn ) can form a Cauchy sequence. Thus, B(0; 1) is not compact. ¥ Proof of Riesz’s Lemma: Take x1 ∈ / M . Put d = d(x1 , M ) = inf m∈M km − x1 k. Then d > 0 since M is closed. For any ε > 0, by definition of the infimum, there exists m1 ∈ M such that 0 < km1 − x1 k < d + ε. Set x = x1 −m1 kx1 −m1 k . Then kxk = 1 and kx − mk = ° ° 1 °x1 − (m1 + kx1 − m1 km)° {z } | kx1 − m1 k ∈M This implies that d(x, M ) = inf kx − mk = m∈M By choosing ε > 0 small, d d+ε inf m∈M kx1 − mk d ≥ . kx1 − m1 k d+ε can be arbitrary close to 1. ¥ Problem 17. Let X be a Banach space and M a closed subspace of X. Prove that the quotient space X/M is also a Banach space under the quotient norm. Solution. We use criterion established above (in problem 15). Suppose that ([xn ]) is any www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 18 CHAPTER 2. BANACH SPACES sequence in X/M such that P∞ n=1 k[xn ]k < ∞. We show that ∃[x] ∈ X/M : k X [xn ] → [x] as k → ∞. n=1 For each n, k[xn ]k = inf z∈M kxn + zk, and therefore there is zn ∈ M such that kxn + zn k ≤ k[xn ]k + 1 2n by definition of the infimum. Hence ∞ X kxn + zn k ≤ n=1 ∞ X n=1 k[xn ]k + 1 < ∞. 2n But (xn + zn ) is a sequence in the Banach space X, and so ∞ X (xn + zn ) = x for some x ∈ X. n=1 Then we have ° k ° ° k ° °X ° °X ° ° ° ° ° [x ] − [x] = [x − x] ° ° ° ° n n ° ° ° ° n=1 n=1 ° ° k °X ° ° ° = inf ° (xn − x + z)° z∈E ° ° n=1 ° ° k °X ¢° ° ¡ ° ≤ ° (xn − x) + zn ° ° n=1 ° ° ° k °X ° ° ° = ° (xn + zn ) − x° → 0 as k → ∞. ° n=1 ° Hence Pk n=1 [xn ] → [x] as k → ∞. ¥ www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 19 SPACE `p (Only properties concerning to norms and completeness will be considered. Other properties such as duality will be discussed later.) Problem 18. Show that `p , 1 ≤ p < ∞ equipped with the norm k.kp is a Banach space. Solution. (i) (i) Let x(i) = (x1 , ..., xk , ...) for i = 1, 2, ... be a Cauchy sequence in `p . Then kx(i) − x(j) kp → 0 as i, j → ∞. (i) (j) Since kx(i) − x(j) kp ≥ |xk − xk | for every k, it follows that (i) (j) |xk − xk | → 0 for every k as i, j → ∞. ¡ (i) ¢ This tells us that the sequence xk is a Cauchy sequence in F, which is complete, ¡ (i) ¢ so that xk converges to xk ∈ F as i → ∞ for each k. Set x = (x1 , ..., xk , ...). We will show that (∗) kx(i) − xkp → 0 as i, j → ∞ and x ∈ `p . Given ε > 0, for any M ∈ N, there exists N ∈ N such that ! p1 à ∞ ! p1 ÃM X (i) X (j) (i) (j) |xk − xk |p |xk − xk |p < ε if i, j > N. ≤ k=1 k=1 Letting j → ∞, for i > N we get ÃM X (∗∗) ! p1 (i) |xk − xk |p < ε. k=1 By Minkowski’s inequality, ÃM ÃM ! p1 à M ! p1 ! p1 X X (N ) X (N ) ≤ + |xk |p |xk − xk |p |xk |p k=1 à ≤ k=1 M X (N ) |xk − x k |p k=1 < ε+ k=1 ! p1 Ã∞ X + k=1 ! p1 (N ) |xk |p Ã∞ X . k=1 www.MATHVN.com ! p1 (N ) |xk |p