Đang tải... (xem toàn văn)
Toán cao cấp Bách khoa TPHCM luyện thi Cao học
Functional Analysis Problems with Solutions ANH QUANG LE, Ph.D September 3, 2013 www.MATHVN.com - Anh Quang Le, Ph.D Contents Contents 1 Normed and Inner Product Spaces Banach Spaces 15 Hilbert Spaces 27 3.1 Hilbert spaces 27 3.2 Weak convergence 40 Linear Operators - Linear Functionals 45 4.1 Linear bounded operators 45 4.2 Linear Functionals 63 Fundamental Theorems 73 Linear Operators on Hilbert Spaces 87 www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D 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 Normed and Inner Product Spaces Problem Prove that any ball in a normed space X is convex Solution Let B(x0 ; r) be any ball of radius r > 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 Consider the linear space C[0, 1] equipped with the norm Z kf k1 = |f (x)|dx Prove that there is no inner product on C[0, 1] agreed with this norm www.MATHVN.com www.MATHVN.com - Anh Quang Le, Ph.D CHAPTER NORMED AND INNER PRODUCT SPACES Solution We show that the norm k.k1 does not satisfy the parallelogram law Let f (x) = and g(x) = 2x Then Z Z kf k1 = while Z kf − gk1 = 1.dx = 1, kgk1 = |2x|dx = 1, |1 − 2x|dx = , kf + gk1 = Thus, kf − gk21 + kf + gk21 = Z |1 + 2x|dx = 17 6= 2(kf k21 + kgk21 ) = 4 ¥ Problem 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) = − 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, and kf − gk = max | − + 2t| = t∈[0,1] t∈[0,1] Thus, kf − gk21 + kf + gk21 = 6= 2(kf k21 + kgk21 ) = ¥ Problem 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 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 = for any a ∈ [0, 1] Choose a = 1/2 then we get k 12 (x + y)k = 1, that is kx + yk = Since x and y belong to the unit sphere, we have kxk = kyk = Hence kx + yk = kxk + kyk Let us show that x, y are linearly independent Assume y = βx for some β ∈ C We have = kax + (1 − a)βxk = |a + (1 − a)β| For a = we get |β| = and for a = 1/2 we get |1 + β| = These imply that β = 1, and so x = y, which is a contradiction ¥ Problem 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 > 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 CHAPTER NORMED AND INNER PRODUCT SPACES That is mkxk2 ≤ kxk ≤ M kxk2 Hence, the two norms are equivalent ¥ Problem 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 > 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 > being arbitrary, it follows that kxk = 0, so x = 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 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 (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) → as n, m → ∞ kx(n) − x(m) k = |ak − ak |2 k=1 Hence, for every k = 1, , d, (n) (m) |ak − ak | → 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 = Then, for any x ∈ X, Pd k=1 k k ¯ k ek , we obtain fk (x) = α ¯ k , and x = k=1 α 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 = k=1 Hence, αk = 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 CHAPTER NORMED AND INNER PRODUCT SPACES Problem ( 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 ) (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 hx, yi = (kx + yk2 − kx − yk2 ) (∗) We use the symmetry of the inner product and compute the right hand side of (): Ô 1Ê (kx + yk2 − kx − yk2 ) = hx + y, x + yi − hx − y, x − yi 4 Ô 1Ê = hx, yi + hy, xi = hx, yi For the complex field case, we again expand the right hand side, using the relation we just established: i 1h 2 2 (kx + yk − kx − yk ) − i(kx + iyk kx iyk ) Ô iÊ Ô 1Ê = hx, yi + hy, xi − hx, iyi + hiy, xi 2 1 i2 i2 = hx, yi + hy, xi − hx, yi + hy, xi 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 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 and we have hx, yi + hx, zi = + = − = − = − = ¢ 1¡ kx + yk2 + kx − yk2 − kx − yk2 − kx − zk2 ¢ i¡ kx + iyk2 + kx + izk2 − kx − iyk2 − kx − izk2 4à ° ° °2 ° ° °2 ° °! ° ° y + z °2 ° ° ° y − z °2 y + z y + z ° ° ° °x + ° +° ° ° ° ° ° − °x − ° − ° ° ° ° ð °2 ° ° ° °2 ° °! ° ° y + z °2 ° ° ° y − z °2 i ° y + z y + z °x + i ° + °i ° ° ° ° ° ° ° − °x − i ° − °i ° ° ° ð °2 ° ° ° °2 ! ° ° ° ° ° y + z °2 ° y − z °2 ° ° y + z y + z °x + ° +° ° −° ° − °x − ° ° ° ° ° ° ° ° ° ð °2 ° ° ° °2 ! ° ° ° ° ° y + z °2 ° y − z °2 ° i ° y + z y + z °x + i ° + °i ° − °i ° − °x − i ° ° ° ° ° ° ° ° ° ¢ 1¡ kx + y + zk2 + kxk2 − kx − (y + z)k2 − kxk2 ¢ i¡ kx + i(y + z)k2 + kxk2 − kx − i(y + z)k2 − kxk2 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 = ihx, yi hx, iyi = Combining these results we have hx, αyi = αhx, yi for α ∈ Q + iQ www.MATHVN.com ... well-defined Indeed, if [f ] = [g], then (f − g)(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)|... 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. .. k.k1 does not satisfy the parallelogram law Let f (x) = and g(x) = 2x Then Z Z kf k1 = while Z kf − gk1 = 1.dx = 1, kgk1 = |2x|dx = 1, |1 − 2x|dx = , kf + gk1 = Thus, kf − gk21 + kf + gk21 = Z |1