Exam 2 Study Guide 2.4 -‐ 3.1 2.4-1 Lemma + proof Let {x1, x2, ..., xn} 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, α1, ..., αn we have α1x1 +  + αnxn ≥ c(|α1| +  + |αn|). Proof: If αi = 0 for every i ∈ {1, 2, ..., n} then any c will suffice. Suppose ∃ αi ≠ 0. Then 0 < |α1| +  + |αn|. Let βi = αi/(|α1| +  + |αn|). Then Σ ni=1 |βi| = α1 αn = 1. + + α1 + + α n α1 + + α n Define F:n →  by F(β1, ..., βn) = β1x1 +  + βnxn. Let M = {β = (β1, ..., βn) ∈ n | Σ ni=1 |βi| = 1}. € Note t€ hat M ⊂ n, M is bounded, and M is closed, hence M is compact. Then, by the continuity of •, F(M) is compact, hence attains a minimum on M, say c. Since {x1, x2, ..., xn} is linearly independent and αi ≠ 0 for some i ⇒ βi ≠ 0 € for some i, we have that 0 < β1x1 +  + βnxn = F(β) ∀ β ∈ M. ⎛ ⎞ α1 αn ⎟ = c(|β1| +  + |βn|) = c ≤ So then c ⎜⎜ + + α1 + + α n ⎟⎠ ⎝ α1 + + α n α1 αn x1 + + x α1 + + α n α1 + + α n n β1x1 +  + βnxn = € Multiplying both sides by |α1| +  + |αn|, we have c(|α1| +  + |αn|) < α1x1 +  + αnxn. € proof 2.4-2 Theorem + Every finite dimensional subspace Y of a normed space X is complete. In particular, every finite dimensional normed space is complete. € Proof: Let {e1, ..., en} be a basis for Y. Let J = {1, 2, ..., n}. Choose c > 0 c(|α1| +  + |αn|) < α1e1 +  + αnen. Let (x(m)) ⊂ X such that (x(m)) is Cauchy. Let ε > 0. Let i ∈ J. Then ∃ N ∈  ∀ m, k ≥ N, x(m) – x(k)  < ε/c. So then c|αi(m) – αi(k)| ≤ c Σ ni=1 |(αi(m) – αi(k))| ≤  Σ ni=1 (αi(m) – αi(k))ei =  Σ ni=1 αi(m)ei – Σ ni=1 αi(k)ei = x(m) – x(k)  < c • ε/c = ε. Thus, ∀ i ∈ J, (αi(m)) is Cauchy, hence, αi(m) → αi. Let y = α1e1 +  € + αnen. Let A = max ei. 1≤i≤n Then for each i € ∈ J, ∃ Nα i ∈€  |αi(m) – αi| < ε/A. Let N ' = max{ Nα i , ..., Nα n }. So then x(m) – y =  Σ ni=1 αi(m)ei – Σ ni=1 αiei =  Σ ni=1 (αi(m) – αi)ei ≤ Σ€ni=1 |(αi(m) – αi)| ei ≤ A • Σ ni=1 |αi(m) – αi| < A • ε/A = ε. ∴ x(m) → y ∈ € Y, hence Y is complete. € € € € € € € Exam 2 Study Guide 2.4 -‐ 3.1 2.4-4 Definition A norm • on a vector space X is said to be equivalent to a norm •0 on X if there are positive numbers a and b such that for all x ∈ X we have ax0 ≤ x ≤ bx0. 2.4-5 Theorem + proof On a finite dimensional vector space X, any norm • is equivalent to any other norm •0. Proof: Let x = α1e1 +  + αnen ∈ X. Let a = max ei0. Let c > 0 c(|α1| +  + |αn|) < α1x1 +  + αnxn. 1≤i≤n Then x0 = α1e1 +  + αnen ≤ α1e1 +  + αnen = |α1|e1 +  +|αn|en € ≤ a•(|α1| +  + |αn|) ≤ a • (1/c)α1x1 +  + αnxn = (a/c)x. Similarly, for b = max ei and 1≤i≤n c* > 0 c*(|α1| +  + |αn|) < α1x1 +  + αnxn0, we have x ≤ (b/c*)x0. ∴ Together, € (c/a)x0 ≤ x ≤ (b/c*)x0. 2.5-1 Definition A metric space X is said to be compact if every sequence in X has a convergent subsequence. A subset M of X is said to be compact if M is compact considered as a subspace of X, that is, if every sequence in M has a convergent subsequence whose limit is an element of M. 2.5-2 Lemma A compact subset M of a metric space is closed and bounded. 2.5-3 Theorem In a finite dimensional normed space X, any subset M ⊂ X is compact ⇔ M is closed and bounded. Exam 2 Study Guide 2.4 -‐ 3.1 2.5-4 Lemma + proof (Riesz's Lemma) Let Y and Z be subspaces of a normed space X (of any dimension), and suppose that Y is closed and is 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, z – y ≥ θ for all y ∈ Y. Proof: Let θ ∈ (0, 1). Let v ∈ Z \Y. Then 0 < a = inf v − y as Y is closed. y∈Y By definition infemum, ∃ y0 ∈ Y a ≤ v – y0 < a/θ. 1 Let z = (v − y 0 ) . v − y0 € 1 (v − y 0 ) − y = v − y0 1 1 € v − y 0 ) − v − y 0 • y = v − y 0 + v − y 0 • y . ( v − y0 v − y0 1 θ − y 0 • y ∈ Y, then z – y ≥ • a > • a = θ. Since y 0 + v € v − y0 a € € 2.5-5 Theorem + proof € If a normed space X has the property that the €closed unit ball M = {x: x ≤ 1} € is compact, then X is finite dimensional. Then ∀ y ∈ Y, z – y = ( ( ) ) Proof: We will prove the contrapositive; if X has infinite dimension, then the closed unit ball is not compact. Assume X has infinite dimension. Let x1 ∈ X x1 = 1. Let Y = span{x1}. Then Y is a proper subspace of X of dimension 1. Let θ = 1/2. Then ∃ x2 ∈ X \ Y x2 = 1 and x1 – x2 ≥ 1/2. Let Y2 = span{x1, x2}. Then Y2 is a proper subspace of X of dimension 2. And ∃ x3 ∈ X \ Y2 x3 = 1 for which x3 – x1 ≥ 1/2 and x3 – x1 ≥ 1/2. Continuing by induction, we have (xn) ⊂ X where ∀ n, m ∈ , xn – xm ≥ 1/2. Thus, (xn) does not converge. So then the closed unit ball is not compact. 2.5-6 Theorem Let X and Y be metric spaces and T:X → Y a continuous mapping. Then the image of a compact subset M of X under T is compact. 2.5-7 Corollary A continuous mapping T of a compact subset M of a metrics space X into  assumes a maximum and a minimum at some points of M. Exam 2 Study Guide 2.4 -‐ 3.1 2.6-1 Definition A linear operator T is an operator such that (i) the domain D (T) of T is a vector space and the range R (T) lies in a vector space over the same field, (ii) for all x, y ∈ D (T) and scalars α, T(x + y) = Tx + Ty and T(αx) = αTx. 2.6-2 through 2.6-8 Examples 2.6-2 Identity Operator 2.6-3 Zero operator 2.6-4 Differentiation 2.6-5 Integration 2.6-6 Multiplication by t. Tx(t) = tx(t). 2.6-7 Elementary vector algebra (cross product and dot product) 2.6-8 Matrices 2.6-9 Theorem Let T be a linear operator. Then: (a) The range R (T) is a vector space. (b) If dim D (T) = n < +∞, then dim R (T) ≤ n. (c) The null space N (T) is a vector space. 2.6-10 Theorem Let X, Y be vector spaces, both real or both complex. Let T:D (T) → Y be a linear operator with domain D (T) and range R (T). Then: (a) The inverse T –1:R (T) → D (T) exists ⇔ Tx = 0 ⇒ x = 0. (b) If T –1 exists, it is a linear operator. (c) If dim D (T) = n < +∞ and T –1 exists, then dim R (T) = dim D (T). 2.6-11 Lemma Let T:X → Y and S: Y → Z be bijective linear operators, where X, Y, Z are vector spaces. Then the inverse (ST)–1:Z → X of the product (composite) ST exists, and (ST)–1 = T–1S–1. 2.7-1 Definition Let X and Y be normed space and T:D (T) → Y a linear operator, where D (T) ⊂ X. The operator T is said to be bounded if there is a real number c such that for all x ∈ D (T), Tx ≤ cx. Exam 2 Study Guide 2.4 -‐ 3.1 2.7-2 Lemma + proof Let T be a bounded linear operator. Then (a) An alternative formula for the norm of T is T = sup Tx. x∈D(T ) x =1 (b) The norm T = sup x∈D(T ) x ≠0 Tx satisfies (N1) to (N4). x Proof: ⎛ x ⎞ Tx (a) T = sup = sup T⎜ ⎟ = sup Ty . x x∈X x∈X y∈X ⎝ x ⎠ € x ≠0 x ≠0 y =1 (b) € € x =1 Hence, T = 0. The converse, 0 = 0, is evident. (N3): αT = sup αTx = |α| sup Tx = αT. x =1 € x =1 (N4): T1 + T2 = sup (T1 + T2)x = sup T1x + T2x € (N1): T ≥ 0 is evident. (N2): Assume T = sup Tx = 0, then Tx = 0 for every x ∈ X. x∈D(T ) € x∈D(T ) x∈D(T ) x∈D(T ) x =1 x∈D(T ) x =1 sup T1x +€ sup T2x = T1 + T2. € ≤ x∈D(T ) x∈D(T ) x =1 x =1 2.7-3 through 2.7-7 E€xamples € 2.7-3 Identity operator, IX is bounded and IX = 1. 2.7-4 Zero operator, 0X is bounded and 0X = 0 € € 2.7-5 Differentiation operator is not bounded. 2.7-6 Integral operator is bounded. 2.7-7 Matrix is bounded. 2.7-8 Theorem + proof Let X, Y be normed spaces. Let T:X → Y be linear. Suppose dim X = n < +∞. Then T is bounded. Proof: Let {e1, ..., en} be a basis for X. Let x = Σ ni=1 αiei ∈ X. Let c > 0 c(|α1| +  + |αn|) ≤ α1e1 +  + αnen = x. Then Tx =  Σ ni=1 αiTei ≤ Σ ni=1 |αi| Tei € ≤ max Tek• Σ ni=1 |αi| € ≤ max Tek•(1/c)x. k∈N € Thus, T is bounded. € € k∈N € Exam 2 Study Guide 2.4 -‐ 3.1 2.7-9 Theorem + proof Let T:D (T) → Y be a linear operator, where D (T) ⊂ X and X, Y are normed spaces. Then: (a) T is continuous ⇔ T is bounded. (b) If T is continuous at a single point, it is continuous. Proof: (b) Suppose T is continuous at x0 ∈ X. Let ε > 0. Then ∃ δ > 0 x – x0 < δ ⇒ Tx – Tx0 < ε. Consider x ∈ X x < δ. Let y = x + x0. Then y – x0 < δ, hence Tx = T(x + x0 – x0) = Ty – Tx0 ≤ ε. Thus, T is continuous at Θ. Let x1 ∈ X. ∀ x ∈ X x1 – x < δ, we have Tx1 – Tx = T(x1 – x) < ε. ∴ T is continuous at x1. (a) ⇐: Assume T is bounded, then Tx ≤ T x, ∀ x ∈ X. Let ε > 0. Let δ = ε/T. Then Tx < T δ = ε. Thus, T is continuous at Θ, hence, by (a), T is continuous everywhere. ⇒: Assume T is continuous. Let ε = 1. Then ∃ δ > 0 x < δ ⇒ Tx < 1. Let u ∈ X u = 1. Then δu < 1, hence T(δu) < 1, which implies that Tu = (1/δ)T(δu) = (1/δ)T(δu) < 1/δ. Thus, T = sup Tu < 1/δ. u =1 2.7-10 Corollary Let T be a bounded linear operator. Then: €x → x [where x , x ∈ D (T)] implies Tx → Tx. (a) n n n (b) The null space N (T) is closed. 2.8-1 Definition A linear functional f is a linear operator with domain in a vector space X and range in the scalar field K of X; thus, f:D (f ) → K, where K =  if X is real and K =  if X is complex. 2.8-2 Definition A bounded linear functional f is a bounded linear operator with range in the scalar field of the normed space X in which the domain D (f ) lies. Thus there exists a real number c such that for all x ∈ D (f ), |f (x)| ≤ cx. 2.8-4 through 2.8-7 Examples 2.8-4 Norm, •, is not a linear functional. 2.8-5 Dot product, f (x) = α • x = α1ξ1 +  + αnξn on n is a linear functional. b 2.8-6 Definite integral, f (x) = 2.8-7 Space C [a, b] . f1(x) = x(t0) on C [a, b], t0 ∈ [a, b] is a linear functional. € ∫ a x(t)dt on C [a, b] is a linear functional. Exam 2 Study Guide 2.4 -‐ 3.1 2.10-1 Theorem The vector space B(X, Y) of all bounded linear operators from a normed space X into a normed space Y is itself a normed space with norm defined by Tx T = sup . x∈D(T ) x x ≠0 2.10-2 Theorem + proof If Y is a Banach space, then B(X, Y) is a Banach space. € Proof: Let (Tn) be a Cauchy sequence in B(X, Y). Let ε > 0. And let x ∈ X. Then ∀ n, m ∈ , Tnx – Tmx ≤ Tn – Tm x as Tn, Tm ∈ B(X, Y). ε Since (Tn) is Cauchy, ∃ N ∈  n, m ≥ N ⇒ Tn – Tm < . x ε This gives us that Tnx – Tmx ≤ x = ε. Thus, (Tnx) is Cauchy in Y. x Since Y is complete, Tnx → y ∈ Y. This is true €for every x ∈ X. So then we can define T: X → Y by T(x) = lim Tnx. n→∞ To show T ∈ B(X, Y), w€e must show T is linear and bounded. To show T is linear, let α ∈ K, x, y ∈ X and note that T(αx + y) = lim Tn(αx + y) = lim (αTnx + Tny) = α lim Tnx + lim Tny = αTx + Ty. n→∞ n → ∞€ n→∞ n→∞ To show T is bounded, we note that by continuity of the norm we have Tnx → Tx ⇒ Tnx → Tx, ∀ x ∈ X. So then Tnx – Tx = Tnx – lim Tmx = lim Tnx – Tmx ≤ lim Tn – Tm x. € € € € m→∞ m→∞ m→∞ We can find N1 ∈  n, m ≥ N1 ⇒ Tn – Tm < 1, hence Tn – T is bounded. Since T = Tn – (Tn – T) and both Tn and Tn – T are bounded, then T is bounded as well, hence T ∈ B(X, Y). € € € Since an arbitrary Cauchy sequence (Tn) ⊂ B(X, Y) converges in B(X, Y), then B(X, Y) is a Banach space. 2.10-3 Definition Let X be a normed space. Then the set of all bounded linear functionals on X fx constitutes a normed space with norm defined by f  = sup which is x∈D(T ) x x ≠0 called the dual space of X and is denoted by X '. 2.10-4 Theorem € space The dual space X ' of a normed space X is a Banach (whether or not X is). Exam 2 Study Guide 2.4 -‐ 3.1 2.10-5 through 2.10-7 Examples 2.10-5 Space n. The dual space of n is n. The dual space of (n, •2) is (n, •2) The dual space of (n, •1) is (n, •∞) The dual space of (n, •∞) is (n, •1) The dual space of (n, •p) is (n, •q) 2.10-6 Space l1. The dual space of l1 is l∞. But, the dual space of l∞ is not l1, yet the dual space of c0 is l1. 2.10-7 Space lp. The dual space of lp is lq. 3.1-1 Definition An inner product space is a vector space X with an inner product defined on X. An inner product on X is a mapping of X × X into the scalar field K of X; that is, with every pair of vectors x and y there is a associated a scalar which is written 〈x, y〉 and is called the inner product of x and y, such that for all vectors x, y, z and scalar α we have (IP1) 〈x + y, z〉 = 〈x, z〉 + 〈y, z〉 (IP2) 〈αx, y〉 = α〈x, y〉 (IP3) 〈x, y〉 = 〈 y, x 〉 (IP4) 〈x, x〉 = 0 ⇔ x = 0. The inner product on X defines a norm on X given by x = x, x and a metric € on X given by d(x, y) = x – y. 3.1-2 Definition An element x of an inner product space X is said to be orthogonal to an € element y ∈ X if 〈x, y〉 = 0. 3.1-3 through 3.1-8 Examples 3.1-3 Euclidean space n. 〈x, y〉 = ξ1η1 +  + ξnηn, x = 〈x, x〉1/2, d(x, y) = x – y. 3.1-4 Unitary space n. 〈x, y〉 = ξ1 η 1 +  + ξn η n, x = 〈x, x〉1/2, d(x, y) = x – y. 3.1-5 Space L2[a, b]. b ( b 2 1/ 2 ) Over reals, 〈x, y〉 = ∫ a x(t)y(t)dt , x = ∫ a x(t) dt . € € 1/ 2 b b 2 Over complex, 〈x, y〉 = ∫ a x(t)y(t)dt , x = ∫ a x(t) dt . ( ) 3.1-6 Hilbert sequence space l2. € € 2 1/ 2 ∞ ⎛ ∞ ⎞ 〈x, y〉 = ∑ ξ j η j , x = ⎜∑ ξ j ⎟ ⎝ j =1 ⎠ j =1 € € p 3.1-7 Space l with p ≠ 2 is not an inner product space as it does not satisfy the parallelogram equality. € 3.1-8 Space C [a, b] € is not an inner product space as it does not satisfy the parallelogram equality.