204Lecture32009

# 204Lecture32009 - Economics 204 Lecture 3–Wednesday...

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Unformatted text preview: Economics 204 Lecture 3–Wednesday, July 29, 2009 Revised 7/29/09, Revisions Indicated by ** and Sticky Notes Section 2.1, Metric Spaces and Normed Spaces Generalization of distance notion in R n Definition 1 A metric space is a pair ( X, d ), where X is a set and d : X × X → R + , satisfying 1. ∀ x,y ∈ X d ( x, y ) ≥ , d ( x, y ) = 0 ⇔ x = y 2. ∀ x,y ∈ X d ( x, y ) = d ( y, x ) 3. ( triangle inequality ) ∀ x,y,z ∈ X d ( x, y ) + d ( y, z ) ≥ d ( x, z ) y % & x → z Definition 2 Let V be a vector space over R . A norm on V is a function k · k : V → R + satisfying 1. ∀ x ∈ V k x k ≥ 2. ∀ x ∈ V k x k = 0 ⇔ x = 0 3. ( triangle inequality ) ∀ x,y ∈ V k x + y k ≤ k x k + k y k x x % & y → x + y y & % x y 1 4. ∀ α ∈ R ,x ∈ V k αx k = | α |k x k A normed vector space is a vector space over R equipped with a norm. Theorem 3 Let ( V, k · k ) be a normed vector space. Let d : V × V ⇒ R + be defined by d ( v, w ) = k v − w k Then ( V, d ) is a metric space. Proof: We must verify that d satisfies all the properties of a metric. 1. d ( v, w ) = k v − w k ≥ d ( v, w ) = 0 ⇔ k v − w k = 0 ⇔ v − w = 0 ⇔ ( v + ( − w )) + w = w ⇔ v + (( − w ) + w ) = w ⇔ v + 0 = w ⇔ v = w 2. First, note that for any x ∈ V , 0 · x = (0 + 0) · x = 0 · x + 0 · x , so 0 · x = 0. Then 0 = 0 · x = (1 − 1) · x = 1 · x + ( − 1) · x = x + ( − 1) · x , so we have ( − 1) · x = ( − x ). d ( v, w ) = k v − w k = | − 1 |k v − w k = k ( − 1)( v + ( − w )) k = k ( − 1) v + ( − 1)( − w ) k 2 = k − v + w k = k w + ( − v ) k = k w − v k = d ( w, v ) 3. d ( u, w ) = k u − w k = k u + ( − v + v ) − w k = k u − v + v − w k ≤ k u − v k + k v − w k = d ( u, v ) + d ( v, w ) Examples of Normed Vector Spaces • E n : n-dimensional Euclidean space. V = R n , k x k 2 = | x | = v u u u t n X i =1 ( x i ) 2 • V = R n , k x k 1 = n X i =1 | x i | • V = R n , k x k ∞ = max {| x 1 | , . . . , | x n |} • C ([0 , 1]) , k f k ∞ = sup {| f ( t ) | : t ∈ [0 , 1] } • C ([0 , 1]) , k f k 2 = s Z 1 ( f ( t )) 2 dt • C ([0 , 1]) , k f k 1 = Z 1 | f ( t ) | dt 3 Theorem 4 (Cauchy-Schwarz Inequality) If v, w ∈ R n , then ⎛ ⎜ ⎝ n X i =1 v i w i ⎞ ⎟ ⎠ 2 ≤ ⎛ ⎜ ⎝ n X i =1 v 2 i ⎞ ⎟ ⎠ ⎛ ⎜...
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204Lecture32009 - Economics 204 Lecture 3–Wednesday...

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