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Unformatted text preview: ROYDEN, REAL ANALYSIS 3RD ED. CHAPTER 10 Problem 1020. Let P ( * ξ ) = lim ξ n . Take G = { A n } where A n * ξ = * η with η k = ξ n + k . Choose S = C and define F ( * η ) = lim η n for * η ∈ S . Hence F ( * ξ ) ≤ p ( * ξ ) = lim ξ n for any * ξ ∈ ‘ ∞ . Replace * ξ above by * ξ to get F ( * ξ ) ≥ lim ξ n . Problem 1021. Let X = { all bounded functions on R } and S = { all bounded mea surable functions on R } . Take G = { A t ; t ∈ R } where ( A t f )( x ) = f ( x + t ). Define F ( f ) = R f ( x ) dx for f ∈ S and P ( f ) = sup Z φ where the sup is taken over all φ ∈ S with 0 ≤ φ ≤  f  . Finally define μ ( A ) = F ( χ A ). Part(i) follows from the linearity of F , part(ii) from Proposition 10.5 and part(iv) by definition. How to prove part(ii)? Problem 1026. Use Proposition 10.13 to { T n } to get a constant M such that k T n x k ≤ M k x k for all n . Now let n → ∞ . Certainly T is linear. Problem 1028 a. If f n ∈ S converges to f in C [0 , 1], then k f n f k 2 ≤ k f n f k ∞ → and thus f ∈ S which is closed in L 2 [0 , 1]. b. Use Proposition 10.11. c. L x ( f ) = f ( x ) is a continuous linear function on S ⊆ C [0 , 1]. Use HahnBanach Theorem and then Riesz Representation Theorem. Note that  f ( y )  ≤ k f k ∞ ≤ M k f k 2 for f ∈ S . Problem 1030. ⇒ (i) Definion of a base. (ii) If x ∈ U and U is open, there exists a neighborhood W of x such that W ⊆ U . By translation invariance W = x + V for some open V which can be taken from the base B . (iii) θ + θ = θ . Now use continuity of the addition from X × X to X at ( θ,θ ). Typeset by A M ST E X 1 2 (iv) x ∈ αU iff α 1 x ∈ U . Use the continuity of scalar multiplication at (0 ,x ). (v) Follow the hint in part d . (vi) For any x 6 = θ , there exist U and V in B such that x + U ∩ V = φ . That implies x / ∈ ( U ) + V which equals U + V by (v). Since U + V is open, there exists W ∈ B such that W ⊆ U + V . Hence x / ∈ W and then x / ∈ ∩{ U ∈ B } . Conversely, there exists, for any x 6 = θ,W in B such that x / ∈ W . By (iii) and (v) x / ∈ V + V = ( V )+ V for some V ∈ B . Thus x + V ∩ V = φ , which implies x and θ can be seperated. By translation invariance, the same result holds for x 6 = y . Hence X is Hausdorff. ⇐ Assume O is open and x + y ∈ O . By (ii) and (iii) there is V ∈ B such that ( x + V ) + ( y + V ) = ( x + y ) + V + V ⊆ x + y + W ⊆ . This shows addition is continuous. For any V ∈ B ,W = (  α  + 1) 1 V ∈ B by (v)....
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This note was uploaded on 10/18/2010 for the course MATH 2411 taught by Professor Pf.choi during the Spring '10 term at 카이스트, 한국과학기술원.
 Spring '10
 Pf.Choi
 Numerical Analysis

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