Midterm Exam Solutions February 27, 2009
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1 Midterm Exam Solutions February 27, 2009 (24. Deine each o the ollowing statements. You may assume that everyone knows the deinition o a topological space and a linear space. (4 a. X is a compact topological space. Solution. Every open cover o X has a inite subcover. (4 b. X is a locally compact topological space. Solution. Every point in X has a compact neighborhood. (4 c. is a norm on a real linear space X. Solution. : X [ 0. satisies. x + y x + y, or all x, y X, 2. λx = λ x, or all x X, λ, 3. x = 0 x= 0. (4 d. C( X, F is an equicontinuous amily, where X is a topological space. Solution. For every x X and ε > 0, there exists a neighborhood U o x such that ( x ( y < ε or all y U and F. (4 e. X is the dual space o the complex Banach space X. Solution. X is the space o all bounded linear unctionals on X. (4. X is a meager topological space. Solution. X is the countable union o nowhere dense sets. Page o 5
2 (20 2. (0 a. Deine the product topology and state Tychono s Theorem. X α α is a amily o topological spaces, the product topology on X = X α is α A the weak topology generated by the projection maps π : X X. Solution. I { } A α α Alternatively, the product topology is the weakest topology such that the projection maps are continuous. Alternatively, the product topology is the weakest topology generated by the sets π, where α A and U is open in X α. α ( U Tychono s Theorem: I { X α } α A X = is compact in the product topology. X α α A is any amily o compact topological spaces, then (0 b. State the Baire Category Theorem, and give an example o a residual set which is not open. Solution. The Baire Category Theorem. Let X be a complete metric space. U is a sequence o open dense subsets o X, then = U n is dense in X. a. I { n} n b. X is not a countable union o nowhere dense sets. n= Example. Let X = with the usual topology, and let E = \. Then is meager, so E c is residual. E is not open, since E is dense. Page 2 o 5
3 (6 3. (8 a. State the Open Mapping Theorem or Banach spaces. Solution. I X and Y are Banach spaces and i T is a continuous linear surjection rom X to Y, then T is an open mapping. (8 b. Let X and Y be Banach spaces, and assume that T : X Y is a continuous linear bijection. Show that T is an isomorphism. Solution. I T is a continuous linear bijection, then T is a continuous linear surjection, so the Open Mapping Theorem implies that T is open. Since T is a bijection, T : Y X exists. Since T is open, the inverse image under T o every open set is open, so is continuous, hence T is an isomorphism. T Page 3 o 5
4 (20 4. Let be a seminorm on the linear space X, and let M { x X : x 0} subspace o X and that the map x + M x is a norm on X M. Solution. First we show that M is a subspace o X. Let x, y M and let a and b be scalars. Then ax + by ax + by = a x + b y = 0 so ax + by M, so M is a linear subspace o X. Next we show that the map x + M = x = =. Show that M is a is well-deined on X M. I x and y are in the same equivalence class, then x y M, so x y = 0. Thereore, x y x y = 0, so x = y, and hence x + M = x = y = y+ M Finally, we show that is a norm on X M. [ x+ M = x 0,, x + y+ M = x+ y x + y = x+ M + y+ M, ( x M x M x x x M λ + = λ + = λ = λ = λ +, x + M = 0 x = 0 x M x+ M = 0 in X M. Page 4 o 5
5 (20 5. Recall that, i ( X, ρ and (, continuous i there exists a contant λ 0 such that Y σ are metric spaces, a unction : X Y is called Lipschitz ( ( ( λρ( σ x, y x, y or all x, y X. Let X be a compact metric space, and let Y =, Show that the Lipschitz continuous unctions are dense in the space o continuous unctions with the uniorm norm. Solution. Let Lip ( X, denote the set o Lipschitz continuous real-valued unctions rom X to. First we show that the Lip ( X, is a linear subspace o C( X,. Since Lipschitz continuity implies continuity, we have that Lip ( X, C( X, ( X, g Lip,, and let ab,. Let Lipschitz constant or g. Then. To show that it is a linear subspace, let λ be the Lipschitz constant or, and let λ g be the ( a + bg ( x ( a + bg ( y = a( ( x ( y + b( g ( x g ( y a ( x ( y + b g( x g( y ( ( ( ( aλρ xy, + bλρ xy, = aλ + bλ ρ xy,, g g so ag + bg is Lipschitz continuous with Lipschitz constant a λ b λg +. Thereore, Lip ( X, is a linear subspace o C( X,. Next we show that Lip ( X, is a subalgebra o C( X,. Let, g Lip ( X, and λ g be as beore. Since X is compact, and g are bounded, so let ( x M g( x Mg, or all x X. Then ( ( = ( ( ( ( + ( ( ( ( ( x g( x g( y + g( y ( x ( y g x g y x g x x g y x g y y g y ( ( ( ( M λρ xy, + Mλρ xy, = Mλ + Mλ ρ xy,, g g g g so g is Lipschitz continuous with Lipschitz constant M λg + Mgλ., and let λ and Constant unctions are Lipschitz continuous with Lipschitz constant 0, so it remains only to Lip X, separates points. Let p and q be distinct points in X, and let show that ( Then ( p = 0 and ( q 0 that ( = ρ (, x x p, so separates p and q. Since the triangle inequality implies ( ( = ρ(, ρ(, ρ(, x y x p y p x y it ollows that is Lipschitz continuous with Lipschitz constant. The Stone-Weierstrass Theorem now implies that Lip ( X, is dense in (, C X. Page 5 o 5
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