ANSWERS TO MIDTERM EXAM # 2

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1 MATH 03, FALL 003 ANSWERS TO MIDTERM EXAM # PENN STATE UNIVERSITY Problem 1 (18 pts). State ad prove the Itermediate Value Theorem. Solutio See class otes or Theorem from our textbook. Problem (18 pts). Let f be a cotiuous fuctio from the closed disk D = {(x, y) R : x + y 3} ito R. Show that f caot be oe-to oe. Solutio Let g : D R be the fuctio defied by g(u) = f(u) f( u), u = (x, y) D. The g is cotiuous at every poit u D sice g is a liear combiatio of cotiuous fuctios. Moreover g(u) = g( u). Now, cosider the poits p = (3, 0) ad q = p = ( 3, 0) i D. Either g(p) = 0 or g(p) 0. Case 1: If g(p) = 0 the f(p) = f(q). Thus, f is ot oe-to-oe. Case : If g(p) > 0 the g(q) < 0. I this case we cosider the path γ : [0, 1] D defied by γ(t) = (3 cos(πt), 3 si(πt)). The image of γ is the circular arc from p to q. Applyig the Itermediate Value Theorem to the fuctio g γ, we coclude that there is a poit r o the path such that g(r) = 0. This gives f(r) = f( r). Sice r r, f is ot oe-to-oe. Case 3: Similarly, oe shows that f is ot oe-to-oe whe g(p) < 0. Problem 3 (18 pts). Cosider the sequece (f ) =3 of fuctios i C([0, 1]) give by { x for 0 x f (x) = for x Show that f m f 1 for ay m. 3.. Use Questio 3.1 to show that the ball B = {f C([0, 1]) : f } is NOT compact Show that B = {f C([0, 1]) : f } is closed ad bouded. Solutio: 3.1. For ay m, we have f m (x) f (x) = (m )x for 0 x m x for x m 0 for x 1 1

2 Therefore, f m f = sup f m (x) f (x) = x [0,1] m. Sice m, we get f m f 1 = By defiitio, B is compact if every sequece i B has a subsequece that coverges to some fuctio f B. However, f = sup f (x) =, x [0,1] so (f ) =3 is a sequece i B. But, oe of the subsequeces of (f ) =3 is coverget. Ideed, ay coverget subsequece is Cauchy, which cotradicts the fact that f m f 1, for ay m. Therefore, B is ot compact. 3.. B is bouded by costructio. If g is a limit poit of B the there is a sequece (g ) =1 i B such that lim g g =0. Sice g g g + g g g +. Takig the limit, we get g. This shows that g B Problem (16 pts). Cosider the sequece of fuctios give by f (x) = x(1 x ) o [0, 1]. a. Show that f coverges poitwise to the zero fuctio. b. Show that f is ot uiformly coverget. Solutio: a. Observe that f (0) = f (1) = 0, for all N. For 0 < x < 1, we have lim f (x) = lim xe Log(1 x ) = 0, sice Log(1 x ) < 0. Therefore, (f ) =1 coverges poitwise to f 0. b. I order to determie the uiform covergece, we compute f f. The derivative of f is ( f (x) = (1 x ) 1 1 (1 + )x ). The fuctio f attais its maximum at x = 1/ 1 +.

3 Hece [ lim f f = lim f = lim 1 + = e 1/ lim This shows that f is ot uiformly coverget. ( 1 1 ) ] = + Problem 5 (1 pts). Let (V, ) be a ormed vector space Show that x y x y, for all x, y V. 5.. Give x 1, x,, x V, show by iductio that x 1 + x + + x x 1 x x. Solutio: 5.1. By the triagle iequality, we get x = x y + y x y + y, for all x, y V. This implies x y x y. Iterchagig the role of x ad y, we get y x x y. Multiplyig by 1, we get It follows x y x y. x y x y x y, for all x, y V. 5.. Case =: Let x = x 1 ad y = x. From Questio 5.1, we get x 1 + x x 1 x. Assume that, for all x 1, x,, x V x 1 + x + + x x 1 x x. Let x 1, x,, x +1 be vectors i V. By the previous case, we get x 1 +x + +x +1 = (x 1 +x + +x )+x +1 (x 1 +x + +x ) x +1. The iductio hypothesis gives x 1 + x + + x +1 x 1 x x x +1. 3

4 Problem 6 (18 pts). Let K be a compact subset of R ad let f : K C R m, ad g : C R k be two fuctios such that f is surjective ad cotiuous. a. Show that C is compact. b. Now suppose that g f is cotiuous. Prove that g is cotiuous. Solutio: a. C = f(k) is compact as the image of a compact subset of R uder a cotiuous fuctio. b. Suppose g is discotiuous at y. There a sequece (y ) =1 i C so that lim y = y but lim g(y ) g(y). The, there is a positive real umber ɛ such that g(y ) g(y) ɛ for ifiitely may. Thus, we ca extract a subsequece (y i ) i=1 such that ( ) g(y i ) g(y) ɛ, i N. Because f is surjective, there is a poit x i K such that f(x i ) = y i, for every y i. Sice K is compact, there is a subsequece (x ij ) j=1 of (x i ) that coverges to some x K. Deote y ij = f(x ij ). By the cotiuity of f, we have lim f(x i j ) = f(x). j By the uiqueess of the limit, we get y = f(x). It follows from the cotiuity of g f that lim j g f(x i j ) = g f(x). I other words, lim j g(y ij ) = g(y). This cotradicts the iequality ( ). Problem 7: Bous. a. (3 pts) Show that a closed subset of a Baach space is complete. b. ( pts) Cosider the fuctio defied o R by { 1 if x 0 or x y f(x) = cos( πx ) if 0 < x < y y Show that f is ot cotiuous at the origi. Solutio:

5 a. By defiitio, a Baach space is a complete ormed vector space. Let C be a closed subset of a Baach space (V, ). Let (v ) be a Cauchy sequece i C. Because V is complete, the sequece coverges to some vector v V. But C is closed, so it cotais all its limit poits. I particular v C. b. Pick x = 1/ ad y = /. The x < y, lim (x, y ) = (0, 0) but lim f(x, y ) = cos( π ) 1. 5