# M597K: Solution to Homework Assignment 7

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1 M597K: Solution to Homework Assignment 7 The following problems re on the specified pges of the text book by Keener (2nd Edition, i.e., revised nd updted version) Problems 3 nd 4 of Section 2.1 on p.94; Problem 1 of Section 3.1 on p.128; Problem 1 of Section 3.2 on p Show tht the sequence (x n ), x n = n 1 k=1 k! is cuchy sequence using the mesure of distnce d(x, y) = x y. A sequence of points x n IR is Cuchy if for every ɛ>thereexistsnn, such tht m>n>n implies x n x m <ɛ Let ɛ>begiven,choosen = 1 ɛ.thenform>n>n,wehve x n x m = 1 (n+1)! + 1 (n+2)! m! n(n+1) + 1 n(n+1)(n+2) +... n(n+1) + 1 n(n+1)(n+1) +... n(n+1) (1 + 1 n ) (n+1) 2 n(n+1) 1 = n = ɛ. 2 n+1 Therefore the sequence is Cuchy. 2. Show tht the continues functions on [, 1] form n infinite dimensionl vector spce. Find set of linerly independent vectors which is not finite. To prove the continues functions on [, 1] form vector spce is trivil, omitted. We show tht the polynomils re lredy infinite dimensionl. Functions 1,x,x 2,x 3,... is infinite nd none is liner combintion of the others. They re linerly independent becuse the only wy we cn write c 1 x n + c 2 x m + c 3 x l +... = (ssuming n>m>l> ) is if ll the coefficients re zero. Becuse otherwise we will hve polynomil of finite degree n which hs t most n roots. Thus the continues functions form n infinite dimensionl vector spce. nd 1,x,x 2,x 3,... re independent.

2 3. Verify tht the solution of u = f(x),u() = u(1) = is given by u(x) = 1 k(x, y)f(y)dy where y(x 1), y<x 1, k(x, y) = x(y 1), x<y. We should first check the boundry: u() = k(,y)f(y)dy = u(1) = k(1,y)f(y)dy = And u(x) = 1 k(x, y)f(y)dy = x y(x 1)f(y)dy + 1 x x(y 1)f(y)dy = 1 xyf(y)dy x yf(y)dy 1 x xf(y)dy u (x) = 1 yf(y)dy xf(x)+ x 1 f(y)dy + xf(x) = x 1 f(y)dy u (x) = f(x) 4. Show T (f) = f(), defined for ll continuous functions on [ 1, 1], is not bounded liner functionl in the L 2 norm, but it is bounded liner functionl with the uniform norm. For T to be bounded on continuous functions in the L 2 norm, there must exist constnt C such tht Tf C f L 2, for ll continuous f(x). The converse: i.e., for T not bounded, the sttement is: For ny finite number C, there is lwys continuous function f, which my depend on C, thusweuse f C (x) to denote it, such tht Tf >C f L 2. (1) 2

3 We prove tht such n f C (x) exists. So let C be positive number. Tke f C (x) be such tht f C (x) =C, x [ d, d], nd f C (x) = for ll other x, whered is smll number. Thus, we hve Tf C (x) = f C () = C, nd f C (x) L 2 = C 2d. If we tke d = 1 8C, then inequlity (1) 2 holds like C>C 1 2. Thus this T is not bounded liner functionl on the set of continuous functions mesured in the norm L 2. (The function f C (x) cnbemde continuous esily without chnging is primry property.) Since f() mx f(x), T is bounded with uniform norm (tke C =1). 5. Let l 2 denote ll the sequences (x 1,x 2,x 3,,x n, ) of rel numbers. Let x denote such sequence. Use vector ddition nd sclr multipliction. Then it is vector spce (no proof needed). Use the inner product <x,y>= x i y i i=1 where y =(y 1,y 2,,y n, ). Show tht this inner product is well defined in l 2 nd it stisfies the four properties of the definition of inner product. (It is clled the little l two spce. (Reference: p. 59 of text book) By Cuchy-Schwrz inequlity: <x,y> = x i y i x 2 i yi 2 =<x,x>1 2 <y,y> 1 2 i=1 i=1 i=1 Thus <x,y> <, <x,y>is well defined in l 2. And (1) <x+ y,z >= i=1 (x i + y i )z i = i=1 x i z i + i=1 y i z i =<x,z>+ <y,z> (2) <αx,z>= α i=1 x i z i = α<x,z>. (3) <x,y>=< y,x> (4) <x,x> nd<x,x>=iffx = This <.,.>is n inner product. 6. Let f(t) be equl to 1 for t between 19 nd 2, nd equl to zero for ll other times t. This my represent Miss Universe s sttus score history of her lifetime. 3

4 Let g(t) be equl to 85 for t between 35 nd 116, nd equl to zero for ll other times t. This might represent nother person s socil sttus score history, who t ge 35 invented perpetul mchine nd enjoyed the fme he got throughout his lifetime. (Unfortuntely he lived only to 116. Thus his perpetul mchine ws just somewht perpetul, nd tht explins the score 85 insted of higher number.) Now pnel wnt to select one from the two to put into Hll of Fme nd wnt you to give the pnel single mesurement number from ech of the two so tht the pnel cn decide. Do the mximum norm nd the L 2 norm clcultion (serious prt) nd decide which norm you wnt to use to give to the pnel ( decision tht is purely up to you). 1, 19 t 2, f(t) =, else, 85, 35 t 16, g(t) =, else, f(t) = 1, g(t) =85 f(t) L 2=( dx) 1 2 = 1, g(t) L 2=( dx) 1 2 = Solve Problem 5, p. 94, of the text book, where L 2 is replced by L 1. As f n (t) =when t< n,ndf n(t) =1whent> n Suppose n<m, f n (t) f m (t) =when t< n or t> n nd f n (t) f m (t) f n (t) + f m (t) 2 Thus f n (t) f m (t) L 2=( 1 (f n(t) f m (t)) 2 ) 1 2 ( n 1 (f n (t) f m (t)) 2 ) 1 2 ( n 1 2 (2) 2 ) = 8 n sn n Tht implies it is Cuchy in L n 8. Show tht the functionl T on the Bnch spce C[, 1] defined by is liner nd bounded. Tf = 1 f(x) x dx, f(x) C[, 1] 4

5 T (αf + βg) = 1 αf(x)+βg(x) x dx = α 1 f(x) x dx + β 1 f(x) x dx = αt (f)+βt(g). Thus T is liner. And Tf = 1 Thus T is bounded. f(x) x dx mx f(x) 1 1 x dx =2 mx f(x) 9. Let nd b be two points in the intervl [, 1]. Show tht the functionl T on the Bnch spce C[, 1] defined by Tf = f() f(b), f(x) C[, 1] is liner nd bounded. (This functionl is generlly written s δ(x ) δ(x b).) (Hint: Lecture notes might help.) T (αf + βg) =α(f()) + βg() (α(f(b)) + βg(b)) = α(f() f(b)) + β(g() g(b)) = αt (f)+βt(g). Thus T is liner. And Tf = f() f(b) f() + f(b) 2 mx f(x) Thus T is bounded. 1. Find the djoint opertor T for the opertor T defined on the Hilbert spce L 2 [, b] by b Tu(x) =w(x) k(x, y)u(y)dy where k(x, y) L 2 ([, b] [, b]) nd w(x) C[, b] re both given functions nd u is n rbitrry member in L 2 [, b]. (Hint: Red text book p.17.) Strting with the definition of Tu(x) =w(x) b k(x, y)u(y)dy nd the definition of T we formulte <Tu,v> = b (w(x) b k(x, y)u(y)dy)v(x)dx = b b w(x)k(x, y)v(x)u(y) dx dy = b ( b k(x, y)w(x)v(x) dx)u(y) dy =< T v, u >. Thus T v(y) = b k(x, y)w(x)v(x)dx or equivlently but more stndrd: T u(x) = b k(y,x)w(y)u(y) dy. 5

6 11. Verify tht λ =(nπ) 2 nd u =cos(nπx) re eigenvlues nd corresponding eigenfunctions for the Sturm-Liouville eigenvlue problem: d 2 u dx 2 + λu =, ( <x<1); u () = u (1) = for ll positive integer n. From λ =(nπ) 2 nd u =cos(nπx), we cn get du dx = nπ sin(nπx), nd d 2 u + λu =, ( <x<1) dx2 And s u (x) = nπ sin(nπx), u () =, u (1) =. 6

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