McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected Exercises. g(x) 2 dx 1 2 a

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1 McGill University Mth 354: Honors Anlysis 3 Fll 2012 Assignment 1 Solutions to selected Exercises Exercise 1. (i) Verify the identity for ny two sets of comlex numers { 1,..., n } nd { 1,..., n } ( n ) 2 ( n k k = ) ( n 2 k 2 k k=1 k=1 k=1 ) 1 2 n n ( i j i j ) 2. i=1 j=1 ii) Let f(x) nd g(x) e continuous functions on [, ]. Prove tht ( 2 f(x)g(x)dx) = f(x) 2 dx g(x) 2 dx 1 2 [f(x)g(y) g(x)f(y)] 2 dxdy. Solution (i) : The RHS exnds to n 2 i 2 j 1 n ( 2 i 2 j + 2 i 2 j 2 i i j j ) = 2 i,j=1 i,j=1 ( n n ) 2 i i j j = k k i,j=1 k=1 Solution (ii) : Becuse f, g C([, ]), we my very sfely invoke Fuini s theorem to chnge the order of integrtion. The RHS exnds to f(x) 2 dx g(x) 2 dx 1 2 [ f(x) 2 g(y) 2 + g(x) 2 f(y) 2 2f(x)g(x)f(y)g(y) ] dxdy = = = (f(x)g(x)f(y)g(y)) dxdy ( ) f(x)g(x)f(y)g(y)dx dy ( ) f(x)g(x)dx f(y)g(y)dy ( ) ( ) = f(x)g(x)dx f(y)g(y)dy

2 Exercise 2. (i) Strting from the ineulity xy x / + y /, where x, y,, > 0 nd 1/ + 1/ = 1, deduce Hölder s integrl ineulity for continuous functions f(t), g(t) on [, ]: ( / ( 1/ f(t)g(t)dt f(t) dt g(t) dt) ; (ii) Use (i) to rove Minkowski s integrl ineulity for continuous functions f(t), g(t) on [, ] nd 1: ( 1/ ( 1/ ( 1/ f(t) + g(t) dt) f(t) dt) + g(t) dt). Solution (i) : If f or g is identiclly 0 on [, ], then the ineulity holds. So we my ssume tht f, g > 0. Tke Then Rerrnge to get f(t) x = ( ) 1/ = f(t) nd y = f f(t) dt f(t) g(t) dt f g f(t)g(t)dt ( f(t) f g(t) ( g(t) dt ) + g(t) g dt = 1. f(t)g(t) dt f g. / = g(t) g Young s Ineulity (i) : For ny x, y,, > 0, with, conjugte exonents, xy x / + y /. Proof: xy = e ln xy = e 1 ln x + 1 ln y 1 eln x + 1 eln y = x + y y convexity of the exonentil function. Solution (ii) : Minkowski s ineulity holds for = 1, so we my ssume > 1. f(t)+g(t) dt = f(t)+g(t) f(t)+g(t) 1 dt ( ( f(t) dt f(t) + g(t) ( 1) dt ( = f(t) dt Since ( 1) =, rerrnge to get ( f(t) + g(t) dt ( + g(t) dt ( = f(t) + g(t) dt f(t) f(t)+g(t) 1 dt+ ( + g(t) dt 1 ( ( f(t) + g(t) ( 1) dt ( f(t) dt g(t) f(t)+g(t) 1 dt f(t) + g(t) ( 1) dt ( + g(t) dt.

3 Exercise 3. Prove tht the set of ll oints x = (x 1, x 2,..., x k,...) with only finitely mny nonzero coordintes, ech of which is rtionl numer, is dense in the sce l 2 of seuences. Solution: Let x = (x 1, x 2,..., x n,...) l 2. Then i=1 x2 i <, so for ny ɛ > 0, there exists N N such tht i>n x2 i < ɛ. For ech 1 j N, choose rtionl numer y i such tht (x i y i ) 2 < ɛ/n. Let y = (y 1, y 2,..., y N, 0, 0,...). Then y hs only N nonzero rtionl coordintes, nd N (d 2 (x, y)) 2 = i y i ) j=1(x 2 + x 2 i < ɛ + N(ɛ/N) = 2ɛ. i>n Since ɛ ws ritrry, we hve roved the density.

4 Exercise 4 (extr credit). i) Suose φ C([, ]) (which need not e differentile) stisfies φ((x + y)/2) (φ(x) + φ(y))/2, x, y [, ]. Prove tht for ll x, y [, ], nd for ny t [0, 1], we hve i.e. tht φ is convex on [, ]. φ(tx + (1 t)y) tφ(x) + (1 t)φ(y), (1) ii) Assume tht function φ (tht is not ssumed to e continuous on n oen intervl (, )), stisfies (1). Prove tht φ is then ctully continuous on (, ). iii) Prove tht if φ C 2 ([, ]), nd φ (x) > 0, x [, ], then φ is convex on [, ]. iv) Prove tht if x 1,..., x n [, ], nd t 1,..., t n > 0 stisfy t t n = 1, nd if φ is convex on [, ], then φ(t 1 x t n x n ) t 1 φ(x 1 ) t n φ(x n ). Solution i): Let x, y (, ) nd define f(λ) = ϕ((1 λ)x + λy) for 0 λ 1. Note first tht f() (1 )f(0) + f(1) for ll dydic rtionls 0 1. To see this, suose tht the ineulity holds for dydic rtionls of the form = k 2 for 1 n N; then if 0 k < 2 N, we hve n ( ) 2k + 1 f 2 N+1 1 ( ( ) ( )) k k + 1 f 2 2 N + f 2 N 1 (( 1 k ) 2 2 N f(0) + k ( 2 N f(1) + 1 k + 1 ) 2 N f(0) + k + 1 ) 2 N f(1) ( = 1 2k + 1 ) 2 N+1 f(0) + 2k + 1 f(1) 2N+1 Consider now the seuence of dydic rtionls otined y tking successively ccurte roximtions to the inry exnsion of λ. Then λ n = 2 n λ2 n is seuence converging to λ, nd since f is continuous the ineulity holds in the limit; tht is, f(λ) (1 λ)f(0) + λf(1) for ll 0 λ 1, nd hence ϕ is convex. ii): We find tht the definition is convexity is euivlent to reuiring tht for ll < s < t < u <, we hve φ(t) φ(s) t s φ(u) φ(t). (2) u t Fix t (, ); we shll rove tht φ is continuous t t. Let r(t, s 0 ) denote the rtio in the left-hnd side of (2) for some fixed < s 0 < t; similrly, denote r(t, u 0 ) denote the rtio in the right-hnd side of (2) for some fixed t < u 0 <. Suose now tht s (s 0, t). Then it follows from (2) tht φ(t) r(t, u 0 )(t s) φ(s) φ(t) r(s 0, t)(t s), i.e. the grh of φ lies in etween two stright lines tht intersect t the oint (t, φ(t)). The continuity s s t from the left follows. The roof of continuity s u t from the right follows similrly from the ineulity φ(t) + r(s 0, t)(u t) φ(u) φ(t) + r(t, u 0 )(u t), tht holds for u (t, u 0 ). iii): By the rgument in ii), it suffices to rove (2), which for continuously differentile functions is euivlent to sying tht φ is nondecresing, nd tht follows from the ssumtion φ (t) > 0, t [, ].

5 iv): The roof is y induction, strting with n = 2 which is the ssumtion of continuity. The induction ste is roved s follows: φ(t 1 x t n x n +t n + 1x n+1 ) = φ(t 1 x (t n +t n+1 )y) t 1 φ(x 1 )+...+(t n +t n+1 )φ(y), (3) where y = (t n x n + t n+1 x n+1 )/(t n + t n+1 ) nd where we hve used induction hyothesis. On the other hnd, y convexity φ(y) t nφ(x n ) + t n+1φ(x n+1 ). t n + t n+1 t n + t n+1 Sustituting into (3), we comlete the roof.

6 Exercise 5. Let X e metric sce, A X suset of X, nd x oint in X. The distnce from x to A is denoted y d(x, A) nd is defined y Prove tht d(x, A) = inf d(x, ). A i) If x A, then d(x, A) = 0, ut not conversely; ii) For fixed A, d(x, A) is continuous function of x; iii) d(x, A) = 0 if nd only if x is contct oint of A (i.e. every neighorhood of x contins oint from A); iv) The closure A stisfies A = A {x : d(x, A) = 0}. Solution (i) : If x A, then 0 = d(x, x) = inf A d(x, ). Converse is not true, for exmle if A = (0, 1], then d(0, A) = 0 while 0 / A. Solution (ii) : Let x, y X. Given ɛ > 0, choose A such tht d(x, ) d(x, A) + ɛ. By tringle ineulity we hve d(y, ) d(x, ) + d(x, y) d(x, y) + d(x, A) + ɛ. Since ɛ ws ritrry, nd since d(y, A) d(y, ), we get d(y, A) d(x, A) + d(x, y). Reversing the roles of x nd y we get d(x, A) d(y, A) + d(x, y). It follows tht which imlies continuity of d(, A). d(x, A) d(y, A) d(x, y). Solution (iii) : If x is contct oint of A, then for every r > 0, B(x, r) contins oint of A, hence inf A d(x, ) < r. Since r ws ritrry, d(x, A) = 0, roving the if rt. Now, suose ll B(x, r) doesn t contin oints from A for some r > 0. Then d(x, A) r > 0, finishing the roof of the only if rt of the sttement. Solution (iv) : The set A is union of A nd the set of ll limit oints of A. d(x, A) = 0 for ny limit oint tht doesn t elong to A. By rt (iii), Remrk : To show tht the function d : X [0, ), d : x d(x, A) ws continuous, some of you cme u with intriguing ineulities such s d(x, A) d(y, A) = inf d(x, ) A inf d(y, ) inf (d(x, ) d(y, )). The following counterexmle shows tht it is wrong in A A generl even if y cn e mde ritrrily close to x. Consider the metric sce of ounded seuences l (N) with entries in R euied with the metric d(x, y) = su x i y i. Let x = i N (0, 0,...), y 1 = ( 1 3, 0, 0,...), y 2 = (0, ( 1 3 )2, 0,...), y 3 = (0, 0, ( 1 3 )3, 0,...), etc... nd A = {ω l (N) : ll comonents of ω re 0 excet exctly one which is eul to 1}. Then y n x nd d(x, A) = 1, d(y n, A) = 1 ( 1 3 )n so tht we hve inf d(x, ) inf d(y, ) = ( 1 A A 3 )n > 0 = inf (d(x, ) d(y, )). A

7 Exercise 6. Let (X, d) e metric sce, nd f : X R continuous function. The nodl set of f, denoted y Z(f), is the set {x X : f(x) = 0}. i) Prove tht Z(f) is closed suset of X. Next, let A, B e two closed nonemty susets of X, A B =. Let d(x, A) (res. d(x, B)) denote the distnce from x X to A (res. B), defined in Exercise 5 in Assignment 1. Define function F : X R y the formul F (x) = d(x, A) d(x, A) + d(x, B). Prove tht ii) F is continuous; iii) F (x) = 0 iff x A, nd F (x) = 1 iff x B. Solution (i) : Let x n Z(f), nd let x n y s n. By continuity of f, 0 = f(x n ) f(y), therefore f(y) = 0 nd so y Z(f). Solution (ii) nd (iii) : By the results roved in Exercise 5, Assgmt 1, d(x, A) = 0 iff x A = A, since A is closed, nd similrly for B. It ws lso shown in Exercise 5, Assgmt 1, tht d(x, A) d(y, A) d(x, y). These results re used undntly in the following demonstrtion. There re 3 ossile different cses: x A: then F (x) = 0. Let = d(x, B) > 0, nd let ɛ <. Suose tht y X is such tht d(x, y) < ɛ. Then d(y, A) d(x, y) < ɛ, nd ɛ d(y, B) + ɛ. It follows tht s ɛ 0, so F is continuous t x. F (y) ɛ/( ɛ) 0 = F (x) x B: then d(x, B) = 0 so F (x) = 1. Let = d(x, A) > 0. Choose ɛ < nd suose y X is such tht d(x, y) < ɛ. By n rgument similr to the rgument ove, we find tht d(y, B) < ɛ nd ɛ d(y, A) + ɛ. Accordingly, F (x) = 1 F (y) = s ɛ 0, roving tht F is continuous t x d(y, B)/d(y, A) ɛ/( ɛ) 1, x / A nd x / B: Let = d(x, A) > 0 nd let = d(x, B) > 0. We hve 0 < F (x) = /( + ) < 1. Choose ɛ < min(, ), nd suose y X is such tht d(x, y) < ɛ. It follows tht ɛ < d(y, A) < + ɛ, nd ɛ < d(y, B) < + ɛ. Then ( + ɛ)/( ɛ) F (y) ( ɛ)/( + ɛ). Both sides of the ineulity converge to /( + ) s ɛ 0, roving the continuity of F t x.

8 Exercise 7. Let Mt n denote the sce of n n rel mtrices. For A Mt n, define the norms A 1 s follows: Ax A 1 = su 0 x R n x, where x is the usul Eucliden norm. Next define nother norm A 2 y Prove tht A 2 = i) Prove tht A 1,2 defines norm on Mt n ; mx A ij. 1 i,j n ii) Prove tht there exists constnt C n > 1 such tht 1/C n A 1 / A 2 C n. Solution (i) : The only nontrivil roerty is the tringle ineulity, A+B 1,2 A 1,2 + B 1,2 ; the other roerties re very esy. Now, (A+B) 1 = su Ax+Bx su ( Ax + Bx ) su Ax + su Bx = A 1 + B 1. x =1 x =1 x =1 x =1 (A + B) 2 = mx i,j (A + B) ij mx i,j ( A ij + B ij ) mx i,j A ij + mx B ij = A 2 + B 2. i,j Solution (ii) : Let x = (x 1, x 2,..., x n ) e n ritrry unit vector in (R n, ), where stnds for the Eucliden 2 norm. Then y Cuchy-Schwrtz (or Hölder s ineulity), n n Ax = ( A ij x j ) 2 n n n = A 2 ij A 2 2 = n A 2. i=1 j=1 i=1 j=1 A 2 ij j=1 x 2 j 1 i,j n 1 i,j n Denote e j = (0,..., 1, 0,..., 0) where the 1 occurs in the j th osition. Then Ae j = (A 1j, A 2j,..., A nj ) nd it follows tht A 1 = su Ax Ae j = n A 2 ij A ij 1 i, j n x =1 So A 1 A 2. Therefore i=1 1 n 1 A 1 A 2 n. which shows tht 1 nd 2 re euivlent norms on Mt n.

9 Exercise 8 (extr credit). Let e rime numer ( ositive integer tht is only divisile y 1 nd itself, e.g. = 2, 3, 5, 7, 11 etc). Define -dic distnce d on the set Q of rtionl numers s follows: given 1, 2 Q, let 1 2 = Q. If 1 = 2, = 0, then we set d ( 1, 2 ) = 0. If 0, we cn write s = m, where m Z, GCD(, ) = 1, GCD(, ) = GCD(, ) = 1. Here GCD(, ) is the gretest common divisor of two nturl numers nd. Then we define the -dic distnce y d ( 1, 2 ) = m. Plese, note the minus sign in the definition. Exmles: d 2 (5/2, 1/2) = 1/2; d 3 (17, 8) = 1/9; d 5 (4/15, 1/15) = 5. Prove tht d stisfies ll the roerties of distnce. The only nontrivil rt is the tringle ineulity: d ( 1, 2 ) + d ( 2, 3 ) d ( 1, 3 ). You my use without roof ll stndrd roerties of the gretest common divisor, rime decomosition etc. Solution : Let us introduce the -dic norm on the vector sce Q over itself defined y x = m, for x = m (/), where GCD(, ) = GCD(, ) = GCD(, ) = 1 nd 0 = 0. x = 0 iff x = 0. For x, y Q, x = m, y = k c d, then x y m+k c = d = m k = m k = x y For the tringle ineulity we will show tht x + y mx{ x, y }. Assume without loss of generlity tht mx{ x, y } = x := m, i.e. tht x = m (/), y = m+k (c/d), where GCD(, ) = 1 = GCD(, ) = GCD(c, ) = GCD(d, ), nd where k 0. Then x + y = m (k d + c) d Since GCD(, d) = 1, we see tht x+y m. The norm could e smller, if GCD(, k d+ c) =. Hence defines norm on Q over Q. Now if 1, 2 Q re distinct, let 1 2 = m e the uniue decomosition. Then d ( 1, 2 ) = m = m = 1 2 = 1 2. If 1 = 2 then d ( 1, 2 ) = 0 = 0 = 1 2. So the -dic norm induces the -dic distnce d.

10 Exercise 9 (extr credit). Denote y P the set of olygons in R 2, not necessrily convex. A olygon P with vertices x 1, x 2,..., x n is the set of oints in R 2 ounded y simle closed curve tht is union of line segments [x 1, x 2 ], [x 2, x 3 ],..., [x n 1, x n ], [x n, x 1 ]. The oundry curve is denoted P nd is sometimes clled olyline or roken line. We reuire tht different line segments do not intersect excet t common endoints. A symmetric difference of two sets A, B is denoted y A B nd is defined y A B = (A\B) (B\A), where A\B = A B c is the set of oints {x A, x / B}. Given two olygons P 1, P 2 R 2, define the distnce etween them y d(p 1, P 2 ) = Are(P 1 P 2 ). Prove tht d stisfies ll the roerties of distnce. Hint: if X Y, then Are(X) Are(Y ). Solution: d(p 1, P 2 ) 0 nd d(p 1, P 1 ) = 0 P 1, P 2 P. d(p 1, P 2 ) = Are(P 1 P 2 ) = Are(P 2 P 1 ) = d(p 1, P 2 ). As shown in the lemm elow, for ny sets P 1, P 2, P 3 we hve (P 1 P 2 ) (P 1 P 3 ) (P 2 P 3 ). In rticulr the reltion holds for olygons in R 2. Tking res, we find tht Are(P 1 P 2 ) Are((P 1 P 3 ) (P 2 P 3 )) Are(P 1 P 3 ) + Are(P 2 P 3 ). Lemm: For ritrry sets P 1, P 2, P 3, (P 1 P 2 ) (P 1 P 3 ) (P 2 P 3 ). Proof: P 1 P c 2 = (P 1 P c 2 P 3 ) (P 1 P c 2 P c 3 ). The first set in renthesis is contined in P c 2 P 3 (P 2 P 3 ), while the second set in renthesis is contined in P 1 P c 3 (P 1 P 3 ). So, P 1 P c 2 (P 1 P 3 ) (P 2 P 3 ). Reversing the roles of P 1 nd P 2, we see tht P 2 P c 1 (P 1 P 3 ) (P 2 P 3 ). Therefore (P 1 P 2 ) = (P 1 P c 2 ) (P 2 P c 1 ) (P 1 P 3 ) (P 2 P 3 ).

k and v = v 1 j + u 3 i + v 2

k and v = v 1 j + u 3 i + v 2 ORTHOGONAL FUNCTIONS AND FOURIER SERIES Orthogonl functions A function cn e considered to e generliztion of vector. Thus the vector concets like the inner roduct nd orthogonlity of vectors cn e extended