Math Solutions to homework 1


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1 Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht stisfies the prllelogrm lw ie + b + b = + b ) Consider x, y) := x + y x y ) Show tht x + y, z) = x, z) + y, z) Hint: consider pplying the prllelogrm lw to {, b} = {x + z, y}, {, b} = {x z, y}, {, b} = {y + z, x}, nd {, b} = {y z, x}) b) Show tht nx, z) = nx, z) for n N c) Show tht qx, z) = qx, z) for q Q d) Show tht, ) is liner in the first slot e) Conclude tht, ) is n inner product Solution: Let X be rel vector spce ) Applying the prllelogrm lw to {, b} = {x + z, y}, {, b} = {x z, y}, {, b} = {y + z, x}, nd {, b} = {y z, x}, one gets the following four equtions: x + y + z + x y + z = x + z + y ) x + y z + x y z = x z + y ) x + y + z + x + y + z = y + z + x 3) x + y z + x + y z = y z + x ) By the property N in the book, pge 3, it is true for ny vector w tht w = w = w Therefore, we hve x + y + z = x y z) = x y z nd x + y z = x y + z) = x y + z
2 Using these two identities to replce the second terms in the equtions 3) nd ), we get: x + y + z + x y + z = x + z + y 5) x + y z + x y z = x z + y 6) x + y + z + x y z = y + z + x 7) x + y z + x y + z = y z + x 8) Finlly, we dd together the equtions 5) nd 7), nd substrct from the result the equtions 6) nd 8) We get x + y + z x + y z = x + z x z + y + z y z, which rewrites s fter dividing by 8): x + y + z x + y z ) = x + z x z ) + y + z y z ), which is x + y, z) = x, z) + y, z) b) The result is cler for n = 0: we hve 0, z) = 0 + 0, z) = 0, z) + 0, z) by prt ), nd this lst eqution implies tht 0, z) = 0 For n, let us prove it by induction For n =, the result is obvious Now, ssume tht the result is true for n To prove it for n +, we use prt ) gin: n + )x, z) = nx + x, z) = nx, z) + x, z) = nx, z) + x, z) = n + )x, z), where we use the induction hypothesis t the third equlity c) Let us first prove it for negtive integers Let n N; we wnt to prove it for n) To do this, write 0 = 0, z) = nx + n)x, z) = nx, z) + n)x, z) = nx, z) + n)x, z); this implies tht n)x, z) = nx, z) Now, we prove it for n rbitrry rtionl number q Write q s b, where Z nd 0 b N We hve ) b b x, z = b ) b x, z = x, z) = x, z), where we used the fct tht the result is true for integers in both the first nd the lst equlities Dividing by b, one get ) b x, z = x, z), b hence the result for ny rtionl number q d) Since we lredy know tht, ) is dditive in the first slot ie x + y, z) = x, z) + y, z)), we only hve to prove tht rx, z) = rx, z) for ny rel number r We lredy know tht this is true if r is rtionl The ide here is to use continuity rgument: the inner product is continuous, nd Q is dense in R, hence if the equlity rx, z) = rx, z) is true for r rtionl, it must lso be true for r rel Let us see how this works precisely
3 By the exercise nd the theorem 5 in the book, we know tht the norm function : X R is continuous, s well s the ddition X X X nd the sclr multipliction R X X Therefore, for ny vectors x, z X, the composition f : R X R : t tx tx + z tx z ) = tx, z) must lso be continuous Since we know tht tx, z) = tx, z) when t Q, we know tht the continuous function bove coincides with the liner function tx, z) recll tht x nd z re fixed) when t is rtionl number Let us recp: we hve function f : R R tht is continuous nd tht extends to R the liner function g : Q R : t tx, z) defined on Q But we lredy know function tht is continuous nd extends g to R: it is the liner function g : R R : t tx, z) Recll tht Q is dense in R Since continuous function defined on dense set of topologicl spce hs t most one continuous extension to the whole spce, we conclude tht f must be equl to g In other words, we hve tx, z) = ft) = gt) = tx, z) for every t R e) Let us verify tht, ) is n inner product we refer to the definition in the book) The properties ii) nd iii) were respectively proved in d) nd ) bove For the property i), we hve x, y) = x + y x y ) = y + x y x) ) = y + x y x ) = y, x) = y, x) We use the definition of, ) t the first nd fourth equlities; the third one follows from w = w s we lredy remrked bove, nd the lst one follows from the fct tht y, x) R, which is cler from its definition Finlly, we check iv): x, x) = x + x x x ) = x = x > 0 when x 0, by the property N of the norm in the book definition ) 3
4 Problem, prt : Let X be complex vector spce ) Let, ) be n inner product on X with inner product norm Show tht x, y) = x + y x y + i x + iy x iy )) b) Use the previous prt of the problem to show tht if stisfies the prllelogrm lw, then there is n inner product with x, x) = x Solution: Let X be complex vector spce ) We simply expnd the righthnd side: x + y x y + i x + iy x iy ) = x + y, x + y) x y, x y) + i x + iy, x + iy) x iy, x iy)) = x, x) + x, y) + y, x) + y, y) x, x) x, y) y, x) y, y) +ix, x) + ix, iy) + iiy, x) + iiy, iy) ix, x) ix, iy) i iy, x) i iy, iy) = x, y) + y, x) + x, y) y, x) = x, y) We used the fct tht λx, y) = λx, y), while x, λy) = λx, y) b) Define x, y) by the formul bove Write x, y) = x, y) 0 + ix, iy) 0, where x, y) 0 is the inner product defined in the first prt, on the vector spce X seen s rel vector spce Let us verify the four properties of the inner product i) Notice tht y +ix = ix iy) nd y ix = ix+iy), hence y + ix = i x iy = x iy nd y ix = i x + iy = x + iy Therefore, y, x) = y + x y x + i y + ix y ix )) = x + y x y + i x iy x + iy )) x + y x y i x + iy x iy )) = = x, y) iii) We proved bove tht, ) 0 is dditive in the first slot Therefore, x + y, z) = x + y, z) 0 + ix + y, iz) 0 = x, z) 0 + ix, iz) 0 + y, z) 0 + iy, iz) 0 = x, z) + y, z) ii) We wnt to prove tht for every + bi C, we hve + bi)x, y) = + bi)x, y) We lredy know tht + bi)x, y) = x, y) + bix, y) = x, y) 0 + ix, iy) 0 + bix, y) 0 + ibix, iy) 0 = x, y) 0 + ix, iy) 0 + bix, y) 0 + biix, iy) 0 = x, y) + bix, y)
5 If we cn prove tht ix, y) = ix, y), then this is equl to +bi)x, y), nd we re done Let us therefore prove tht ix, y) = ix, z) Using s in i) the trick tht iw = i w = w, we compute: ix, y) = ix + y ix y + i ix + iy ix iy )) = i x + y x y + i x + iy x iy )) = ix, y) Therefore, we proved tht λx, y) = λx, y) for every λ C iv) Finlly, we hve x, x) = x, x) 0 + ix, ix) 0 If this is equl to 0, then x, x) 0 = 0, which implies x = 0 we cn do this becuse, ) 0 tkes only rel vlues) Problem in the book): Let < b in R Show tht the spce W [, b] of continuously differentible functions on [, b], with vlues in C, is n inner product spce with respect to pointwise ddition nd sclr multipliction, nd inner product f, g) W = ft)gt) + f t)g t) dt Solution: We hve to check the four properties of the definition in the book Let f, g, h be functions in W [, b], nd let λ C i) ii) iii) iv) f, g) W = λf, g) W = = ft)gt) + f t)g t) dt = λft)gt) + λf t)g t) dt = λ f + g, h) W = ft)ht) + f t)h t) dt + f, f) W = ft)gt) + f t)g t) dt = g, f) W ft)gt) + f t)g t) dt = λf, g) W ft) + gt))ht) + f t) + g t))h t) dt ft)ft) + f t)f t) dt = gt)ht) + g t)h t) dt = f, h) W + g, h) W ft) + f t) dt Since f is continously, differentible, the function ft) + f t) is continuous If f is not identiclly 0, then ft) + f t) is lso not identiclly zero Therefore, by continuity, there exists n intervl of length α > 0 on which the function ft) + f t) is lrger thn β > 0 Since this function is lso nonnegtive, this implies tht f, f) W > 0 5
6 Problem 3: Show tht W [0, ] from the previous problem is not Hilbert spce ie it is not complete with respect to the inner product norm) Solution: Consider the sequences of functions see digrm 3 pge 8 in the book) 0 if 0 x f nx) n = nx + n if n x if x Their indefinite integrls re f n x) = 0 if 0 x n + n ) x + n 8 + n if n x n + x if x nx It is somewht long, but esy to check tht this is Cuchy sequence: compute f n f m, f n f m ) W, nd show tht for every ε, there exists N N such tht this quntity is smller thn ε if m, n N Now in W [0, ], if sequence g n ) n of functions converges to function g, then g n) n converges to g for the norm on the spce of continuous functions given by g C = 0 g x) dx Indeed, g g n C = g x) dx 0 0 g g n ) x) + g g n) x) dx = g g n W, so if g g n W is smller thn ε for n lrge, then so is g g n C The only function tht the functions f n bove converge to, in the norm C is the function f x) = { 0 if 0 x if x Therefore, if the functions f n bove converge to something in W [0, ], then their derivtives must converge to the function f bove But tht function is not continuous, hence f / W [0, ] So, W [0, ] is not complete Problem 7 in the book): Let c 0 denote the subspce of l comprising ll sequences x n ) n which tend to zero s n Prove tht c 0 is closed in l with respect to Solution: We hve to prove tht the limit in the topology given by ) of sequence in c 0 tht converges in l is ctully in c 0 Let x m n ) n ) m be such sequence there re two indices becuse we re considering sequence of elements of c 0, which lredy re sequences) Let us cll x n ) n the limit of tht sequence in l Let ε > 0 Since lim m x m n ) n = x n ) n this is limit in l ), there exists M N such tht for every m M, we hve x m n ) n x n ) n < ε By definition of the norm, this mens tht for every m M nd for every n N, we hve x m n x n < ε 6
7 Now, pick ny m M Since x m n ) n c 0, we hve tht lim n x m n = 0 Therefore, there exists number N m) N which depends on m) such tht for every n N m), we hve x m n < ε By the tringle inequlity, for every n N m), we hve tht x n x n x m n + x m n < ε + ε = ε, which proves tht lim n x n = 0, nd hence tht the sequence x n ) n is in c 0 Problem 5 8 in the book): The open nd closed unit blls in normed spce E, ) re defined to be respectively the sets E 0 = {x E : x < }, E = {x E : x } Show tht the ltter is the closure of the former Solution: We hve to show tht closuree 0 ) = E, which we prove by showing the two inclusions seprtely closuree 0 ) E : let x be in the closure of E 0, this mens tht there must exist sequence x n ) n in E 0, tht converges to x Since we know tht the norm is continuous, the sequence x n ) n must converge in R to x But this sequence only contins terms in [0, ) becuse it is in E 0 ), hence x must be in [0, ], which proves tht x E closuree 0 ) E : let x be in E For every t [0, ), the vector tx is in E 0, since tx = t x < x Tke ny sequence t n ) n in [0, ) tht converges to Then, by continuity of the sclr multipliction, the sequence t n x) n converges to x in E Since this sequence is in E 0, this shows tht x is in the closure of E 0 Problem 6 38 in the book): Prove tht the closure of convex set in normed spce is convex Solution: Let us cll C our convex set, nd C its closure Tke two elements x nd y in C; this mens tht there re sequences x n ) n nd y n ) n in C tht converge to x nd y respectively We wnt to show tht the segment in E between x nd y lies in C; in other words, we wnt to show tht for every t [0, ], the vector tx + t)y is in C Fix t [0, ], nd consider w n = tx n + t)y n Since x n nd y n re both in C, nd since C is convex, w n is lso in C And the limit of the sequence w n ) n is tx + t)y, which proves tht tx + t)y is in the closure of C Since this holds for every t [0, ], we conclude tht C is convex 7
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