Hilbert Spaces. Chapter Inner product spaces

Size: px
Start display at page:

Download "Hilbert Spaces. Chapter Inner product spaces"

Transcription

1 Chpter 4 Hilbert Spces 4.1 Inner product spces In the following we will discuss both complex nd rel vector spces. With L denoting either R or C we recll tht vector spce over L is set E equipped with ddition, i.e., mp (x, y) x + y from E E to E nd multipliction by sclrs in L, i.e., mp (λ, x) λx from L E to E stisfying the well known vector spce xioms. Exmples of rel vector spces re known from liner lgebr they include R k nd F(M, R) of rel functions defined on set M with pointwise ddition nd multipliction by rel numbers. Likewise we hve the complex vector spces C k nd F(M, C) of complex vlued functions on set M with pointwise ddition nd multipliction by complex numbers. If M is subset of R k then the set C(M, C) of continuous functions is subspce of F(M, C), since ddition of continuous functions results in continuous function nd multipliction of continuous function by number is lso continuous function. We will encounter mny other interesting exmples of subspces of F(M, C). We turn to the definition of inner product spces. Definition 4.1 (Inner product spces). Let E be vector spce over L (= R or C). An inner product (lso clled (sclr product) on E is mp (, ) : E E L, stisfying the following conditions (where 0 denotes the null-vector in E): i) x E \ {0} : (x, x) > 0, ii) x, y E : (x, y) = (y, x), iii) x, y, z E : (x + y, z) = (x, z) + (y, z), iv) λ L x, y E : (λx, y) = λ(x, y). If (, ) is n inner product on E we cll the pir (E, (, )) n inner product spce. Note tht (x, x) > 0 in i) mens tht (x, x) is rel positive number. In the cse L = R, complex conjugtion in ii) is of course superfluous. 1

2 Chp. 4 Opertors on Hilbert Spces Version of The lst two conditions bove express tht the mp x (x, y) from E into L is liner for ech fixed y E. If we combine this with ii) we see tht (x, y + z) = (x, y) + (x, z), (4.1) (x, λy) = λ(x, y), (4.2) for ll x, y, z E nd λ L. We sy tht the inner product is conjugte liner in the second vrible. In the cse L = C mp from E E C, which is liner in the first vrible nd conjugte liner in the second vrible is often sid to be sesqui-liner form on E. Thus we hve seen tht n inner product on complex vector spce E is sesquiliner form on E. Conversely, ny sesqui-liner form, stisfying the property i), which is referred to s sying tht the form is positive definite, is n inner product. To see this we only hve to show tht ii) is stisfied. Note first tht by the linerity in the first vrible we hve (0, x) = 0 for ll x E. In prticulr, we obtin (0, 0) = 0, nd thus from i) we find x E : (x, x) = 0 x = 0. (4.3) Together with i)this shows tht (x, x) R for ll x E. Applying (x + y, x + y) = (x, x) + (y, y) + (x, y) + (y, x) we obtin tht (x, y) + (y, x) R, i.e., Im(x, y) = Im(y, x) for ll x, y E. Replcing x by ix nd using the sesqui-linerity we find i(x, y) i(y, x) R, i.e., Re(x, y) = Re(y, x), which is wht we wnted to show. Given n inner product we define the norm x of x E by x = (x, x). From i) bove it follows tht x > 0 for x 0, nd from iv) nd (4.2) it follows tht λx 2 = (λx, λx) = λλ(x, x) = λ 2 x 2. i.e., λx = λ x. (4.4) Below, s consequence of the Cuchy-Schwrz inequlity in Theorem 4.10 we will show the tringle inequlity x + y x + y, x, y E. (4.5) Positivity nd the two previous reltions re the chrcteristic properties of the norm. The norm defines nturl generliztion of the Eucliden distnce in R 3 (see (1.1) in Chpter 1) by d(x, y) = x y. (4.6) Of specil importnce in wht follows is tht the norm defines notion of convergence of sequences in E:

3 Chp. 4 Opertors on Hilbert Spces Version of Definition 4.2 (Convergence of sequences). Let (x n ) n N be sequence in n inner product spce E, i.e., mp n x n from N into E. We sy tht the sequence converges to x E, or hs the limit x, nd write x n x for n or lim n x n = x, if x n x 0 for n. Finlly, we shll need the importnt notion of Cuchy sequence. Definition 4.3 (Cuchy sequence). We sy tht the sequence (x n ) n N in E is Cuchy sequence, if for ll ε > 0 there is n N N such tht x n x m < ε for ll n, m > N, (which is sometimes written s x n x m 0 s n, m ). Proposition 4.4. Every convergent sequence in n inner product spce E is Cuchy sequence Proof. Exercise 4.1 If the converse holds we sy tht the spce E is complete or tht it is Hilbert spce. Definition 4.5 (Hilbert Spce). If every Cuchy sequence in nd inner product spce E is convergent we sy tht E is complete inner product spce or tht it is Hilbert spce. It is fundmentl property of the rel numbers (equivlent to the supremum property, tht every non-empty set of rel numbers with n upper bound hs supremum, i.e., lest upper bound) tht R is complete wrt. the usul inner product, i.e., the usul product. It is simple consequence of this tht R k s well s C k re complete for ll k wrt. the usul inner products defined in the next exmple (see Exercise 4.2). Exmple The usul inner product on C k (corresponding to the usul inner product on R k ) is defined by ((x 1,..., x k ), (y 1,..., y k )) = x 1 y x k y k. It is left s n exercise for he reder to show tht the requirements i) iv) re stisfied. 2. On the vector spce C([, b]) of continuous complex functions on the intervl [, b] we define n inner product by (f, g) = b f(x)g(x)dx (4.7) for f, g C([, b]). In generl, for ny positive continuous function ρ : [, b] R + = ]0, + [ we get n inner product (, ) ρ by (f, g) ρ = b f(x)g(x)ρ(x)dx. (4.8)

4 Chp. 4 Opertors on Hilbert Spces Version of We remind the reder tht the integrl of continuous complex function is obtined by integrting the rel n imginry prt, i.e., if f = Ref + iimf then b f(x)dx = b Ref(x)dx + i b Imf(x)dx. (4.9) Then f b f(x)dx defines liner mp from C([, b]) into C, which immeditely implies tht (, ) ρ stisfies iii) nd iv) in Definition 4.1. Property ii) follows from b f(x)dx = b f(x)dx, nd i) is consequence of the fct tht if f is continuous nd f 0 then b f(x)dx = 0 if nd only if f = 0. Note tht if ρ nd ρ denote the mx nd the min of the positive convex function ρ on [, b] such tht 0 < ρ ρ ρ < +, then ρ f f ρ ρ f for f C([, b]), where is the norm given by the inner product (4.7)nd ρ is the norm given by (4.8). 3. Let l 0 (N) denote the set of complex sequences (x n ) n N tht vnish from certin point, i.e., for which only finite number of elements re different from 0. The set l 0 (N) is then subspce of the vector spce of ll complex sequences, which with our previous nottion is the spce F(N, C). On l 0 (N) we define n inner product by ((x n ), (y n )) = x n y n, where the sum on the right hs only finitely mny terms different from 0 (nd thus obviously is convergent). Tht this defines n inner product on l 0 (N) is seen s in the cse of C k. 4.2 Orthogonlity Let now (E, (, )) be n inner product spce. We sy tht two vectors x, y E re orthogonl nd write x y if (x, y) = 0. We generlize this notion nd sy tht vector x E is orthogonl to subset A E, nd write x A, if x is orthogonl to ll vectors in A. The orthogonl complement A of A is defined to be the set of ll vectors orthogonl to A, i.e., A = {x E (x, y) = 0 for ll y A}. (4.10) We note tht by iii) nd iv) in Definition 4.1 A is subspce of E for ny subset A E. For the sme reson A = (spna). (4.11) Here spna is the subspce of E spnned by ll elements in A, i.e., the spce consisting of ll liner combintions of vectors from A.

5 Chp. 4 Opertors on Hilbert Spces Version of A fmily (x i ) i I of vectors from E, where I is ny index set is sid to be n orthogonl fmily if (x i, x j ) = 0, whenever i j, i.e., if the vectors in the set re pirwise orthogonl. If moreover x i = 1 for ll i I we sy tht the fmily is orthonorml. A fmily (x i ) i I of vectors from E is sid to be linerly independent if ny finite subfmily of (x i ) i I is linerly independent. Lemm 4.7. Let (x i ) i I be n orthogonl fmily in E, such tht x i 0 for ll i I. Then (x i ) i I is linerly independent fmily. Proof. Let (x i1,..., x in ) be finite subfmily of (x i ) i I nd ssume tht the sclrs λ 1,..., λ n stisfy λ 1 x i λ n x in = 0. If we for ll j {1,..., n} clculte the inner product with x ij on both sides of this eqution we rrive t λ 1 (x i1, x ij ) λ n (x in, x ij ) = 0. By the ssumption tht (x ik, x ij ) = 0 for k j we rrive t λ j (x ij, x ij ) = 0. Hence λ j = 0 becuse (x ij, x ij ) 0, since x ij 0. We lso hve the following generliztion of the Pythgoren Theorem. Theorem 4.8 (Pythgors). Let (x 1,..., x n ) be finite orthogonl fmily. Then x i 2 = x i 2. Proof. We hve x i 2 = x i, = j=1 (x i, x i ) = x j = (x i, x j ) i,j=1 x i 2, where the third equlity used tht only the digonl terms corresponding to i = j contribute to the sum by the orthogonlity ssumption. The next theorem is lso well known from elementry liner lgebr. Theorem 4.9 (Orthogonl projections nd Bessel s inequlity). Let (e 1,..., e n ) be finite orthonorml fmily in E. For ll vectors x E there is unique vector u spn{e 1,..., e n } such tht x u {e 1,..., e n }. It is given by u = (x, e i )e i. (4.12) nd my be chrcterized s the unique vector in spn{e 1,..., e n } with the shortest distnce to x wrt. the norm. The vector u given in (4.12) is clled the orthogonl projection of x on the subspce spn{e 1,..., e n }.

6 Chp. 4 Opertors on Hilbert Spces Version of Moreover, we hve Bessel s inequlity (x, e i ) 2 x 2 (4.13) for ll x E. Proof. All vectors u spn{e 1,..., e n } cn be written in the form u = λ 1 e λ n e n, where λ 1,..., λ n L. Clculting the inner product with e i on both sides of this eqution we rrive t (u, e i ) = λ i (e i, e i ) = λ i since (e i, e i ) = 1. Hence u = (u, e i )e i (4.14) for u spn{e 1,..., e n }. But x u {e 1,..., e n } is equivlent to (x u, e i ) = 0 for ll i = 1,..., n. Thus (x, e i ) = (u, e i ) for i = 1,..., n. Together with (4.14) this gives the first prt of the theorem. If v spn{e 1,..., e n } then x v = (x u) + (u v) where (x u) (u v). Thus x v 2 = x u 2 + u v 2 x u 2, with equlity if nd only if u = v. Bessel s inequlity follow from Theorem 4.8 if we note tht x = u + (x u) where u (x u), since then x 2 = u 2 + x u 2 u 2 = (x, e i ) 2. In the lst step we hve used tht (x, e i )e i 2 = (x, e i ) 2. From Bessel s inequlity we shll now derive the Cuchy-Schwrz inequlity Theorem 4.10 (Cuchy-Schwrz inequlity). If E is n inner product spce then for ll x, y E we hve (x, y) x y. (4.15) Proof. In fct, if y = 0 then both sides of the inequlity re zero nd if y 0 then 1 y y = 1 nd it follows from (4.13) with n = 1 nd e 1 = 1 y y tht ( ) 1 x, y y x, from which (4.15) follows fter multipliction by y on both sides. From the Cuchy-Schwrz inequlity we derive the tringle inequlity (4.5) x + y 2 = (x + y, x + y) = x 2 + (x, y) + (y, x) + y 2 = x 2 + 2Re(x, y) + y 2 x (x, y) + y 2 x x y + y 2 = ( x + y ) 2.

7 Chp. 4 Opertors on Hilbert Spces Version of Continuity of the inner product In the following we will repetedly mke use of the fct tht the inner product (, ) : E E L is continuous. This is seen s follows. Let x 0, y 0 E be given nd choose x, y E such tht x x 0 δ nd y y 0 δ where δ > 0 is given. Then (x, y) (x 0, y 0 ) = (x, y y 0 ) + (x x 0, y 0 ) (x, y y 0 ) + (x x 0, y 0 ) x y y 0 + x x 0 y 0 δ( x + y 0 ) δ( x 0 + δ + y 0 ). (4.16) Here we hve used the Cuchy-Schwrz inequlity nd tht x = (x x 0 ) + x 0 x x 0 + x 0 x 0 + δ. Since the lst expression in (4.16) tends to 0 s δ 0 we conclude tht for ll ε > 0 there is δ > 0 such tht (x, y) (x 0, y 0 ) < ε if x x 0 δ nd y y 0 δ. This is exctly the continuity of the inner product. The continuity my equivlently be stted s sying tht for ll sequences (x n ) nd (y n ) in E we hve tht (x n, y n ) (x 0, y 0 ) s n, (4.17) if x n x 0 nd y n y 0 s n. Definition 4.11 (Convergence of series). A series x n with terms in normed vector spce E is sid to be convergent with sum x = x n in E if the sequence of prtil sums (s k ) k N defined by converges to x s k. s k = k x n (4.18) In Exercise 4.19 n lterntive definition of Hilbert spce is given using the concept of bsolute convergence of series. From (4.17) nd iii) in Definition 4.1 it follows tht ( ) ( k ) k x n, y = lim x n, y = lim (x n, y) = (x n, y) (4.19) k k nd likewise tht ( y, ) x n = (y, x n ) (4.20) for ny convergent series x n in E nd ll y E. An other consequence of the continuity of the inner product is tht A = (A) for ny subset A E. Here A denotes the closure of the set A, i.e., the set of ll limit points of

8 Chp. 4 Opertors on Hilbert Spces Version of sequences from A. In fct, if x A nd y A there is sequence (y n ) in A such tht y = lim y n. Thus we conclude tht (x, y) = lim(x, y n ) = 0, i.e., x y. Since y A ws rbitrry we see tht A (A). The opposite inclusion follows immeditely from A A. Together with (4.11) this implies tht A = (spna) = (spna). (4.21) We cll spna the closed subspce spnned by A. Likewise it it seen in Exercise 4.5 tht A is closed subspce of E (i.e., A = A for ny subset A E). 4.4 Hilbert spces It is well known from liner lgebr tht ny finite dimensionl inner product spce E hs orthonorml bses. If we let (e 1,..., e n ) denote such bsis. If x E we denote by x = (x 1,..., x n ) L n the coordintes of x wrt. this bsis, i.e., x = x 1 e x n e n, According to Pythgors Theorem 4.8 we hve x = ( x x n 2 ) 1/2. It follows tht the mp (x 1,..., x n ) x 1 e x n e n is liner isometry from L n onto E nd the inverse mp is x ((x, e 1 ),..., (x, e n )). Since the vector spce L n is known to be complete wrt. the usul norm we conclude from tht E is lso complete, i.e., Hilbert spce. For generl Hilbert spces we will see tht the requirement of completeness will ensure tht severl properties of finite dimensionl spces will generlize to infinite dimensionl spces. We will first consider importnt exmples of Hilbert spces. Exmple We hve seen bove tht ny finite dimensionl inner product spce is Hilbert spce. This is in prticulr true for R k nd C k (see Exercise 4.2). 2. The spce l 0 (N) (see Exmple 4.6) is not complete. In fct, let x n l 0 (N) be given by x n = (1, 1 2, 1 3,..., 1 n, 0, 0,... ). Then for m n we see tht x n x m 2 = k=m+1 nd hence the sequence (x n ) n N is Cuchy sequence in l 0 (N) since the series is convergent. The sequence (x n ) n N is however obviously not convergent in l 0 (N). Thus l 0 (N) cnnot be Hilbert spce. 1 k 2 k=1 1 k 2

9 Chp. 4 Opertors on Hilbert Spces Version of If we insted consider the lrger subspce l 2 (N) of F(N, C) consisting of sequences of complex numbers ( n ) n N tht re squre summble, i.e, n 2 < +. Tht l 2 (N) is subspce of F(N, C) follows from the well-known inequlity +b 2 2( 2 + b 2 ) for complex number, b (this inequlity is ctully the Cuchy-Schwrz inequlity for the vectors (1, 1) nd (, b)). For two sequences ( n ) nd (b n ) in l 2 (N) we conclude tht n + b n 2 2 n b n 2 < +. Thus ( n ) + (b n ) l 2 (N). Since we lso hve λ( n ) = (λ n ) l 2 (N) if λ C nd ( n ) l 2 (N) we hve shown tht l 2 (N) is subspce of F(N, C). Using the inequlity b 1 2 ( 2 + b 2 ) for, b C we see tht (( n ), (b n )) = n b n defines mp (, ) : l 2 (N) l 2 (N) C. In fct, the series bove is bsolutely convergent (see the definition of bsolute convergence in Exercise 4.19). It is now cler tht the bove mp defines n inner product on l 2 (N). We will now show tht l 2 (N) with this inner product is Hilbert spce. Thus ssume tht (x n ) n N is Cuchy sequence in l 2 (N). We write x n = ( n 1, n 2,... ). For every k N we hve tht n k m k xn x m. Thus the sequence ( n k ) n N is Cuchy sequence in C for ech k N. Since C is complete we conclude tht this sequence converges, i.e., there is k such tht k = lim n n k for ll k N. Let x = ( 1, 2,... ). We will now show tht x l 2 (N) nd tht x n x s n. Given ε > 0 we cn find n N N, such tht K n k m k 2 k=1 n k m k 2 = x n x m 2 ε 2 k=1 for ll n, m N nd ll K N. For m we see from this tht K n k k 2 ε 2 for n N nd ll K N. For K we get x n x 2 = n k k 2 ε 2 k=1 for n N. This shows tht x N x l 2 (N) nd hence tht x = x N (x N x) l 2 (N) nd lso tht x n x s n s ws climed. k=1

10 Chp. 4 Opertors on Hilbert Spces Version of The spce C([, b]) with inner product given by (4.8) is not complete s we shll now discuss. The norm on this spce is given by f 2 ρ = b f(x) 2 ρ(x)dx. Let f n denote the function on [0, 2] which is equl to 0 on [0, 1], grows linerly from 0 to 1 on [1, n ] nd is equl to 1 on [1 + 1 n, 2] (drw the grph!), It is esy to see tht (f n ) is Cuchy sequence in C([0, 2]) w.r.t. the norm ρ, but tht it is not convergent in C([0, 2]). As in the discussion of l 0 bove we cn extend C([, b]) to Hilbert spce L 2 ([, b]) consisting of squre integrble functions on [, b], i.e., functions f : [, b] C, such tht b f(x) 2 dx <. We here refer to generliztion of the Riemnn integrl, clled the Lebesgue integrl. We lso generlize wht is ment by functions or rther wht is ment by two functions being equl. The detils of this is beyond the scope of these notes. We shll only need the following three fcts: i) Two functions f nd g in L 2 ([, b]) re considered to be equl if b f(x) g(x) 2 dx = 0 i.e., if f g = 0, nd we sy tht f is equl to g lmost everywhere on [, b], or tht f(x) = g(x) for lmost ll x [, b]. This is, in prticulr, true for ny two functions tht differ t only finitely mny points of the intervl. This strictly speking mens tht L 2 ([, b]) is not spce of functions, but of clsses of functions tht gree lmost everywhere. The proof tht L 2 ([, b]) is vector spce with inner product given by (4.7) cn now be completed s for l 2 (N) bove. Tht L 2 ([, b]) is Hilbert spce is one of the most fundmentl results in mesure theory known s the Riesz-Fischer Theorem. ii) L 2 ([, b]) is miniml extension of C([, b]) in the sense tht the closure (C([, b]) in L 2 ([, b]) is equl to ll of L 2 ([, b]). In other words to every f L 2 ([, b]) there is sequence (f n ) n N in C([, b]), such tht f n f 0 s n. iii) In contrst to the Riemnn integrl the Lebesgue integrl is defined on n rbitrry intervl I R nd for continuous positive functions it grees with the improper Riemnn integrl. We my thus define the Hilbert spce L 2 (I) for ll intervls I, in prticulr, we my consider L 2 (R). For this spce we hve insted of (ii) bove tht the closure of the subspce C 0 (R), consisting of ll continuous functions vnishing outside bounded intervl, is ll of L 2 (R). In Section 5.5 we will need tht the sme holds for the subspce C0 (R) consisting of C -functions vnishing outside bounded intervl. For more detils on mesure theory nd Lebesgue integrtion the interested reder my consult the book M.Reed nd B. Simon: Methods of modern mthemticl physics, Vol I, Acdemic press 1972.

11 Chp. 4 Opertors on Hilbert Spces Version of In the following we will let H denote Hilbert spce with inner product (, ). Any subspce X of H is n inner product spce with the inner product defined s the restriction of (, ) to X X. Then X H is Hilbert spce if nd only if X is closed subspce of H, i.e., if X = X (see Exercise 4.6). In prticulr, ll finite-dimensionl subspces of H re closed. We will need the following extension of Pythgors Theorem. Theorem 4.13 (Infinite Pythgoren Theorem). Let (x i ) i N be n orthogonl fmily in Hilbert spce H. Then x i is convergent in H if nd only if nd x i 2 < +, x i 2 = x i 2. (4.22) Proof. Since H is Hilbert spce x i is convergent in H if nd only if the sequence of prtil sums (s n ) is Cuchy sequence. This mens tht for ll ε > 0 there is n N N such tht s n s m 2 = x i 2 = x i 2 ε 2 (4.23) i=m+1 i=m+1 for ll n > m N where we hve used Theorem 4.8. Since R is lso complete we hve on the other hnd tht x i 2 is convergent if nd only if the sequence (r n ) given by r n = x i 2 is Cuchy sequence in R. This mens tht for ll ε > 0 there is n N N such tht r n r m = x i 2 < ε 2 (4.24) i=m+1 for ll n > m N. The first clim in the theorem follows by compring (4.23) nd (4.24). Finlly, we see from the continuity of x x nd Theorem 4.8 tht x i 2 = lim x i 2 = lim x i 2 = lim x i 2 = x i 2, n n n which implies (4.22). We wnt to extend the notion of orthonorml bsis to the infinite dimensionl cse. We my chrcterize n orthonorml bsis in finite dimensionl inner product spce H s n orthonorml fmily tht spns H. In the infinite dimensionl cse we will only require tht n orthonorml bsis spns dense subspce, i.e., set whose closure is the whole spce. Since s we just sw ny finite dimensionl subspce in n inner product spce is closed the following notion of orthonorml bsis spce is, indeed, n extention of the finite dimensionl notion.

12 Chp. 4 Opertors on Hilbert Spces Version of Definition 4.14 (Orthonorml bsis). An orthonorml bsis for A Hilbert spce H is n orthonorml fmily (e i ) i I in H such tht spn{e i i I} = H. We emphsize tht this notion of orthonorml bsis is different from the purely lgebric notion of bsis, where it is required tht the bsis spns the whole set. Since H = {0} it follows tht from (4.21)tht {e i i I} = {0}. for ny orthonorml bsis (e i ) i I for H. This mens ny orthonorml bsis is mximl orthonorml fmily in H, i.e., tht there is no vector e H with e = 1, such tht e together with (e i ) i I is n orthonorml fmily. The converse tht ny mximl orthonorml fmily in H is n orthonorml bsis will be shown in Theorem 4.16 below. Note tht the definition of orthonorml bsis implies tht ny orthonorml fmily (e i ) i I is n orthonorml bsis for spn{e i i I}, i.e., for the closure of the subspce spnned by the fmily. Any Hilbert spce hs n orthonorml bsis. This clim however relies on the wht is known s the xiom of choice, which we will not discuss here. In the following we shll restrict ttention to seprble Hilbert spces, i.e., Hilbert spces tht hve n orthonorml bsis which is either finite or countble, i.e., cn be indexed by N. We know from elementry liner lgebr tht in finite dimensionl vector spces ll bses hve the sme number of elements equl to the dimension of the spce. In Exercise 4.21 we show tht in n infinite dimensionl seprble Hilbert spce ll orthonorml bses re infinite nd hve countbly mny elements. Exmple In the spce l 2 (N) discussed in Exmple 4.12 we cn get n orthonorml bsis (e i ) i N by letting e i be the sequence with ll elements equl to 0 except element number i which we choose to be 1, i.e., { 1 for j = i (e i ) j = 0 for j i. Tht this is n orthonorml bsis follows by observing tht it is obviously n orthonorml fmily nd tht spn{e i i N} = l 0 (N) whose closure is l 2 (N) (see Exercise 4.15) We cll (e i ) i N the nturl orthonorml bsis for l 2 (N). In Section 4.6 we shll determine n orthonorml bsis for the spce L 2 ([ L, L]). It will in this cse turn out to be more convenient to use ll integers Z s n index ser for the bsis. Of course if one cn use Z one could lso hve used N (why?). In Exercise 4.18 n lterntive construction of n orthonorml bsis for L 2 ([0, 1]) is given 4.5 Orthonorml expnsions Our gol in this section is to generlize the expnsion of vectors in orthonorml bses known in finite dimensionl Hilbert spces to the cse of infinite dimensionl seprble Hilbert spces. Thus we will in this section ssume tht H is n infinite dimensionl seprble Hilbert spce. All results in this section generlize to non-seprble spces but we shll not discuss this here.

13 Chp. 4 Opertors on Hilbert Spces Version of Let (e i ) i N be n orthonorml fmily in H (ssumed to be infinite dimensionl) nd consider vector x H. From Bessel s inequlity (4.13) we hve tht n (x, e i ) 2 x 2 for ll n N. This implies the generlized Bessel inequlity (x, e i ) 2 x 2. (4.25) By Theorem 4.13 together with (4.25) we therefore conclude tht the series (x, e i )e i is convergent in H. If we write u = (x, e i )e i (4.26) it is cler tht u spn{e i i N}. From Definition 4.11 for convergence of series the vector u could in principle depend on the order in which we perform the sum in (), i.e., on the order we chose for the bsis e 1, e 2, e 3,.... Tht this is not the cse follows from the next theorem. Theorem 4.16 (Projections in infinite dimensionl spces). Let (e i ) i N be n orthonorml fmily in the seprble Hilbert spce H. To ny vector x H there is unique vector u spn{e i i N} such tht x u {e i i N}, nd u is given by (4.5). In prticulr, u does not depend on the order of the summtion nd we lso write u = i N(x, e i )e i. Among ll vectors in spn{e i i N} the vector u hs the shortest distnce to x. Proof. The vector u given by (4.5) stisfies tht for ech j N ( ) (u, e j ) = (x, e i )e i, e j = (x, e i )(e i, e j ) = (x, e j ). Thus, (x u, e j ) = 0 for ll j N nd hence x u {e j j N}. This proves the existence of u we show the uniqueness below. If v spn{e i i N} then u v spn{e i i N} nd thus Pythgors Theorem gives x v 2 = x u 2 + u v 2 x u 2. This shows the lst clim of the theorem. If v moreover, ssumes tht x v {e i i N} then u v = (x v) (x u) {e i i N} = (spn{e i i N}). In prticulr, (u v, u v) = 0, i.e., u = v showing tht u is unique. We now hve the following importnt result on orthonorml bses.

14 Chp. 4 Opertors on Hilbert Spces Version of Theorem 4.17 (Properties of Orthonorml Bsis). For n orthonorml fmily (e i ) i N in seprble Hilbert spce H the following sttements re equivlent (i) (e i ) i N is n orthonorml bsis in H. (ii) {e i i N} = {0}. (iii) The orthonorml expnsion x = i N(x, e i )e i holds for ll x H. (iv) Prsevl s identity holds for ll x H. x 2 = (x, e i ) 2 (4.27) Proof. We hve previously seen tht (i) (ii). The impliction (ii) (iii) follows from Theorem 4.16 since with the nottion used there we hve x = u. Tht (iii) (iv) follows immeditely from (4.22). If we finlly ssume tht (iv) holds we find for ll x H x (x, e i )e i 2 = x 2 (x, e i ) 2 0 for n. Thus spn{e i i I} is dense in H nd hence (iv) (i). As further consequence of Theorem 4.16 we obtin the projection theorem. Theorem 4.18 (Projection Theorem). Let X be closed subspce of seprble Hilbert spce H. Then for ech x H there exist unique vectors u X nd v X such tht x = u + v. (4.28) Proof. Since X is closed subspce of the seprble Hilbert spce H then X is seprble Hilbert spce (see Exercise 4.17). We my thus choose nd orthonorml bsis (e i ) i I for X, where I is finite or equl to N. Hence X = spn{e i i I}. The clim is now n immedite consequence of Theorems 4.9 nd As we hve seen u in (4.28) my lso be chrcterized s the vector in X with the shortest distnce to x. In generl if two subspces V nd W of vector spce E stisfy tht ny vector x E cn be uniquely written s x = v + w, where v V nd w W we sy tht V nd W re complementry nd write E = V W. We lso express this s sying tht E is the direct sum of V nd W. We my thus rephrse Theorem 4.18 s H = X X

15 Chp. 4 Opertors on Hilbert Spces Version of for ny closed subspce X of H. The vector u in (4.28) is clled the orthogonle projection of x onto X nd s we hve seen it cn be clculted s u = i I (x, e i )e i, where (e i ) i I is n orthonorml bsis for X. Since X is lso closed subspce of H we hve H = X X. If we use tht ny vector x X my be written uniquely s x = u + v, with u X nd v X nd the obvious inclusion X X (why?) we see tht, in fct, u = x nd v = 0. Thus for every closed subspce X of H. projection of x onto X. X = X It then follows tht v in (4.28) is the orthogonl Proposition Let X be closed subspce in seprble Hilbert spce H. Assume we hve orthonorml bses (e i ) i I for X nd (e j ) j J for X. Then the combined fmily (e i ) i I J is n orthonorml bsis for H Proof. Exercise Fourier series The celebrted theory of Fourier series which hs its origin in J. Fourier s 1807 nlysis of the het eqution my be conveniently formulted in terms of orthonorml expnsions. We shll now discuss this very briefly. The relevnt Hilbert spce is H = L 2 ([ L, L]) with inner product (f, g) = 1 2L L L f(θ)g(θ) dθ. 1 We hve included the fctor 2L for convenience. The key result bout Fourier series is stted in the next theorem. We will give proof bsed on the Stone-Weierstrss Theorem (see, e.g., W. Rudin: Functionl nlysis, Chpter 5). Theorem 4.20 (Fourier bsis). Let the functions e n C([ L, L]) be given by e n (θ) = e in π L θ, θ [ L, L], n Z. Then (e n ) n Z is n orthonorml bsis in H = L 2 ([ L, L]). Proof. Tht (e n ) n Z is n orthonorml fmily is seen from the clcultion (e n, e m ) = 1 2L L L L e n (x)e m (x)dx = 1 e i(n m) π L θ dθ 2L. L [ 1 L = 2L i(n m)π ei(n m) π θ] L L = 0 for n m L 1 2L [x]l L = 1 for n = m,

16 Chp. 4 Opertors on Hilbert Spces Version of Here we hve used tht e n is periodic function of θ R with period 2L. It remins to prove tht spn{e n n Z} = H. To conclude this we note tht it is enough to show tht C([ L, L]) spn{e n n Z} since C([ L, L]) = H s lredy remrked. It is thus enough to show tht for ll functions f C( L, L) nd ll ε > 0 there is n f 1 spn{e n n Z} such tht f f 1 < ε. (4.29) To see this we first choose function (why is this possible?) f 2 C([ L, L]) such tht f 2 ( L) = f 2 (L) = 0 nd f f 2 < ε 2. (4.30) Then f 2 my be extended to continuous periodic function on R with period 2L. It the follows from the Stone-Weierstrss Theorem tht there is function f 1 spn{e n n Z}, such tht f 2 (θ) f 1 (θ) < ε, θ [ L, L]. 4L Hence f 2 f 1 < ε 2. (4.31) From (4.30), (4.31), nd the tringle inequlity we rrive t (4.29). If for f L 2 ([ L.L]) we define the Fourier coefficients c n (f) by the bove theorem nd Theorem 4.17 imply tht c n (f) = 1 L f(θ)e in π L θ dθ, (4.32) 2L L f(θ) = n Z c n (f)e in π L θ. (4.33) This series is clled the Fourier series for f. We emphsize tht it converges in sense of the Hilbert spce L 2 ([ L, L]), i.e., with respect to the L 2 -norm. More precisely, this mens tht N c n (f)e n f 0 for N. n= N It is importnt to relize tht this is not the sme s uniform or pointwise convergence. For discussion on uniform nd pointwise convergence of Fourier series we refer the reder to more dvnced texts on the subject. Exercises Opgve 4.1. Proof Proposition 4.4. Hint: Use the tringle inequlity (4.5).

17 Chp. 4 Opertors on Hilbert Spces Version of Opgve 4.2. () Show tht sequence (x n ) n N in R k (or C k ) is convergent or Cuchy sequence wrt. the stndrd norm if nd only if ech coordinte sequence (x i n) n N, i = 1,..., k, is convergent or Cuchy sequence respectively (in R or C). nottion x n = (x 1 n,..., x k n). We hve used the (b) Use () to conclude tht R k nd C k re complete, i.e., Hilbert spces, for ll k 2. The completeness of R is ssumed known. Opgve 4.3. Consider the finite dimensionl complex Hilbert spce H = C k, with the inner product from Exmple 4.6 (1). Show tht ((1, 0,..., 0), (0, 1, 0..., 0),..., (0,..., 0, 1)) is n orthonorml bsis for H. Opgve 4.4. Show tht n inner product on complex Hilbert spce Hilbert stisfies the polriztion identity (x, y) = 1 4 ( x + y 2 x y 2 + i x + iy 2 i x iy 2 ). Opgve 4.5. Let H be Hilbert spce. Show tht for ny subset A H the subset A is closed subspce of H. Opgve 4.6. Let H be Hilbert spce. Show tht subspce X of H is Hilbert spce if nd only if X is closed. Show lso tht the closure of subspce of H is itself subspce. (here you should use the continuity of ddition nd of sclr multipliction). Opgve 4.7. Show tht (sin nθ) n N is n orthogonl fmily in C([0, π]) with the inner product given in (4.7). Opgve 4.8. Determine 1, 2, 3 C such tht is s smll s possible. π 0 cos θ 3 n sin nθ 2 dθ Opgve 4.9. Let the polynomils p 0 (x), p 1 (x),... be given such tht p n (x) is polynomil of degree n in the vrible x with the coefficient of x n being 1 nd (p 0 (x), p 1 (x),...) being n orthonorml fmily in L 2 ([0, 1]). Find p 0 (x), p 1 (x) nd p 2 (x). Opgve Show tht (sin(n 1 2 )θ) n N is n orthogonl fmily in C([0, π]) with inner product given by (4.7). Opgve Let H be n infinite dimensionl seprble Hilbert spce nd let (e n ) n N be n orthonorml fmily in H. (1) Show tht the series 1 n e n is convergent in H nd determine for which α R the series nα e n is convergent in H.

18 Chp. 4 Opertors on Hilbert Spces Version of (2) Determine the orthogonl projection of the vectors e 1 ± 2e 2 on the subspce spnned by the vector n 1 e n. You my use tht 1 = π2 n 2 6. π Opgve Show tht lim n 0 log θ sin nθdθ = 0. Hint: This cn be seen s corollry to Bessel s inequlity (4.25) used on the the orthogonl fmily in Exercise 4.7. Opgve Let (e i ) i N be n orthonorml bsis for the Hilbert spce H. Show tht the following generliztion of Prsevl s eqution holds for ll x, y H: (x, y) = (x, e i )(y, e i ). Opgve Consider the inner product spce l 0 (N) defined in Exmple 4.6 (3). Let { } 1 X = (x n ) n N l 0 (N) x n n = 0. Show tht X is closed subspce of l 0 (N) nd tht X X l 0 (N). Opgve Argue tht the closure of l 0 (N) is equl to l 2 (N), i.e., tht ny sequence x in l 2 (N) is the limit of convergent sequence (x n ) n N in l 0 (N). Hint: Let x n be the sequence which is equl to x up to the n-th term nd zero fterwrds. Opgve Let H be complex Hilbert spce nd let n N, C be such tht n = 1 nd 2 1. Show the generlized polriztion identity (x, y) = 1 n 1 ν x + ν y 2. n ν=0 Opgve () Show tht the Hilbert spce H is seprble if nd only if it hs countble dense subset ( countble set whose closure is the whole spce). Hint to the if-prt: Construct n orthonorml bsis using the Grm-Schmidt procedure. (b) Let H be seprble Hilbert spce. Show tht ny subset M of H hs n t most countble dense subset (n t most countble subset of M whose closure is M) Hint: Assume tht M is not finite. Let {x i i N} be countble dense subset of H. For i, n N let x i,n denote ny rbitrrily chosen point in {x M x x i 1 n } ssuming tht this set is non-empty. Otherwise let x i,n be n rbitrrily chosen point in M. Now show tht {x i,n i, n N} is countble dense subset of M. (c) Show tht closed subspce X of seprble Hilbert spce is itself seprble Hilbert spce. Hint: Use the chrcteriztion of seprble in question () nd the result of (b). Opgve (Difficult) Consider the Hilbert spce H = L 2 ([0, 1]) with the inner product (f, g) = f(θ)g(θ)dθ. Let F 0 : R R be given by F 0 (t) = 1 if 0 t < 1 nd F (t) = 1 if 1 t < 2. Define f n, n = 0, 1, 2,... by F n (t) = 2 n 1 F 0 (2 n t).

19 Chp. 4 Opertors on Hilbert Spces Version of Let f n,k H, n = 0, 1, 2,..., k = 0, 1,..., 2 n 1 1 be given by f n,k (θ) = f n ( θ k2 n 1 ), θ [0.1] () Show tht the fmily (f n,k n = 0, 1, 2,..., k = 0, 1,..., 2 n 1 1) is countble orthonorml fmily. (b) Show tht the subspce spn{f n,k n = 0, 1, 2,..., N, k = 0, 1,..., 2 n 1 1} consists of ll functions tht re constnt on the intervls [k2 N, (k + 1)2 N ), k = 0, 1,..., 2 N 1. (c) Show tht 1 C 1 ([, b]) spn{f n,k n = 0, 1, 2,..., k = 0, 1,..., 2 n 1 1}. Hint: Use the Men Vlue Theorem. (d) Conclude from the informtion L 2 ([0, 1]) = C 1 ([, b]) tht (f n,k n = 0, 1, 2,..., k = 0, 1,..., 2 n 1 1) is n orthonorml bsis for L 2 ([0, 1]). Opgve Show tht n inner product spce E is Hilbert spce if nd only if ny series x n with terms in E stisfying x n < is convergent. On Hilbert spce series x n stisfying x n < is sid to be bsolutely convergent. Opgve Consider the Hilbert spce H = L 2 ([ 1, 1]) with the inner product (f, g) = 1 1 f(t)g(t)dt nd the elements f 1, f 2, g H given by f 1 (t) = 3/2 t, f 2 (t) = 5/2 t 2, nd g(t) = 1. Show tht f 1, f 2 re orthonorml nd find the closest vector to g in spn{f 1, f 2 }. Opgve Show tht if H is n infinite dimensionl seprble Hilbert spce then ll orthonorml bses re infinite nd hve countbly mny elements. Hint: From the chrcteriztion of seprbility in Exercise 4.17 you know tht H hs countble dense set D. Given n orthonorml bsis show tht the blls centered t bsis vectors with rdius less thn 1 re disjoint. Argue tht these blls hve non-empty intersection with D nd then tht the orthonorml bsis is is t most countble. Opgve Prove Proposition It is lso possible to show tht C([, b]) spn{f n,k n = 0, 1, 2,..., k = 0, 1,..., 2 n 1 1} but this is most esily done by fist showing tht continuous functions on [, b] re uniformly continuous. Uniform continuity mens tht for ny ε > 0 there is δ > 0, such tht x y < δ f(x) f(y) < ε, for ll x, y [0, 1].

The Regulated and Riemann Integrals

Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

More information

Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n

More information

Theoretical foundations of Gaussian quadrature

Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of

More information

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further

More information

MATH 174A: PROBLEM SET 5. Suggested Solution

MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous rel-vlued function on I), nd let L 1 (I) denote the completion

More information

Abstract inner product spaces

WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the

More information

Inner-product spaces

Inner-product spces Definition: Let V be rel or complex liner spce over F (here R or C). An inner product is n opertion between two elements of V which results in sclr. It is denoted by u, v nd stisfies:

More information

Chapter 6. Infinite series

Chpter 6 Infinite series We briefly review this chpter in order to study series of functions in chpter 7. We cover from the beginning to Theorem 6.7 in the text excluding Theorem 6.6 nd Rbbe s test (Theorem

More information

g i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f

1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where

More information

Fourier series. Preliminary material on inner products. Suppose V is vector space over C and (, )

Fourier series. Preliminry mteril on inner products. Suppose V is vector spce over C nd (, ) is Hermitin inner product on V. This mens, by definition, tht (, ) : V V C nd tht the following four conditions

More information

2 Fundamentals of Functional Analysis

Fchgruppe Angewndte Anlysis und Numerik Dr. Mrtin Gutting 22. October 2015 2 Fundmentls of Functionl Anlysis This short introduction to the bsics of functionl nlysis shll give n overview of the results

More information

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence

More information

Math Solutions to homework 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

More information

Review of Riemann Integral

1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

More information

Best Approximation in the 2-norm

Jim Lmbers MAT 77 Fll Semester 1-11 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the -norm Suppose tht we wish to obtin function f n (x) tht is liner combintion

More information

Best Approximation. Chapter The General Case

Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

More information

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

More information

STUDY GUIDE FOR BASIC EXAM

STUDY GUIDE FOR BASIC EXAM BRYON ARAGAM This is prtil list of theorems tht frequently show up on the bsic exm. In mny cses, you my be sked to directly prove one of these theorems or these vrints. There

More information

Lecture 3. Limits of Functions and Continuity

Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

More information

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,

More information

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)

More information

SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set

SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σ-finite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such

More information

Analytical Methods Exam: Preparatory Exercises

Anlyticl Methods Exm: Preprtory Exercises Question. Wht does it men tht (X, F, µ) is mesure spce? Show tht µ is monotone, tht is: if E F re mesurble sets then µ(e) µ(f). Question. Discuss if ech of the

More information

Lecture 19: Continuous Least Squares Approximation

Lecture 19: Continuous Lest Squres Approximtion 33 Continuous lest squres pproximtion We begn 31 with the problem of pproximting some f C[, b] with polynomil p P n t the discrete points x, x 1,, x m for

More information

NOTES AND PROBLEMS: INTEGRATION THEORY

NOTES AND PROBLEMS: INTEGRATION THEORY SAMEER CHAVAN Abstrct. These re the lecture notes prepred for prticipnts of AFS-I to be conducted t Kumun University, Almor from 1st to 27th December, 2014. Contents

More information

DEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1.

398 CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 11.1 ORTHOGONAL FUNCTIONS REVIEW MATERIAL The notions of generlized vectors nd vector spces cn e found in ny liner lger text. INTRODUCTION The concepts

More information

Notes on length and conformal metrics

Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued

More information

MAA 4212 Improper Integrals

Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

More information

A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int

A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure

More information

FUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (

FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions

More information

Presentation Problems 5

Presenttion Problems 5 21-355 A For these problems, ssume ll sets re subsets of R unless otherwise specified. 1. Let P nd Q be prtitions of [, b] such tht P Q. Then U(f, P ) U(f, Q) nd L(f, P ) L(f, Q).

More information

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner

More information

Functional Analysis I Solutions to Exercises. James C. Robinson

Functionl Anlysis I Solutions to Exercises Jmes C. Robinson Contents 1 Exmples I pge 1 2 Exmples II 5 3 Exmples III 9 4 Exmples IV 15 iii 1 Exmples I 1. Suppose tht v α j e j nd v m β k f k. with α j,

More information

Sturm-Liouville Eigenvalue problem: Let p(x) > 0, q(x) 0, r(x) 0 in I = (a, b). Here we assume b > a. Let X C 2 1

Ch.4. INTEGRAL EQUATIONS AND GREEN S FUNCTIONS Ronld B Guenther nd John W Lee, Prtil Differentil Equtions of Mthemticl Physics nd Integrl Equtions. Hildebrnd, Methods of Applied Mthemtics, second edition

More information

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

More information

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

More information

Rudin s Principles of Mathematical Analysis: Solutions to Selected Exercises. Sam Blinstein UCLA Department of Mathematics

Rudin s Principles of Mthemticl Anlysis: Solutions to Selected Exercises Sm Blinstein UCLA Deprtment of Mthemtics Mrch 29, 2008 Contents Chpter : The Rel nd Complex Number Systems 2 Chpter 2: Bsic Topology

More information

8 Laplace s Method and Local Limit Theorems

8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved

More information

Math 61CM - Solutions to homework 9

Mth 61CM - Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ

More information

ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2.

ACM 05: Applied Rel nd Functionl Anlysis. Solutions to Homework # 2. Andy Greenberg, Alexei Novikov Problem. Riemnn-Lebesgue Theorem. Theorem (G.F.B. Riemnn, H.L. Lebesgue). If f is n integrble function

More information

Prof. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf

Knpp, Chpter, Section, # 4, p. 78 Homework For ny two sequences { n } nd {b n} in R, prove tht lim sup ( n + b n ) lim sup n + lim sup b n, () provided the two terms on the right side re not + nd in some

More information

Math 270A: Numerical Linear Algebra

Mth 70A: Numericl Liner Algebr Instructor: Michel Holst Fll Qurter 014 Homework Assignment #3 Due Give to TA t lest few dys before finl if you wnt feedbck. Exercise 3.1. (The Bsic Liner Method for Liner

More information

Definite integral. Mathematics FRDIS MENDELU

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the

More information

Orthogonal Polynomials and Least-Squares Approximations to Functions

Chpter Orthogonl Polynomils nd Lest-Squres Approximtions to Functions **4/5/3 ET. Discrete Lest-Squres Approximtions Given set of dt points (x,y ), (x,y ),..., (x m,y m ), norml nd useful prctice in mny

More information

A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction

Ttr Mt. Mth. Publ. 44 (29), 159 168 DOI: 1.2478/v1127-9-56-z t m Mthemticl Publictions A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES Miloslv Duchoň Peter Mličký ABSTRACT. We present Helly

More information

1. On some properties of definite integrals. We prove

This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.

More information

arxiv:math/ v2 [math.ho] 16 Dec 2003

rxiv:mth/0312293v2 [mth.ho] 16 Dec 2003 Clssicl Lebesgue Integrtion Theorems for the Riemnn Integrl Josh Isrlowitz 244 Ridge Rd. Rutherford, NJ 07070 jbi2@njit.edu Februry 1, 2008 Abstrct In this pper,

More information

ODE: Existence and Uniqueness of a Solution

Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =

More information

Properties of the Riemann Integral

Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2

More information

MA Handout 2: Notation and Background Concepts from Analysis

MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,

More information

Definite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30

Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion - re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function

More information

Problem Set 4: Solutions Math 201A: Fall 2016

Problem Set 4: s Mth 20A: Fll 206 Problem. Let f : X Y be one-to-one, onto mp between metric spces X, Y. () If f is continuous nd X is compct, prove tht f is homeomorphism. Does this result remin true

More information

Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.

Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points

More information

NOTES ON HILBERT SPACE

NOTES ON HILBERT SPACE 1 DEFINITION: by Prof C-I Tn Deprtment of Physics Brown University A Hilbert spce is n inner product spce which, s metric spce, is complete We will not present n exhustive mthemticl

More information

440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam

440-2 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP

More information

1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers...

Contents 1 Sets 1 1.1 Functions nd Reltions....................... 3 1.2 Mthemticl Induction....................... 5 1.3 Equivlence of Sets nd Countbility................ 6 1.4 The Rel Numbers..........................

More information

Review of basic calculus

Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

More information

Chapter 28. Fourier Series An Eigenvalue Problem.

Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why

More information

Convex Sets and Functions

B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line

More information

Entrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim

1. Let n be positive integers. ntrnce xm, Rel Anlysis September 1, 29 Solve exctly 6 out of the 8 problems. Sketch the grph of the function f(x): f(x) = lim e x2n. Compute the following nd justify your

More information

ON THE C-INTEGRAL BENEDETTO BONGIORNO

ON THE C-INTEGRAL BENEDETTO BONGIORNO Let F : [, b] R be differentible function nd let f be its derivtive. The problem of recovering F from f is clled problem of primitives. In 1912, the problem of primitives

More information

Math 1B, lecture 4: Error bounds for numerical methods

Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

More information

II. Integration and Cauchy s Theorem

MTH6111 Complex Anlysis 2009-10 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.

More information

The Henstock-Kurzweil integral

fculteit Wiskunde en Ntuurwetenschppen The Henstock-Kurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft

More information

MAT 215: Analysis in a single variable Course notes, Fall Michael Damron

MAT 215: Anlysis in single vrible Course notes, Fll 2012 Michel Dmron Compiled from lectures nd exercises designed with Mrk McConnell following Principles of Mthemticl Anlysis, Rudin Princeton University

More information

The Banach algebra of functions of bounded variation and the pointwise Helly selection theorem

The Bnch lgebr of functions of bounded vrition nd the pointwise Helly selection theorem Jordn Bell jordn.bell@gmil.com Deprtment of Mthemtics, University of Toronto Jnury, 015 1 BV [, b] Let < b. For f

More information

Lecture Notes on Functional Analysis. Kai-Seng Chou Department of Mathematics The Chinese University of Hong Kong Hong Kong

Lecture Notes on Functionl Anlysis Ki-Seng Chou Deprtment of Mthemtics The Chinese University of Hong Kong Hong Kong My 29, 2014 2 Contents 1 Normed Spce: Exmples 5 1.1 Vector Spces of Functions...................................

More information

1.9 C 2 inner variations

46 CHAPTER 1. INDIRECT METHODS 1.9 C 2 inner vritions So fr, we hve restricted ttention to liner vritions. These re vritions of the form vx; ǫ = ux + ǫφx where φ is in some liner perturbtion clss P, for

More information

Chapter 0. What is the Lebesgue integral about?

Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous

More information

Appendix to Notes 8 (a)

Appendix to Notes 8 () 13 Comprison of the Riemnn nd Lebesgue integrls. Recll Let f : [, b] R be bounded. Let D be prtition of [, b] such tht Let D = { = x 0 < x 1

More information

A Convergence Theorem for the Improper Riemann Integral of Banach Space-valued Functions

Interntionl Journl of Mthemticl Anlysis Vol. 8, 2014, no. 50, 2451-2460 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/10.12988/ijm.2014.49294 A Convergence Theorem for the Improper Riemnn Integrl of Bnch

More information

Riemann is the Mann! (But Lebesgue may besgue to differ.)

Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >

More information

Discrete Least-squares Approximations

Discrete Lest-squres Approximtions Given set of dt points (x, y ), (x, y ),, (x m, y m ), norml nd useful prctice in mny pplictions in sttistics, engineering nd other pplied sciences is to construct curve

More information

Series of functions. Chapter lim sup and lim inf

Chpter 4 Series of functions In this chpter we shll see how the theory in the previous chpters cn be used to study functions. We shll be prticulrly interested in how generl functions cn be written s sums

More information

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.

Chpter 6 Integrtion In this chpter we define the integrl. Intuitively, it should be the re under curve. Not surprisingly, fter mny exmples, counter exmples, exceptions, generliztions, the concept of the

More information

11 An introduction to Riemann Integration

11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in

More information

INTRODUCTION TO INTEGRATION

INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide

More information

CMDA 4604: Intermediate Topics in Mathematical Modeling Lecture 19: Interpolation and Quadrature

CMDA 4604: Intermedite Topics in Mthemticl Modeling Lecture 19: Interpoltion nd Qudrture In this lecture we mke brief diversion into the res of interpoltion nd qudrture. Given function f C[, b], we sy

More information

1 The Lagrange interpolation formula

Notes on Qudrture 1 The Lgrnge interpoltion formul We briefly recll the Lgrnge interpoltion formul. The strting point is collection of N + 1 rel points (x 0, y 0 ), (x 1, y 1 ),..., (x N, y N ), with x

More information

Elementary Linear Algebra

Elementry Liner Algebr Anton & Rorres, 1 th Edition Lecture Set 5 Chpter 4: Prt II Generl Vector Spces 163 คณ ตศาสตร ว ศวกรรม 3 สาขาว ชาว ศวกรรมคอมพ วเตอร ป การศ กษา 1/2555 163 คณตศาสตรวศวกรรม 3 สาขาวชาวศวกรรมคอมพวเตอร

More information

arxiv: v1 [math.ca] 7 Mar 2012

rxiv:1203.1462v1 [mth.ca] 7 Mr 2012 A simple proof of the Fundmentl Theorem of Clculus for the Lebesgue integrl Mrch, 2012 Rodrigo López Pouso Deprtmento de Análise Mtemátic Fcultde de Mtemátics, Universidde

More information

1 The Riemann Integral

The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

More information

1 A quick look at topological and functional spaces

1 A quick look t topologicl nd functionl spces The unified chrcter of mthemtics lies in its very nture; indeed, mthemtics is the foundtion of ll exct nturl sciences. Dvid Hilbert (1862-1943) Nowdys, functionl

More information

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

More information

Chapter 3. Inner Products and Norms

Chpter 3 Inner Products nd Norms The geometry of Eucliden spce relies on the fmilir properties of length nd ngle. The bstrct concept of norm on vector spce formlizes the geometricl notion of the length

More information

A PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES

INROADS Rel Anlysis Exchnge Vol. 26(1), 2000/2001, pp. 381 390 Constntin Volintiru, Deprtment of Mthemtics, University of Buchrest, Buchrest, Romni. e-mil: cosv@mt.cs.unibuc.ro A PROOF OF THE FUNDAMENTAL

More information

AMATH 731: Applied Functional Analysis Fall Additional notes on Fréchet derivatives

AMATH 731: Applied Functionl Anlysis Fll 214 Additionl notes on Fréchet derivtives (To ccompny Section 3.1 of the AMATH 731 Course Notes) Let X,Y be normed liner spces. The Fréchet derivtive of n opertor

More information

Lecture 1: Introduction to integration theory and bounded variation

Lecture 1: Introduction to integrtion theory nd bounded vrition Wht is this course bout? Integrtion theory. The first question you might hve is why there is nything you need to lern bout integrtion. You

More information

The Bochner Integral and the Weak Property (N)

Int. Journl of Mth. Anlysis, Vol. 8, 2014, no. 19, 901-906 HIKARI Ltd, www.m-hikri.com http://dx.doi.org/10.12988/ijm.2014.4367 The Bochner Integrl nd the Wek Property (N) Besnik Bush Memetj University

More information

Math Theory of Partial Differential Equations Lecture 2-9: Sturm-Liouville eigenvalue problems (continued).

Mth 412-501 Theory of Prtil Differentil Equtions Lecture 2-9: Sturm-Liouville eigenvlue problems (continued). Regulr Sturm-Liouville eigenvlue problem: d ( p dφ ) + qφ + λσφ = 0 ( < x < b), dx dx β 1 φ()

More information

Chapter 22. The Fundamental Theorem of Calculus

Version of 24.2.4 Chpter 22 The Fundmentl Theorem of Clculus In this chpter I ddress one of the most importnt properties of the Lebesgue integrl. Given n integrble function f : [,b] R, we cn form its indefinite

More information

Continuous Random Variables

STAT/MATH 395 A - PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is rel-vlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht

More information

Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties

More information

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.

MATH 409 Advnced Clculus I Lecture 19: Riemnn sums. Properties of integrls. Drboux sums Let P = {x 0,x 1,...,x n } be prtition of n intervl [,b], where x 0 = < x 1 < < x n = b. Let f : [,b] R be bounded

More information

CHAPTER 4 MULTIPLE INTEGRALS

CHAPTE 4 MULTIPLE INTEGAL The objects of this chpter re five-fold. They re: (1 Discuss when sclr-vlued functions f cn be integrted over closed rectngulr boxes in n ; simply put, f is integrble over iff

More information

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

More information

Lecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)

Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of

More information

REPRESENTATION THEORY OF PSL 2 (q)

REPRESENTATION THEORY OF PSL (q) YAQIAO LI Following re notes from book [1]. The im is to show the qusirndomness of PSL (q), i.e., the group hs no low dimensionl representtion. 1. Representtion Theory

More information

Riemann Sums and Riemann Integrals

Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct

More information