115VII. Inner Product SpacesAn inner product space ( =: ips) is a nls in which the norm derives from an innerproduct, i.e., from a hermitian positive definite form. (Here, form is used i n the senseof functional, i.e., scalar-valued map, particularly a scalar-valued map on the cartesianproduct of the space with itself.) In such a space, one has a notio n of angle in addition to(translation- and scale-invariant) distance and so is much closer to the familiar Euclideann-space ℓ2(n) than in other nls’s.We have occasion to use the complex scalar field C. Recall the agreement t hat Fstands for either R or C.** definit ion **X ls. X × X → F : (x, y) 7→ hx, yi is an inner product :=∀{y ∈ X} h·, yi is linear (linearity)∀{x ∈ X\0} hx, xi > 0 (positive definite)∀{x, y ∈ X} hy, xi =hx, yi (skew-symmetric or hermitian)The bar denotes formation of the complex conjugate. If F = R, then the skew-symmetry becomes just symmetry, hy, xi = hx, yi.Note that hx, ·i is skew-linear, i.e., additive but skew homogeneous, sincehx, y + zi = hy + z, xi = hy, xi + hz, xi = hx, yi+ hx, zibuthx, αyi =hαy, xi = ¯αhy, xi = ¯αhx, yi.The model example is X = Fnwith h, i the scalar product, i.e.,hx, yi := ycx :=Xix(i)y(i) = xty = ytx.(Unfortunately, it is tradit ional to have the inner product skewlinear in the second slotrather than the first, therefore t he switch in the order here.) The corresp onding continuousexample is: X some ls of functions on some T ⊂ Rmwith the inner product given by theint egra lhf, gi :=ZTf(t)g(t) dt.** bilinear forms **Each lm A on the ips X into X gives rise to a (skew)bilinear form by the ruleh,iA: X × X → F : (x, y) 7→ hAx, yi,and A is called hermitian or positive definite or w hat ever in case h,iAis hermitian or positivedefinite or whatever. In particular, arguments concerning inner products are applicable toa more general bilinear form h,iAto the extent that it shares the properties of the ip h,iused in those arguments.bilinear formsc2002 Carl de Boor116 VII. Inner Product SpacesH.P.(1)(i) Prove: If X ips over F = C, A ∈ L(X), |x|A:= hAx, xi, then4hAx, yi =P4i=1ii|x + iiy|A.This means that the bilinear form h, iAcan be reconstructed from its values on the “di agonal”, i.e., on{(x, x) : x ∈ X}. Thi s is called polarization.(ii) Prove: If F = C and A ∈ L(X) is positive definite, then h, iAis an ip. Explain why this conclusioncannot be drawn when F = R.** ips is nls **The “norm” on the ips X is defined bykxk := hx, xi1/2.It is positive definite and positive homogeneous, by insp ection. A proof of the triangleinequality can be obtained along the following lines. Computekx ± yk2= hx ± y, x ± yi = hx, xi ± hx, yi ± hy, xi + hy, yi,using the bi-additi vity of h, i. Therefore(1) kx ± yk2= kxk2± 2 Rehx, yi + kyk2.Taking a cue from ℓ2(n), one calls x, y ∈ X orthogonal and writ esx ⊥ yin case hx, yi = 0. This gives(2) Pythagoras. x ⊥ y =⇒ kx ± yk2= kxk2+ kyk2.Perhaps t he most important aspect of an ips X is the possibili ty of using the elementsof X to represent linear functionals on X. This possibility exists since∀{y ∈ X} yt:= h·, yi ∈ X′.Note that ker yt= {x ∈ X : x ⊥ y} =: {y}⊥, the orthogonal complement of y. Also,for y 6= 0, ycy = hy, yi 6= 0. Therefore (recall (IV.5)elimination), for y 6= 0,∀{x ∈ X}x −hx, yihy, yiy⊥ y.So, by Pythagoras,kxk2= kx −hx, yihy, yiyk2+ khx, yihy, yiyk2≥ |hx, yihy, yi|2kyk2= |h x, yi|2/kyk2with equality iff x =hx,yihy,yiy. This provesips is nlsc2002 Carl de BoorBasics 117(3) CBS (Cauchy-Bunyakovski-Schwarz). |hx, yi| ≤ kxkkyk with equality iff [x, y] isnot 1-1.As a consequence of the CBS inequality, kyck = kyk, hence the map y 7→ ycembedsX i sometri cally, but only skewlinearly, in X∗. (Since X∗is complete (being a dual space),this map cannot be onto unless X is complete. The Riesz-Fischer Theorem below saysthat the map is onto in that case.) Also, yc||y (for y 6= 0). Further, with (1),kx + yk2≤ kxk2+ 2kxkkyk+ kyk2.This proves the triangle inequalitykx + yk ≤ kxk + kyk,and so finishes the proof of the fact that an ips is a nls wrto the norm k · k := h·, ·i1/2.This nls is called Hilbert space ( =: Hs) in case it is complete.H.P.(2) Prove that the CBS inequality holds (though equality is more complicated) even when X is onlya semi-inner product space, meaning that h, i is onl y p ositive semidefinite, i. e., still kxk ≥ 0 for all x ∈ X,but equality is possible even for some nonzero x. (Hint: Show that only the case kxk = 0 = kyk needs to beconsidered; then cons ider it.)H.P.(3) Prove that, for any A ∈ BL(X), with X ips, supkxk,kyk≤1|hAx, yi| = kAk.H.P.(4) (a) Prove that any weakly convergent sequence in an ips converges in norm iff its norms convergeto the norm of its weak limit. (b) Show that the sequence (en) in ℓ2converges weakly, but not strongly, i.e.,not in norm, to 0.** parallelogram law **(1) implies the(4) Parallelogram Law. kx + yk2+ kx − yk2= 2kxk2+ kyk2.xykykkx+ykkxkkx−yk(5) Figure. Parallelogram lawVon Neumann showed that this law is characteristic of a norm derived from an ip,i.e., the parallel ogram law implies that(x, y) 7→4Xi=1iikx + iiyk2/4is an inner product and it gives rise to the norm from which we started. (Here, the twopurely imaginary terms are omitted in case F = R.)parallelogram lawc2002 Carl de Boor118 VII. Inner Product SpacesH.P.(5) Prove von Neumann’s assertion. (Hint: additivity of hx, ·i follows by two applications of the Par-allelogram Law, whence (with some H.P. from Chapter I) homogeneity wrto rationals; now use continuity andeffect of substitution of iy for y; etc.)(6) Corollary 1. An ips is strictly convex.Proof: kxk = kyk = 1 = k(x + y)/2k =⇒ 4 = 4k(x + y)/2k2= kx + yk2=2kxk2+ kyk2− kx − yk2= 4 − kx − yk2, i.e., x = y.** ba from convex sets **(7) Corollary 2. If K i s a complete convex subset of the ips X, then every x ∈ X hasexactly one ba from K.Proof: After a shift by x, we may assume that x = 0. Let ( kn) be a minimizingsequence in K, i.e., li m kknk = d(0, K). By the Parallelogram law,0 ≤ kkn− kmk2= 2kknk2+ kkmk2− 4k(kn+ km)/2k2≤ 2kknk2+ kkmk2− 4d( 0 , K)2n,m→∞−−−−−−→ 0since (kn+ km)/2 ∈ K and kknk2→ d(0, K)2. This shows (kn) to be Cauchy. Since K iscomplete, this gi ves lim kn= k for some k ∈ K, while kkk = lim kknk = d(0, K).This proves existence of a ba. Uniqueness follows from the strict convexity or