63IV. The (continuous) dualWe callX∗:= bL(X, IF) = {λ ∈ X0: λ is bounded}the continuous dual of the nls X. It is at times useful to know that X∗can be identifiedin a natural way with a closed lss of b(B), the complete metric space of bounded functionson the unit ball B of X.(1) Lemma. For any nls X, the restriction map(2) r : X∗→ b(B) : λ 7→ λBis linear, an isometry (i.e., ∀{λ} kr(λ)k = kλk), and its range is closed.Proof: Since λ is bounded, i.e., λ(B) is a bounded subset of IF, the restrictionλBof λ to B is a bounded function. Moreover,kλk = sup |λ(B)| = kλBk∞.This shows that r is a linear isometry into b(B). To see that its range is closed, let g ∈ b(B)be the limit of some sequence (λn|B). Then g is necessarily linear on B. This makes itpossible to extend g to a fnl λ on all of X by the recipeλx := αg(x/α), all α > kxk, all x ∈ Xand to verify that λ ∈ X0, hence ∈ X∗since kλk = kgk∞. Therefore, g = r(λ), i.e., ran ris closed.It follows that ran r is complete (as a closed subset of the complete ms b(B)), and,since r is an isometry, this implies that X∗is a Banach space (=: Bs), i.e., a completenls.(3) Proposition. The continuous dual X∗of any nls X is complete.A second advantage that X∗has over X is that its closed unit ball is compact in anatural topology.(4) Alaoglu’s theorem. The closed unit ball of X∗is compact in the topology of point-wise convergence.Proof: Let Y := B−X∗. Then Y is a subset of ×x∈X{z ∈ IF : |z| ≤ kxk}, and thelatter, by (II.41) Tykhonov’s Theorem, is compact in the topology of pointwise conver-gence. Hence, by (II.25)Lemma, it is sufficient to prove that Y is closed in this topology.For this, let f be an element in the pointwise closure of Y . Then, for arbitrary α, β ∈ IFand arbitrary x, y ∈ X, and arbitrary ε > 0, there is some λ ∈ Y with |f(s) − λs| < ε fors ∈ {x, y, αx + βy}. Consequently, |f(x)| < |λx| + ε ≤ kxk + ε and|f(αx+βy)−αf(x)−βf(y)| = |(f −λ)(αx+βy)−α(f −λ)(x)−β(f −λ)(y)| < (1+|α|+|β|)ε.Since ε > 0 is arbitrary, this shows that f ∈ Y .c2002 Carl de Boor64 IV. The (continuous) dualx xzyλxλyyd(x, ker λ)H(λ, λx) H(λ, λy) H(λ, λx)ker λ ker λ(6) Figure. A lfl is constant on hyperplanes parallel to its kernel. Thedistance of such a hyperplane from the kernel is proportionalto the lfl’s value on that hyperplane.** hyperplanes and lfl’s **The action of a nontrivial (bounded or unbounded) lfl λ on a ls X is easy to visualize:Each element of X lies in exactly one of the hyperplanesH(λ, t) := {x ∈ X : λx = t},and each such hyperplane is a translate of any other such hyperplane, since each is atranslate of ker λ, i.e.,∀{x ∈ H(λ, t)} H(λ, t) = ker λ + x.This reflects the fact that H(λ, 0) = ker λ has codimension 1, i.e.,X = ker λ˙+ ran[y],for any y 6∈ ker λ; hence the term “hyperplane”. To see this, observe that, for any x ∈ Xand any y 6∈ ker λ, x − (λx/λy)y ∈ ker λ.** elimination **This observation is the basic step of the well known numerical linear algebra processcalled(5) Elimination. To convert x into something in ker λ, pick y 6∈ ker λ and computez := x − (λx/λy)y ∈ ker λ.eliminationc2002 Carl de BoorBasics 65H.P.(1) Prove that x 7→ (λx/λy)y is a linear projector. What is its range, what are its interpolationfunctionals?** a useful formula **Elimination provides the following useful formula:(7) ∀{λ ∈ X0} ∀{x ∈ X\ ker λ}d(x, ker λ)|λx|=0 (ker λ)−= X;1/kλk otherwise.Indeed, if there is some y ∈ X\(ker λ)−, then, by elimination, for any x ∈ X, d(x, ker λ) =d((λx/λy)y, ker λ) = |λx/λy|d(y, ker λ), hence (ker λ)−= B−0(ker λ) = ker λ and further,for any x 6∈ ker λ,kλk = supy6∈ker λ|λy|kyk=|λx|d(x, ker λ)supy6∈ker λd(y, ker λ)kyk=|λx|d(x, ker λ),the last equality by (III.7)Riesz’ Lemma. The fact that d(x, ker λ) = 0 for all x ∈ X iffker λ is dense in X, is trivial.** λ is continuous iff ker λ is closed **We infer from (7) that λ is bounded in case (ker λ)−6= X. Put differently, it says that,for λ ∈ X0\X∗, (ker λ)−= X, i.e., the kernel of a discontinuous lfl is dense. Since such alfl is necessarily nontrivial, it says that ker λ 6= (ker λ)−for a discontinuous lfl. Conversely,if λ is continuous, then ker λ = λ−1{0} is closed as the pre-image of a closed set under acontinuous map. This proves:(8) Proposition. Let X be a ls, and λ ∈ X0. Then: λ is continuous iff ker λ is closed.(9) Corollary. For λ ∈ X0, ker λ is either closed or dense (with ker λ both closed anddense iff λ = 0).** error estimates **(7) proves the useful identity(10) Lemma. ∀{λ ∈ X∗, x ∈ X} |λx| = kλkd(x, ker λ).This identity contains all basic Numerical Analysis error estimates, in the followingway.In applications of (10), λ is an error functional, and ker λ is not completely known.Rather, one knows some set F contained in ker λ, and so obtains the bound|λx| ≤ kλkd(x, F ).(11) Example. λ :=Rba· − hPn−11δa+jh− (h/2)(δa+ δb) (with h := (b − a)/n) isthe error in the composite trapezoidal rule, and this rule is exact for all linear polynomials.Further, on C([a . . b]), kλk ≤ 2|b − a|. Hence (with d(f, Y ) ≤ kfk for any lss Y )|λf| ≤ 2|b − a|d(f, Π1) = O(|b − a|kfk∞).Actually, the composite trapezoidal rule is exact for Π01,h:= all broken lines on [a . . b] withbreakpoints a + jh, all j, and d(f, Π01,h) ≤18h2kD2fk∞. Hence |λf| ≤ kλkd(f, Π01,h) =O(|b − a|h2kD2fk∞), a much better estimate.existence of ba from a hyperplanec2002 Carl de Boor66 IV. The (continuous) dual** existence of ba from a hyperplane **(12) Corollary. λ ∈ X∗\0 takes on its norm at x iff 0 is a ba to x from ker λ.Indeed, since kλkkxk ≥ |λx| = kλkd(x, ker λ), we have kλkkxk = |λx| iff kxk =d(x, ker λ), and, since 0 ∈ ker λ, this last equality can only hold if 0 is a ba.Since (III.13)Example provides an example of a bounded lfl that does not take on itsnorm, while its kernel is closed by (8)Proposition, this provides the illustration, promisedearlier, of a closed lss that fails to provide ba’s (since, if y is a ba from the lss Y to x, then0 is a ba from Y to x − y).H.P.(2) Let x, k ∈ X nls, λ ∈ X∗\0. Prove: |λx−λk| ≤ kλkkx−kk with equality iff B−kx−kk(x)∩H(λ, λk) 6={} = Bkx−kk(x) ∩ H(λ, λk). Draw the picture.One says that λ