REAL ANALYSIS II HOMEWORK 4
                                                      ˙
                                                    CIHANBAHRAN
                                                Folland, Chapter 5
                   1. If X is a normed vector space over K (= R or C), then addition and scalar
                   multiplication are continuous from X × X and K ×X to X. Moreover, the norm
                   is continuous from X to [0,∞); in fact, |kxk − kyk| ≤ kx − yk.
                 Since X has a metric topology, to show that a map into X is continuous it suffices to
                 show that the inverse image of balls are open.
                 If we write α : X × X → X for the addition map, then for z ∈ X and r > 0 we have
                                        α−1(Br(z)) = {(x,y) : kx+y −zk < r} .
                 Now, given (x ,y ) ∈ α−1(B (z)) there exists ε > 0 such that
                               0  0          r
                                                 kx +y −zk 0 such that
                               0  0          r
                                         kc x −zk 0, if we choose δ = ε then kx −yk < δ implies |kxk − kyk| < ε. Thus the norm
                 map x 7→ kxk is continuous.
                   12. Let X be a normed vector space and M a proper closed subspace of X.
                        a. kx+Mk=inf{kx+yk:y∈M}isanormonX/M.
                                                            1
                                                   REAL ANALYSIS II HOMEWORK 4                                      2
                           b. For any ε > 0 there exists x ∈ X such that kxk = 1 and kx +Mk ≥ 1−ε.
                           c. The projection map π(x) = x+M from X to X/M has norm 1.
                           d. If X is complete, so is X/M. (Use Theorem 5.1.)
                           e. The topology defined by the quotient norm is the quotient topology as
                              defined in the exercise 28 in §4.2.
                   a.  First of all note that the definition of kx + Mk only depends on the coset, not
                   the representative x. Indeed, kx + Mk = inf{kzk : z ∈ x + M}. Observe that given
                   x,y ∈ X and z ∈ M we have
                                        k(x+y)+Mk≤k(x+y)+zk≤kx+zk+kyk.
                   So k(x+y)+Mk−kyk is a lower bound for the set {kx+zk : z ∈ M}. Therefore
                   k(x+y)+Mk−kyk≤kx+Mkandhence
                                                 k(x+y)+Mk≤kx+Mk+kyk.
                   For every z ∈ M, replacing y with y +z above yields
                                  k(x+y)+Mk=k(x+y+z)+Mk≤kx+Mk+ky+zk.
                   So k(x+y)+Mk−kx+Mkisalowerboundfortheset{ky+zk:z ∈M}. Therefore
                   k(x+y)+Mk−kx+Mk≤ky+Mkandhence
                                k(x+M)+(y+M)k=k(x+y)+Mk≤kx+Mk+ky+Mk.
                   This proves the triangle inequality.
                   Second, let x ∈ X and λ ∈ K. If λ = 0, then
                                kλ(x+M)k=k0+Mk=inf{kyk:y∈M}=0=|λ|kx+Mk
                   since 0 ∈ M. If λ 6= 0, then λM = M so
                                              kλ(x+M)k=kλx+Mk
                                                            =inf{kλx+yk:y ∈M}
                                                            =inf{kλx+λyk:y ∈M}
                                                            =inf{|λ|kx+yk:y ∈M}
                                                            =|λ|inf{kx+yk:y ∈M}
                                                            =|λ|kx+Mk.
                   Thus we have a seminorm on X/M. To see that this is a norm, suppose kx+Mk = 0.
                   This yields that
                                               d(x,M) = inf{kx−yk : y ∈ M} = 0.
                   Since M is closed this implies that x ∈ M, so x + M = 0 +M.
                   c. Let x ∈ X −M, so kx+Mk > 0. Given α ∈ (0,1), since α−1kx+Mk > kx+Mk
                   there exists y ∈ M such that
                                                      α−1kx+Mk>kx+yk.
                   Write z = x+y. Then since x+M = z +M = π(z), we get
                                                             kπ(z)k > α,
                                                               kzk
                                            REAL ANALYSIS II HOMEWORK 4                             3
                 therefore kπk > α. As this holds for every α ∈ (0,1) we get kπk ≥ 1. On the other
                 hand, for every x ∈ X we have kπ(x)k = kx+Mk ≤ kxk by choosing y = 0. Thus
                 kπk = 1.
                 b. By (c) we have
                                         1 = kπk = sup{kx+Mk : kxk = 1}.
                 So for every ε > 0 there exists x with kxk = 1 such that kx + Mk > 1 − ε.
                                               ∞
                 d. Assume X is complete. Let X(xn +M) be an absolutely convergent series in X/M.
                                               n=1
                 For every n, choose y ∈ M such that
                                      n
                                             kx +y k 1 kf k. Consider the closed
                                           n         n                     n  n      2   n
                   subspace
                                                      M=span{xn:n∈N}
                   of X. We want to show M = X, so suppose not. Then there exists x ∈ X − M so
                   by Hahn-Banach theorem there exists f ∈ X∗ such that f(x) 6= 0 and f|               =0. Since
                                                                                                    M
                   {f }∞ is dense in X∗ there exists a subsequence {f } with
                      n 1                                                   n
                                                                             k
                                                            lim f    =f.
                                                            k→∞ nk
                   But |(f    −f)(x )| = |f (x )| > 1 kf k so kf              −fk > 1kf k thus we conclude
                           n          n        n    n           n          n               n
                            k          k         k   k      2     k         k          2    k
                   that
                                                       kfk = lim kf k = 0
                                                               k→∞ nk
                   so f = 0, contradicting f(x) 6= 0.
                     27. There exist meager subsets of R whose complements have Lebesgue measure
                     zero.
                   Let {qk} be an enumeration of the rational numbers and let
                                                      ∞ Ç                           å
                                               E = [ q − 1 ,q + 1                      .
                                                n           k     k−1    k     k−1
                                                     k=1        2    n        2   n
                   Consider the set E = T∞ E . Observe that
                                             n=1   n
                                                                  ∞
                                                       m(E )≤X 1 = 1
                                                            n          k
                                                                  k=1 2 n    n
                   hence m(E) = 0.
                   On the other hand, we claim that for every n the complement Ec is a nowhere dense
                                                                                            n
                   subset of R. Indeed, since E is open Ec is closed; so it suffices to show that Ec has
                                                   n            n                                            n
                   empty interior. This is the case because otherwise Ec would contain an interval and
                                                                               n