Solutions to Exercises

Home , Absorbing set, Compact operator, Sequence space, Unitary operator

Chapter 2 2.1. U is convex: p (1− t)x + ty <(1 − t) + t = 1ifx, y ∈ U and t ∈ (0, 1). U is absorbing: p x/2p(x) = 1/2 < 1ifx ∈ X and p(x) = 0. U is balanced: p(kx) =|k|p(x)<1ifx ∈ U and |k|≤1.

2.2. p j (x) + p j (y) ≤ max{p1(x), . . . , pm (x)}+max{p1(y),...,pm (y)} if j = 1,...,m and x, y ∈ X. Hence p(x + y) ≤ p(x) + p(y) for x, y ∈ X. 2 Let X := K , and let p1(x) := |x(1)|+2|x(2)|, p2(x) = 2|x(1)|+|x(2)| for x := (x(1), x(2)) ∈ X. Then p1 and p2 are norms on X.Butq is not a seminorm on X:Ifx := (1, 0) and y := (0, 1), then q(x + y)>q(x) + q(y). p 2.3. Let p ∈{1, 2, ∞}, x ∈ , xn := (x(1), . . . , x(n), 0, 0,...)∈ c00, n ∈ N. If p := 1orp := 2, then xn − xp → 0. If p := ∞ and x ∈ c0, then xn − x∞ → 0. Conversely, let (yn) be a ∞ sequence in c0 such that yn → x in , and let > 0. There is n0 ∈ N − < / ∈ ∈ N such that yn0 x ∞ 2. Since yn0 c0, there is j0 such that | ( )| < / ≥ | ( )|≤| ( ) − ( )|+| ( )|≤ yn0 j 2 for all j j0. Then x j x j yn0 j yn0 j x − y ∞ +|y ( j)| < /2 + /2 = for all j ≥ j . Hence x ∈ c . n0 n0 √ 0 0 1 2.4. For t ∈ (0, 1],letx(t)√:= 1/ t, and let x(0√) := 0. Then x ∈ L ([0, 1]),but x ∈/ L2([0, 1]).Also, x ∈ L2([0, 1]),but x ∈/ L∞([0, 1]). Let y(t) := x(t) if t ∈[0, 1] and y(t) := 0 if either√ t < 0ort > 1. Then y ∈ L1(R),buty ∈/ L2(R)∪ L∞(R).Letz(t) := x(t) if t ∈[0, 1], z(t) := 0 if t < 0, and z(t) := 1/t if t > 1. Then z ∈ L2(R),butz ∈/ L1(R) ∪ L∞(R). Let u(t) := 1ift ∈ R. Then u ∈ L∞(R),butu ∈/ L1(R) ∪ L2(R).

2.5. For x ∈ C([0, 1]), x1 ≤x2 ≤x∞.Forn ∈ N,letxn(t) := 1 − nt if 0 ≤ t ≤ 1/n, and xn(t) := 0if(1/n)

by the mean value theorem, and so |x(t)|≤|x(a)|+(b − a)x ∞. Hence

x1,∞ = max{x∞, x ∞}≤max{1, b − a}x , x ≤x1,∞ and x∞ ≤x1,∞ for all x ∈ X. n n Also, if for n ∈ N,weletxn(t) := (t − a) /(b − a) , t ∈[a, b], then ∈ , = = /( − ) →∞ xn X xn ∞ 1, and xn ∞ n b a . 2.7. For x ∈ X, let us write |||x + Z(F)||| := inf{x + z:z ∈ Z(F)}.Lety ∈ Y . There is x ∈ X such that F(x) = y, and then |||x + Z(F)||| = inf{u: u ∈ X and u − x ∈ Z(F)}=inf{u:u ∈ X and F(u) = y}=q(y).If k ∈ K, then F(kx) = ky. Also, if y1, y2 ∈ Y and F(x1) = y1, F(x2) = y2, then F(x1 + x2) = y1 + y2. As in the proof of Proposition 2.5(i), q(ky) = |||kx + Z(F)||| = |k| |||x + Z(F)||| = |k|q(y), and q(y1 + y2) = |||x1 + x2 + Z(F)||| ≤ |||x1 + Z(F)||| + |||x2 + Z(F)||| = q(y1) + q(y2). Suppose Z(F) is a closed subset of X.Lety ∈ Y be such that q(y) = 0. If x ∈ X and F(x) = y, then |||x + Z(F)||| = q(y) = 0, and so x ∈ Z(F), that is, y = F(x) = 0. Hence q is a norm on Y . Conversely, suppose q is a norm on Y .Let(xn) be a sequence in Z(F) such that xn → x in X. Then |||x + Z(F)||| = |||x −xn + Z(F)||| ≤ x −xn→0, and so |||x + Z(F)||| = 0. Let y := F(x). Then q(y) = |||x + Z(F)||| = 0, so that F(x) = y = 0, that is, x ∈ Z(F). Thus Z(F) is a closed subset of X.

2.8. Suppose there are x0 ∈ X and r > 0 such that U(x0, r) ⊂ E. Since E is compact, it is closed in X, and so U(x0, r) ⊂ E. But then the closed unit ball of X is a closed subset of the compact set s(E − x0), where s := 1/r. 2.9. The closed unit ball E of 2 is not compact since the normed space 2 is inﬁnite dimensional. Let (xn) be a sequence in the Hilbert cube C. There is a subsequence xn,n of (xn) such that for each j ∈ N, the sequence xn,n( j) converges in K to x( j), say. Let x := (x(1), x(2),...). Then x ∈ 2 since |x( j)|≤1/j for each j ∈ N.Also,

m ∞ 2 2 2 2 x , − x ≤ |x , ( j) − x( j)| + for every m ∈ N. n n 2 n n j j=1 j=m+1

Let > 0. Choose m ∈ N such that the second term above is less than 2/2, 2 2 and then choose n0 ∈ N such that |xn,n( j) − x( j)| < /2m for all n ≥ n0 and j = 1,...,m. It follows that xn,n − x2 < for all n ≥ n0. 2.10. Suppose a norm ·on a linear space X is induced by an inner product. If x, y ∈ X, x=1 =y and x = y, then

x + y2 = 2x2 + 2y2 −x − y2 < 2x2 + 2y2 = 4.

2 Thus (X, ·) is strictly convex. The norm ·2 on , and the norm ·2 on L2([0, 1]) are induced by inner products. Hence they are strictly convex. On the other hand, let x := e1 and y := e2. Then x1 = 1 =y1 and x + y1 = 2, 1 and x + y∞ = 1 =x − y∞ and (x + y) + (x − y)∞ = 2. Hence and ∞ are not strictly convex. Similarly, if x and y denote the characteristic functions of [0, 1/2] and (1/2, 1], then x1 = 1/2 =y1 and x +y1 = 1, Solutions to Exercises 205

and x + y∞ = 1 =x − y∞ and (x + y) + (x − y)∞ = 2. Hence L1([0, 1) and L∞([0, 1]) are not strictly convex. Thus if p ∈{1, ∞}, then the norms on p and L p([0, 1]) are not induced by an inner product. 2.11. Suppose | x, y|2 = x, x y, y, and deﬁne z := y, yx − x, yy. Pro- ceeding as in the proof of Proposition 2.13(i), we see that z, z=0, and so z = 0, that is, y, yx = x, yy. Conversely, if y, yx = x, yy, then we can readily check that | x, y| = xy. Suppose x + y2 = (x+y)2. Then Re x, y=xy. If either x = 0 or y = 0, then clearly, yx =xy.Nowletx = 0 and y = 0, and deﬁne u := x/x,v:= y/y. Then u=1 =v, and u + v=2 since

2Re x, y u + v2 =u2 +v2 + 2Re u,v=2 + = 4. xy

By Exercise 2.10, (X, ·) is strictly convex, and so u = v, that is, yx = xy. Conversely, if yx =xy, then we can readily check that x + y= x+y. 2.12. Let x, y, z ∈ X. By the parallelogram law,

x + y + z2 +x + y − z2 +x − y + z2 +x − y − z2 = 2x + y2 + 2z2 + 2x − y2 + 2z2 = 4(x2 +y2 +z2).

2.13. Let x ∈ Z.Ifk ∈ K, then clearly kx ∈ Z.Lety ∈ X, and let > 0. Then

0 ≤ x + y, x + y= x, x+ 2 y, y+2 Re x, y=2 y, y+2 Re x, y,

and so 0 ≤ y, y+2Re x, y.Let → 0 to obtain 0 ≤ Re x, y. Replace x by −x to obtain Re x, y≤0, and so Re x, y=0. Replace x by ix to obtain Im x, y=0, and so x, y=0. In particular, if x, y ∈ Z, then x + y, x + y= x, x+ y, y+2Re x, y= 0 + 0 + 0 = 0, that is, (x + y) ∈ Z. Thus Z is a subspace of X. Suppose x + Z = x1 + Z and y + Z = y1 + Z. Then x1 − x, y1 − y ∈ Z, and so x1, y1= (x1 − x) + x,(y1 − y) + y, which is equal to

(x1 − x), (y1 − y)+ (x1 − x), y+ x,(y1 − y)+ x, y= x, y.

Let x + Z, y + Z := x, y for x, y ∈ X.If x + Z, x + Z = 0, then x ∈ Z, that is, x +Z = 0+Z. It follows that ·, · is an inner product on X/Z. Now | x, y|2 =| x + Z, y + Z|2 ≤ x + Z, x + Z y + Z, y + Z = x, x y, y for all x, y ∈ X. 2.14. Let x, y ∈ X. Then x + y, x + y− x − y, x − y=4Re x, y. Replacing y by iy, x + iy, x + iy− x − iy, x − iy=4Re x, iy=4Im x, y. Hence the right side is equal to 4 Re x, y+i 4Im x, y=4 x, y. 206 Solutions to Exercises

2 2 2 2 2 2.15. x − y =x +y − 2Re x, y=x +y − 2 xy cos θ , . √x y 2.16. In Example 2.12(i), replace n by mn.Also,I = (1 +···+1)1/2 = n. n F ∈ 2 2 = ∞ w( )| ( )|2 ≤w 2 v ∈ ∞ 2.17. For x , x w j=1 j x j ∞ x 2.If , then w( ) ≥ /v ∈ N 2 ≥2/v j 1 ∞ for all j , and so x w x 2 ∞. Conversely, 2 suppose there is α > 0 such that x2 ≤ αxw for all x ∈ . Then 1 ≤ α w( j) by considering x := e j for j ∈ N. Hence v∞ ≤ α. { +···+ , − ,..., − } 2.18. The set e1 em e1 e2 e1 em is linearly independent. Next,√ y1 := 0, z1 := x1 = e1 +···+em , u1 := z1/z12 = (e1 +···+em )/ m and y2 :=√ x2, u1u1 = 0, z2 := x2 − y2 = e1 − e2, u2 := z2/z22 = ( − )/ ∈{,..., − } := e1 e2 2, as desired. Let √n 2 m 1 and assume that u j (e1 +···+e j−1 − ( j − 1)e j )/ ( j − 1) j for j = 2,...,n. Then

n n n u j yn+1 := xn+1, u j u j = xn+1, u j u j = √ , ( j − 1) j j=1 j=2 j=2 √ := − = − − n / ( − ) and zn+1 xn+1 yn+1 e1 en+1 j=2 u j j 1 j.Now

n n n u j e1 +···+e j−1 − ( j − 1)e j 1 e1 − √ = e1 − = e j . ( j − 1) j ( j − 1) j n j=2 j=2 j=1

= ( +···+ )/ − = ( +···+ − )/ Hence zn+1 e1 en n en+1 e1 √ en nen+1 n, and un+1 := zn+1/zn+12 = (e1 +···+en − nen+1)/ n(n + 1) as desired. m 2.19. For j = 1,...,n, denote the jth column of A by x j ∈ K . As in Theorem 2.17, obtain u1,...,un by the Gram–Schmidt orthogonalization of x1,...,xn. Then x j = y j +x j − y j u j , where y1 := 0, and

y j := x j , u1u1 +···+ x j , u j−1u j−1 for j = 2,...,n.

Let Q denote the m × n matrix whose jth column is u j for j = 1,...,n, and let R := [ri, j ] denote the n ×n matrix, where ri, j := x j , ui if 1 ≤ i ≤ j −1, r j, j := x j − y j and ri, j := 0if j + 1 ≤ i ≤ n for j = 1,...,n. Then for j = 1,...,m, x j = x j , u1u1 +···+ x j , u j−1u j−1 +x j − y j u j = r1, j u1 +···+r j−1u j−1 + r j u j , that is, A =[x1,...,xn]=[u1,...,un]R = QR. = ,..., Uniqueness: Suppose A Q R , where the columns u1 un of Q form an Km := [ ] orthonormal subset of , and R ri, j is upper triangular with positive = = = = diagonal entries. Then x1 r1,1u1. Hence r1,1 x1 r1,1, and u1 / = ∈{,..., } = x1 r1,1 u1. Next, let j 2 n , and suppose we have shown u1 ,..., = = + ··· + + = u1 u j−1 u j−1. Then x j r1, j u1 r j−1, j u j−1 r j, j u j + ··· + + = , = = r1, j u1 r j−1, j u j−1 r j, j u j . Hence ri, j x j ui ri, j for i ,..., − = − −···− = 1 j 1 and r j, j x j r1, j u1 r j−1, j u j−1 r j, j , so that = ( − −···− )/ = = = u j x j r1, j u1 r j−1, j u j−1 r j, j u j . Thus Q Q and R R. Solutions to Exercises 207

If A is an infinite matrix whose columns form a linearly independent subset of 2, then the above arguments hold. ∈ N m | 1 ( ) ( ) |2 ≤ 1 | ( )|2 2.20. Let m . Then n=0 −1 x t pn t dt −1 x t dt by the Bessel inequality. Also, equality does not hold here since x ∈/ span {p , p ,...,p }. 0 1 m 2.21. If r := 0, then T u (t)dt = 2T , and if r = 0, then T u (t)dt = −T r −T r 1 T (2sinrT)/r for all T > 0. Hence lim →∞ u (t)dt is equal to 1 if T 2T −T r 1 T = = →∞ ( ) r 0, and it is equal to 0 if r 0. As a result, limT 2T −T x t dt exists for every x ∈ X.Letp, q ∈ X. Then p q ∈ X, and so p, q is well-defined. Clearly, the function · , · : X × X → K is linear in the first variable, it is conjugate symmetric, and p, p≥0 for all p ∈ X. := + ··· + ,..., Let p c1ur1 cnurn , where r1 rn are distinct real numbers, 2 2 and c1,...,cn ∈ C. Then p, p=|c1| + ··· + |cn| . It follows that = , = · , · p 0 whenever p p 0. Thus is an inner product on X.Also, 1 T , = →∞ − ( ) { : ∈ R} ur us limT 2T −T ur s t dt. Hence ur r is an uncountable orthonormal subset of X. (In particular, it is a linearly independent subset of X.)

2.22. Since the map F : X → Y is linear, and since · , ·Y is an inner product on Y , the function · , ·X : X × X → K is linear in the ﬁrst variable, it is conjugate-symmetric, and x, xX ≥ 0 for all x ∈ X. (i) · , ·X is an inner product on X if and only if x = 0 whenever x, xX =

F(x), F(x)Y = 0, that is, F(x) = 0. (ii) Since F is one-one, if F(uα) = F(uβ), then uα = uβ.Also, uα, uβX =

F(uα), F(uβ)Y for all α, β. (iii) Since F is one-one, {F(uα)} is an orthonormal subset of Y . Also, since

F is onto, if {F(uα)} is a proper subset of an orthonormal subset E of Y , then −1 {uα} is a proper subset of the orthonormal subset F (E) of X. 2.23. There are α > 0 and β > 0 such that βx≤x ≤ αx for all x ∈ X.

Hence (xn) is a Cauchy sequence in (X, ·) if and only if (xn) is a Cauchy

sequence in (X, ·), and (xn) is a convergent sequence in (X, ·) if and only if (xn) is a convergent sequence in (X, ·). ∞ 2.24. c0 is the closure of c00 in the Banach space (Exercise 2.3). ∞ To show that c is a closed subspace of the Banach space ,let(xn) be a ∞ sequence in c such that xn → x in , and let > 0. There is n0 ∈ N such − < / K ∈ N that xn0 x ∞ 3. Since xn0 is a Cauchy sequence in , there is j0 | ( ) − ( )| < / , ≥ such that xn0 i xn0 j 3 for all i j j0. Then

2 |x(i) − x( j)|≤|(x − x )(i) − (x − x )( j)|+|x (i) − x ( j)| < + = n0 n0 n0 n0 3 3

for all i, j ≥ j0. Hence (x( j)) is Cauchy sequence in K, and so x ∈ c. To show that C0(T ) is a closed subspace of the Banach space C(T ),let(xn)

be a sequence in C0(T ) such that xn → x in C(T ), and let > 0. There 208 Solutions to Exercises

∈ N − < / ∈ ( ) is n0 such that xn0 x ∞ 2. Since xn0 C0 T , there is a | ( )| < / ∈ \ compact subset T of T such that xn0 t 2 for all t T T. Then | ( )|≤| ( ) − ( )|+| ( )|≤ − +| ( )| < / + / = x t x t xn0 t xn0 t x xn0 ∞ xn0 t 2 2 for all t ∈ T \T, and so x ∈ C0(T ). 2.25. Since X has a denumerable basis, X is not a Banach space. 2.26. (i) Let Y be an m dimensional subspace of X. There is an orthonormal basis u1,...,um of Y .Let(xn) be a Cauchy sequence in Y . Then xn = xn, u1u1 + ···+ xn, um um for n ∈ N.Fix j ∈{1,...,m}. For all n, p ∈ N, | xn, u j − x p, u j |≤xn − x p. Hence the Cauchy sequence ( xn, u j ) converges in K to k j , say. Then xn → k1u1 +···+kmum in Y . (ii) Let X be infinite dimensional. By Theorem 2.17, there is a denumerable orthonormal subset {u1, u2,...} of X. The sequence (un) in the closed√ unit ball of X does not have a convergent subsequence since un − u p= 2for all n = p. Hence the closed unit ball of X is not compact. Conversely, suppose X is finite dimensional. By Theorem 2.23(iii), there is an isometry from X onto an Euclidean space whose closed unit ball is compact by the classical Heine–Borel theorem. (iii) Suppose X is complete, and it has a denumerable (Hamel) basis. By { , ,...} Theorem 2.17, there is an orthonormal subset u1 u2 of X which is a := ∞ / (Hamel) basis for X. Then x n=1 un n belongs to X by the Riesz– Fischer theorem, but it does not belong to span {u , u ,...}. 1 2 := ( ( ), ( ),...) ∈ ∈ N ( ) := m ( ) . 2.27. Let x x 1 x 2 c0, and for m ,letsm x n=1 x n en ( ) − = {| ( )|: = + , + ,...}→ →∞ Then sm x x ∞ sup x n n m 1 m 2 0asm . = ∞ ( ) , ,... K = Thus x n=1 x n en. Also, if there are k1 k2 in such that x ∞ = ∞ ( ) = ( ) ∈ N n=1 knen, then k j n=1 knen j x j for each j . ∈ ( ) → := − ∈ = ∞ ( ) Let x c, and x n x . Then y x x e0 c0, and y n=1 y n en. = + ∞ ( ( ) − ) = ∞ Hence x x e0 n=1 x n x en. Also, ifx n=0 knen, where , , ,... K := − = ∞ k0 k1 k2 are in , then y0 x k0e0 n=1 knen belongs to the ∞ ∈ = = closure of c00 in , that is, y0 c0. It follows that k0 x and k j ∞ ( ) = ( ) − ( ) = ( ) − ∈ N n=1 knen j x j x e0 j x j x for each j . 1 Let x ∈ . Define kn := x(2n − 1) + x(2n) and n := x(2n − 1) − x(2n) for n ∈ N.Form ∈ N,lets2m (x) := k1u1 + 1v1 +···+kmum + mvm and s2m−1(x) := k1u1 +1v1 +···+km−1um−1 +m−1vm−1 +km um . Then for all m in N, s2m(x) = x(1)e1+···+x(2m)e2m and s2m−1(x) = x(1)e1+···+x(2m− 2)e2m−2 +kmum . Since km → 0, we obtain x = k1u1+1v1+k2u2+2v2+···. = + v + + v +··· ( − ) = ( + )/ Also, if x k1u1 1 1 k2u2 2 2 , then x 2n 1 kn n 2 ( ) = ( − )/ = ( − ) + ( ) = = and x 2n kn n 2, that is, kn x 2n 1 x 2n kn and n x(2n − 1) − x(2n) = n for n ∈ N. As ∞ and L∞([a, b]) are not separable, they do not have Schauder bases. 1/2 2.28. For j ∈ N and x( j) ∈ X j , define x( j) j := x( j), x( j) . The function · , · : X × X → K is well-defined since for x, y ∈ X, Solutions to Exercises 209

∞ ∞ ∞ 1/2 ∞ 1/2 | ( ), ( ) |≤ ( ) ( ) ≤ ( )2 ( )2 . x j y j j x j j y j j x j j y j j j=1 j=1 j=1 j=1

If x, y ∈ X, then x + y ∈ X since x( j) + y( j) j ≤x( j) j +y( j) j and

∞ 1/2 ∞ 1/2 ∞ 1/2 ( ) + ( ) 2 ≤ ( )2 + ( )2 . x j j y j j x j j y j j j=1 j=1 j=1

Also, it is easy to see that if x ∈ X and k ∈ K, then kx∈ X. Further, it follows that · ,, · is an inner product on X. Suppose X is complete. Fix j ∈ N.If (xn( j)) is a Cauchy sequence in X j , deﬁne xn := (0,...,0, xn( j), 0, 0,...), where xn( j) is in the jth entry, and note that (xn) is Cauchy sequence in X. If xn → x in X, then xn( j) → x( j) in X j . Thus X j is complete. Conversely, suppose X j is complete for each j ∈ N. Then the completeness of X follows exactly as in the proof of the completeness of 2 given in Example 2.24(ii). 2.29. The case k := 1 is treated in Examples 2.24(iii), (v) and 2.28(iv). We consider here the case k := 2. The cases k > 2 are similar. ( ) ( 2([ , ]), · ) ( ), ( ) (i) Let xn be a Cauchy sequence in C a b 2,∞ . Then xn xn and ( ) ( ([ , ]), · ) xn are Cauchy sequences in the Banach space C a b ∞ . A well- known result in Real Analysis shows that there are x and y in C1([a, b]) such − → − → − → − → that xn x ∞ 0, xn x ∞ 0, and xn y ∞ 0, xn y ∞ 0. 1 2 Then x = y ∈ C ([a, b]), that is, x ∈ C ([a, b]), and xn − x2,∞ = − + − + − → xn x ∞ xn x ∞ xn x ∞ 0. ( ) ( 2,1([ , ]), · ) ( ), ( ) (ii) Let xn be a Cauchy sequence in W a b 2,1 . Then xn xn ( ) ( 1([ , ]), · ) and xn are Cauchy sequences in the Banach space L a b 1 .Aswe have seen in Example 2.24 (v), there are absolutely continuous functions x and [ , ] − → − → − → y on a b such that xn x 1 0, xn x 1 0, and xn y 1 0, − → = [ , ] xn y 1 0. Then x y is absolutely continuous on a b , that is, ∈ 2,1([ , ]) − = − + − + − → x W a b , and xn x 2,1 xn x 1 xn x 1 xn x 1 0. ( ) ( 2,2([ , ]), · ) ( ), ( ) (iii) Let xn be a Cauchy sequence in W a b 2,2 . Then xn xn ( ) ( 2([ , ]), ·) and xn are Cauchy sequences in the Hilbert space L a b 2 .As we have seen in Example 2.28(iv), there are absolutely continuous functions x [ , ] − → ∈ 2([ , ]) − → and y on a b such that xn x 2 0, x L a b , xn x 2 0, and − → ∈ 2([ , ]) − → = xn y 2 0, y L a b , xn y 2 0. Then x y is absolutely continuous on [a, b] and x = y ∈ L2([a, b]), that is, x ∈ W 2,2([a, b]), and − = − + − + − → xn x 2,2 xn x 2 xn x 2 xn x 2 0. =⇒ := ∞ , = , = 2 ∈ N 2.30. (i) (ii): Let s n=1 xn. Then s xn xn xn xn for n . =⇒ := = := / = (ii) (iii): Let un 0ifxn 0, and un xn xn if xn 0. Then , = ∈ N ∞ 2 = ∞ | , |2 ≤ 2 s un xn for n , and n=1 xn n=1 s un s . =⇒ := m , ∈ N > − 2 = (iii) (i): Let sm n=1 xn m . Then for m p, sm sp 2 2 2 x p+1 +···+xm =x p+1 +···+xm . Thus (sm) is a Cauchy sequence in the Hilbert space H, and so it converges in H. 210 Solutions to Exercises

2 2.31. For each n ∈ N, En := {un,vn,wn} is an orthonormal subset of and span En = span {e3n−2, e3n−1, e3n}.(Letm := 3 in Exercise 2.18.) Hence E := {un : n ∈ N}∪{vn : n ∈ N}∪{wn : n ∈ N} is an orthonormal subset of 2 2 and span E = span {e j : j ∈ N}, which is dense in . 2 2.32. For each n ∈ N, En := {un,vn,wn} is an orthonormal subset of and span En = span {e3n−2, e3n−1, e3n}. (Compare Exercise 4.35.) Hence E := 2 {un : n ∈ N}∪{vn : n ∈ N}∪{wn : n ∈ N} is an orthonormal subset of 2 and span E = span {e j : j ∈ N}, which is dense in .

2.33. It is easy to see that E := {u0, u1,v1, u2,v2,...} is an orthonormal subset of L2([−π, π]).Letx ∈ E⊥.Fork ∈ Z,

1 π 1 π xˆ(k) = x(t)e−iktdm(t) = x(t)(cos kt − i sin kt)dm(t) = 0 2π −π 2π −π

since x, u0=0 and x, un= x,vn=0 for all n ∈ N. Hence x = 0a.e. ⊥ on [−π, π]. Thus√E ={0}. √ √ , = , = , ,v = ∈ N Since x u0 2π a0 and x un π an x n π bn for all n , = , + ∞ , + ,v v the Fourier expansion x x u0 u0 n=1 x un un x n n and the 2 =| , |2 + ∞ | , |2 +| ,v |2 Parseval formula x 2 x u0 n=1 x un x n yield the desired results. 2 2.34. The subsets E := {u0, u1, u2,...,} and F := {v1,v2,...} of L ([0, 1]) are orthonormal. Given x ∈ L2([0, 1]),lety(t) := x(t/π) if t ∈[0, π], and y(t) := x(−t/π) if t ∈[−π, 0). Then y is an even function on [−π, π]. Also, y ∈ L2([−π, π]). Further, π y(t)dm(t) = 2π 1 x(s)dm(s) = 2πa . −π 0 0 ∈ N π ( ) ( ) = 1 ( ) ( ) = For n , −π y t cos nt dm t 2π 0 x s cos nπsdms πan and π ( ) ( ) = ( ) = + ∞ −π y t sin nt dm t 0. By Exercise 2.33, y t a0 n=1 an cos nt, [− , ] ( ) = + the series converging in the mean square on π π . Hence x s a0 ∞ [ , ] n=1 an cos nπs, the series converging in the mean square on 0 1 . Similarly, given x ∈ L2([0, 1]),letz(t) := x(t/π) if t ∈[0, π], and z(t) := −x(−t/π) if t ∈[−π, 0). Then z is an odd function on [−π, π]. As above, we ( ) = ∞ may use Exercise 2.33 to obtain z t n=1 bn sin nt, the series converging [− , ] ( ) = ∞ in the mean square on π π , and so x s n=1 bn sin nπs, the series converging in the mean square on [0, 1]. 3 2.35. The subspace G := span {v0,v1,...} of H consists of all polynomials in t defined on [−1, 1].Letx be a continuous function on [−1, 1], and define y ∈ X by y(t) := x(t1/3), t ∈[−1, 1].Let > 0. By the Weierstrass theorem, there is a polynomial q defined on [−1, 1] such that y − q∞ < . Define p(s) := 3 3 3 q(s ) for s ∈[−1, 1]. Then p ∈ G and x − p∞√= sup{|y(s ) −√q(s )|: s ∈[−1, 1]} = y−q∞ < . Further, x − p2 ≤ 2 x − p∞ < 2 .By Proposition 1.26(ii), G is dense in H. The calculations of v ,v v are routine. 0 1 2 ∈ | , |2 ≤ 2 < ∞ 2.36. If x H, then by Bessel’s inequality, α x uα x .Bythe , v ∈ : → Riesz–Fischer theorem, α x uα φ(α) G. Thus the map F H G is := , ∈ well-defined. It is easy to see that F is linear. Let x j α x j uα uα H Solutions to Exercises 211

:= , { } {v } , = for j 1 2. By the orthonormality of the sets uα and φ(α) , x1 x2 , , = ( ), ( ) α x1 uα x2 uα F x1 F x2 . In particular, F is continuous. To ∈ = ,v v show that Fis onto, let y G. Then y β y β β. By Bessel’s | ,v |2 ≤2 < ∞ inequality, β y β y . By the Riesz–Fischer theorem, − := ,v − ∈ := 1( ) ( ) = v x β y β uφ 1(β) H.Ifα φ β , then F uα β. Hence ( ) = ,v ( − ) = ,v v = F x β y β F uφ 1(β) β y β β y by the continuity and the linearity of F. 2.37. Let G denote the closure of span E. Since E⊥ = G⊥, we obtain E⊥⊥ = G⊥⊥ = G, as in the Projection Theorem. 2.38. Since y ∈ G, we obtain x − y≥d(x, G).Letw ∈ G. Then (y − w) ∈ G. Since (x − y) ∈ G⊥, we obtain (x − y) ⊥ (y − w), so that

x − w2 =(x − y) + (y − w)2 =x − y2 +y − w2 ≥x − y2.

Thus x − y=d(x, G).Ifw ∈ G and x − w=d(x, G), then x − w= x − y, and so y − w=0 by the above inequality, that is, w = y. 2.39. (i) Let x ∈ X\Y , and Z := span {x, Y }.NowY = Z, and since Y is ﬁnite dimensional, Y is a closed subspace of Z. By the lemma of Riesz, there is zn ∈ Z such that zn=1 and (1 − 1/n)

2.40. The function · , · : H/G × H/G → K is well-deﬁned: Suppose x j + G = + = , = + = + , ∈ x j G for j 1 2. Let x j y j z j and x j y j z j , where y j y j G , ∈ ⊥ = , − = ( − ) + ( − ) and z j z j G for j 1 2. Since z j z j x j x j y j y j , where ( − ) ∈ ( − ) ∈ ( − ) ∈ = , x j x j G and y j y j G, we see that z j z j G for j 1 2. ( − ) ⊥ ( − ) ⊥ Hence z1 z1 z2 and z2 z2 z1, and so + , + = , = , = , = + , + . x1 G x2 G z1 z2 z1 z2 z1 z2 x1 G x2 G · , · / It is easy to see that is an inner product on H G. Also, by Exercise 2 2 2 2.38, x1 + G, x1 + G = z1, z1=z1 =x1 − y1 = d(x1, G) = 2 |||x1 + G||| . Further, since H is complete, so is H/G.

Chapter 3

3.1. Suppose L := {x1, x2,...} is an inﬁnite linearly independent subset of X, and let B be a (Hamel) basis for X containing L. Deﬁne f (xn) := nxn for n ∈ N, 212 Solutions to Exercises

and f (b) := 0forb ∈ B\L.Let f : X → K denote the linear extension of this function. Then f is not continuous. Similarly, deﬁne F(xn) := nxn for n ∈ N, and F(b) := b for b ∈ B\L.LetF : X → X denote the linear extension of this function. Then F is one-one and R(F) = X,butF is not continuous. Let Y be a normed space, and y0 ∈ Y, y0 = 0. If G(x) := f (x)y0, x ∈ X, then G is linear but not continuous.

3.2. (i) If x := k1x1 +···+kn xn, then x ≥ |||k j x j + X j ||| = |k j | |||x j + X j ||| for j = 1,...,n, and so F(x)≤|k1|y1+···+|kn|yn≤αx. (ii) If y := 1 y1 +···+m ym , then y ≥ |||i yi + Yi ||| = |i | ||| yi + Yi ||| for i = 1,...,m, and if x := 1x1 + ··· + m xm , then F(x) = y and x≤|1|x1+···+|m|xm ≤γy. 3.3. Suppose Y is a subspace of X such that Z( f ) ⊂ Y and Y = Z( f ).Let y0 ∈ Y \Z( f ). Consider x ∈ X.Letk := f (x)/f (y0) and y := x − ky0. Then x = y + ky0, where y ∈ Z( f ) ⊂ Y , and so x ∈ Y . Next, let Y denote the closure of Z( f ). Suppose f is continuous. Then Y = Z( f ).IfZ( f ) is dense in X, then Z( f ) = Y = X. Conversely, suppose Z( f ) is not dense in X. Then Z( f ) ⊂ Y and Y = X, and so Y = Z( f ), that is, Z( f ) is a closed subspace of X. Hence f is continuous. · := ∈{ , , ∞} 3.4. Consider the norm p on X c00, where p 1 2 . = ≥ | ( )|≤ ∞ | ( )|= ∈ p 1: If r 0, then fr x j=1 x j x 1 for all x X, and −r fr (e1) = 1, so that f =1. Conversely, if r < 0, then fr (en) = n →∞. / = > / := ∞ −2r 1 2 < ∞ | ( )|≤ p 2: If r 1 2, then α j=1 j , and fr x α x 2 −r −r for all x ∈ X. Further, let xn := (1, 2 ,...,n , 0, 0 ...) ∈ X and αn := / n −2r 1 2 ∈ N = ( ) = 2 ∈ N j=1 j for n . Then xn 2 αn and fr xn αn for n .If > / ( / ) → ≤ / ( / ) →∞ r 1 2, then fr xn αn α, and if r 1 2, then fr xn αn . =∞ > := ∞ −r < ∞ | ( )|≤ p :Ifr 1, then β j=1 j , and fr x βx ∞ for all ∈ := ( ,..., , , ,...) ∈ := n −r x X. Further, let xn 1 1 0 0 X and βn j=1 j for n ∈ N. Then xn∞ = 1 and fr (xn) = βn for each n ∈ N.Ifr > 1, then f (x ) → β, and if r ≤ 1, then f (x ) →∞. r n r n ∈ 1 ( ) ≤ ∞ ∞ | ( )| / 2 ≤ ∞ ∞ | ( )| / 2 3.5. For x , F x 1 i=1 j=i x j i i=1 j=1 x j i = 2 / ( ) =( , / 2,..., / 2, , ,...) → 2/ π x 1 6. Also, F en 1 1 1 2 1 n 00 1 π 6. ∈ 1 ( ) = ∞ | ( )| < ∞ | ( )| Further, if x and x n 0, then j=n+1 x j j=1 x j , and so 2 F(x)1 < π x1/6. 3.6. (i) If P = 0, then there is y ∈ R(P) with y=1, and so P≥P(y)= y=1. Suppose X is an inner product space. If P is an orthogonal projection, then x2 =P(x)2 +x − P(x)2 ≥P(x)2, x ∈ X, and so P≤1. Conversely, let P=0orP=1. Then P≤1. Let y ∈ R(P) and z ∈ Z(P).Ifz := 0, then clearly y, z=0. Let z = 0, and assume z=1 without loss of generality. Deﬁne x := y − y, zz. Then x2 = y2 −| y, z|2 =P(x)2 −| y, z|2 ≤x2 −| y, z|2, and so y, z=0, that is y ⊥ z. Thus P is an orthogonal projection operator. (ii) ||| Q(x)||| = |||x + Z||| ≤ x for all x ∈ X, and so Q≤1. If Z = X, then X/Z ={0 + Z}, and so Q=0. On the other hand, if Z = X Solutions to Exercises 213

and > 0, then by the Riesz lemma, there is x ∈ X such that x=1 and ||| Q(x)||| = d(x, Z)>1−, and so Q > 1−, which shows that Q=1.

3.7. Suppose M := [ki, j ] deﬁnes a map F from c00 to itself. Then the jth column F(e j ) := (k1, j , k2, j ,...) is in c00 for each j ∈ N. Conversely, suppose the jth column (k1, j , k2, j ,...) of M is in c00 for each j ∈ N. Then ∈ N ∈ N = > for each j , there is m j such that ki, j 0 for all i m j . := ( ( ),..., ( ), , ,...) ∈ ( ) := ∞ ( ) = If x x 1 x n 0 0 c00, then y i j=1 ki, j x j n ( ) ∈ K ∈ N ( ) = > { ,..., } j=1 ki, j x j for all i , and so y i 0ifi max m1 mn . Thus y := (y(1), y(2), . . .) ∈ c00. The result for the norm ·1 follows as in Example 3.14(i) since e j ∈ c00 for j ∈ N, and the result for the norm ·∞ follows as in Example 3.14(ii) by := ( ,..., , , ,...)∈ , ∈ N considering xi,m sgn ki,1 sgn ki,m 0 0 c00 for i m . ∈ ∈ N ∞ | , | ≤ ( ) ( ) := 3.8. (i) Let x X.Fori , j=1 ki, j x u j x β1 i , where β1 i ∞ | | ( ) := ∞ , | , | = j=1 ki, j , and so let fi x j=1 ki, j x u j . Also, writing ki, j x u j 1/2 1/2 |ki, j | |ki, j | | x, u j | for i, j ∈ N, Bessel’s inequality shows that

∞ ∞ ∞ ∞ 2 2 2 | fi (x)| ≤ |ki, j | |ki, j || x, u j | ≤ β1α1x < ∞. i=1 i=1 j=1 j=1

(ii) Let x ∈ X.Fori ∈ N,

∞ ∞ 1/2 ∞ 1/2 2 2 2 |ki, j x, u j | ≤ |ki, j | | x, u j | ≤ γ2,2x , j=1 j=1 j=1 ( ) := ∞ , and so let fi x j=1 ki, j x u j . Also, Bessel’s inequality shows that

∞ ∞ ∞ ∞ | ( )|2 ≤ | |2 | , |2 ≤ 2 2. fi x ki, j x u j γ2,2 x i=1 i=1 j=1 j=1 := ∞ ( )v In both cases, y i=1 fi x i belongs to Y by the Riesz–Fischer theorem ( ) := ( )2 ≤ ∞ | ( )|2 for the Hilbert space Y .LetF x y. Then F x i=1 fi x again by the Bessel inequality. √ Hence the matrix M deﬁnes F ∈ BL(X, Y ), and F≤ α1β1 in case (i), ≤ while F γ2,2 in case (ii). = ∈ N ( ) = ∞ / 2 2 = 2/ 2 = = 3.9. p 1: For j , α1 j i=1 1 i j π 6 j , and so F α1 2 π /6. =∞ ∈ N ( ) = ∞ / 2 2 = 2/ 2 = = p :Fori , β1 i j=1 1 i j π 6i , and so F β1 2 π /6. = 2 = ∞ ∞ / 4 4 = ( 4/ )2 ≤ = 4/ p 2: γ2,2 i=1 j=1 1 i j π 90 , and so F γ2,2 π√90. j 2 2 2 Also, if we let x( j) := (−1) /j for j ∈ N, then x ∈ , and x2 = π / 90, 6 3/2 4 whereas F(x)2 = π /(90) . Hence F=π /90. 214 Solutions to Exercises

3.10. Let p ∈{1, 2, ∞}, and (1/p) + (1/q) = 1. As in Example 3.13, fy is a continuous linear map on (X, ·p) and fy≤yq . 2 (i) p = 1: Let t0 ∈ (a, b), and for t ∈[a, b],letxn(t) := n − n |t − t0| if |t − t0|≤1/n and xn(t) := 0 otherwise. Clearly, xn ∈ X, xn ≥ 0 and xn1 ≤ 1forn ∈ N.LetAn := {t ∈[a, b]:t0 − (1/n) ≤ t ≤ t0} and Bn := {t ∈[a, b]:t0 ≤ t ≤ t0 + (1/n)} for n ∈ N. Then

2 2 fy(xn) = n + n (t − t0) y(t)dt + n − n (t − t0) y(t)dt A n Bn 2 2 = n y(t)dt + n (t − t0)y(t)dt − n (t − t0)y(t)dt. An ∪Bn An Bn 2 Now n y(t)dt → y(t0), n y(t)dt → y(t0), n (t − t0)y(t)dt → An Bn An −y(t )/2, and n2 (t − t )y(t)dt → y(t )/2 by the continuity of y at t . 0 Bn 0 0 0 Hence fy(xn) → 2y(t0) − y(t0)/2 − y(t0)/2 = y(t0). Thus y∞ ≤fy. (ii) p =∞:Letxn := ny/(1 + n|y|), n ∈ N. Then xn ∈ X and xn∞ ≤ 1. Since n|y|2/(1 + n|y|) →|y| pointwise and monotonically on [a, b],

b n|y|2 b fy(xn) = dm → |y|dm =y1. a 1 + n|y| a

Thus y1 ≤fy. = := ∈ = ( ) =2 (iii) p 2: If x y X, then x 2 y 2 and fy x y 2. Thus y2 ≤fy. 3.11. If x ∈ X, then clearly F(x) ∈ X. (i) Let x ∈ X. Using the Fubini theorem, we obtain F(x)1 ≤ αx1. Hence F ∈ BL(X) and F≤α1. On the other hand, let t0 ∈ (a, b), and ( ) ∈[ , ] consider the sequence xn given in Exercise 3.10(i). For s a b , deﬁne ( ) := ( , ), ∈[ , ] ∈ b ( ) ( ) → ( ) ys t k s t t a b . Then ys X, and a xn t ys t dt ys t0 for each s ∈[a, b]. By the bounded convergence theorem,

b b b

F(xn)1 = k(s, t)xn(t)dt ds → |k(s, t0)|ds. a a a b |k(s, t )|dt ≤F t ∈ (a, b) Thus a 0 for every 0 . Also, note that the function −→ b | ( , )| [ , ] ≤ t a k s t ds is continuous on a b . Hence α1 F . (ii) For s ∈[a, b], deﬁne ys (t) := k(s, t), t ∈[a, b]. Then ys ∈ X, and β1 = { : ∈[ , ]} sup ys 1 s a b . Consider the linear functional fys on X considered = { : ∈[ , ]} ( )( ) = in Exercise 3.10(ii). Then β1 sup fys s a b . Further, F x s ( ) ∈ ∈[ , ] fys x for all x X and s a b . It follows that

( ) = {| ( )|: ∈[ , ]} ≤ ∈ . F x ∞ sup fys x s a b β1 x ∞ for all x X

Hence F ∈ BL(X) and F≤β1. On the other hand, for each s ∈[a, b], Solutions to Exercises 215

= {| ( )|: ∈ ≤ }≤ fys sup fys x x X and x ∞ 1 F

| ( )|≤ ( ) ∈ ≤ since fys x F x ∞ for all x X. Hence β1 F . (iii) Let x ∈ X and s ∈[a, b]. By the Schwarz inequality,

b b |F(x)(s)|2 ≤ |k(s, t)|dt |k(s, t)||x(t)|2dt a a b 2 ≤ β1 |k(s, t)||x(t)| dt . a ( )2 ≤ b b | ( , )| | ( )|2 ≤ By the Fubini theorem, F x 2 β1 a a k s t ds x t dt β1α1 x2. Hence F ∈ BL(X) and F≤(α β )1/2. 2 1 1 , ∈ N ∞ ( , ) ( , ) K 3.12. (i) For i j , the series n=1 k1 i n k2 n j converges in since ∞ | ( , ) ( , )|≤( ∞ | ( , )|2)1/2( ∞ | ( , )|2)1/2 < ∞ n=1 k1 i n k2 n j n=1 k1 i n n=1 k2 n j . 2 2 Consider x ∈ , and let y := F2(x) ∈ . Then for i ∈ N, ∞ ∞ ∞ F(x)(i) = F1(y)(i) = k1(i, n)y(n) = k1(i, n) k2(n, j)x( j) n=1 n=1 j=1 ∞ ∞ ∞ = k1(i, n)k2(n, j) x( j) = k(i, j)x( j). j=1 n=1 j=1

Interchanging the order of summation is justiﬁed since

∞ ∞ 2 ∞ ∞ | ( , ) ( , ) ( )| ≤ | ( , ) ( )|2 2 k1 i n k2 n j x j k1 i n x j γ2,2 n=1 j=1 n=1 j=1 ∞ = | ( , )|2 2 2 , k1 i n x 2γ2,2 n=1 2 := ∞ ∞ | ( , )|2 where γ2,2 n=1 j=1 k2 n j . (See, for example, [13, Proposition 7.21].) Hence the matrix M := [k(i, j)] deﬁnes the map F.Also,

∞ ∞ ∞ ∞ ∞ 2 2 2 F ≤ |k(i, j)| ≤ |k1(i, n)k2(n, j)| i=1 j=1 i=1 j=1 n=1 ∞ ∞ ∞ ∞ 2 2 ≤ |k1(i, n)| |k2(n, j)| i=1 j=1 n=1 n=1 ∞ ∞ ∞ ∞ 2 2 = |k1(i, n)| |k2(n, j)| . i=1 n=1 n=1 j=1 216 Solutions to Exercises

(ii) Replace i, j, n ∈ N by s, t, u ∈[a, b] respectively, and replace summation by Lebesgue integration in (i) above. n 3.13. The seminorm p is discontinuous on X:Letxn(t) := t /n for n ∈ N and ∈[ , ] ∈ = ( ) = − →∞ t 0 1 . Then xn X and xn 1,∞ 1, but p xn n 1 . := ∞ The seminorm pis countably subadditive on X:Lets k=1 xk in X with ∞ ( ) = ∞ < ∞ := ([ , ]) k=1 p xk k=1 xk ∞ .LetY C 0 1 with the sup norm. ∞ Since Y is a Banach space, the absolutely summable series k=1 xk of terms := ∞ ∈ := n in Y is summable in Y .Lety k=1 xk Y . Deﬁne sn k=1 xk ∈ N → ( ) 1([ , ]) for n . Since sn s in X, the sequence sn in C 0 1 converges ∈ 1([ , ]) ( ) (uniformly) to the function s C 0 1 , and the derived sequence sn , = n , ∈ N where sn k=1 xk n , converges uniformly to the functiony.Bya = ( ) = = ∞ well-known theorem in Real Analysis, y s s . Thus s k=1 xk , ( ) = ≤ ∞ = ∞ ( ) and so p s s ∞ k=1 xk ∞ k=1 p xk . 3.14. (i) If xn := kn,1 y1 +···+kn,m ym → x := k1 y1 +···+km ym in X, then

|p(xn)− p(x)|≤p(xn −x) ≤|kn,1 −k1|p(y1)+···+|kn,m −km |p(ym ) → 0.

(ii) Let p be a lower semicontinuous seminorm on a Banach space X. With → ( ) ≤ ∞ ( ) notation as in Lemma 3.18, sn s in X, and p sn k=1 p xk for ∈ N. ( ) ≤ { ( ) : ≥ }≤ ∞ ( ) n Hence p s limn→∞ inf p sm m n k=1 p xk . Thus p is countably subadditive. By the Zabreiko theorem, p is continuous. r 3.15. If F ∈ BL(X, ) and j ∈ N, then | f j (x)|=|F(x)( j)|≤F(x)r ≤ Fx for all x ∈ X, and so f j ∈ BL(X, K). Conversely, suppose f j ∈ BL(X, K) for all j ∈ N. r = 1: For n ∈ N,letpn(x) := | f1(x)|+···+|fn(x)|, x ∈ X. Then each pn is a continuous seminorm on X, and for each x ∈ X, pn(x) ≤F(x)1 for all n ∈ N. By Corollary 3.22, there is α > 0 such that pn(x) ≤ αx for all n ∈ N and x ∈ X, and so F(x)1 ≤ αx for all x ∈ X. Hence F ∈ BL(X,r ). A similar argument holds if r ∈{2, ∞}: / 2 2 1 2 r = 2: For n ∈ N,letpn(x) := | f1(x)| +···+|fn(x)| , x ∈ X. r =∞:Forn ∈ N,letpn(x) := | fn(x)|, x ∈ X. r r Aliter: The map F : X → is closed: Let xn → 0inX and F(xn) → y in .

Then F(xn)( j) → y( j), and also F(xn)( j) = f j (xn) → 0, so that y( j) = 0 for each j ∈ N. By the closed graph theorem, F is continuous.

3.16. Let E be a totally bounded subset of X, and let > 0. Find x1,...,xm in E such that E ⊂ U(x1, ) ∪···∪U(xm, ). Deﬁne F(x) := limn→∞ Fn(x) for x ∈ X. There is n0 such that Fn(x j ) − F(x j ) < for all n ≥ n0 and j = 1,...,m.Letx ∈ E, and choose x j such that x − x j < . By Theorem 3.24, there is α > 0 such that Fn≤α for all n ∈ N, and F≤α. Hence

Fn(x) − F(x)≤Fn(x − x j )+Fn(x j ) − F(x j )+F(x j − x)

≤Fnx − x j +Fn(x j ) − F(x j )+Fx − x j ≤ (2α + 1) Solutions to Exercises 217

for all n ≥ n0. Thus (Fn(x)) converges to F(x) uniformly for x ∈ E.

3.17. Suppose Fn≤α for all n ∈ N, and there is E ⊂ X with span E dense in X and (Fn(x)) is Cauchy in Y for each x ∈ E.LetX0 := span E. Since Y is a Banach space, (Fn(x)) converges in Y for each x ∈ E, and hence for each x ∈ X0. Deﬁne F0(x) := limn→∞ Fn(x) for x ∈ X0. Then F0(x)≤ limn→∞ Fn(x)≤αx for all x ∈ X0. Thus F0 ∈ BL(X0, Y ) and F0≤ α. By Proposition 3.17(ii), there is F ∈ BL(X, Y ) satisfying F(x0) = F0(x0) for all x0 ∈ X0 and F=F0≤α.Letx ∈ X and > 0. Since X0 is dense in X, there is x0 ∈ X0 such that x−x0 < .Also,Fn(x0) → F0(x0) = F(x0) in Y , and so there is n0 ∈ N such that Fn(x0) − F(x0) < for all n ≥ n0. Hence

Fn(x) − F(x)≤Fn(x − x0)+Fn(x0) − F(x0)+F(x0 − x)

≤Fnx − x0+Fn(x0) − F(x0)+Fx0 − x ≤ (2α + 1)

for all n ≥ n0. Thus Fn(x) → F(x) in Y for all x ∈ X. Conversely, let F ∈ BL(X, Y ) be such that Fn(x) → F(x) in Y for each

x ∈ X. Then (Fn) is bounded by the Banach–Steinhaus theorem. ∈ N ∈ ( ) ( ) ⊂ ( ) ( ) = 3.18. If n0 and x R Pn 0 , then R Pn0 R Pn , and so Pn x x for ≥ := ∞ ( ) ( ) → ∈ all n n0. Thus if E n=1 R Pn , then Pn x x for each x E.In Exercise 3.17, let Y := X, Fn := Pn for n ∈ N, and F = I . n 3.19. Deﬁne x0(t) := 1 and xn(t) := t for n ∈ N and t ∈[a, b]. In Polya’s theorem, let E := {x0, x1, x2,...}. Then span E is the linear space of all polynomial functions on [a, b], which is dense inC([a, b]) by the Weierstrass theorem. mn |w |= mn w = ( ) → If all weights are nonnegative, then j=1 n, j j=1 n, j Qn x0 Q(x0) = b − a.

3.20. (i) xX ≤xF for all x ∈ X.Also,F is continuous if and only if there is α > 0 with F(x)Y ≤ αxX , that is, xF ≤ (1 + α)xX for all x ∈ X. (ii) By the closed graph theorem, F is continuous, and so ·F is equivalent to ·X by (i) above. Hence (X, ·F ) is a Banach space (Exercise 2.23). (iii) Let (xn) be a sequence in X such that xnX → 0 and there is y ∈ Y with F(xn) − yY → 0. Then (xn) is a Cauchy sequence in (X, ·F ) since (xn) is a Cauchy sequence in (X, ·X ) and (F(xn)) is a Cauchy sequence in (Y, ·Y ). Since (X, ·F ) is a Banach space, there is x ∈ X such that xn − xF → 0, that is, xn − xX → 0 and F(xn) − F(x)Y =F(xn − x)Y → 0. Hence x = 0 and F(x) = y, so that y = F(0) = 0. (iv) The comparable norms ·X and ·F are equivalent by the two-norm theorem. Hence F is continuous by (i) above.

3.21. F is linear: Let F(x1) := x2 and F(x˜1) :=x ˜2. Then F1(x1) = F2(x2) and F1(x˜1) = F2(x˜2). Since F1 and F2 are linear, F1(x1 +˜x1) = F2(x2 +˜x2), that is, F(x1 +˜x1) = x2 +˜x2 = F(x1) + F(x˜1). Similarly, F(kx1) = kF(x1). 218 Solutions to Exercises

F is a closed map: Let x1,n → x1 in X1 and F(x1,n) → x2 in X2. Define x2,n := F(x1,n) for n ∈ N. Since F1 is continuous, F1(x1,n) → F1(x1), and since F2 is continuous, F2(x2,n) → F2(x2).ButF1(x1,n) = F2(x2,n) since F(x1,n) = x2,n for all n ∈ N. Hence F1(x1) = F2(x2), that is, F(x1) = x2. Since X and X are Banach spaces, F is continuous. 1 2 ∈ q = ∞ | ( ) ( )|≤ ∈ 1 3.22. (i) If y , then xy 1 j=1 x j y j x p y q , and so xy p 1 p for all x ∈ . Conversely, suppose xy ∈ for all x ∈ .LetM := [ki, j ], where k1, j := y( j) for j ∈ N, and ki, j := 0 otherwise. Then M defines p 1 q a map from to , and so its first row y = (y(1), y(2),...) is in by ( ) := ∞ ( ) ( ) ∈ p Corollary 3.26. Also, if we let fy x j=1 x j y j for x , then F=fy=yq . r r ∞ (ii) If y ∈ , then xyr ≤x∞yr , and so xy ∈ for all x ∈ , and r ∞ F≤yr . Conversely, suppose xy ∈ for all x ∈ .Letx := (1, 1,...). ∞ r Since x ∈ , y = xy ∈ and yr =xyr =F(x)r ≤F. ∞ (iii) If y ∈ and p ≤ r, then xyr ≤xr y∞ ≤xpy∞, and so r p xy ∈ for all x ∈ , and F≤y∞. Conversely, suppose p ≤ r and xy ∈ r for all x ∈ p.LetM := diag (y(1), y(2), . . .). Then M defines the matrix transformation F from p to r , and so F is continuous by Proposition 3.30. Hence |y( j)|=F(e ) ≤F for j ∈ N, and so y∞ ≤F. j r ∈ q = 1 | ( ) ( )| ( ) ≤ ∈ 1 3.23. (i) If y L , then xy 1 0 x t y t dm t x p y q , and so xy L for all x ∈ L p. Conversely, suppose xy ∈ L1 for all x ∈ L p.Forn ∈ N, ( ) := ( ) | ( )|≤ ( ) := ∈ ∞ ⊂ let yn t y t if y t n and yn t 0 otherwise. Then yn L Lq n ∈ N f (x) := 1 x(t)y (t)dm(t), x ∈ L p .For , define n 0 n , and note that = ∈ p ( ) → 1 ( ) ( ) ( ) fn yn q .Ifx L , then fn x 0 x t y t dm t by the dominated convergence theorem, and so there is α > 0 such that ynq =fn≤α for all n ∈ N by Theorem 3.24.Ifp = 1, then the set

∞ {t ∈[0, 1]:|y(t)| > α}= {t ∈[0, 1]:|yn(t)| > α} n=1

= | |≤ = 2 → is of measure zero, and so y ∞ ess sup y α.Ifp 2, then yn 2 1 | ( )|2 ( ) ≤ 0 y t dm t by the monotone convergence theorem, and so y 2 α.If =∞ ( ) := ∈[ , ] = ∈ 1 p , then letting x t 1fort a b , we see that y xy L . Thus ∈ q ( ) := 1 ( ) ( ) ( ) ∈ p y L in all cases. Also, if we let fy x 0 x t y t dm t for x L , then F=fy=yq . r r ∞ (ii) If y ∈ L , then xyr ≤x∞yr , and so xy ∈ L for all x ∈ L , and r ∞ F≤yr . Conversely, suppose xy ∈ L for all x ∈ L .Letx(t) := 1for ∞ r t ∈[a, b]. Since x ∈ L , y = xy ∈ L and yr =xyr =F(x)r ≤ F. ∞ 2 2 (iii) If y ∈ L , then xy2 ≤x2y∞, and so xy ∈ L for all x ∈ L , 2 2 and F≤y∞. Conversely, suppose xy ∈ L for all x ∈ L . We show that 2 2 F is a closed map. Let xn → x in L and F(xn) = xn y → z in L . Since Solutions to Exercises 219

xy−z1 ≤(x − xn)y1 +xn y −z1 ≤x − xn2y2 +xn y −z2 → 0, we see that z = xy = F(x). By the closed graph theorem, F is continuous. Let > 0, and let E := {t ∈[0, 1]:|y(t)| > F+}. Assume for a moment thatm(E)>0. If x denotes the characteristic function of E, then |x| F+ ≤|xy| on [0, 1], and F+ x2 ≤xy2 =F(x)2 ≤ Fx2, where x2 = 0. This is impossible. Hence |y|≤F+ a.e. on [0, 1]. Since this holds for every > 0, y∞ = ess sup|y|≤F.

3.24. Let X := C([a, b]). Then (X, ·∞) is a Banach space, and xn −x∞ → 0if and only if (xn) converges to x uniformly on [a, b]. We show that the identity map I from (X, ·∞) to (X, ·) is a closed map. Let xn∞ → 0 and xn − y=I (xn) − y→0. Then y(t) = limn→∞ xn(t) = 0 for every t ∈[a, b], that is, y = 0. By the closed graph theorem, I is continuous, and so there is α > 0 such that x≤αx∞ for all x ∈ X, that is, the norm ·∞ is stronger than the norm ·. The norms ·∞ and ·on X are complete and comparable, and so they are equivalent. 3.25. Let W := Y × Z, and (y, z) := y+z for (y, z) ∈ W. Then (W, · ) is a Banach space. Define F : W → X by F(y, z) := y + z for (y, z) ∈ W. Then F is linear, onto, and F(y, z)=y + z≤y+z=(y, z) for (y, z) ∈ W. By the open mapping theorem, F is an open map. Hence there is γ > 0 such that for every x ∈ X, there is (y, z) ∈ W with x = F(y, z) = y+z and y+z=(y, z) ≤ γx. 3.26. Since F is an open map, there is γ > 0 such that for each n ∈ N, there is zn ∈ X with F(zn) = yn − y and zn≤γyn − y.Letxn := x + zn for n ∈ N. Then F(xn) = F(x) + F(zn) = yn and xn − x=zn→0. 3.27. Note: In this exercise, the bounded inverse theorem and the open mapping theorem are deduced directly from the Zabreiko theorem, and then closed graph theorem follows. (i) Since F is continuous, one-one and onto, F and F −1 are closed maps. In Remark 3.29, replace X, Y, F and p by Y, X, F −1 and q respectively. If both X ad Y are Banach spaces, the continuity of q follows from the Zabreiko theorem. := ( ) ∈ ( / , ) (ii) Let Z Z F . Then F BL X Z Y is one-one and onto. Also, q(y) = (F)−1(y) for y ∈ Y .IfX is a Banach space, then q is a countably subadditive seminorm on Y , and if Y is also a Banach space, then q is continuous, as in (i) above, and so there is γ > 0 such that q(y)<γy for every y ∈ Y . By the definition of an infimum, for every y ∈ Y , there is x ∈ X satisfying F(x) = y and x≤γy. Hence F is an open map by Proposition 3.41. (iii) is continuous since (x, F(x))=x≤x+F(x) for x ∈ X. If X and Y are Banach spaces, and F is a closed map, then Gr(F) is a closed subspace of the Banach space X × Y , and so −1 ∈ BL(X, Gr(F)) by (i) above. Hence there is α > 0 such that F(x)≤x+F(x)≤αx for all x ∈ X. 220 Solutions to Exercises

3.28. (i) Let p ≤ p and r ≤ r . Then xp ≤xp and F(x)r ≤F(x)r for p p r p r all x ∈ . Hence BL( , ) ⊂ BL( , ), and Fp ,r = sup{F(x)r : p p x ∈ and xp ≤ 1}≤sup{F(x)r : x ∈ and xp ≤ 1}=Fp,r for all F ∈ BL(p,r ). p r p Let F ∈ CL( , ), and let (xn) be a bounded sequence in . Then (xn) p ( ) is a bounded sequence in , and so there is a subsequence xnk such that r r p r (F(xn )) converges in , and hence in . Thus F ∈ CL( , ). k (ii) Let p ≥ p and r ≥ r . Replace p,r ,p ,r by L p, Lr , L p , Lr respectively in (i) above. n 3.29. Let F := k1 F +···+kn F . Then F ∈ CL(X). Hence G = k0 I + F belongs to CL(X) if and only if k0 I belongs to CL(X), that is, k0 = 0. 3.30. Since P2 = P and P is a closed map, R(P) is a closed subspace of X.The desired result follows from Theorem 3.42. 3.31. (i) Suppose M defines a map F from p to r . Then F is continuous. Since ( ) = ( ) ∈ N = { ( ) : ∈ N}≤ αr j F e j r for j , αr sup F e j r j F .Next, ∈ N ( ) := ∞ ( ), ∈ p | ( )|≤ let i , and define fi x j=1 ki, j x j x . Then fi x p F(x)r ≤Fxp for all x ∈ , and so fi ≤F. Since βq (i) =fi for i ∈ N, βq = sup{ fi :i ∈ N}≤F. (See Corollary 3.26.) (ii) Let p = 1, and r ∈{1, 2, ∞}. 1 r First suppose αr < ∞. We show that M defines a map from to . r = 1: This is worked out in the text (Example 3.14(i)). 1 ∞ r = 2: Let x ∈ and i ∈ N. Then |k , x( j)|≤α x .Lety(i) := j=1 i j 2 1 ∞ ( ) | ( )|= | || ( )|1/2 | ( )|1/2 j=1 ki, j x j . Writing ki, j x j ki, j x j x j , we obtain

∞ 2 ∞ ∞ ∞ 2 2 |ki, j x( j)| ≤ |ki, j | |x( j)| |x( j)|= |ki, j | |x( j)| x1. j=1 j=1 j=1 j=1 ∞ | ( )|2 ≤ ∞ ∞ | |2 | ( )|≤ 2 2 := Hence i=1 y i x 1 j=1 i=1 ki, j x j α2 x 1, and y ( ( ), ( ),...)∈ 2 1 2 ≤ y 1 y 2 . Thus M defines a mapF from to , and F α2. 1 ∞ r =∞:Letx ∈ and i ∈ N. Then |k , x( j)|≤α∞x .Let j=1 i j 1 ( ) := ∞ ( ) | ( )|≤ := ( ( ), ( ),...)∈ ∞ y i j=1 ki, j x j . Since y i α∞ x 1, y y 1 y 2 . 1 ∞ Thus M defines a map F from to , and F≤α∞. 1 r Conversely, if M defines a map F from to , then αr ≤F < ∞.

Lastly, let r ∈{1, 2, ∞}, and assume that αr ( j) → 0. For n ∈ N, let Mn denote the infinite matrix whose first n columns are the same as those of the matrix M, and the remaining columns are zero. Then the matrices Mn 1 r and M − Mn define maps Fn and F − Fn from to respectively, and F − Fn=sup{αr ( j) : j = n + 1, n + 2,...} for each n ∈ N. Since each 1 r Fn is of finite rank and F − Fn→0, F ∈ CL( , ). (iii) Let r =∞, and p ∈{1, 2, ∞}. p ∞ First suppose βq < ∞. We show that M defines a map from to . Solutions to Exercises 221

p =∞: This is worked out in the text (Example 3.14(ii)). 2 ∞ p = 2: Let x ∈ and i ∈ N. Then |k , x( j)|≤β (i)x .Lety(i) := j=1 i j 2 2 ∞ ( ) | ( )|≤ := ( ( ), ( ),...) ∈ ∞ j=1 ki, j x j . Hence y i β2 x 2, and y y 1 y 2 . 2 ∞ ≤ Thus M defines a map F from to , and F β2. 1 ∞ p = 1: Let x ∈ and i ∈ N. Then |k , x( j)|≤β∞(i)x .Lety(i) := j=1 i j 1 ∞ ( ) | ( )|≤ := ( ( ), ( ), . . .) ∈ ∞ j=1 ki, j x j . Hence y i β∞ x 1, and y y 1 y 2 . 1 ∞ Thus M defines a map F from to , and F≤β∞. p ∞ Conversely, if M defines a map F from to , then βq ≤F < ∞.

Lastly, let p ∈{1, 2, ∞}, and assume that βq (i) → 0. For n ∈ N, let Mn denote the infinite matrix whose first n rows are the same as those of the matrix M, and the remaining rows are zero. Then the matrices Mn and M − Mn define p ∞ maps Fn and F − Fn from to respectively, and F − Fn=sup{βq (i) : i = n + 1, n + 2,...} for each n ∈ N. Since each Fn is of finite rank and p ∞ F − Fn→0, F ∈ CL( , ). 3.32. We use Exercise 3.31. √ (i) For j ∈ N, α1( j) = j, α2( j) = j, and α∞( j) = 1, while for i ∈ N, p r β∞(i) = 1 and β2(i) = β1(i) =∞.IfM defines a map from to , then αr < ∞ and βq < ∞, and so p = 1 and r =∞. Conversely, suppose p = 1 and r =∞. Then M defines a map F ∈ BL(1,∞), and F= 1 ∞ α∞ = β∞ = 1. But F ∈/ CL( , ), since e j 1 = 1 for all j ∈ N, and F(e j ) − F(ek )∞ =ek+1 +···+e j ∞ = 1 for all j > k in N. / ∈ N ( ) = j / , ( ) = j / 2 1 2 ( ) = (ii) For j , α1 j k=1 1 k α2 j k=1 1 k , and α∞ j 1, while for i ∈ N, β∞(i) = 1/i, β2(i) = β1(i) =∞.IfM defines a map p r from to , then αr < ∞ and βq < ∞, and so p = 1 and r ∈{2, ∞}. Conversely, suppose√p = 1 and r ∈{2, ∞}. Then M defines F in BL(1,r ), and F=α2 = π/ 6ifr = 2, and F=α∞ = β∞ = 1ifr =∞.Tosee 1 2 that F ∈ CL( , ),letMn denote the infinite matrix whose first n rows are the same as those of the matrix M, and the remaining rows are zero. Then the 1 2 matrices Mn and M −Mn define maps Fn and F −Fn from to respectively, / − = ∞ / 2 1 2 → . each Fn is of finite rank, and F Fn k=n+1 1 k 0 Also, 1 ∞ F ∈ CL( , ) since β∞(i) → 0. 3.33. (i) Let p = 1. By Exercise 3.31(ii), the converse of Corollary 3.31 holds. Let p ∈{2, ∞}. Define M := [ki, j ], where k1, j := 1 for all j ∈ N, and := ∈{ , , ∞} ( ) = ∈ N ki, j 0 otherwise. Let r 1 2 . Then αr j 1 for allj , and so = ( ) := / ∈ N ∈ p ∞ ( ) αr 1. If x j 1 j for j , then x , but the series j=1 k1, j x j does not converge in K. Hence M does not define a map from p to r . (ii) Let r =∞. By Exercise 3.31(iii), the converse of Corollary 3.26 holds.

Let r ∈{1, 2}. Deﬁne M := [ki, j ], where ki,1 := 1 for all i ∈ N, and ki, j := 0 ∈{ , , ∞} ( ) = ∈ N = otherwise. Let p 1 2 . Then βq i 1 for all i , and so βq 1. := ∈ p ∞ ( ) = ∈ N If x e1, then x , and j=1 ki, j x j 1 for all i . However, (1, 1,...)/∈ r . Hence M does not deﬁne a map from p to r . (iii) Let p ∈{2, ∞} and r ∈{1, 2}.Ifp > r, then there is x ∈ p\r , and so 222 Solutions to Exercises

the identity matrix I does not define a map from p to r , although the r-norm of each column of I and the q-norm of each row of I is equal to 1. If p = 2 = r,letM denote√ the matrix which has an n × n diagonal block with all entries equal to 1/ n for each n = 1, 2,...in that order, and whose all other entries are equal to 0. Then the 2-norm of each column as well as each row of M is equal to 1. Assume for a moment that M defines an 2 2 operator F on . Then√ F ∈ √BL( ) by Proposition √3.30.Forn ∈ N,let xn := (0,...,0, 1/ n,...,1/ n, 0, 0,...),where 1/ n occurs only in the ( − ) / + ,..., ( + )/ = n places numbered√ n 1 n 2 1 n n 1 2. Then xn 2 1, but F(xn)2 = n →∞, which is impossible. p r 3.34. Let p ∈{2, ∞}, r ∈{1, 2, ∞}. Suppose M := [ki, j ] defines F ∈ CL( , ). Assume that F(e j )r = αr ( j) → 0. Then there are j1 < j2 < ··· in N and p r there is δ > 0 such that αr ( jk ) ≥ δ for all k ∈ N. Since F ∈ CL( , ), ( ) ( ) ∈ r there is a subsequence em of the sequence e jk , and there is y such r that F(em) → y in .Fixi ∈ N.Nowβq (i) ≤ βq ≤F (Exercise 3.31(i)), where q ∈{1, 2}. Hence F(em)(i) = ki,m → 0, and so y(i) = limm→∞ F(em )(i) = 0. Thus y = 0. But yr = limm→∞ F(em )r ≥ δ. Hence αr ( j) → 0. Let p = 1. If M := [ki, j ], where k1, j := 1for j ∈ N, and ki, j := 0 otherwise, 1 r then M defines a map in CL( , ),butαr ( j) = 1 for all j ∈ N.

3.35. (i) Define a0 := c0 := 0. For j ∈ N, α1( j) =|a j |+|b j |+|c j−1| and 2 2 2 2 α2( j) =|a j | +|b j | +|c j−1| , while for i ∈ N, β1(i) =|ai−1|+|bi |+|ci |. 1 Now M defines F ∈ BL( ) if and only if α1 := sup{α1( j) : j ∈ N} < ∞, ∞ and M defines F ∈ BL( ) if and only β1 := sup{β1(i) : i ∈ N} < ∞.Also, 2 if α1 < ∞ and β1 < ∞, then M defines F ∈ BL( ), and conversely, if M 2 defines F ∈ BL( ), then α2 < ∞ (Exercise 3.31(i)). All these statements hold if and only if (a j ), (b j ), (c j ) are bounded sequences. Let a j → 0, b j → 0 and c j → 0. Then α1( j) → 0 and β1(i) → 0, so that F ∈ CL(p) for p ∈{1, 2, ∞}. To prove the converse, note that for j ∈ N, α∞( j) = max{|a j |, |b j |, |c j−1|}, while for i ∈ N, β∞(i) = 1 max{|ai−1|, |bi |, |ci |}.IfF ∈ CL( ), then β∞(i) → 0 (Exercise 4.21), if 2 ∞ F ∈ CL( ), then α2( j) → 0 (Exercise 3.34), and if F ∈ CL( ), then α∞( j) → 0 (Exercise 3.34). In each case, a j → 0, b j → 0 and c j → 0. (ii) Let a j := c j := 0 and b j := k j for all j ∈ N in (i) above.

(iii) Let a j := w j and b j := c j := 0 for all j ∈ N in (i) above.

3.36. Let p ∈{2, ∞} and let r ∈{1, 2}.Forn ∈ N, let Mn denote the infinite matrix whose first n rows are the same as those of the matrix M, and the remaining rows are zero. =∞ = ∈ ∞ ∞ ∞ | ( )|≤ p and r 1: For x , i=1 j=1 ki, j x j γ1,1 x ∞. Hence M ∈ (∞,1) ≤ ∈ (∞,1) defines F BL , and F γ1,1.Also,Mn defines Fn CL , ∞ ∞ ∞ 1 and F − F ≤ |k , |→0. Hence F ∈ CL( , ). n i=n+1 j=1 i j =∞ = ∈ ∞ ∞ ∞ | ( )| 2 ≤ 2 2 p and r 2: For x , i=1 j=1 ki, j x j γ1,2 x ∞. Hence Solutions to Exercises 223

∈ (∞,2) ≤ M defines F BL , and F γ1,2. Also, the matrix Mn defines ∈ (∞,2) − 2 ≤ ∞ ∞ | | 2 → Fn CL , and F Fn i=n+1 j=1 ki, j 0. Hence ∞ 2 F ∈ CL( , ). = = ∈ 2 ∞ ∞ | ( )|≤ p 2 and r 1: For x , i=1 j=1 ki, j x j γ2,1 x 2. Hence M ∈ (2,1) ≤ ∈ (2,1) defines F BL , and F γ2,1.Also,Mn defines Fn CL , − ≤ ∞ ( ) → ∈ (2,1) and F Fn i=n+1 β2 i 0. Hence F CL . The case p = 2, r = 2 is treated in the text (Example 3.14(iii)). ∞ 1 ∞ 1 Let γ1,1 < ∞. Then M defines F ∈ CL( , ), and CL( , ) is contained in CL(p,r ) for all p, r ∈{1, 2, ∞} (Exercise 3.28(i)). Note that γ2,2 ≤ γ1,2, γ2,1 ≤ γ1,1.

3.37. Let n ∈ N. Then kn(· , ·) ∈ C([0, 1]×[0, 1]), and for x ∈ X and s ∈[0, 1], |Fn(x)(s)|≤kn(· , ·)∞x1, so that Fn(x)∞ ≤kn(· , ·)∞x1. Thus i n−i Fn ∈ BL(X, Y ).Fori = 0, 1,...,n,letyi (s) := s (1 − s) , s ∈[0, 1]. Then yi ∈ Y for each i, and

n n n i j n 1 F (x)(s)= c y (s), where c := k , t j (1 − t)n− j dt n i i i i n n j i=0 j=0 0

for all x ∈ X and s ∈[0, 1]. Hence each Fn is of ﬁnite rank. Also, it follows that Fn − F≤kn(· , ·) − k(· , ·)∞ → 0. (See [6, p.10].) Since Y is a Banach space, F ∈ CL(X, Y ). 3.38. Let E denote the closure of F(U). Suppose F ∈ CL(X, Y ), and let (y˜n) be a sequence in E. Then there is yn ∈ F(U) such that yn −˜yn < 1/n, and there is xn ∈ U such that ( ) = ∈ N ( ) ( ) F xn yn for each n .Let ynk be a subsequence of yn such that → ˜ → ∈ ynk y in Y . Then ynk y, and y E. Hence E is a compact subset of Y . In particular, every sequence in F(U) has a Cauchy subsequence, that is, F(U) is totally bounded. Conversely, suppose E is a compact subset of Y .Let(xn) be a bounded

sequence in X. There is α > 0 such that xn/α belongs to U, and let := ( / ) ∈ N ( ) ( ) yn F xn α for n .Let ynk be a subsequence of yn such that → ( ) → ∈ ( , ) ynk y in E. Then F xnk αy in Y . Hence we see that F CL X Y . This conclusion also holds if we assume that F(U) is totally bounded and Y is a Banach space, since a Cauchy subsequence in F(U) converges in Y . 3.39. By Exercise 3.38, F(U) is a totally bounded subset of Y , and by Exercise 3.16, (Fn(y)) converges uniformly to F(y), y ∈ F(U).Now(Fn − F)F= sup{(F − F)(F(x)):x ∈ U}=sup{(F − F)(y):y ∈ F(U)}→0. n n ( ˜ )2 = ∗( ˜ )2 = ( )2 3.40. By Exercise 4.31(iii), k A uk k A uk j A u j . (i) Let A ∈ BL(H, G) be a Hilbert–Schmidt map, and let {u , u ,...} be a 1 2 ( )2 < ∞ countable orthonormal basis for H such that j A u j . Consider = , ∈ ( ) = , ( ) ∈ N x j x u j u j H. Then A x j x u j A u j .Forn , deﬁne 224 Solutions to Exercises ( ) := n , ( ), ∈ ∈ ( , ) An x j=1 x u j A u j x H. Since An BL H G is of ﬁnite rank, it is a compact linear map for each n ∈ N. Also, for all x ∈ X,

2 2 2 2 A(x) − An(x) ≤ | x, u j |A(u j ) ≤ A(u j ) x . j>n j>n − 2 ≤ ( )2 → Hence A An j>n A u j 0, and so A is a compact map. 2 (ii) Let A ∈ BL( ). Suppose A is defined by a matrix M := [k , ]. Since ij ( )( ) = , ∈ N ∞ ( )2 = ∞ ∞ | |2 = 2 A e j i ki, j for i j , j=1 A e j 2 j=1 i=1 ki, j γ2,2. If γ2,2 < ∞, then clearly A is a Hilbert–Schmidt map. Conversely, suppose 2 A is a Hilbert–Schmidt map on . Then there is a denumerable orthonormal {˜ , ˜ ,...} 2 ∞ (˜ )2 < ∞ := ( )( ) basis e1 e2 for such that k=1 A ek 2 .Letki, j A e j i , ∈ N 2 = ∞ ( )2 = ∞ (˜ )2 < ∞ for i j . Then γ2,2 j=1 A e j 2 k=1 A ek 2 , and so A is defined by the matrix M := [ki, j ] satisfying γ2,2 < ∞. 2 (iii) Let A ∈ BL(L ).Let{u1, u2,...} be a denumerable orthonormal basis for L2 consisting of continuous functions on [a, b].Fori, j ∈ N,let wi, j (s, t) := ui (s)u j (t). Then {wi, j : i, j ∈ N} is a denumerable orthonormal basis for L2([a, b]×[a, b]). Suppose A is a Fredholm integral operator defined by a kernel k(· , ·) in L2([a, b]×[a, b]). Define

b b ci, j := k(s, t)wi, j (s, t)dm(s)dm(t) for i, j ∈ N. a a

Then A(u j ), ui =ci, j for all i, j ∈ N. By Parseval’s formula,

∞ ∞ ∞ ∞ ∞ ( )2 = | ( ), |2 = | |2 = (· , ·)2. A u j 2 A u j ui ci, j k 2 j=1 j=1 i=1 j=1 i=1

Hence A is a Hilbert–Schmidt map. Conversely, suppose A is a Hilbert– := ( ), , ∈ N Schmidt map. Deﬁne ci, j A u j ui for i j . Arguing as in (ii) ∞ ∞ | |2 = ∞ ( )2 < ∞ above, we obtain i=1 j=1 ci, j j=1A uj 2 . The Riesz– ∞ ∞ w Fischer theorem shows that the double series i=1 j=1 ci, j i, j converges in L2([a, b]×[a, b]),tosay,k(· , ·).LetB denote the Fredholm integral opera- 2 tor on L deﬁned by the kernel k(· , ·). Then B(u j ), ui =ci, j = A(u j ), ui for all i, j ∈ N. Hence A = B.

Chapter 4

4.1. (i) Let a := (1, 0). Clearly, g is linear, continuous, and g=1 = g(a).A function f : K2 → K is a Hahn–Banach extension of g to K2 if and only if Solutions to Exercises 225

2 f is linear on K and f =1 = f (a), that is, there are k1, k2 ∈ K such that 2 f (x) = k1x(1) + k2x(2) for all x := (x(1), x(2)) ∈ K , f =|k1|+|k2|= 2 1, and k1 = 1. Hence the only Hahn–Banach extension of g to K is given by f (x) := x(1) for x := (x(1), x(2)) ∈ K2. (ii) Let b := (1, 1). Clearly, h is linear, continuous, and h=1 = h(b).A function f : K2 → K is a Hahn–Banach extension of h to K2 if and only 2 if f is linear on K and f =1 = f (b), that is, there are k1, k2 ∈ K 2 such that f (x) = k1x(1) + k2x(2) for all x := (x(1), x(2)) ∈ K , f = |k1|+|k2|=1, and k1 + k2 = 1. But for k1 ∈ K, |k1|+|1 − k1|=1 if and 2 only if k1 ∈[0, 1]. Hence the Hahn–Banach extensions of h to K are given f (x) := tx( ) +( −t)x( ) x := (x( ), x( )) ∈ K2 t ∈[ , ] by t 1 1 2 for 1 2 , where 0 1 . > ≤ 4.2. Suppose there is α 0 such that s cs ks α s cs xs as stated. Let := { : ∈ } := ∈ ( ) := Y span xs s S , and for y s cs xs Y , deﬁne g y s cs ks. Then g ∈ Y and g≤α, and so there is f ∈ X such that f =g≤α and f (xs ) = g(xs ) = ks for all s ∈ S. The converse holds with α := f . 4.3. By the Hahn–Banach extension theorem, E =∅. E is convex: Suppose f1, f2 ∈ E, t ∈ (0, 1), and f := (1 − t) f1 + tf2. Then

f (y) = (1 − t) f1(y) + tf2(y) = (1 − t)g(y) + tg(y) = g(y) for all y ∈ Y , and so g≤f ≤(1 − t) f1+t f2=(1 − t)g+tg=g.

E is closed: Suppose ( fn) is in E, and f ∈ X such that fn − f →0.

Then f (y) = limn→∞ fn(y) = limn→∞ g(y) = g(y) for all y ∈ Y , and f =limn→∞ fn=limn→∞ g=g. E is bounded and E contains no open ball: E ⊂{f ∈ X :f =g}. E may not be compact: Let X := (C([0, 1]), ·∞), and let Y denote the subspace of X consisting of all constant functions. Deﬁne g(y) := y(0) for

y ∈ Y . Then g ∈ Y and g=1. Given t ∈[0, 1], deﬁne ft (x) := x(t) for x ∈ X. Then each ft is a Hahn–Banach extension of g, and ft − fs ≥1 if t = s, since there is x ∈ X such that x∞ = 1, x(t) = 0 and x(s) = 1. Hence the sequence ( f1/n) in E does not have a convergent subsequence. 4.4. Suppose X is strictly convex. Let Y be a subspace of X, g ∈ Y with g=1, and let f1 and f2 be Hahn–Banach extensions of g to X. Then f1=f2= g=1. Also, ( f1 + f2)/2 is a Hahn–Banach extension of g to X, and so ( f1 + f2)/2=g=1. Hence f1 = f2.

Conversely, suppose there are f1 = f2 in X such that f1=1 =f2 and f1 + f2=2. Let Y := {x ∈ X : f1(x) = f2(x)}, and deﬁne g : Y → K by g(y) := f1(y) for y ∈ Y . Then g≤1. It can also be shown that g≥1. (See [11].) Hence f1 and f2 are distinct Hahn–Banach extensions of g. 4.5. If Y is a Banach space, then BL(X, Y ) is a Banach space (Proposition 3.17(i)). If BL(X, Y ) is a Banach space, then its closed subspace CL(X, Y ) is a Banach space. Now suppose CL(X, Y ) is a Banach space. Let a ∈ X be nonzero,

and let f ∈ X be such that f (a) =a and f =1. Let (yn) be a Cauchy sequence in Y , and for n ∈ N, deﬁne Fn : X → Y by Fn(x) := f (x)yn, x ∈ X. Then Fn ∈ CL(X, Y ) and Fn − Fm=yn − ym for 226 Solutions to Exercises

all n, m ∈ N. Hence there is F ∈ CL(X, Y ) such that Fn − F→0. In particular, ayn = Fn(a) → F(a). Hence Y is a Banach space.

4.6. For j ∈{1,...,m}, let us deﬁne g j (y) := G(y)( j), y ∈ Y . Then G(y) =

(g1(y), . . . , gm (y)) for y ∈ Y . By Lemma 2.8(ii), g j ∈ Y , and so there is a m Hahn–Banach extension f j ∈ X of g j for j = 1,...,m. Deﬁne F : X → K by F(x) := ( f1(x),..., fm (x)) for x ∈ X. Then F(y) = G(y) for y ∈ Y , and by Lemma 2.8(ii), F ∈ BL(X, Km ). Note that F≥G, and |F(x)( j)|= | f j (x)|≤f j x=g j x for all x ∈ X and j = 1,...,m. m Consider the norm ·∞ on K . We show that F≤G.Forx ∈ X,

F(x)∞ = max{|F(x)(1)|,...,|F(x)(m)|} ≤ max{g1,...,gm }x,

while for y ∈ Y , |g j (y)|≤max{|g1(y)|,...,|gm (y)|} = G(y)∞, and so g j ≤G for each j = 1,...,m. Thus F(x)∞ ≤Gx for all x ∈ X. ∞ Finally, suppose G ∈ BL(X, ).For j ∈ N, deﬁne f j ∈ X as above, and

let F(x) := ( f1(x), f2(x),...) for x ∈ X. Since |F(x)( j)|=|f j (x)|≤ ∞ f j x=g j x≤Gx for x ∈ X and j ∈ N, F(x) ∈ . Clearly, F : X → ∞ is linear and F(y) = G(y) for all y ∈ Y . Also, on replacing ‘max’ by ‘sup’ in the earlier argument, it follows that F=G. 3 4.7. Suppose F : K → Y is linear, and F(y) = y for all y ∈ Y .LetF(e1) := (k1, k2, k3), where k1 + k2 + k3 = 0. Then F(e2) = F(e2 − e1) + F(e1) = e2 −e1 +(k1, k2, k3) = (k1 −1, k2 +1, k3), and F(e3) = F(e3 −e1)+ F(e1) = e3 −e1 +(k1, k2, k3) = (k1 −1, k2, k3 +1). Hence F(e1)1 =|k1|+|k2|+|k3|, F(e2)1 =|k1 −1|+|k2 +1|+|k3| and F(e3)1 =|k1 −1|+|k2|+|k3 +1|. We show that at least one of F(e1)1, F(e2)1, and F(e3)1 is greater than 1. Suppose F(e3)1 ≤ 1. Then k3 = 0, for otherwise |k1 − 1|+|k2|≤0, that is, k1 = 1 and k2 = 0, and so k1 + k2 + k3 = 0. If F(e1)1 ≤ 1also, then F(e2)1 ≥ 1 −|k1|+1 −|k2|+|k3|≥1 + 2|k3| > 1. Thus F > 1. 4.8. Suppose the stated condition holds. Assume for a moment that a ∈/ E. Then there is r > 0 such that U(a, r) ∩ E =∅.NowU(a, r) and E are disjoint convex subsets of X, and U(a, r) is open. By the Hahn–Banach separation theorem, there are f ∈ X and t ∈ R such that Re f (a) 0, then Re g(a)<1 and Re g(x) ≥ 1 for all x ∈ E, while if t < 0, then Re g(a)>1 and Re g(x) ≤ 1 for all x ∈ E, contrary to the stated condition. If t = 0, there is s ∈ R such that Re f (a)~~0 for all x + Y ∈ E.Let f := f˜ ◦ Q.~~

~~4.10. Let r := inf{x1 − x2:x1 ∈ E1 and x2 ∈ E2}. Assume for a moment that r = 0. Then there is a convergent sequence (x1,n) in E1 and a sequence (x2,n) Solutions to Exercises 227~~

in E2 such that x1,n − x2,n→0. Also, if x1,n → x1, then x2,n → x1 as ∈ ∩ > well, and so x1 E1 E2, contrary to the hypothesis. Hence r 0. Deﬁne Er := E1 + U(0, r). Then Er = {x1 + U(0, r) : x1 ∈ E1} is an open convex subset of X.Ifx1 ∈ E1, x ∈ U(0, r), and x2 := x1 + x ∈ E2, then r ≤x1 − x2=x < r, a contradiction. Hence Er ∩ E2 =∅.Bythe

~~Hahn–Banach separation theorem, there are f ∈ X and t2 ∈ R such that Re f (x1)~~

for x := (x(1), x(2),...) ∈ c0. Clearly, fy ∈ (c0) , and fy≤y1. ∈ N := ( ( ),..., ( ), , ,...) ∈ For n , define xn sgn y 1 sgn y n 0 0 c0.Now ≤ ≥| ( )|= n | ( )| ∈ N xn ∞ 1, and fy fy xn j=1 y j for all n . Hence 1 1 fy≥y1. Define (y) : → (c0) by (y) := fy, y ∈ . ∈ ( ) Then is a linear isometry. To show that is onto, let f c0 . Then ( ) = ∞ ( ) = ∞ ( ) ( ) ∈ := f x f j=1 x j e j j=1 x j f e j for x c0. Define y ( ( ), ( ), . . .) n | ( )|= n ( ) ( ) = ( ) ≤ f e1 f e2 . Then j=1 y j j=1 xn j y j f xn f 1 for all n ∈ N. Thus y ∈ , and f = fy. (Compare the case p := ∞ of Example 3.12, and also Example 4.18(iii).) p (iii) If p ∈{1, 2}, then c00 is a dense subspace of , and if p := ∞, then c00 is a dense subspace of c0 (Exercise 2.3). By Proposition 4.13(i), the dual p q space of (c00, ·p) is linearly isometric to ( ) ,thatis,to if p ∈{1, 2}, 1 and to (c0) , that is, to if p := ∞.

~~4.12. (i) It is clear that : c → (c0) is linear, and (x )≤x for all x ∈ c .~~

~~Since c0 is a closed subspace of c, and e0 := (1, 1,...)∈ c\c0, there is x ∈ c~~

~~such that x (y) = 0 for all y ∈ c0 and x (e0) =e0∞ = 1. Then x = 0, but~~

(x ) = 0. Hence is not an isometry. ∈ 1 | ( )|≤ | ( )|+ ∞ | ( + )| = (ii) Let y . Then fy x y 1 j=1 y j 1 x ∞ y1x∞ for all x ∈ c, and so fy≤y1. On the other hand, deﬁne xn := (sgn y(2),...,sgn y(n), sgn y(1), sgn y(1),...) for n ∈ N. Then xn ∈ c, xn∞ ≤ 1, and

~~n−1 ∞~~

fy≥|fy(xn)|= |y(1)|+ |y( j + 1)|+ sgn y(1)y( j + 1) j=1 j=n n ∞ ≥ |y( j)|− |sgn y(1)||y( j)| for all n ∈ N. j=1 j=n+1 228 Solutions to Exercises ∞ | ( )|→ ≥ Since j=n+1 y j 0, we see that fy y 1. To show that is

onto, let f ∈ c .Forn ∈ N,letun := sgn f (e1)e1 +···+sgn f (en)en. Then ∈ , ≤ | ( )|+···+| ( )|= ( ) ≤ ∈ N un c un ∞ 1, and f e1 f en f un f for all n . ∞ ( ) K ∈ K Hence the series j=1 f e j converges in .Letsf denote its sum. Con- ∈ ( ) → = + ∞ ( ( ) − ) sider x c, and let x j x . Then x xe0 j=1 x j x e j (Exer- ( ) = ( ) + ∞ ( ) − ( ) = ( ) − cise 2.27), and so f x x f e0 j=1 x j x f e j x f e0 + ∞ ( ) ( ) := ( ( ) − , ( ), ( ),...) sf j=1 x j f e j . Deﬁne y f e0 s f f e1 f e2 . Then n | ( )|=| ( ) − |+ ∞ | ( )|≤| ( ) − |+ < ∞ j=1 y j f e0 s f j=1 f e j f e0 s f f . 1 1 Thus y ∈ , and f = fy. Hence is a linear isometry from onto c . 1,2 4.13. Let H := W .Fixy ∈ H. Then fy(x) = x, y1,2 for x ∈ H. (See Example

~~2.28(iv).) As in the proof of Theorem 4.14, fy ∈ H and fy=y2 =y2.~~

~~Thus the map (y) := fy, y ∈ H, gives a linear isometry from H to H .~~

Next, if f ∈ H , then by Theorem 4.14, f = fy, where y := y f . Hence is a linear isometry from H onto H . p 4.14. Let p ∈{1, 2}. Then (C([a, b]), ·p) is a dense subspace of L ([a, b]). Hence the dual space of (C([a, b]), ·p) is linearly isometric to the dual space of L p([a, b]), that is, to Lq ([a, b]), where (1/p) + (1/q) = 1. 4.15. Let X be a reﬂexive normed space, and let Y := X . Since J is a linear isometry from X onto X , and since X = BL(Y, K) is a Banach space, X is a Banach space. Suppose X is separable as well. Then X = Y is separable, and so Y = X is separable. Let X := 1. Then X is a separable Banach space. But X is not reﬂexive. Otherwise X would be separable, but X is isometric to ∞ which is not separable. w w 4.16. Suppose xn → x and xn →˜x in X. Then x (x˜) = limn→∞ x (xn) = x (x) for all x ∈ X . Hence x˜ = x by Proposition 4.6(i). w (i) Let E := {xn : n ∈ N}. Suppose xn → x in X. Then x (E) is a bounded subset of K for every x ∈ X . Hence E is a bounded subset of X. Conversely, suppose there is α > 0 such that xn≤α for all n ∈ N, and

~~there is a subset D of X whose span is dense in X and x (xn) → x (x)~~

~~for every x ∈ D, that is, J(xn)(x ) → J(x)(x ) for every x ∈ D. Since~~

~~J(xn), J ∈ BL(X , K), and J(xn)=xn≤α for n ∈ N, Exercise 3.17 w shows that J(xn)(x ) → J(x)(x ) for every x ∈ X , that is, xn → x in X.~~

(ii) If xn → x, then |x (xn) − x (x)|≤x xn − x→0 for every x ∈ X , w w and so xn → x in X. Suppose X is an inner product space, xn → x in X, and 2 2 2 xn→x. Then xn − x =xn − 2Re xn, x+x → 0. (iii) Let (xn) be a bounded sequence in X. Then the bounded sequence

( x1, xn) in K has a convergent subsequence ( x1, x1,n) by the Bolzano– Weierstrass theorem for K. Next, the bounded sequence ( x2, x1,n) has a convergent subsequence ( x2, x2,n), and so on. Deﬁne un := xn,n for n ∈ N. The diagonal sequence (un) is a subsequence of (xn). For each ﬁxed m ∈ N,the sequence ( xm , un) is convergent, and so the sequence ( y, un) is convergent for every y ∈ span {x1, x2,...}.LetY denote the closure of span {x1, x2,...}. Solutions to Exercises 229

It can be seen that for every y ∈ Y , ( y, un) is a Cauchy sequence, and hence ⊥ it is convergent in K. Further, if z ∈ Y , then z, un=0 for all n ∈ N. Since ⊥ X = Y ⊕ Y , it follows that the sequence ( x, un) is convergent for every

x ∈ X. Deﬁne f (x) := limn→∞ x, un for x ∈ X. Then f ∈ X , and so, there is u ∈ X such that f (x) = x, u for all x ∈ X. Thus x, un→ x, u for w every x ∈ X, that is, x (un) → x (u) for every x ∈ X . Thus un → u in X. Suppose ( xn, x˜) is convergent in K for every x˜ ∈ X.Forn ∈ N,let

~~fn(x˜) := x ˜, xn, x˜ ∈ X. Then fn ∈ X and fn=xn. Deﬁne f (x˜) := limn→∞ fn(x˜), x˜ ∈ X. By the Banach–Steinhaus theorem, ( fn) is~~

bounded, that is, (xn) is bounded, and so f ∈ X . As above, there is x ∈ X w such that xn → x in X. w (iv) Let xn → 0inX.IfF ∈ BL(X, Y ), then y ◦F(xn) → 0 for every y ∈ Y , w and so F(xn) → 0inY .NowletF ∈ CL(X, Y ), and assume for a moment that F(xn) → 0inY . By passing to a subsequence, if necessary, we may assume that there is δ > 0 such that F(xn)≥δ for all n ∈ N. Since (xn) is a bounded sequence and F is a compact linear map, there is a subsequence ( ) ( ) → ( ) →w xnk such that F xnk y in Y . But since F xnk 0inY , we see that = ( )≥ ∈ N y 0. This is impossible since F xnk δ for all k . (v) Let (un) be an orthonormal sequence in the Hilbert space X. As a conse- w quence of the Bessel inequality, x, un→0 for every x ∈ X, that is, un → 0 in X.LetF ∈ CL(X, Y ). Then F(un) → 0inY by (iv) above. In particular, let X = Y := 2, and let an infinite matrix M define a map 2 t F ∈ CL( ). Then F(e j ) → 0, that is, α2( j) → 0. Also, the transpose M t 2 t of M defines a map F ∈ CL( ), and so F (ei ) → 0, that is, β2(i) → 0. 1 1 4.17. (i) |x (x)|≤x1 for all x ∈ , and so x ∈ ( ) . Clearly, x (en) = 1 → 0. 2 (ii) The span of the set E := {e j : j ∈ N} is dense in , which is linearly isometric to (2) . Hence the result follows from Exercise 4.16(i). (iii) Without loss of generality, we let x := 0. Assume for a moment that w xn→ 0inX. Then there is x ∈ X such that x (xn) → 0, and so there is > < < ··· N | ( )|≥ δ 0 and there are n1 n2 in such that x xnk δ for all ∈ N ∈ N := ( ) +···+ ( ) . k .Form , define um sgn x xn1 xn1 sgn x xnm xnm Since | ( )|≤| ( )|+···+| ( )|≤ ∞ | ( )|≤ ∈ N um j xn1 j xnm j n=1 xn j α for all j ,we ≤ ∈ N ≤| ( )|+···+| ( )|= see that um ∞ α for all m , and so m δ x xn1 x xnm | ( )|≤ ∈ N →w ∞ x um α x for all m , which is impossible. Thus xn 0in .In := ∈ N ∞ | ( )|= particular, let xn en for n , and observe that n=1 en j 1 for all w ∞ j ∈ N. Hence en → 0in . (iv) Let x ∈ X . By Exercise 4.11(ii), there is y ∈ 1 such that x (x) = ∞ ( ) ( ) ∈ ( ) = n ( ) → ∞ ( ) j=1 x j y j for all x c0, and so x xn j=1 y j j=1 y j . w Assume for a moment that xn → x in c0.Fix j ∈ N. Then xn( j) → x( j).But xn( j) = 1 for all n ≥ j, and so x = (1, 1,...)which is not in c0. Thus there w is no x ∈ c such that x → x. 0w n (v) Let xn → x in C([a, b]). Then (xn) is bounded in (C([a, b]), ·∞) by 230 Solutions to Exercises

∈[ , ] ( ) := ( ), ∈ ∈ Exercise 4.16(i). For t a b , deﬁne xt x x t x X, so that xt X . ( ) = ( ) → ( ) = ( ) ∈[ , ] Hence xn t xt xn xt x x t for each t a b . Conversely, suppose (xn) is uniformly bounded on [a, b], and xn(t) → x(t) ∈[ , ] ∈ ∈ ([ , ]) ( ) = for each t a b .Letx X . There is y BV a b such that x x b xdyfor all x ∈ X. There are nondecreasing functions y , y , y , y deﬁned a 1 2 3 4 [ , ] = − + ( − ) b → b on a b such that y y1 y2 i y3 y4 , and further, a xn dyi a xdyi = ,..., for i 1 4 by the bounded convergence theorem for the Riemann– ( ) = b → b = ( ) Stieltjes integration. Thus x xn a xn dy a xdy x x . 4.18. (i)=⇒ (ii)=⇒ (iii) by the projection theorem. (iii)=⇒ (iv): If Y ⊥ ={0}, then Y = Y ⊥⊥ ={0}⊥ = X. (iv)=⇒ (v): Let Y := Z( f ) in the proof of Theorem 4.14.

~~(v)=⇒ (i): Let (yn) be a Cauchy sequence in X.Forn ∈ N, deﬁne fn(x) :=~~

~~x, yn, x ∈ X. Then fn ∈ X , and fn − fm =yn − ym for all n, m ∈ N.~~

Since X is a Banach space, there is f ∈ X such that fn − f →0. Let y ∈ X be such that f (x) := x, y, x ∈ X. Then yn − y=fn − f →0. 4.19. Let H denote the completion of X. { } { } (i) Let span uα be dense in X. Since X is dense in H, we see that span uα = , is dense in H. By Theorem 2.31, x n x un un for every x ∈ H, and in particular for every x ∈ X. The converse follows as in the proof of (iv)=⇒ (i) of Theorem 2.31. (ii) Let G denote the closure of Y in H. Then G ∩ X = Y since Y is closed in X.Let · , · be the inner product on H/G which induces the quotient norm on H/G (Exercise 2.40). Let x1 + Y, x2 + Y := x1 + G, x2 + G for x1, x2 ∈ X.Ifx ∈ X and x + Y, x + Y = x + G, x + G = 0, then x ∈ G ∩ X = Y , that is x + Y = 0 + Y . It follows that · , · is an inner product on X/Y , and x + Y, x + Y = x + G, x + G = |||x + G|||2 = d(x, G)2 = d(x, Y )2 = |||x + Y |||2 for all x ∈ X. (iii) Let · , · denote the inner product on H which induces the norm on H ,

~~as in Corollary 4.16(i). Let f0, g0 ∈ X . By Proposition 3.17(ii), there are~~

~~unique f, g ∈ H such that f (x) = f0(x) and g(x) = g0(x) for all x ∈ X,~~

f =f0 and g=g0. Define f0, g0 := f, g . Then · , · is an inner product on X , and f , f = f, f =f 2 =f 2. 0 0 0 ,..., ,..., ( ) = m ( ) 4.20. If there are x1 xm in X and y1 ym in Y such that F x i=1 xi x yi for x ∈ X, then R(F) ⊂ span {y1,...,ym }, and so F is of finite rank. Conversely, suppose F is of finite rank. Let y1,...,ym be a basis for R(F). Then there are unique functions f1,..., fm from X to K such that F(x) = f1(x)y1 +···+ fm (x)ym for x ∈ X.Fixi ∈{1,...,m}. Clearly, fi is linear. Let Yi := span {y j : j = 1,...,m and j = i} and di := d(yi , Yi ). Since di > 0 and | fi (x)|di ≤F(x)≤Fx for all x ∈ X, fi is contin- := ∈ = ,..., uous. Let xi fi X for i 1 m. ( ) := m ( ) ∈ ∈ Suppose F x i=1 xi x yi for x X as above, and let y Y . Then ( )( ) = ( ( )) = m ( ) ( ) = m ( ) ( ) F y x y F x i=1 xi x y yi i=1 y yi xi x for all ∈ ( ) = m ( ) x X, that is, F y i=1 y yi xi . Solutions to Exercises 231

4.21. Let p, r ∈{1, 2}, and (1/p)+(1/q) = 1 and (1/r)+(1/s) = 1. The transpose Mt of M defines a map F t ∈ BL(s ,q ), which can be identified with F . Since F is compact, F is compact, and so is F t . Since s, q ∈{2, ∞},the sequence of the columns of Mt , which is the sequence of the rows of M, tends to 0 in q by Exercise 3.34. Let p ∈{1, 2, ∞} and r := ∞.fM := [ki, j ], where ki,1 := 1fori ∈ N, and p ∞ ki, j := 0 otherwise, then M defines F ∈ CL( , ),butβr (i) = 1, i ∈ N. p p 4.22. If p ∈{1, 2, ∞}, then A(x)p ≤xp for all x ∈ L , and so A ∈ BL(L ). Now let p ∈{1, 2}, and (1/p) + (1/q) = 1. For y ∈ Lq ,let

~~∞ (y)(x) := x(t)y(t)dm(t), x ∈ L p. 0~~

Now : Lq → (L p) is a linear isometry, and it is onto. (Compare Examples 4.19(ii) and 4.24(ii).) Deﬁne At : Lq → Lq by At := ()−1 A . Thus A ∈ BL((L p) ) can be identiﬁed with At ∈ BL(Lq ).Fixy ∈ Lq . Let z(s) := 0ifs ∈[0, 1), and z(s) := y(s − 1) if s ∈[1, ∞). Then z ∈ Lq , and

~~(At (y))(x) = A ((y))(x) = (y)(A(x)) ∞ ∞ = x(t + 1)y(t)dm(t) = x(s)y(s − 1)dm(s) 0 1 ∞ = x(s)z(s)dm(s) = (z)(x). 0~~

for all x ∈ L p. Hence (At (y)) = (z), and in turn, At (y) = z, as desired. ∈ N ∈ q ( )( ) := ∞ ( ) ( ), ∈ p 4.23. Let n .Fory ,let y x j=1 x j y j x . Then is q (p) t := ()−1 . ∈ ((p) ) an isometry from onto .LetPn Pn Thus Pn BL t ∈ (q ) ( t )2 = t ∈ q can be identiﬁed with Pn BL . Clearly, Pn Pn .Fixy , and let q yn := (y(1),...,y(n), 0, 0,...). Then yn ∈ , and

~~n ( t ( ))( ) = (( ))( ) = ( )( ( )) = ( ) ( ) = ( )( ) Pn y x Pn y x y Pn x x j y j yn x j=1~~

∈ p ( t ( )) = ( ) t ( ) = for x . Hence Pn y yn , and in turn, Pn y yn, as desired. 4.24. For x ∈ X ,lety := P (x ). Then (P )2(x )(x) = P (y )(x) = y (P(x)) = x (P(P(x))) = x (P(x)) = P (x )(x) for all x ∈ X. Hence (P )2 = P . Also, R(P ) ={x ∈ X : P (x ) = x }={x ∈ X : x (P(x) − x) = 0forx ∈ X}=Z 0 and Z(P ) ={x ∈ X : P (x )(x) = 0forx ∈ X}= {x ∈ X : x (P(x)) = 0forx ∈ X}=Y 0. 4.25. Let x ∈ X and y ∈ Y . Then

~~F JX (x) (y ) = JX (x) F (y ) = F (y )(x) = y (F(x)) = JY F(x) (y ). 232 Solutions to Exercises~~

~~Hence F JX (x) = JY F(x) for all x ∈ X, that is, F JX = JY F.Also, F =(F ) =F =F.~~

~~Let Xc denote the closure of JX (X) in X , and let Yc denote the closure of~~

~~JY (Y ) in Y .Letx ∈ Xc. We show that F (x ) ∈ Yc.Let(xn) be a sequence~~

~~in X such that JX (xn) → x in X . Then JY F(xn) = F JX (xn) → F (x )~~

~~in Y , where JY F(xn) ∈ JY (Y ). Thus F (x ) ∈ Yc. Deﬁne Fc : Xc → Yc by~~

~~Fc(x ) = F (x ), x ∈ Xc. Then Fc is linear, and Fc(JX (x)) = F (JX (x)) =~~

JY F(x) for all x ∈ X, that is, Fc JX = JY F.Also,F≤Fc≤F = F. The uniqueness of Fc ∈ BL(Xc, Yc) follows by noting that JX (X) is dense in Xc. 4.26. Suppose F ∈ CL(Y , X ). Then F ∈ CL(X , Y ). By Exercise 4.25,

~~F JX = JY F, and so JY (F(U)) ⊂{F (x ) : x ∈ X and x ≤1}. The latter set is a totally bounded subset of Y since F = (F ) is compact~~

~~(Exercise 3.38), and so JY (F(U)) is also a totally bounded subset of Y . Since JY is an isometry, F(U) is a totally bounded subset of Y . Since Y is a Banach space, F ∈ CL(X, Y ) (Exercise 3.38).~~

4.27. Let xn → x0 in H and A(xn) → y0 in G. Then for every y ∈ Y , A(xn), y→ y0, y on one hand, and on the other hand, A(xn), y= xn, B(y)→ x0, B(y)= A(x0), y, and so A(x0), y= y0, y. It follows that y0 = A(x0). Thus A is a closed map. By the closed graph theorem, A ∈ BL(H, G). Aliter: Let E := {A(x) : x ∈ H and x≤1}⊂G. Consider y ∈ G . There

is y0 ∈ G such that y (y) = y, y0 for all y ∈ G. Hence |y (A(x)|= | A(x), y0| = | x, B(y0| ≤ B(y0) for all x ∈ H satisfying x≤1. By the resonance theorem, E is a bounded subset of G, that is, A ∈ BL(H, G). Similarly, B ∈ BL(G, H). By the uniqueness of the adjoint, B = A∗. ∗ ∗ 4.28. R(A ) ⊥ R(B) if and only if A (y1), B(y2)=0 for all y1, y2 ∈ G.But ∗ A (y1), B(y2)= y1, AB(y2) for y1, y2 ∈ G, and y1, AB(y2)=0 for all y1, y2 ∈ G if and only if AB(y2) = 0 for all y2 ∈ G, that is, AB = 0. 4.29. A∗(G⊥) ⊂ G⊥ if and only if A∗(x), y=0 for all x ∈ G⊥ and y ∈ G.But A∗(x), y= x, A(y) for x ∈ G⊥ and y ∈ G = (G⊥)⊥, and x, A(y)=0 for all x ∈ G⊥ and y ∈ G⊥⊥ if and only if A(G) = A(G⊥⊥) ⊂ (G⊥)⊥ = G. 4.30. Suppose A ∈ BL(H, G) is one-one and onto. Replacing A by A∗ in Theorem 4.27(i), we see that R(A∗) is dense in H. Also, by Theorem 4.27(ii), A∗ is bounded below. Let β > 0 be such that βy≤A∗(y) for all y ∈ G. To show A∗ is onto, consider x ∈ H. Since R(A∗) is dense in H, there is a ∗ ∗ sequence (yn) in G such that A (yn) → x in H. Then (A (yn)) is a Cauchy ∗ ∗ sequence in H, and βyn −ym ≤A (yn)− A (ym ) for all n, m ∈ N. Hence (yn) is a Cauchy sequence in G. Since G is complete, there is y ∈ Y such that ∗ ∗ ∗ ∗ ∗ yn → y in G. Then A (yn) → A (y). Hence x = A (y) ∈ R(A ). Thus A is onto. Since A∗ is also one-one, consider B := (A∗)−1 : H → G. Then B is linear. Also, if x ∈ H and x = A∗(y), then B(x)=y≤A∗(y)/β = x/β, and so B ∈ BL(H, G). Hence A−1 = B∗ ∈ BL(G, H). Open mapping theorem: Suppose A ∈ BL(H, G) is onto. Let Z := Z(A), Solutions to Exercises 233

and deﬁne A : H/Z → G by A(x + Z) := A(x) for x ∈ H. Then A is one-one and onto. As we saw above, A−1 ∈ BL(G, H/Z). Hence A is an open map, and so is A = A◦ Q since Q : H → H/Z is an open map. The closed graph theorem can be deduced from the open mapping theorem as in Exercise 3.27(iii). 4.31. (i) Let A ∈ CL(H, G). Then A∗ ∈ BL(G, H), and so A∗ A ∈ CL(H). ∗ Conversely, let A A ∈ CL(H). Consider a bounded sequence (xn) in H, and ∗ let α > 0 be such that xn≤α for all n ∈ N. Since A A is compact, there ( ) ( ∗ ( )) is a subsequence xnk such that A A xnk converges in G.But

~~ ( ) − ( )2 =| ∗ ( − ), − | A xnk A xn j A A xnk xn j xnk xn j ≤ ∗ ( ) − ∗ ( ) , ∈ N. 2α A A xnk A A xn j for all k j~~

( ( )) ∈ ( , ) Now the Cauchy sequence A xnk converges in G. Thus A CL H G . (ii) Let A ∈ CL(H, G). Then (A∗)∗ A∗ = AA∗ ∈ CL(G), and by (i) above, A∗ ∈ CL(G, H). (Note: Theorem 4.21 of Schauder is not used.) (iii) Let A ∈ BL(H, G) be a Hilbert–Schmidt map. Let {u , u ,...} be 1 2 ( )2 < ∞ a countable orthonormal basis for H such that n A un .Let {v1,v2,...} be an countable orthonormal basis for G. Then

∗ 2 ∗ 2 2 A (vm) = | A (vm ), un| = | vm, A(un)| m m n n m ( )2 ∗ which is equal to n A un . Hence A is a Hilbert–Schmidt map. {˜ , ˜ ,...} (This proof shows that if u1 u2 is another countable orthonormal basis ( ˜ )2 = ∗(v )2 = ( )2 for H, then n A un m A m n A un .) 4.32. Let A ∈ BL(H). Deﬁne B := (A + A∗)/2 and C := (A − A∗)/2. Then B is hermitian, C is skew–hermitian, and A = B + C. Suppose A = B1 + C1, ∗ where B1 is hermitian and C1 is skew–hermitian. Then A = B1 − C1, and so ∗ ∗ B1 = (A + A )/2 = B and C1 = (A − A )/2 = C. Note that BC = (A2 − (A∗)2 − AA∗ + A∗ A)/4 and CB = (A2 − (A∗)2 − A∗ A + AA∗)/4. Hence BC = CB if and only if A∗ A = AA∗, that is, A is normal. Also, C = 0 if and only if A∗ = A, that is, A is hermitian, and B = 0 if and only if A∗ =−A, that is, A is skew–hermitian. Finally, note that B2 − C2 = (AA∗ + A∗ A)/2. Hence BC = CBand B2 − C2 = I if and only if A∗ A = AA∗ and A∗ A + AA∗ = 2I , that is, A∗ A = I = AA∗. 4.33. Let A ∈ BL(H). Clearly, A∗ A − AA∗ is self-adjoint. Hence A is hyponormal if and only if A∗ A(x), x≥ AA∗(x), x, that is, A(x)2 ≥A∗(x)2 for all x ∈ H. Since A is normal if and only if A∗(x)=A(x) for all x ∈ H, and A∗ is hyponormal if and only if A∗(x)≥A(x) for all x ∈ H, A is normal if and only if A and A∗ are both hyponormal. Let A denote the right shift operator on 2. Then A∗ is the left shift operator 234 Solutions to Exercises

on 2. Hence A∗ A = I and AA∗(x) = (0, x(2), x(3), . . .) for all x ∈ 2. Thus A∗ A ≥ AA∗,butA∗ A = AA∗. 4.34. For x in H, let B(x)( j):= x( j + 1) for all j ∈ Z. Then x, B(y)= ∞ ( ) ( + ) = ∞ ( − ) ( ) = ( ), , ∈ j=−∞ x j y j 1 j=−∞ x j 1 y j A x y for all x y H. Hence A∗ = B, the left shift operator on H. Also, it is easy to see that A∗ A(x) = x = AA∗(x) for x ∈ H. Hence A is a unitary operator on H. n = p = = ,..., − | |= = −1 4.35. Note: ωn 1, but ωn 1forp 1 n 1, ωn 1 and ωn ωn . n We show that the n columns of Mn form an orthonormal subset of C .Let ,∈{ ,..., } j 1 n . Then the inner product of the jth and th columns is equal to n (p−1)( j−1) −(p−1)(−1) / = n (p−1)( j−) / p=1 ωn ωn n p=1 ωn n, which is equal to (1 +···+1)/n = 1if j = , and which is equal to 0 if j = since j− j− j− (n−1)( j−) ( j−)n ωn = 1 and (1 − ωn )(1 + ωn +···+ωn ) = 1 − ωn = 0. t t t Hence M Mn = I = Mn M , and so the operator A is unitary. Also, M = Mn n √n n ( j−1)(p−1) since k j,p := ωn / n = k p, j for p, j = 1,...,n.

4.36. (i) Let x ∈ H.Forn ∈ N,letan(x) := An(x), x. Then (an(x)) is a monotonically increasing sequence in R, and it is bounded above by α x, x. Hence it is a Cauchy sequence in R.Form ≥ n, deﬁne Bm,n := Am − An. Then 0 ≤ Bm,n ≤ αI − A1 for all m ≥ n, and

~~Bm,n=sup{ Bm,n(x), x:x ∈ H and x≤1}≤|α|+A1.~~

~~By the generalized Schwarz inequality, for all m ≥ n,~~

~~ ( )= ( ), 1/4 2 ( ), ( )1/4 Bm,n x Bm,n x x Bm,n x Bm,n x 1/4 3/4 1/2 ≤ Bm,n(x), x Bm,n x 1/4 3/4 1/2 ≤ Bm,n(x), x (|α|+A1) x .~~

It follows that (An(x)) is a Cauchy sequence in H.LetAn(x) → y in H, and deﬁne A(x) := y. Clearly, A : H → H is linear. Also, since A1 ≤ An ≤ αI , we see that An≤|α|+A1 for all n ∈ N, and so A(x)≤ (|α|+A1)x for all x ∈ H. Thus A ∈ BL(H). Since An(x) → A(x), we see that the monotonically increasing sequence ( An(x), x) converges to A(x), x for each x ∈ H. Thus A is self-adjoint, and An ≤ A for all n ∈ N. Finally, suppose A is self-adjoint, and An ≤ A for all n ∈ N. Then A(x), x=limn→∞ An(x), x≤ A(x), x for all x ∈ H. Hence A ≤ A. The uniqueness of A is obvious. − := − − ≤ − ≤− ∈ N (ii) Let An An, so that An An+1 βI for all n . Then the desired result follows from (i) above.

4.37. For x, y ∈ H, define x, yA := A(x), y. Then · , ·A : H × H → K is linear in the first variable, is conjugate symmetric, and satisfies x, xA ≥ 0 for all x ∈ H since A is a positive operator. In Exercise 2.13, let X := H and replace · , · by · , ·A. Then G ={x ∈ H : x, xA = 0}, and G is a subspace Solutions to Exercises 235

of H. In fact, G is closed since A is continuous. For x + G, y + G ∈ H/G, let x + G, y + G := x, yA. It follows that · , · is an inner product 2 2 on H/G. In particular, | x, yA| ≤ x, xA y, yA, that is, | A(x), y| ≤ A(x), x A(y), y for all x, y ∈ H.

4.38. Let n, m ∈ N, m = n. Since Pn Pm = 0, we see that R(Pm ) ⊂ Z(Pn), and since Pn is an orthogonal projection operator, Z(Pn) ⊥ R(Pn). Hence for all m = n, R(Pn) ⊥ R(Pm ).Form ∈ N,letQm := P1 +···+ Pm . Then ∗ ∗ ∗ Qm = P1 +···+Pm = P1 +···+Pm = Qm, and so Qm is an orthogonal projection operator. By Exercise 3.6(i), Qm =0orQm=1. Let x ∈ H. Then the Pythagoras theorem shows that for each m ∈ N,

2 2 2 2 2 P1(x) +···+Pm (x) =P1(x) +···+ Pm (x) =Qm (x) ≤x . ( ) ( ) := ∞ ( ) By Exercise 2.30, n Pn x is summable in H.LetP x n=1 Pn x . Clearly, P is linear, and P(x)≤x for all x ∈ H. Hence P≤1. ∈ ( ( )) = ( ) ∈ N ( ( )) = Let x H. SincePn P x Pn x for all n , we obtain P P x ( ( )) = ( ) = ( ) n Pn P x n Pn x P x . Thus P is an orthogonal projection operator on H (Exercise 3.6(i)).

∞ ( ) ∈ Let G denote the closure of the linear span of n=1 R Pn .Letx H. Then Pn(x) ∈ R(Pn) ⊂ G for all n ∈ N. Since G is a closed subspace of H,it follows that P(x) ∈ G. Hence R(P) ⊂ G. Conversely, Pn(x) = P(Pn(x)) ∈ R(P) for all n ∈ N. Hence R(Pn) ⊂ R(P). Since R(P) is a closed subspace of H,itfollowsthatG ⊂ R(P). Thus R(P) = G.Ifx ∈ H and Pn(x) = 0 for all n ∈ N, then clearly P(x) = 0. Conversely, if x ∈ H, and P(x) = 0, ∞ then P (x) = P (P(x)) = P (0) = 0. Thus Z(P) = Z(P ). n n n n=1 n √ ∈ N := ( , −iθ, −2iθ,..., −(n−1)iθ, , ,...)/ 4.39. Let n , and let xn 1 e e e 0 0 n, where√ −iθ −2iθ −(n−1)iθ θ ∈ (−π, π]. Then A(xn) = (0, 1, e , e ,...,e , 0, 0,...)/ n. iθ Clearly, xn ∈ X and xn2 = 1. Hence A(xn), xn=(n − 1)e /n ∈ ω(A). Letting n = 1, we see that 0 ∈ ω(A). Next, let k ∈ K satisfy 0 < |k| < 1. Then k = reiθ, where 0 < r < 1 and θ ∈ (−π, π]. There is n ∈ N such that r <(n − 1)/n. Since 0 and (n − 1)eiθ/n belong to ω(A), and since ω(A) is a convex subset of K, we see that k ∈ ω(A). Thus {k ∈ K :|k| < 1}⊂ω(A). Since A=1, it follows that ω(A) ⊂{k ∈ K :|k|≤1}. We show that if k ∈ K and |k|=1, then k ∈/ ω(A).Letx ∈X. Then | ( ), | ≤ ∞ | ( )|| ( + )|≤ 1 ∞ (| ( )|2 +| ( + )|2). A x x j=1 x j x j 1 2 j=1 x j x j 1 Thus 2 | A(x), x| ≤ x . Assume for a moment that x2 = 1 =| A(x), x|. 2 2 Since |x( j)||x( j + 1)|≤(|x( j)| +|x( j + 1)|) /2 for every j ∈ N, |x( j)|= | ( + )| ∈ N < ∞ | ( )|2 < ∞ x j 1 for each j . This is impossible since 0 j=1 x j . Thus ω(A) ={k ∈ K :|k| < 1}. 4.40. For x, y ∈ X, A(x + y), x + y− A(x − y), x − y=2 A(x), y+ 2 A(y), x and A(x + iy), x + iy− A(x − iy), x − iy=−2i A(x), y+ 2i A(y), x. Multiplying the second equality by i and adding it to the ﬁrst, we obtain the generalized polarization identity. 236 Solutions to Exercises

Let H ={0} be a Hilbert space over C, A ∈ BL(H), and ω(A) ⊂ R. Then 4 A(y), x= A(y + x), y + x− A(y − x), y − x+i A(y + ix), y + ix −i A(y − ix), y − ix for x, y ∈ H. Note that A(z), z∈R for every z ∈ H. Hence 4 x, A(y)=4 A(y), x=4 A(x), y. Thus A is self-adjoint.

Chapter 5 − − − 5.1. Suppose k ∈ K, k = 0. If k ∈/ σ(A), then (A 1 − k 1 I ) (A − kI) 1 A = −k−1(I − kA−1)(I − kA−1)−1 =−k−1 I and (A − kI)−1 A (A−1− k−1 I ) = (I − kA−1)−1(−k−1)(I − kA−1) =−k−1 I , and so k−1 ∈/ σ(A−1). Replacing A by A−1, and k by k−1, we see that if k−1 ∈/ σ(A−1), then k ∈/ σ(A). Hence σ(A−1) ={λ−1 : λ ∈ σ(A)}. n n−1 5.2. Let p(t) := ant + an−1t +···+a1t + a0, where n ∈ N, an,...,a0 ∈ K, and an = 0. Suppose λ ∈ σ(A). Assume for a moment that p(λ)/∈ σ(p(A)), m−1 that is, p(A) − p(λ)I is invertible. For m ∈{1,...,n},letqm (A) := A + m−2 m−2 m−1 m m λA +···+λ A + λ I , so that A − λ I = (A − λI )qm (A) = qm (A)(A − λI ). Deﬁne q(A) := anqn(A) +···+a1q1(A). It follows that p(A)− p(λ)I = (A−λI )q(A) = q(A)(A−λI ). Hence A−λI is invertible. This contradiction shows that p(λ) ∈ σ(p(A)). Next, let K := C, and μ ∈ C. Then there are λ1,...,λn ∈ C such that

p(t)−μ = an(t −λ1) ···(t −λn). Suppose μ ∈ σ(p(A)). Then p(A)−μI = an(A − λ1 I ) ···(A − λn I ).IfA − λ j I is invertible for each j = 1,...,n, ( ) − ∈ ( ) = ( ) then so would be p A μI . Hence there is λ j σ A such that μ p λ j . − 5.3. Let I − AB be invertible. Then (I − BA) I + B(I − AB) 1 A = I − BA+ − − − B (I − AB) 1 − AB(I − AB) 1 A = I − BA+ B(I− AB)(I − AB) 1 A = I − BA+ BA = I . Similarly, I + B(I − AB)−1 A (I − BA) = I . − (This formula is conceived as follows: (I − BA) 1 = I + BA+ BABA+ BABABA+···= I + B I + AB+ ABAB+··· A = I + B(I − AB)−1 A.) Let k ∈ K be nonzero, and let A := A/k. Then AB − kI =−k(I − AB ) is invertible if and only if −k(I − B A) = BA− kI is invertible. n 5.4. Let λ ∈ σ(A) = σe(A). Then there is nonzero x := (x(1),...,x(n)) ∈ K such that A(x) − λx = 0. Suppose x∞ =|x(i)|. We show that λ ∈ Di . Now ki,1x(1) +···+(ki,i − λ)x(i) +···+ki,n x(n) = 0, that is,

~~x(1) x(i − 1) x(i + 1) x(n) k , − λ =−k , −···−k , − − k , + −···−k , . i i i 1 x(i) i i 1 x(i) i i 1 x(i) i n x(i)~~

Since |x( j)|/|x(i)|≤1 for all j ∈{1,...,n}, we obtain |k , − λ|≤r . i i i ( ) = = ( ( ),..., ( )) n ( ) = 5.5. Suppose Ax λx and u g1 x gn x . Then j=1 g j x x j λx, n ( ) ( ) = ( ) n ( ) ( ) = ( ) and hence j=1 g j x gi x j λgi x , that is, j=1 u j gi x j λu i for = ,..., t = t = n ( ) / each i 1 n. Thus Mu λu .Also,x j=1 u j x j λ. Solutions to Exercises 237 Conversely, suppose Mut = λut and x = n u( j)x /λ. Then for each j=1 j = ,..., n ( ) ( ) = ( ) i 1 n, j=1 gi x j u j λu i , and so

n 1 n n A(x) = g (x)x = u( j)g (x ) x i i λ i j i i=1 i=1 j=1 1 n n = λu(i)x = u(i)x = λx. λ i i i=1 i=1 ( ) = n ( ) ( ) = n ( ) = ( ) = ( ) Also, λu i j=1 gi x j u j gi j=1 u j x j gi λx λgi x , and so u(i) = gi (x) for all i = 1,...,n, that is, u = (g1(x), . . . , gn(x)). Also, x = 0 if and only if u = 0. Thus x is an eigenvector of A corresponding to λ if and only if ut is an eigenvector of M corresponding to λ.

5.6. Suppose λ ∈ σe(A). Then there is nonzero x ∈ X such that A(x) = λx, that is, x0(t) − λ x(t) = 0 for all t ∈[a, b].Lett0 ∈[a, b] be such that x(t0) = 0. Since x is continuous at t0, there is δ > 0 such that x(t) = 0for all t ∈ I := [a, b]∩(t0 − δ, t0 + δ). Hence x0(t) = λ for all t ∈ I . Conversely, suppose λ ∈ K, and x0(t) = λ for all t in a nontrivial subinterval I

of [a, b]. Then there is t0 ∈[a, b], and there is δ > 0 such that (t0 −δ, t0 +δ) ⊂ I .Fort ∈[a, b], deﬁne x(t) := 1if|t − t0|≤δ/2, x(t) := 2(t − t0 + δ)/δ if t0 − δ < t < t0 − δ/2, x(t) := 2(t0 − t + δ)/δ if t0 + δ/2 < t < t0 + δ, and x(t) := 0, if |t − t0|≥δ. Then x ∈ X and x = 0. Since x0(t) − λ = 0 for all t ∈[a, b] satisfying |t − t0| < δ, and x(t) = 0 for all t ∈[a, b] satisfying |t − t0|≥δ, we see that x0(t) − λ x(t) = 0 for all t ∈[a, b], that is, A(x) = λx.

5.7. Let E denote the essential range of x0. Suppose λ ∈ E.Forn ∈ N,let Sn := {t ∈[a, b]:|x0(t) − λ| < 1/n}, and let xn denote the characteristic 2 = ( )> function of Sn. Then xn 2 m Sn 0, and

~~m(S ) x 2 A(x ) − λx 2 = |x − λ|2|x |2dm ≤ n = n 2 . n n 2 0 n 2 2 Sn n n~~

~~Hence A − λI is not bounded below, that is, λ ∈ σa(A). Thus E ⊂ σa(A). Conversely, suppose k ∈/ E, that is, there is > 0 such that m({t ∈[a, b]:~~

|x0(t) − k| < }) = 0. Then |x0(t) − k|≥ for almost all t ∈[a, b], and the function 1/(x0 −k) belongs to X.Fory ∈ X, deﬁne B(y) := y/(x0 −k). Then B ∈ BL(X) and (A − kI)B = I = B(A − kI). This shows that σ(A) ⊂ E. Since σa(A) ⊂ σ(A), we obtain σa(A) = E = σ(A). Next, suppose λ ∈ σe(A). Then there is nonzero x ∈ X such that A(x) = λx,

that is, (x0(t) − λ)x(t) = 0 for almost all t ∈[a, b]. Since x is nonzero, m({t ∈[a, b]:x(t) = 0})>0, and so m({t ∈[a, b]:x0(t) = λ})>0. Conversely, let λ ∈ K, S := {t ∈[a, b]:x0(t) = λ}, and suppose m(S)>0. 238 Solutions to Exercises

Let x denote the characteristic function of the set S. Then x ∈ X and x = 0. Since (x0(t) − λ)x(t) = 0 for all t ∈[a, b], λ ∈ σe(A). 5.8. For n ∈ N, An −λn I = B(A−λI ), where B := An−1+λAn−2+···+λn−2 A + n−1 λ I .Letλ ∈ σa(A). Then there is a sequence (xn) in X such that xn=1 n n for every n ∈ N, and A(xn) − λxn → 0. Consequently, A (xn) − λ xn = n n n n B(A(xn) − λxn) → 0, and so λ ∈ σa(A ). Hence |λ |≤A , and so |λ|≤inf An1/n : n ∈ N . 5.9. (i) Let α > 0 be such that A(x)≤αx for all x ∈ X.Ifk ∈ K and |k| > α, then A(x) − kx≥|k|x−A(x)≥(|k|−α)x for all x ∈ X, and so k ∈/ σa(A). Thus σa(A) ⊂{k ∈ K :|k|≤α}. (ii) Let β > 0 be such that A(x)≥βx for all x ∈ X.Ifk ∈ K and |k| < β, then A(x) − kx≥A(x)−|k|x≥(β −|k|)x for all x ∈ X, and so k ∈/ σa(A). Thus σa(A) ⊂{k ∈ K :|k|≥β}. (iii) If A is an isometry, let α = β := 1 in (i) and (ii) above. 1 5.10. For x ∈ , A(x)≤2x1, and A(e2)1 =2e31 = 2. Hence A=2. 2 2 Further, A (x) = (0, 0, 2x(1), 2x(2), . . .), and A (e1)1 =√2e3=2. Hence A2=2. Since 1 is a Banach space, |λ|≤A21/2 = 2 for every λ ∈ σ(A).

5.11. Let α := sup{|λn|:n ∈ N}. Then A(x)p ≤ αxp for all x ∈ X. Hence A≤α. In fact, A=α since A(e j ) = λ j e j for each j ∈ N.Thisalso shows that λ j ∈ σe(A) for each j ∈ N. Conversely, let λ ∈ σe(A), and A(x) = λx for a nonzero x ∈ X. Then there is j ∈ N such that x( j) = 0. Since λ j x( j) = λx( j), we obtain λ = λ j . Thus σe(A) ={λ j : j ∈ N}. Let E denote the closure of {λ j : j ∈ N}. Since σa(A) is closed in K,

E ⊂ σa(A). Conversely, suppose k ∈ K\E, and let δ := d(k, E)>0. Deﬁne B(y) := (y(1)/(k − λ1), y(2)/(k − λ2),...)for y ∈ X.Fory ∈ X, |B(y)( j)|=|y( j)|/|k − λ j |≤|y( j)|/δ for all j ∈ N, and so B(y) ∈ X, and B(y)p ≤yp/δ. Thus B ∈ BL(X). It is easy to check that (A − kI)B = I = B(A − kI). Hence σ(A) ⊂ E. Thus σa(A) = σ(A) = E.

5.12. Since K is a separable metric space, so is its subset E.Let{λ j : j ∈ N} be a countable dense subset of E.Forx := (x(1), x(2),...) in 2, define A(x) := (λ1x(1), λ2x(2),...). Then σ(A) = E by Exercise 5.11. Suppose (λ j ) is a sequence in K such that λ j → 0, and define A as above. 2 Also, for n ∈ N, define An(x) := (λ1x(1),...,λn x(n), 0, 0,...), x ∈ . 2 Then An ∈ BL( ) is of finite rank for each n ∈ N. Since A − An= 2 sup{|λ j |: j = n + 1, n + 2,...}→0, we see that A ∈ CL( ). Further, σe(A) ={λ j : j ∈ N}, and σ(A) = E ={λn : n ∈ N}∪{0}. 5.13. Let k ∈ K with |k| > A, and define A := A/k. Then A < 1. Hence − ( − )−1 =−( − )−1/ =− ∞ n / = I A is invertible, and A kI I A k n=0 A k − ∞ n/ n+1. ( − )−1≤ /| |( −) = /(| |− ) n=0 A k Also, A kI 1 k 1 A 1 k A . Further, for n ∈ N, and x ∈ p, An(x) = (0,...,0, x(1), x(2),...), where the ( − )−1( )( ) =− ∞ n( )( )/ n+1 = first n entries are equal to 0. Hence A kI y j n=0 A y j k −y( j)/k −···−y(1)/k j for y ∈ p, and j ∈ N. Solutions to Exercises 239

5.14. Clearly, B=1. Hence σe(B) ⊂ σa(B) ⊂ σ(B) ⊂{λ ∈ K :|λ|≤1}.Let x ∈ X.Forλ ∈ K, B(x) = λx if and only if x( j + 1) = λx( j) for all j ∈ N, 2 1 2 2 that is, x := x(1)(1, λ, λ ,...).IfX := ,,orc0, then (1, λ, λ ,...)∈ X if and only if |λ| < 1; if X := ∞, then (1, λ, λ2,...) ∈ X if and only if |λ|≤1, and if X := c, then (1, λ, λ2,...) ∈ X if and only if |λ| < 1or λ = 1. In all cases, σa(B) ={λ ∈ K :|λ|≤1} since σa(B) is a closed subset of K.Itfollowsthatσ(B) ={λ ∈ K :|λ|≤1}.

5.15. Clearly, A=1. Hence σe(A) ⊂ σa(A) ⊂ σ(A) ⊂{λ ∈ K :|λ|≤1}.Let x ∈ L p.Forλ ∈ K, A(x) = λx if and only if x(t + 1) = λx(t) for almost all ∈[ , ∞) ( + ) = j ( ) ∈[ , ) ∈ N t 0 , that is, x t j λ x t for almost all t 0 1 and all j . ∈{, } j+1 | ( )|p ( ) =| j |p 1 | ( )|p ( ) Let p 1 2 . Then j x t dm t λ 0 x s dm s by the p = translation invariance of the Lebesgue measure. It follows that x p 1 | ( )|p ( ) ∞ | p| j ∞ | p| j 0 x s dm s j=0 λ . The series j=0 λ is convergent if and only if |λ| < 1. Hence σe(A) ={λ ∈ K :|λ| < 1}. Next, let p =∞. Since {|λ| j : j ∈ N} is a bounded subset of K if and only if |λ|≤1, we see that σe(A) ={λ ∈ K :|λ|≤1}. In all cases, σa(A) ={λ ∈ K :|λ|≤1} since σa(A) is a closed subset of K. It follows that σ(A) ={λ ∈ K :|λ|≤1}.

~~5.16. Let x ∈ X and x = 0. Then rA(x), x= A(x), x−qA(x) x, x=0, that 2 2 is, rA(x) ⊥ x.Itfollowsthatfork ∈ K, A(x)−kx =rA(x) +|qA(x)− 2 2 2 k| x ≥rA(x) .~~

~~5.17. Since A ∈ CL(X), σ(A) = σa(A), which is a closed and bounded subset of n 1/n K by Proposition 5.5. Also, if λ ∈ σa(A), then |λ|≤inf{A : n ∈ N}. (See Exercise 5.8.)~~

5.18. For x := (x(1), x(2),...) ∈ X, deﬁne D(x) := (w1x(1), w2x(2),...), R(x) := (0, x(1), x(2), . . .) and L(x) := (x(2), x(3), . . .). Then A = RD and B = LD. Since wn → 0, D is a compact operator (Exercise 3.35(ii)). Hence A and B are compact operators. Let x ∈ X and λ ∈ K. Now A(x) = λx if and only if 0 = λx(1) and w j x( j) = λx( j + 1) for all

j ∈ N.Ifλ = 0 and A(x) = λx, then x = 0, and so λ ∈/ σe(A). Since A is compact, σ(A) = σe(A) ∪{0}={0}.LetE0 := {x ∈ X : A(x) = 0}= {x ∈ X : w j x( j) = 0 for all j ∈ N}. Hence E0 ={0} if and only if there is j ∈ N such that w j = 0. Also, e j ∈ E0 if and only if w j = 0. Thus if { j ∈ N : w j = 0} is a ﬁnite set, then E0 = span {e j : j ∈ N and w j = 0}, and otherwise dim E0 =∞. Next, B(x) = λx if and only if w j+1x( j + 1) = λx( j) for all j ∈ N.Ifλ = 0

and B(x) = λx, then x = 0. For if there is j0 ∈ N such that x( j0) = 0, then w j = 0 and x( j) = 0 for all j ≥ j0, and in fact |x( j)|→∞as j →∞. Hence σe(B) ⊂{0}. Since B is compact, σ(B) = σe(B) ∪{0}={0}.Let G0 := {x ∈ X : B(x) = 0}. Then G0 ={x ∈ X : w j x( j) = 0 for all j ≥ 2}. Clearly, e1 ∈ G0, and so 0 ∈ σe(B).Also,for j ≥ 2, e j ∈ G0 if and only if w j = 0. Thus if { j ∈ N : j ≥ 2 and w j = 0} is a ﬁnite set, then G0 ={e1}∪span {e j : j ≥ 2 and w j = 0}, and otherwise dim G0 =∞. 240 Solutions to Exercises

~~5.19. Let X := L2([a, b]). Since k(· , ·) ∈ L2([a, b]×[a, b]), we see that A is a compact operator on X.Letx ∈ X, and deﬁne~~

~~s y(s) := A(x)(s) = x(t)dm(t), s ∈[a, b]. a~~

Since x ∈ L1([a, b]), the fundamental theorem of calculus for Lebesgue integration shows that y is absolutely continuous on [a, b], and y = x almost everywhere on [a, b].Also,y(a) = 0. Let A(x) = 0. Then y(s) = 0 for almost all s ∈[a, b]. In fact, y(s) = 0for all s ∈[a, b] since y is continuous on [a, b]. Hence x(s) = y (s) = 0for almost all s ∈[a, b]. This shows that 0 ∈/ σe(A). Next, let λ ∈ K, λ = 0 and A(x) = λx. Then x = A(x)/λ = y/λ, and so x is absolutely continuous on [a, b]. By the fundamental theorem of calculus for Riemann integration, y is in C1([a, b]), and y (s) = x(s) for all s ∈[a, b]. Hence x = y/λ is in C1([a, b]), and λx (s) = y (s) = x(s) for all s ∈[a, b].Also,x(a) = y(a)/λ = 0. Thus x satisﬁes Bernoulli’s differential equation λx − x = 0, and also the initial condition x(a) = 0. It follows that x = 0, and so λ ∈/ σe(A). Thus σe(A) =∅. Also, σa(A) = σ(A) ={0} since A is compact. 5.20. First, consider X := C([0, 1]). Then A ∈ CL(X) since k(· , ·) is continuous. Let x ∈ X, and deﬁne

~~s 1 y(s) := A(x)(s) = tx(t) dt + s x(t) dt, s ∈[0, 1]. 0 s~~

~~Then y(0) = 0. By the fundamental theorem of calculus for Riemann integration, y ∈ C1([0, 1]), and~~

~~1 1 y (s) = sx(s) − sx(s) + x(t) dt = x(t) dt for all s ∈[0, 1]. s s~~

Then y (1) = 0. Further, y ∈ C1([0, 1]), and y (s) =−x(s) for s ∈[0, 1]. Thus we see that if x ∈ X and y := A(x), then y ∈ C2([0, 1]), y =−x and y(0) = 0 = y (1). Conversely, suppose x ∈ X, and let y ∈ C2([0, 1]) satisfy y =−x and y(0) = 0 = y (1). Integrating by parts,

~~s 1 A(y )(s) = ty (t)dt + s y (t)dt 0 s s = sy (s) − 0 y (0) − y (t)dt + s y (1) − y (s) 0 s =− y (t)dt =−y(s) + y(0) =−y(s) 0 Solutions to Exercises 241~~

~~for all s ∈[0, 1]. Hence A(y ) =−y, that is, A(x) = y. Let x ∈ X be such that A(x) = 0. Then 0 =−x, that is, x = 0. Hence~~

0 ∈/ σe(A). Next, let λ ∈ K and λ = 0. Let x ∈ X be such that A(x) = λx. Then it follows that λx =−x and λx(0) = 0 = λx (1), that is, λx + x = 0 and x(0) = 0 = x (1). Now the differential equation λx + x = 0 has a nonzero solution satisfying x(0) = 0 = x (1) if and only if λ = 4/(2n − 1)2π2, n ∈ N. In this case, the general solution is given by x(s) := cn sin(2n − 1)πs/2, s is in [0, 1], where cn ∈ K. (If K := C,wemustﬁrst show that λ ∈ R, as in the footnote in Example 5.23(ii).) Fix n ∈ N,letλn := 2 2 4/(2n − 1) π , xn(s) := sin(2n − 1)πs/2, s ∈[0, 1], and let yn := λn xn. = =− ( ) = = ( ) ( ) = = Then yn λn xn xn and yn 0 0 yn 1 . Hence A xn yn λn xn. It follows that λn is an eigenvalue of A, and the corresponding eigenspace of A is spanned by the function xn. There are no other eigenvalues of A.It 2 2 follows that σe(A) ={1/(2n −1) π : n ∈ N}. Since A is a compact operator, σa(A) = σ(A) = σe(A) ∪{0}. Next, consider Y := L p([0, 1]), where p ∈{1, 2, ∞}. The arguments in this case are exactly the same as the ones given in Example 5.23(ii). 5.21. Let k ∈ K. Suppose A + B − kI is invertible, and A − kI is one-one. Since A − kI = A + B − kI − B = (A + B − kI)(I − (A + B − kI)−1 B),we see that (I − (A + B − kI)−1 B) is one-one, that is, 1 is not an eigenvalue of (A + B − kI)−1 B. Since B is compact, so is (A + B − kI)−1 B. Hence 1 is not a spectral value of (A + B − kI)−1 B, that is, I − (A + B − kI)−1 B is invertible. It follows that A − kI is invertible. 5.22. Suppose A − kI is one-one, that is, k ∈/ σe(A). Since A is compact, A − kI is invertible. In particular, A − kI is onto. n Conversely, suppose A − kI is onto. For n ∈ N,letZn := Z((A − kI) ). Then Zn is a closed subspace of Zn+1 for all n ∈ N. Assume for a moment that Zn Zn+1 for all n ∈ N.Fixn ∈ N. By the Riesz lemma, there is zn+1 ∈ Zn+1 such that zn+1=1 and d(zn+1, Zn) ≥ 1/2. It is easy to see that (A − kI)(Zn+1) ⊂ Zn and A(Zn) ⊂ Zn. Hence for all z ∈ Zn,

~~A(zn+1) − A(z)=kzn+1 + (A − kI)(zn+1) − A(z)≥|k|/2 > 0.~~

In particular, A(zn+1) − A(zm+1)≥|k|/2 for all n, m ∈ N with n = m. Now (zn+1) is a bounded sequence in X, but the sequence (A(zn+1)) has no convergent subsequence. This contradicts the compactness of A. Hence there is m ∈ N such that Zm+1 = Zm .LetZ0 := {0}. We show that Zm = Zm−1. Let y ∈ Zm . Since A − kI is onto, there is x ∈ X such that y = (A − kI)(x). m+1 m Now (A − kI) (x) = (A − kI) (y) = 0, that is, x ∈ Zm+1 ⊂ Zm. m−1 m Thus (A − kI) (y) = (A − kI) (x) = 0, that is, y ∈ Zm−1. Similarly, Zm−1 = Zm−2,...,Z2 = Z1, and Z1 = Z0, that is, A − kI is one-one. (Compare the proof of Proposition 5.20.)

~~5.23. If k ∈/ σa(A), then the result follows from Lemma 5.19. Now suppose k is in σa(A). Since A is compact and k = 0, we see that k ∈ σe(A), 242 Solutions to Exercises~~

and the corresponding eigenspace Ek is finite dimensional. Let {x1,...,xm } ,..., ( ) = be a basis for Ek , and find x1 xm in X such that x j xi δi, j for , = ,..., := m ( ) i j 1 m. Define Y j=1 Z x j . Then Y is a closed subspace of X, and X = Y ⊕ Z(A−kI). Define B : Y → X by B(y) := (A−kI)(y), y ∈ Y . Then B is one-one. In fact, arguing as in the proof of Proposition 5.18,wesee that B is bounded below. Let β > 0 be such that βy≤(A − kI)(y) for all y ∈ Y . We show that R(B) is a closed subspace of X.Forn ∈ N,letyn ∈ Y be such that (A(yn) − kyn) converges in X to, say, z.Letα > 0 be such that A(yn) − kyn≤α for all n ∈ N. Since βyn≤A(yn) − kyn≤α for all n ∈ N, (yn) is a bounded sequence in Y . By Lemma 5.17, (yn) has a convergent subsequence, and if it converges to y in Y , then A(y) − ky = z. Thus z ∈ R(B), and so R(A − kI) = R(B) is a closed subspace of X. (Compare the proof of Lemma 5.19.)

5.24. Suppose (i) holds. Then 1 ∈/ σe(A).Infact,1∈/ σ(A) since A is compact. In this case, the inverse (I − A)−1 is continuous, that is, x := (I − A)−1(y) depends continuously on y ∈ X. Clearly, (ii) holds if and only if 1 ∈ σe(A). In this case, the eigenspace E1 := {x ∈ X : A(x) = x} of A corresponding to its nonzero eigenvalue 1 is ﬁnite dimensional, since A is compact.

5.25. (i) Since A is compact, 1 ∈ σe(A ) if and only if 1 ∈ σe(A). (ii) Let y ∈ X. Suppose there is x ∈ X such that x − A(x) = y.If x is in Z(A − I ), that is, if A (x ) = x , then x (y) = x (x) − x (A(x)) = ( ) − ( )( ) = { ,..., } ( − ) x x A x x 0. Conversely, let x1 xm be a basis for Z I A , which is ﬁnite dimensional since A is a compact operator, and suppose that ( ) = = ,..., ( ) = ∈ ( − ) x j y 0for j 1 m. Then x y 0 for all x Z I A . Assume for a moment that there is no x ∈ X such that x − A(x) = y, that is, y ∈/ R(I − A). Clearly, y = 0. Since R(I − A) is a closed subspace of X (Exercise 5.23), there is x ∈ X such that x (z) = 0 for all z ∈ R(I − A) and x (y) =y. Then x (x) − A (x )(x) = x (x − A(x)) = 0 for every x ∈ X, that is, x ∈ Z(I − A ),butx (y) =y = 0. This is a contradiction. Next, suppose x0 ∈ X and x0 − A(x0) = y. Then x ∈ X satisﬁes x − A(x) = y

if and only if x −x0 − A(x −x0) = 0, that is, x −x0 ∈ Z(A− I ), which is ﬁnite dimensional since A is a compact operator. This is the same thing as saying x := x0 + k1x1 +···+km xm , where k1,...,km are in K, and {x1,...,xm } is a basis for Z(A − I ). 5.26. Let J denote the canonical embedding of X into X . By Exercise 4.25, A J = JA, and so (A − kI)J = J(A − kI) for k ∈ K. It follows that if A − kI is one-one, then A − kI is one-one, and if A − kI is bounded below, then

A − kI is bounded below. Hence σe(A) ⊂ σe(A ) and σa(A) ⊂ σa(A ). Next, since X is a Banach space, σ(A ) = σ(A ).Also,σ(A ) ⊂ σ(A). 5.27. Let A be a normal operator on a Hilbert space H. By mathematical induction on n ∈ N, we show that if λ ∈ K, x ∈ X and (A − λI )n(x) = 0, then (A −λI )(x) = 0. If n = 1, then this is obvious. Assume this holds for m ∈ N. Solutions to Exercises 243

~~Suppose (A−λI )m+1(x) = 0. Let y := (A−λI )(x). Then (A−λI )m(y) = 0. By the inductive assumption, (A − λI )(y) = 0. Now~~

~~(A − λI )(x)2 = (A − λI )(x), (A − λI )(x)= (A∗ − λI )(A − λI )(x), x ≤(A∗ − λI )(y)x=(A − λI )(y)x=0~~

~~since A is normal. Thus (A − λI )(x) = 0.~~

5.28. Assume for a moment that σe(A) is uncountable. For each λα ∈ σe(A),let uα be a corresponding eigenvector of A with uα=1. Since A is normal,√ {uα} is an uncountable orthonormal subset of H. Since uα − uβ= 2for α = β, no countable subset of H can be dense in H. This is a contradiction to the separability of H. 2 5.29. Let A ∈ BL( ) be defined by an infinite matrix M := [ki, j ].IfM is diagonal, t 2 2 t then it is clear that M M = diag (|k1,1| , |k2,2| ,...) = M M . Hence A is a normal operator (Example 4.28(i)). Conversely, suppose A is a normal operator, and M is upper triangular. Note = > ( ) = ∞ = that ki, j 0 for all i j.First,A e1 i=1 ki,1ei k1,1e1. Since A is ∗ normal, e1 is an eigenvector of A corresponding its eigenvalue k1,1, that is, ∗( ) = t ∗ A e1 k1,1e1. But since the matrix M defines the operator A ,wesee ∗( ) = ∞ = = ··· = ( ) = that A e1 i=1 k1,i ei . Hence k1,2 k1,3 0. Next, A e2 ∞ = + = = i=1 ki,2ei k1,2e1 k2,2e2 k2,2e2, since k1,2 0 as we have just shown. ∗( ) = ∗( ) = ∞ = Again, since A is normal, A e2 k2,2e2.ButA e2 i=1 k2,i ei ∞ = = ··· = i=2 k2,i ei . Hence k2,3 k2,4 0. In this manner, by mathematical induction, we obtain ki,i+1 = ki,i+2 = ··· = 0 for every i ∈ N. Thus M = diag (k1,1, k2,2,...). If A is a normal operator, and M is lower triangular, then the normal operator A∗ is defined by the upper triangular matrix Mt . Hence Mt is a diagonal matrix, that is, M is a diagonal matrix. 5.30. Let A ∈ BL(H) be unitary. Since A is normal, and A is an isometry, σ(A) = σa(A) ⊂{k ∈ K :|k|=1} by Exercise 5.9(iii). Let k ∈ K with |k| = 1. If y ∈ H and y = (A − kI)(x), x ∈ H, then

~~y=A(x) − kx≥||k|−1|x=||k|−1|(A − kI)−1(y)~~

( )= ( − )−1≤ /|| |− | since A x x . Hence A kI 1 k 1 . 5.31. Let k ∈ K\ω(A), and β := d k, ω(A) . Since σ(A) ⊂ ω(A), A − kI is invertible. If x ∈ H and x=1, then A(x) − kx≥| A(x) − kx, x| = | A(x), x−k|≥β. Let y ∈ H.Ify = (A − kI)(x), where x ∈ H, then

~~y=A(x) − kx≥βx=β(A − kI)−1(y).~~

~~Hence (A − kI)−1≤1/β. Let A ∈ BL(H) be self-adjoint. Since (m A, MA) ⊂ ω(A) ⊂[m A, MA], 244 Solutions to Exercises~~

we see that ω(A) =[m A, MA].Letk ∈ K\[m A, MA]. Clearly, β =|Im k| if Re k ∈[m A, MA], β =|k − m A| if Re k < m A, and β =|k − MA| −1 if Re k > MA. Further, if K := R, and k ∈ R\σ(A), then (A − kI) is self-adjoint, and so (A − kI)−1=sup{|μ|:μ ∈ σ((A − kI)−1)}= sup{|(λ − k)−1|:λ ∈ σ(A)}=1/d, where d := d(k, σ(A)) (Exercise 5.1). (Note: If K := C, A is normal, and k ∈ C\σ(A), then (A − kI)−1 is normal, and so the above proof works since B=sup{|μ|:μ ∈ σ(B)} for a normal operator B on a Hilbert space H over C.) 5.32. Let A ∈ BL(H) be self-adjoint. Then ±i ∈/ σ(A). Since A∗ = A,we obtain T (A)∗ = (A − iI)−1(A + iI) = (A + iI)(A − iI)−1. It follows that (T (A))∗T (A) = I = T (A)(T (A))∗, that is, T (A) is unitary. Also, 1 ∈/ σ(T (A)) since T (A)−I = (A−iI− A−iI)(A+iI)−1 =−2i(A+iI)−1 is invertible. Next, let B ∈ BL(H) be unitary, and suppose 1 ∈/ σ(B). Since B∗ = B−1, we obtain S(B)∗ =−i(I − B∗)−1(I + B∗) =−i(I − B−1)−1(I + B−1) = i(I + B)(I − B)−1 = S(B), that is, S(B) is self-adjoint. Further, it can be easily checked that S(T (A)) = i(I +T (A))(I −T (A))−1 = A and T (S(B)) = (S(B) − iI)(S(B) + iI)−1 = B. 5.33. If A ≥ 0, then σ(A) ⊂ ω(A) ⊂[0, ∞). Conversely, suppose σ(A) ⊂[0, ∞). Then inf ω(A) = m A ∈ σa(A) = σ(A) ⊂[0, ∞), and so ω(A) ⊂[0, ∞). If 0 ∈ σe(A), then 0 ∈ ω(A) since σe(A) ⊂ ω(A). Conversely, suppose A ≥ 0, and 0 ∈ ω(A), that is, there is x ∈ H such that x=1 and A(x), x=0. By the generalized Schwarz inequality, A(x)≤ A(x), x1/4 A2(x), A(x)1/4, and so A(x) = 0. Hence 0 ∈ σe(A). 5.34. Define B(x) := (x(1) cos θ + x(2) sin θ, −x(1) sin θ + x(2) cos θ) for x := (x(1), x(2)) ∈ R2. It is easy to check that A(x), y= x, B(y) for all , ∈ R2 ∗ = ( )2 = ( ( ) − ( ) )2 + x y . Hence A B.Also, A x 2 x 1 cos θ x 2 sin θ ( ( ) + ( ) )2 = ( )2 + ( )2 = 2 ∈ R2 x 1 sin θ x 2 cos θ x 1 x 2 x 2 for all x . Hence A : R2 → R2 is a linear isometry. Since R2 is finite dimensional, A is onto. Thus A is a unitary operator. Clearly, A is defined by the 2 × 2matrixM := [ki, j ], where k1,1 := cos θ, k1,2 := − sin θ, k2,1 := sin θ and k2,2 := cos θ. Now det(M − tI) = t2 − 2t cos θ + 1, and it is equal to 0 if and only if 2 1/2 t = cos θ ± (cos θ − 1) ∈ R. Hence σ(A) = σe(A) ={1} if θ := 0, σ(A) = σe(A) ={−1} if θ := π, and σ(A) = σe(A) =∅otherwise. 5.35. Since A is a Hilbert–Schmidt operator on H, it is a compact operator (Exercise 3.40(i)). Hence σe(A) is countable. Also, the eigenspace corresponding to each nonzero eigenvalue of A has a finite orthonormal basis. Further, since A is a normal operator, any two eigenspaces of A are mutually orthogonal. Let (λn) be the sequence of nonzero eigenvalues of A, each eigenvalue being repeated as many times as the dimension of the corresponding eigenspace. Then there is a countable orthonormal subset {u1, u2,...} of H such that A(un) = λnun ∗ for each n ∈ N. By Exercise 4.31(iii), A is a Hilbert–Schmidt operator on H. {˜ , ˜ ,...} ∗( ˜ )2 < ∞ Let u1 u2 be an orthonormal basis for H such that j A u j . Solutions to Exercises 245

~~by the Parseval formula and the Bessel inequality.~~

5.36. Suppose A is normal, and let μ1,...,μk be the distinct eigenvalues of A.For j ∈{1,...,k},letE j := Z(A − μ j I ), and let Pj denote the orthogonal projection operator on H with R(Pj ) = E j .Ifx ∈ X, then Pj (x) ∈ E j , and so APj (x) = μ j Pj (x) for j = 1,...,k. Also, if i = j, then Ei ⊥ E j , and so ( ) = ⊂ ⊥ = ( ) = R Pj E j Ei Z Pi , that is, Pi Pj 0. ⊥ Let G := E1 + ··· + Ek . Then G ={0} as in the proof of Theorem

5.39, and so G = H. Consider x ∈ H. Then x = x1 + ··· + xk , where x j ∈ E j = R(Pj ) for j = 1,...,k. Thus x = P1(x) + ··· + Pk (x) and A(x) = AP1(x) + ··· + APk (x) = μ1 P1(x) + ··· + μk Pk (x), that is, I = P1 +···+ Pk and A = μ1 P1 +···+μk Pk , as desired. The converse follows easily since every orthogonal projection operator is normal, and a linear combination of normal operators is normal.

5.37. Suppose A is a nonzero compact self-adjoint operator on H.Letμ1, μ2,...be the distinct nonzero eigenvalues of A. Since A is self-adjoint, each μ j is real, and since A is compact, either the set {μ1, μ2,...} is finite or μn → 0. For each j,letE j := Z(A − μ j I ), and let Pj denote the orthogonal projection operator on H with R(Pj ) = E j . Since A is compact, each Pj is of finite rank. If x ∈ X, then Pj (x) ∈ E j , and so APj (x) = μ j Pj (x) for each j. Also, if = ⊥ ( ) = ⊂ ⊥ = ( ) = i j, then Ei E j , and so R Pj E j Ei Z Pi , that is, Pi Pj 0. Let G denote the closure of span (∪ E ), and let P denote the orthogonal pro- j j ( ) = ( ) = ( ) jection operator on H with R P G. By Exercise 4.38, P x j Pj x for all x ∈ H.Also,H = Z(A) ⊕ G, as in the proof of Theorem 5.40. ( ) = ( ) Let P0 denote the orthogonal projection operator on Hwith R P0 Z A . ∈ = ( ) + ( ) = ( ) + ( ) ( ) = Then for x H, x P0 x P x P0 x j Pj x and A x ( ) + ( ) = ( ) AP0 x j APj x j μ j Pj x . = { , ,...} In fact, A j μ j Pj . This is obvious if the set μ1 μ2 is finite. Suppose 2 = ( )2 + ( )2 = now that this set is infinite. First note that x P0 x P x ( )2 + ∞ ( )2 n ( )2 ≤ 2 ∈ P0 x j=1 Pj x , and so j=1 Pj x x for all x H ∈ N ∈ N := n > ∈ N and all n .Forn , define An j=1 μ j Pj .Given 0, find n0 such that |μ j | < for all n > n0. Then for all n ≥ n0, 246 Solutions to Exercises ∞ 2 ∞ 2 2 2 A(x) − An(x) = μ j Pj (x) ≤ |μ j | Pj (x) j=n+1 j=n+1 ∞ 2 2 2 2 < Pj (x) ≤ x , x ∈ H. j=n+1 − < ≥ = ∞ ( ) Thus A An for all n n0. Hence A j=1 μ j Pj in BL H . For the converse, note that each P is self-adjoint since it is an orthogonal j ∈ R ∗ = ∗ = = projection operator, each μ j , and so A j μ j Pj j μ j Pj A. := n Also, A is compact since An j=1 μ j Pj is a bounded operator of finite rank for each n and An − A→0. 5.38. The kernel k(s, t) := min{1−s, 1−t}, s, t ∈[0, 1], is a real-valued continuous function on [0, 1]×[0, 1], and k(t, s) = k(s, t) for all s, t ∈[0, 1]. Hence A is a compact self-adjoint operator on H.Also,A = 0. Let λ be a nonzero eigenvalue of A, and let x ∈ H be a corresponding eigenvector of A. Since A(x) ∈ C([0, 1]), x = A(x)/λ is continuous on [0, 1].Now

~~s 1 y(s) := A(x)(s) = (1 − s) x(t) dt + (1 − t)x(t) dt, s ∈[0, 1]. 0 s~~

~~Clearly, y(1) = 0. By the fundamental theorem of calculus for Riemann integration, y ∈ C1([0, 1]), and~~

~~s s y (s) = (1 − s)x(s) − x(t) dt − (1 − s)x(s) =− x(t) dt, s ∈[0, 1]. 0 0~~

Hence y (0) = 0. Further, y ∈ C1([0, 1]), and y (s) =−x(s) for s ∈[0, 1]. Thus y ∈ C2([0, 1]), y =−x and y (0) = 0 = y(1). Since y = A(x) = λx, we see that λx =−x and λx (0) = 0 = λx(1), that is, λx + x = 0 and x (0) = 0 = x(1). Now the differential equation λx + x = 0 has a nonzero solution satisfying x (0) = 0 = x(1) if and only if λ = 4/(2n−1)2π2, n ∈ N. In this case, the general solution is given by x(s) := cn cos(2n − 1)πs/2for s ∈[0, 1], where cn ∈ K. (If K := C, we must ﬁrst show that λ ∈ R, as in the footnote in Example 5.23(ii).) Conversely, ﬁx n ∈ N,letλn := 2 2 4/(2n − 1) π , xn(s) := cos(2n − 1)πs/2, s ∈[0, 1], and let yn := λn xn. = =− ( ) = = ( ) Clearly, yn λn xn xn and yn 0 0 yn 1 . Integrating by parts,

s 1 ( )( ) = ( − ) ( ) + ( − ) ( ) A yn s 1 s yn t dt 1 t yn t dt 0 s = ( − )( ( ) − ( )) − ( − ) ( ) + ( ) − ( ) 1 s yn s yn 0 1 s yn s yn 1 yn s =−yn(s) for s ∈[0, 1]. Solutions to Exercises 247

( ) =− ( ) = = Thus A yn yn, that is, A xn yn λn xn.Itfollowsthatλn is in fact an eigenvalue of A, and the corresponding eigenspace of A is spanned by the ( ) = ∞ , := /( − )2 2 function xn. Hence√ A x n=1 λn x un un, where λn 4 2n 1 π and un(s) := 2 cos(2n − 1)πs/2, s ∈[0, 1], for n ∈ N. 5.39. The kernel k(s, t) := min{s, t}, s, t ∈[0, 1], is a real-valued continuous function on [0, 1]×[0, 1], and k(t, s) = k(s, t) for all s, t ∈[0, 1]. Hence A is a compact self-adjoint operator on L2([0, 1]).Also,A = 0. By Exercise 2 2 5.20, the nonzero eigenvalues of A are given by λn := 4/(2n − 1) π , n ∈ N, and the eigenspace of A corresponding√ to λn is span {xn}, where xn(s) := sin(n −1/2)πs, s ∈[0, 1].Letun(s) := 2sin(n −1/2)πs, s ∈[0, 1]. Then { : ∈ N} 2([ , ]) un n is an orthonormal basis for L 0 1 consisting of eigenvectors ∈ 2([ , ]) ( ) = ∞ , of A.Forx L 0 1 , A x n=1 λn x un un, that is,

∞ s 1 8 s (x) πs tx(t) dm(t) + s x(t) dm(t) = n sin(2n − 1) , π2 (2n − 1)2 2 0 s n=1 ( ) := 1 ( ) ( − / ) ( ), where sn x 0 x t sin n 1 2 πtdm t and the series on the right side converges in L2([0, 1]). We shall use Theorem 5.43. Let y ∈ L2([0, 1]) and μ ∈ K, μ = 0. Consider the integral equation

~~s 1 x(s) − μ s tx(t) dm(t) + s x(t) dm(t) = y(s), s ∈[0, 1]. 0 s~~

:= −1 = ( − )2 2/ ∈ N = ∈ N, Let μn λn 2n 1 π 4forn .Ifμ μn for any n then there is a unique x ∈ L2([0, 1]) satisfying x − μA(x√) = y. In fact, since 2 2 μ/(μn − μ) = 4μ/ (2n − 1) π − 4μ and y, un= 2 sn(y) for n ∈ N,

~~∞ s (y) πs x(s) = y(s) + 8μ n sin(2n − 1) , s ∈[0, 1]. (2n − 1)2π2 − 4μ 2 n=1~~

2 2 Further, x≤αy, where α := 1 + 4|μ|/ minn∈N{|(2n − 1) π − 4μ|}. 2 2 2 Next, suppose μ := (2n1 − 1) π /4, where n1 ∈ N. There is x in L ([0, 1]) − ( ) = ⊥ ( ) = satisfying x μA x y if and only if x un1 , that is, sn1 y 0. In this 2 case, since μ/(μn − μ) = (2n1 − 1) /4(n − n1)(n + n1 − 1) for n = n1,

~~(2n − 1)2 s (y) πs x(s) = y(s) + 1 n sin(2n − 1) 2 (n − n1)(n + n1 − 1) 2 n=n1 πs + k sin(2n − 1) , s ∈[0, 1], where k ∈ K. 1 1 2 1~~

~~5.40. Suppose A is a nonzero compact operator on H. Then A∗ A is a compact self- adjoint operator on H. Also, since A(x)2 = A∗ A(x), x for all x in H,we 248 Solutions to Exercises~~

see that Z(A∗ A) = Z(A). In particular, A∗ A = 0. Then H has an orthonormal ∗ ∗ basis {uα} consisting of eigenvectors of A A.LetA A(uα) = λαuα for each α. By Theorem 5.36(iii), the set S := {uα : λα = 0} is countable. Let S := { , ,...} ∗ ( ) = u1 u2 and A A un λnun for each n.IfS is in fact denumerable,√ then ∗ λn → 0. Also, λn ≥ 0 for each n since A A is positive. Let sn := λn > 0 and vn := A(un)/sn for each n. Then snsm vn,vm = A(un), A(vm )= ∗ A A(un), um =λn un, um for all n, m.Itfollowsthat{v1,v2,...} is an {v ,v ,...} orthonormal subset of H.If 1 2 is denumerable, then so is S, and → ∈ ≤ | |2| , |2 ≤ 2 sn 0. Let x H.Letsn s for all n. Then n sn x un s x , and , v ( ) := , v ∈ so n sn x un n converges in H. Define B x n sn x un n for x H. ( ) = v = ( ) ∈ ( ) ⊥ Then B un sn n A un for all n. Also, if uα Z A , then uα un for ( ) = = ( ) ( ) = ( ) = , v all n, and so B uα 0 A uα . Hence A x B x n sn x un n for all x ∈ H. ( ) = , v ∈ { , ,...} Conversely, let A x n sn x un n for all x H, where u1 u2 and {v1,v2,...} are countable orthonormal subsets of H, and s1, s2,...are positive → {v ,v ,...} numbers such that sn 0iftheset 1 2 is denumerable. First, suppose {v ,v ,...} ∈ N ( ) = m , v 1 2 is finite. Then there is m such that A x n=1 sn x un n for all x ∈ H, and so R(A) = span {v1,...,vm }. Since A is a bounded operator of finite rank, it is compact. Next, suppose {v1,v2,...} is denumerable. Let > ∈ N | | < > ∈ N 0. There is m0 such that sn for all n m0.Form ,let ( ) := m , v , ∈ Am x n=1 λn x un n x H. Then

~~ ∞ ∞ 2 2 2 2 A(x) − Am(x) = sn x, unvn = |sn| | x, un| n=m+1 n=m+1 ∞ 2 2 2 2 ≤ | x, un| ≤ x for x ∈ H and m ≥ m0. n=m+1~~

Thus A − Am ≤ for all m ≥ m0. Since Am → A in BL(H), and each Am is a bounded operator of ﬁnite rank, A is compact. In this case, A∗(x), y= x, A(y)= s x,v u , y= C(x), y for n n n n , ∈ ( ) := ,v ∈ ∗ = all x y H, where C x n sn x n un for x H. Hence A C. ∈ := ( ) ,v = , Let x H and y A x . Then y n sn x un for all n, and so ∗ ( ) = ∗( ) = ( ) = ,v = 2 , A A x A y C y n sn y n un n sn x un un. Also, A(u )2 = A∗ A(u ), u = λ u , u =s2 for all n, and so n n n n n n n 2 = ( )2 n sn n A un . Now suppose H is a separable Hilbert space. Then the orthonormal basis {uα} ∗ ∗ ( ) = for H consisting of eigenvectors of A A is countable. Also, A A uα 0if ( ) = ( )2 = ( )2 and only if A uα 0. Hence α A uα n Aun .ByExercise ( )2 < ∞ 4.31(iii),A is a Hilbert–Schmidt operator if and only if n A un , 2 < ∞ that is, n sn . References

1. M. Ahues, A. Largillier and B.V. Limaye, Spectral Approximation for Bounded Operators, Chapman & Hall/CRC, Boca Raton, Florida, 2001. 2. F. Altomare and M. Campiti, Korovkin-type Approximation Theory and its Applications,de Gruyter, Berlin, New York, 1994. 3. S. Banach,Théorie des Opérations Linéaires, Monografje Matematyczne, Warsaw, 1932. 4. E.A. Coddington, An Introduction to Ordinary Differential Equations, Prentice-Hall, Engle- wood Cliffs, N. J., 1961. 5. R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. I, Interscience, New York, 1953. 6. R.A. Devore and G.G. Lorentz, Constructive Approximation, Springer-Verlag, Berlin, 1991. 7. J. Diestel, Sequences and Series in Banach Spaces, Springer-Verlag, New York, 1984. 8. J. Dieudonné, Foundations of Modern Analysis, Academic Press, New York, 1969. 9. P. Enﬂo, A counter-example to the approximation problem in Banach spaces, Acta Math. 103 (1973), pp. 309–317. 10. G. Fichtenholtz and L. Kantorovich, Sur les opérations linéaires dans l’espace des fonctions bornées, Studia Math. 5 (1934), pp. 69-98. 11. S.R. Foguel, On a theorem of A.E. Taylor, Proc. Amer. Math. Soc. 9 (1958), p. 325. 12. S.R. Ghorpade and B.V.Limaye, A Course in Calculus and Real Analysis, Undergraduate Texts in Mathematics, Springer, New York, 2006. 13. S.R. Ghorpade and B.V. Limaye, A Course in Multivariable Calculus and Analysis, Under- graduate Texts in Mathematics, Springer, New York, 2010. 14. K. Gustafson, The Toeplitz-Hausdorff theorem for linear operators, Proc. Amer. Math. Soc. 25 (1970), pp. 203–204. 15. P.R. Halmos and V. S. Sunder, Bounded Integral Operators on L2 spaces, Springer-Verlag, New York, 1978. 16. E. Hewitt and K. Stromberg, Real and Abstract Analysis, Graduate Texts in Mathematics, Springer-Verlag, New York, 1965. 17. K. Hoffman, Banach Spaces of Analytic Functions, Prentice-Hall, Englewood Cliffs, N. J., 1962. 18. R.A. Horn and C. R. Johnson, Matrix Analysis, Cambridge Univ. Press, Cambridge, 1985. 19. P.P. Korovkin, Linear Operators and Theory of Approximation, Hindustan Publ. Corp., Delhi, 1960. 20. R.G. Kuller, Topics in Modern Analysis, Prentice-Hall, Englewood Cliffs, N.J., 1969. 21. B.V. Limaye, Operator approximation, Semigroups, Algebras and Operator Theory, Springer Proceedings in Mathematics and Statistics, Vol. 142, Springer India, 2015, pp. 135–147. © Springer Science+Business Media Singapore 2016 249 B.V. Limaye, Linear Functional Analysis for Scientists and Engineers, DOI 10.1007/978-981-10-0972-3 250 References

22. J. Lindenstrauss and L. Tzafriri, On the complemented subspaces problem, Israel J. Math. 9 (1971), pp. 263–269. 23. F. Riesz and B. Sz.-Nagy, Functional Analysis, Frederick Ungar, New York, 1955. 24. H.L. Royden, Real Analysis, third ed., Macmillan, New York, 1988. 25. W. Rudin, Principles of Mathematical Analysis, third ed., McGraw-Hill, New York, 1976. 26. W. Rudin, Real and Complex Analysis, third ed., McGraw-Hill, New York, 1986. 27. F.C. Sánchez and J.M.F. Castillo, Isometries of ﬁnite-dimensional normed spaces, Extracta Mathematicae 10 (1995), pp. 146–151. 28. A.E. Taylor and D.C. Lay, Introduction to Functional Analysis, second ed., Wiley, New York, 1980. 29. D.S. Watkins, Fundamentals of Matrix Computations, second ed., Wiley, New York, 2002. 30. R. Whitley, Projecting m onto c0, Amer. Math. Monthly 73 (1966), pp. 285–286. 31. P.P.Zabreiko, A theorem for semiadditive functionals, Functional Analysis and its Applications, 3 (1969), pp. 86–88. (Translated from the original Russian version in Funksional’nyi Analiz Ego Prilozheniya) Index

A C Absolutely continuous function, 24 Canonical embedding, 129 Absolutely summable series, 53 Carathéodory condition, 20 Absorbing set, 69, 126 Cauchy sequence, 12 Adjoint, 141 Cayley transform, 201 Almost everywhere, 20 Characteristic function, 21 Closed ball, 34 Angle, 71 Closed complement, 100 Antisymmetric relation, 2 Closed graph theorem, 97, 116, 157 Approximate eigenspectrum, 160 Closed map, 96 Approximate eigenvalue, 164 Closed set, 10 Approximate solution, 104 Closed unit ball, 34 Arzelà theorem, 19 Closure, 10 Ascoli theorem, 19 Compact linear map, 106 Compact operator, 107 Compact perturbation, 200 Compact set, 14 B Compact support, 36 Baire theorem, 14 Comparable norms, 37 Balanced set, 69 Complete metric space, 13 Banach space, 53 Completion of a normed space, 129 Banach–Steinhaus theorem, 93 Componentwise convergence, 10 Basic inequality for an operator norm, 79 Compression spectrum, 179 Basis, 3 Conjugate exponents, 81 Bernstein polynomial, 17, 118 Conjugate-linear map, 44 Bessel inequality, 50 Conjugate-symmetric function, 44 Best approximation, 74 Conjugate-transpose of a matrix, 142 Bolzano–Weierstrass theorem, 13 Continuous function, 15 Continuously differentiable function, 24 Bounded below, 77 Convergence in the mean, 28 Bounded convergence theorem, 22 Convergent sequence, 10 Bounded inverse theorem, 101, 116, 157 Convergent series, 53 Bounded linear map, 76 Convex set, 69, 125 Bounded operator, 76 Coset, 4 Bounded sequence, 12 Countable set, 2 Bounded set, 12 Countably additive function, 20 Bounded variation, 25, 135 Countably subadditive seminorm, 89 © Springer Science+Business Media Singapore 2016 251 B.V. Limaye, Linear Functional Analysis for Scientists and Engineers, DOI 10.1007/978-981-10-0972-3 252 Index

D G Dense subset, 11 Gelfand–Mazur theorem, 170 Denumerable set, 2 Generalized eigenvector, 200 Diagonal operator, 117, 142, 188, 198 Generalized polarization identity, 158 Dimension, 4 Generalized Schwarz inequality, 151 Direct sum, 68 General solution, 200 Dominated convergence theorem, 22 Gershgorin disk, 198 Dot product, 44 Gershgorin theorem, 198 Dual basis, 128 Gram–Schmidt orthonormalization, 48 Dual space, 128 Graph of a function, 96 Green’s function, 177

E Eigenequation, 161 H Eigenspace, 160 Haar system, 61 Hahn–Banach extension, 122 Eigenspectrum, 160 Hahn–Banach extension theorem, 122, 133 Eigenvalue, 160 Hahn–Banach separation theorem, 127 Eigenvector, 160 Hamel basis, 3 Energy principle, 153 Heine–Borel theorem, 15 Equicontinuity of seminorms, 92 Helly theorem, 154 Equicontinuous set of functions, 18 Helmert basis, 71 Equivalence class, 2 Hermite polynomials, 50 Equivalence relation, 2, 4 Hermitian operator, 157 Equivalent norms, 37 Hilbert cube, 70 Essential bound, 27 Hilbert space, 62 Essentially bounded function, 27 Hilbert space isomorphism, 68, 146 Essentially uniform convergence, 28 Hilbert–Schmidt map, 99, 118, 157, 202 Essential range, 198 Hilbert–Schmidt test, 113 Essential supremum, 27 Homeomorphic spaces, 15 Homeomorphism, 15 Homogeneous equation, 200 F Hyperplane, 6 Fast Fourier transform, 157 Hyperspace, 6 Féjer theorem, 30 Hyponormal operator, 157 Finite dimensional linear space, 4 Finite rank, 5 Finite set, 2 I Fourier coefficient, 29, 30 Infimum, 8 Fourier expansion, 64 Infinite dimensional linear space, 4 Fourier integral, 152 Inner product, 44 Fourier matrix, 157 Inner product space, 44 Fourier-Plancherel transform, 153 Integrable function, 22 Fourier series, 29 Inverse Cayley transform, 201 Fourier transform, 152 Inversion theorem, 31, 152 Fredholm alternative, 200 Invertible operator, 160 Fredholm integral equation, 196 Isometric spaces, 15 Fredholm integral map, 87 Isometry, 15 Frobenius norm, 71, 86 Fubini theorem, 23 Function space, 36 K Fundamental theorem of calculus for Rie- Kernel, 5, 87 mann and Lebesgue integrations, 24 Kronecker symbol, 6 Index 253

L Open ball, 11, 34 Law of cosines, 71 Open map, 102 Lebesgue integral, 21 Open mapping theorem, 103, 116, 157 Lebesgue measurable set, 20 Open set, 10 Lebesgue measure, 20 Open unit ball, 34 Lebesgue outer measure, 20 Operator equation, 104 Left shift operator, 143, 156, 169, 170, 199 Operator norm, 79 Legendre polynomials, 49 Orthogonal complement, 69 Light-like vector, 45 Orthogonal projection, 69 Limit, 10 Orthogonal projection operator, 100, 152 Linear combination, 3 Orthogonal set, 47 Linear functional, 5 Orthonormal basis, 65 Linearly dependent subset, 3 Orthonormal set, 48 Linearly independent subset, 3 Linear map, 5 Linear space, 3 P Lower triangular matrix, 201 Parallelepiped law, 70 Parallelogram law, 47 Parseval formula, 64 M Parseval identity, 68 Matrix deﬁning a linear map, 82 Partially ordered set, 2 Matrix transformation, 82 Partial order, 2 Maximal element, 2 Partial sum, 53 Maximal orthonormal subset, 52 Particular solution, 200 Mean square convergence, 28 Perturbation technique, 105 Measurable function, 20, 22 Pointwise bounded set of functions, 18 Metric, 7 Pointwise convergence, 17 Metric space, 7 Pointwise limit, 17 Minimum residual property, 199 Polarization identity, 70 Minkowski gauge, 126 Polya theorem, 95 Minkowski inequality, 45 Positive-deﬁnite function, 44 Minkowski inequality for functions, 26 Positive operator, 146 Minkowski inequality for numbers, 8 Product norm, 39, 40 Minkowski space, 44 Product space, 5 Monotone convergence theorem, 22 Projection map, 40 Multiplication operator, 165, 198 Projection operator, 99 Mutually orthogonal subspaces, 182 Projection theorem, 68 Pythagoras theorem, 48

N Neumann expansion, 199 Q Node, 95 QR factorization, 71 Norm, 33 Quadrature formula, 95 Normal operator, 146 Quotient map, 40 Normed space, 34 Quotient norm, 38 Norm of the graph, 115 Quotient space, 4 Nullity, 5 Null space, 5 Numerical range, 149 R Range space, 5 Rank, 5 O Rank-nullity theorem, 5 Oblique projection operator, 100 Rayleigh quotient, 171, 199 One-to-one correspondence, 2 Real hyperplane, 127 254 Index

Reflexive normed space, 129 T Reflexive relation, 2 Taylor–Foguel theorem, 154 Relation, 2 Term of a sequence, 10 Representer, 132 Tietze extension theorem, 16 Residual element, 199 Time-frequency equivalence, 153 Resonance theorem, 130 Time-like vector, 45 Riemann–Lebesgue lemma, 31 Toeplitz-Hausdorff theorem, 150 Riesz-Fischer theorem, 63 Tonelli theorem, 23 Riesz representation theorem, 132, 136 Total energy, 153 Right shift operator, 142, 156, 157, 169, 170 Totally bounded set, 12 Ritz method, 184 Totally ordered set, 2 Total variation, 25, 135 Transitive relation, 2 S Transpose, 136 Saw-tooth function, 61 Transposed homogeneous equation, 200 Schauder basis, 60 Triangle inequality, 7 Schauder theorem, 140 Tridiagonal operator, 117 Schur result, 131 Trigonometric polynomials on R, 72 Schur test, 113 Two-norm theorem, 101 Schwarz inequality, 45 Schwarz inequality for functions, 26 Schwarz inequality for numbers, 8 U Second dual space, 129 Uncountable set, 2 Self-adjoint operator, 146 Uniform boundedness principle, 93 Seminorm, 33 Uniform convergence, 10 Separable metric space, 11 Uniform limit, 17 Sequence, 10 Uniformly bounded set of functions, 18 Sequence space, 35 Uniformly continuous function, 16 Signal analysis, 153 Unique Hahn–Banach extension, 133 Signum function, 81 Unit sphere, 34 Simple function, 21 Unitary operator, 146 Singular value, 202 Upper bound, 2 Skew-hermitian operator, 157 Upper triangular matrix, 201 Sobolev space, 58, 62, 73 Urysohn lemma, 16 Space-like vector, 44 Usual inner product, 44 Span, 3 Spanning subset, 3 Spectral radius formula, 170 V Spectral theorem, 189–191, 193 Vanishing at infinity, 36 Spectral value, 165 Volterra integration operator, 199 Spectrum, 160 Square-integrable function, 26 Standard basis, 4 W Strictly convex normed space, 70, 154 Weak convergence, 155 Sturm–Liouville problem, 177 Weak limit, 155 Sublinear functional, 120 Weierstrass theorem, 17 Subsequence, 10 Weight, 95 Subspace, 3 Weighted shift operator, 117, 199 Sum of a series, 53 Summable series, 53 Sup metric, 10 Z Sup norm, 36 Zabreiko theorem, 91 Supremum, 8 Zero space, 5 Symmetric relation, 2 Zorn lemma, 2