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Solutions to Exercises Solutions to Exercises 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)<t ≤ 1.√ Then xn ∈ C([0, 1]) for n ∈ N, xn∞ = 1, xn1 = 1/2n and xn2 = 1/ 3n. 2.6. Let x ∈ X.Ifx = 0, then x is a constant function with x(a) = 0, that is, x = 0. For t ∈ (a, b], there is s ∈ (a, t) such that x(t) − x(a) = (t − a)x (s) by the mean value theorem, and so |x(t)|≤|x(a)|+(b − a)x ∞. Hence © Springer Science+Business Media Singapore 2016 203 B.V. Limaye, Linear Functional Analysis for Scientists and Engineers, DOI 10.1007/978-981-10-0972-3 204 Solutions to Exercises 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.
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