Real Analysis Chapter 5 Solutions Jonathan

4. Note that ‖Tx− Tnxn‖ ≤ ‖Tx− Txn‖+ ‖Txn − Tnxn‖ ≤ ‖T‖‖x− xn‖+ ‖T − Tn‖‖xn‖, and the limit as n→∞ of the right hand side is 0, so limn→∞ Tnxn = Tx. 6. (a) Clearly ‖x‖1 ≥ 0 for all x ∈ X. If ∑n k=1 akek ∈ X is non-zero then am 6= 0 for some m ∈ {1, 2, . . . , n}. This implies that ‖ ∑n k=1 akek‖1 ≥ |am| > 0. Moreover, if ∑n k=1 akek ∈ X and b ∈ K then ∥∥∥∥b n ∑ k=1 akek ∥∥∥∥ 1 = n ∑ k=1 |bak| = |b| ∥∥∥∥ n ∑ k=1 akek ∥∥∥∥ 1 . If ∑n k=1 akek, ∑n k=1 bkek ∈ X then ∥∥∥∥ n ∑ k=1 akek + n ∑ k=1 bkek ∥∥∥∥ 1 = n ∑ k=1 |ak + bk| ≤ n ∑ k=1 |ak|+ n ∑ k=1 |bk| = ∥∥∥∥ n ∑ k=1 akek ∥∥∥∥ 1 + ∥∥∥∥ n ∑ k=1 bkek ∥∥∥∥ 1 . This shows that ‖ · ‖1 is a norm on X. (b) Let (a1, a2, . . . , an) ∈ Kn and suppose that ‖(a1, a2, . . . , an)‖2 = 1. Then ∑n k=1 |ak| = 1, and in particular |ak| ≤ 1 for all k ∈ {1, 2, . . . , n}. It follows that |ak| ≤ 1 for all k ∈ {1, 2, . . . , n}, in which case ∥∥∥∥ n ∑ k=1 akek ∥∥∥∥ 1 = n ∑ k=1 |ak| ≤ n, and hence the linear map (a1, a2, . . . , an) 7→ ∑n k=1 akek is bounded, thus continuous. (c) Define C := {(a1, a2, . . . , an) ∈ Kn | ‖ ∑n k=1 akek‖1 = 1}, so that the image of C under the map from (b) is the given set. Hence, it suffices to show that C is compact. Note that ‖ · ‖1 : X → R is (Lipschitz) continuous, so the map (a1, a2, . . . , an) 7→ ‖ ∑n k=1 akek‖1 from Kn to R is continuous, implying that C (the preimage of {1}) is closed. Moreover C is bounded, thus compact, because for each (a1, a2, . . . , an) ∈ Kn ‖(a1, a2, . . . , an)‖2 ≤ n ∑ k=1 ‖(0, . . . , 0, ak, 0, . . . , 0)‖2 = n ∑ k=1 |ak| = 1. (d) Let ‖ · ‖ : X→ R be a norm on X. If ∑n k=1 akek ∈ X then ∥∥∥∥ n ∑ k=1 akek ∥∥∥∥ ≤ n ∑ k=1 |ak|‖ek‖ ≤ max{‖ek‖}k=1 n ∑ k=1 |ak| = max{‖ek‖}k=1 ∥∥∥∥ n ∑ k=1 akek ∥∥∥∥ 1 . This proves one half of the equivalence and shows that ‖ · ‖ is (Lipschitz) continuous on (X, ‖ · ‖1). It follows that ‖ · ‖ has a minimum at some u ∈ B, where B is the compact set {x ∈ X | ‖x‖1 = 1}. Since 0 / ∈ B, u 6= 0 and hence ‖u‖ > 0. If x ∈ X is non-zero then ‖x‖ = ‖x‖1 ∥∥∥∥ x ‖x‖1 ∥∥∥∥ ≥ ‖x‖1‖u‖, which also holds if x = 0. This proves the other half of the equivalence, so every norm on X is equivalent to ‖ ·‖1. 7. (a) Since X is a Banach space, L(X,X) is also a Banach space. The series ∑∞ k=0(I − T )k is absolutely convergent because ∑∞ k=0 ‖(I − T )k‖ ≤ ∑∞ k=0 ‖I − T‖k, which is a geometric series with ratio ‖I − T‖ < 1. Therefore