Exercise 2.8.3

Define

t mn = i = 1 m j = 1 n | a ij |
(a)
Prove that ( t nn ) converges.
(b)
Now, use the fact that ( t nn ) is a Cauchy sequence to argue that ( s nn ) converges.

Answers

(a)
Note that t nn is monotone increasing; moreover
i = 1 j = 1 | a ij | i = 1 j = 1 n | a ij | i = 1 n j = 1 n | a ij | = t nn

and therefore t nn is bounded; by the Monotone Convergence Theorem t nn converges.

(b)
Since ( t nn ) is a Cauchy sequence, for any 𝜖 > 0 there exists N such that if p > q > N ,
𝜖 > | t pp t qq | = | i = 1 p j = 1 p | a ij | i = 1 q j = 1 q | a ij | | = | i = q + 1 p j = 1 p | a ij | + i = 1 q j = q + 1 p | a ij | + i = 1 q j = 1 q | a ij | i = 1 q j = 1 q | a ij | | = | i = q + 1 p j = 1 p | a ij | + i = 1 q j = q + 1 p | a ij | | | i = q + 1 p j = 1 p a ij + i = 1 q j = q + 1 p a ij | = | i = q + 1 p j = 1 p a ij + i = 1 q j = q + 1 p a ij + i = 1 q j = 1 q a ij i = 1 q j = 1 q a ij | = | i = 1 p j = 1 p a ij i = 1 q j = 1 q a ij | = | s pp s qq |

and therefore s nn is also a Cauchy sequence and thus converges.

User profile picture
2022-01-27 00:00
Comments