Would you know how to show that ||AB||*<=||A||*||B||*? This would be helpful for Problem 2.
Also, how is ||B||* equal to sup{||B'(x)||: ||x||=1}? Notice that in my definition of || ||*, X refers to an nxn matrix.
Let M_n(R) be the n x n matrices over the reals R. Define a norm || || on M_n(R) by ||A||= sum of absolute values of all the entries of A. Further define a new norm || ||* by ||A||* = sup{||AX||/||X||, ||X||!=0}.
Show that
1. M_n(R) under || ||* is complete.
2. If ||A||<1, then I-A is...
I assume it is not TOTALLY bounded by Heine-Borel. Not compact=> not complete or not totally bounded. If not complete, then we are done. Therefore assume not totally bounded but complete.
OK here is an example. Take the set of the natural numbers. Under the usual metric p it is non-compact. But now I can define a new metric p* by p*(n,n)=0 and p*(n,n+1)=1/2^n. It is easy to check that p and p* are equivalent. Further, the sequence {n} used to be unbounded under p, but it is now a...
Challenging Problem! (Equivalent Metrics)
I need to show that if (X,p) is a non-compact metric space, then there exists a metric p* equivalent to p such that (X,p*) is not complete.
I greatly appreciate your help!