Hint: It's enough to show this for such that
Dear Colleagues,
Could you please help me in solving the following problem:
Let be the space of all such that , and are numbers, and be any norm on , and define the norm on to be . We want to show directly that there is such that for any , but without using the fact that all norms on a finite dimensional vector space are equivalent. Instead we want to use the fact that " A continuous mapping of a compact subset of a normed space into the real numbers assumes a maximum and a minimum at some points of ".
I think the continuous mapping in my problem will be the norm but which norms will be used and what is the compact set.
Best Regards.
Raed.
No, the set is closed, and ymar's hint was a good one.
It's probably best to start with the 2-norm, and to look at the set , which is closed and bounded (as a subset of n-dimensional euclidean space) and therefore compact. You need to show that the function is continuous on this space and hence attains its upper and lower bounds. The tricky part is to do this without implicitly assuming the result that you are trying to prove.