-1
$\begingroup$

I've a problem with some exercise, namely:

Show that if X is a finite-dimensional Banach space, then every linear functional f on X is continuous on X.

Hint.
Use Proposition: Every operator T from a finite-dimensional normed space X into a normed space Y is continuous.

I don't even know how to start...

Can someone help?
Thanks and regards!

$\endgroup$

closed as off-topic by Nosrati, Lord Shark the Unknown, Leucippus, user91500, José Carlos Santos Jan 17 at 10:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Lord Shark the Unknown, Leucippus, user91500, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ A linear functional is a linear operator. With codomain the scalar field. So this is a particular case of the proposition you quote. $\endgroup$ – Julien May 10 '13 at 14:04
2
$\begingroup$

Recall that all norms are equivalent in finite dimensional vector space and let $(e_1,\cdots,e_n)$ a basis for $X$ and let $$x=\sum_{i=1}^n x_i e_i\in X$$ then we have $$|f(x)|=\left|\sum_{i=1}^n x_if( e_i)\right|\leq \sum_{i=1}^n |x_i| |f( e_i)|\leq M \sum_{i=1}^n |x_i|=M||x||_1$$ where $$M=\max_{1\le i\le n}|f(e_i)|$$

and then we can deduce.

$\endgroup$
  • $\begingroup$ Thanks! You really helped me. The rest also thanks $\endgroup$ – Dareq May 11 '13 at 14:29
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 May 11 '13 at 16:31
0
$\begingroup$

Hint: A linear functional is a linear operator into the scalar field.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.