# Thread: Continuity and norm space

1. ## Continuity and norm space

Let V a norma space with 2 norms , lets say norm 1 and norm 2

Show that the function identity $\ I_d :\left( {V,\left\| x \right\|_1 } \right) \to \left( {V,\left\| x \right\|_2 } \right)\$ es continuos iff the set
A $\A = \left\{ {x \in V/\left\| x \right\|_1 = 1} \right\}\$ is bounded with norm 2

Please somebody give me a biiiiiiiiiiiiig hint....

Regards...

2. Originally Posted by orbit
Let V a norma space with 2 norms , lets say norm 1 and norm 2

Show that the function identity $\ I_d :\left( {V,\left\| x \right\|_1 } \right) \to \left( {V,\left\| x \right\|_2 } \right)\$ es continuos iff the set
A $\A = \left\{ {x \in V/\left\| x \right\|_1 = 1} \right\}\$ is bounded with norm 2

Please somebody give me a biiiiiiiiiiiiig hint....

Regards...
Come on man--you really need to show some effort. You need remember that linear operators are continuous if and only if they're bounded. So, $\text{id}:V\to V$ will be bounded if and only if there is a constant $M$ such that $\|\text{id}(v)\|_1\leqslant M\|v\|_2$ for all $v\in V$... now why is this equivalent to saying that this inequality holds true for every $v\in B_{\|\cdot\|_1}(0;1)$?

3. Originally Posted by Drexel28
Come on man--you really need to show some effort. You need remember that linear operators are continuous if and only if they're bounded. So, $\text{id}:V\to V$ will be bounded if and only if there is a constant $M$ such that $\|\text{id}(v)\|_1\leqslant M\|v\|_2$ for all $v\in V$... now why is this equivalent to saying that this inequality holds true for every $v\in B_{\|\cdot\|_1}(0;1)$?
Hi, thanl you for answering. unfortunately for me,linear operators havenīt been taught is my course.

4. Originally Posted by orbit
Hi, thanl you for answering. unfortunately for me,linear operators havenīt been taught is my course.
Ok, so then what have you been taught?

5. Originally Posted by Drexel28
Ok, so then what have you been taught?
Here is what Ive been taught.

6. Originally Posted by orbit
Here is what Ive been taught.
I don't mean to be rude but