- #1

- 16

- 0

why the supremum of a dot u subject to the 2-norm of u is less than or equal to r equals r times 2-norm of a, i.e. sup{a.u | ||u||_2 <=r} = r ||a||_2?

How can I work out that?

Thank you!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter peterlam
- Start date

- #1

- 16

- 0

why the supremum of a dot u subject to the 2-norm of u is less than or equal to r equals r times 2-norm of a, i.e. sup{a.u | ||u||_2 <=r} = r ||a||_2?

How can I work out that?

Thank you!

- #2

chiro

Science Advisor

- 4,790

- 132

why the supremum of a dot u subject to the 2-norm of u is less than or equal to r equals r times 2-norm of a, i.e. sup{a.u | ||u||_2 <=r} = r ||a||_2?

How can I work out that?

Thank you!

Can you please give a definition of the two-norm as I haven't encountered it before (1 and infinity norm, and L^p norms but not "2-norm").

- #3

- 16

- 0

Sorry. I mean Euclidean norm.

Thanks!

Thanks!

- #4

Landau

Science Advisor

- 905

- 0

This would be the [itex]\ell^2[/itex] norm, except that here we are simply in a finite-dimensional space. In more conventional notation,Can you please give a definition of the two-norm as I haven't encountered it before (1 and infinity norm, and L^p norms but not "2-norm").

[tex]\sup\{\left<x,y\right>\ :\ \|y\|\leq r\}=r\|x\|.[/tex]

@peterlam: to show LHS [itex]\leq[/itex] RHS use Cauchy-Schwartz .

Share: