Prove that the dual norm is in fact a norm

[Dafe]

Homework Statement

Let ||\cdot || denote any norm on \mathbb{C}^m. The corresponding dual norm ||\cdot ||' is defined by the formula ||x||^=sup_{||y||=1}|y^*x|.
Prove that ||\cdot ||' is a norm.

Homework Equations

I think the Hölder inequality is relevant: |x^*y|\leq ||x||_p ||y||_q, 1/p+1/q=1 with 1\leq p, q\leq\infty

The Attempt at a Solution

Since a norm is a function satisfying three properties, I need to show that they hold.

(1) ||x||'=0 if and only if x=0.
(2) ||\alpha x||'=|\alpha| ||x||^.
(3) ||x+z||'\leq ||x||^+||z||^.

I manage to do (1) and (2) just fine, but the triangle inequality (3) is giving me problems.

I use the Hölder inequality to get the following:

||x+z||'=sup_{||y||=1}|y^*(x+z)|\leq ||y|| ||x+z||=||x+z||

||x||'=sup_{||y||=1}|y^*x|\leq ||y|| ||x|| =||x||

||z||'=sup_{||y||=1}|y^*z|\leq ||y|| ||z|| =||z||

(1) ||x||'\leq ||x||
(2) ||z||'\leq ||z||
(3) ||x||'+||z||' \leq ||x||+||z||

I also know that
(4) ||x+z|| \leq ||x||+||z||

I am unable to show that ||x+z||\leq ||x||'+||z||' which I think I must if I am to prove the triangle inequality.

Any help is appreciated.

 

[dirk_mec1]

||x+z||'=sup_{||y||=1} |y^*(x+z)| = sup_{||y||=1} | y^*x + y^*z | \leq sup_{||y||=1} | y^*x| + sup_{||y||=1} |y^*z | = ||x||' + ||z||'

The supremum-norm is a norm.

 

[Dafe]

Ah, didn't think of it that way. Thank you very much.