Homework Help: Prove that the dual norm is in fact a norm

1. Jul 4, 2010

Dafe

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Jul 4, 2010

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.

Last edited: Jul 4, 2010
3. Jul 4, 2010

Dafe

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