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I'd like to show that the following defines a norm on $\mathbb C^n$:

$||x||=(a_1^2+a_2^2)^{1/2}$

Where $x=(x_1,..,x_n)$, $a_1$ is the maximum of the $|x_i|$'s and $a_2$ the second maximum.

But I have problems with the triangle inequality axiom...

Hope somebody could help!

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What do you mean exactly by second maximum? –  enzotib Oct 1 '12 at 8:03
    
Yes this was unclear. I mean if $a_1=max |x_i|$ is reached for some $k$, then $a_2=max_{i\neq k}|x_i|$ –  tina Oct 1 '12 at 8:06
    
So it could happen $a_1 = a_2$, if the value $a_1$ is reached in more than one coordinate? –  enzotib Oct 1 '12 at 8:07
    
@enzotib: yes it could –  tina Oct 1 '12 at 8:12

1 Answer 1

up vote 1 down vote accepted

Let me denote $x_{(1)}$ the largest element of $x$ in absolute value and $x_{(2)}$ the second one. Then you want to prove that $[(x+y)_{(1)}^2+(x+y)_{(2)}^2]^{1/2}\leq[x_{(1)}^2+x_{(2)}^2]^{1/2}+[y_{(1)}^2+y_{(2)}^2]^{1/2}$, in the square:

$(x+y)_{(1)}^2+(x+y)_{(2)}^2\leq x_{(1)}^2+x_{(2)}^2+y_{(1)}^2+y_{(2)}^2+2\sqrt{x_{(1)}^2y_{(1)}^2+x_{(1)}^2y_{(2)}^2+x{(2)}^2y_{(1)}^2+x_{(2)}^2y_{(2)}^2}$

There are two possibilities. First: $(x+y)_{(1)}=x_{(1)}+y_{(1)}$, then necesserily $(x+y)_{(2)}\leq x_{(2)}+y_{(2)}$. Second: $(x+y)_{(1)}<x_{(1)}+y_{(1)}$, then necessarily (WLOG) $(x+y)_{(1)}\leq x_{(1)}+y_{(2)}$ and $(x+y)_{(2)}\leq x_{(2)}+y_{(1)}$ (it can happen that $x$ and $y$ are to be exchnged here).

Together: $(x+y)_{(1)}^2+(x+y)_{(2)}^2\leq x_{(1)}^2+x_{(2)}^2+y_{(1)}^2+y_{(2)}^2+2A+2B$, where $A+B=x_{(1)}y_{(1)}+x_{(2)}y_{(2)}$ or $A+B=x_{(1)}y_{(2)}+x_{(2)}y_{(1)}$. Both possibilities easily give $A+B\leq\sqrt{x_{(1)}^2y_{(1)}^2+x_{(1)}^2y_{(2)}^2+x_{(2)}^2y_{(1)}^2+x_{(2)}^2y_{(2)}^2}$.

Q.E.D.

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