# Proving that the 1-norm, $||x||_1$ is not generated by inner products on $\mathbb{C}^n$

Proving that the 1-norm, $||x||_1$ is not generated by inner products on $\mathbb{C}^n$.

Is it sufficient to take $x=(1,0)$, $y=(0,1)$ in $\mathbb{C}^2$ and just showing that \begin{align} ||x+y||^2+||x-y||^2=8\\ 2||x||^2+2||y||^2=4 \end{align}

As a counter example to show that it does not satisfy the parallelogram law? Or in proving this must it be an actual formal proof?

• So for showing this for $\mathbb{C}^n$ it suffices to show it for say $\mathbb{C}^2$? – Pablo Sep 16 '14 at 23:05
• It shows it is not "generated by inner products for all $\mathbb C^n$". If you want to show for each $n \geq 2$, just take your vectors with a bunch of zeroes on the end $(1,0,0,\dots)$ and $(0,1,0,\dots)$ and your proof follows the same. – Clinton Bradford Sep 16 '14 at 23:07