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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?

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Yes!

A norm generated by inner products must satisfy the parallelogram law. A counterexample is a sufficient proof.

If you want to be more formal, you can say "Assume for contradiction that there is such an inner product. Then the norm must satisfy the parallelogram law. However, ..."

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  • $\begingroup$ So for showing this for $\mathbb{C}^n$ it suffices to show it for say $\mathbb{C}^2$? $\endgroup$ – Pablo Sep 16 '14 at 23:05
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    $\begingroup$ 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. $\endgroup$ – Clinton Bradford Sep 16 '14 at 23:07

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