Verify or refute: There exists an inner product in $\mathbb{R}^2$ such that the norm of every vector $v=(v_1,v_2)$ is $\|v\|=|v_1|+|v_2|$.

I think this is untrue. So I took $v=(1,0), y=(0,1)$. After some calculations, I got that:

$\|v+y\|^2-\|v-y\|^2= 4-4=0$

and $2\|v\|^2+ 2\|y\|^2=4$.

But this clearly satisfies the parallelogram law since $0\leq4$.

So, does such an inner product exist? Or perhaps it doesn't exist but my election of vectors $v,y$ wasn't the appropriate?

  • 3
    $\begingroup$ I am not sure what you mean by the parallelogram law, but for inner product spaces the parallelogram identity must hold (math.stackexchange.com/questions/21792/…), which is not the case in the example you have picked. Thus, this does indeed not yield an inner product space. $\endgroup$ Commented May 17 at 18:52
  • $\begingroup$ Got it. Thanks! My confusion stems because I was working with an inequality "version" of the parallelogram law. $\endgroup$
    – user926356
    Commented May 17 at 19:00

1 Answer 1


The actual parallelogram law is an identity, not an inequality:

$$\|v+y\|^2+\|v-y\|^2=2\|v\|^2+ 2\|y\|^2$$

and that is not satisfied in your example, hence the norm is not associated with an inner product.


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