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Show that $\mathbb{R}^2$ with norm given by $\|x\|=|x_1|+|x_2|$ is not an inner-product space, that is, for this norm there is no inner product such that $\|x\|=\sqrt{\langle x,y\rangle}$.

I'm lost on what this question is asking. It seems to me that it's intended to lend to a proof that $\sqrt{x^2+y^2}\neq|x|+|y|$, but I'm not sure. Assistance in both interpreting and solving this problem would be greatly appreciated.

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    $\begingroup$ Check out the polarization identity. $\endgroup$ – Cameron Williams Aug 28 '18 at 23:55
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To expand on Cameron Williams' comment:

The parallelogram law is given by

$$2||x||^2+2||y||^2=||x+y||^2+||x-y||^2\text{ f.a. }x,y\in V$$

for a vector space and it can be shown that every norm induced by an inner product satisfies this property.


However, taking your norm, we find that for $x=(1,0),y=(0,-1)$, we have

$$2||x||^2+2||y||^2=2(|1|+|0|)^2+2(|0|+|-1|)^2=2(1)^2+2(1)^2=2+2=4$$

but

$$||x+y||^2+||x-y||^2=(|1+0|+|0-1|)^2+(|1+0|+|0+1|)^2=(|1|+|-1|)^2+(|1|+|1|)^2=2^2+2^2=4+4=8$$


EDIT: I want to give you a proof that every inner-product induced norm satisfies the parallelogram law:

Let $||x||:=\sqrt{\langle x,x\rangle}$ for an inner product $\langle\cdot,\cdot\rangle$ on a real vector space $V$. Thus $||x||^2=\langle x,x\rangle$ and followingly:

$$||x+y||^2=\langle x+y,x+y\rangle=\langle x,x\rangle +\langle x,y\rangle+\langle y,x\rangle+\langle y,y\rangle$$

and

$$||x-y||^2=\langle x-y,x-y\rangle=\langle x,x\rangle -\langle x,y\rangle-\langle y,x\rangle+\langle y,y\rangle$$

Thus

$$||x+y||^2+||x-y||^2=\langle x,x\rangle +\langle x,y\rangle+\langle y,x\rangle+\langle y,y\rangle+\langle x,x\rangle -\langle x,y\rangle-\langle y,x\rangle+\langle y,y\rangle$$

i.e.

$$||x+y||^2+||x-y||^2=2\langle x,x\rangle+2\langle y,y\rangle=2||x||^2+2||y||^2$$


The so called polarization identity establishes the converse, i.e. it shows that for a normed space $(V,||\cdot||)$, if $||\cdot||$ satisfies the parallelogram law, then it comes from an inner product, i.e. there is an inner product $\langle\cdot,\cdot\rangle$ on $V$ s.t. $||x||^2=\langle x,x\rangle$ f.a. $x\in V$.

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