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.

  • 1
    $\begingroup$ Check out the polarization identity. $\endgroup$ – Cameron Williams Aug 28 '18 at 23:55

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




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:

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


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


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


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

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. $x\in V$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.