3
$\begingroup$

For $x = \left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n}\,,\ n \geq 2\ $, define $\left\vert\,x\,\right\vert := \max_i \left\vert\,x_i\,\right\vert\ $. Show that there exists no inner product $\left\langle ,\right\rangle$ on $\mathbb{R}^{n},\ $ for which $\left\langle x,x \right\rangle = \left\vert\,x\,\right\vert^{2}\,, \forall\ x \in \mathbb{R}^{n}$.

I've been unable to find a contradiction, although I think that is key to solving this problem.

$\endgroup$
  • $\begingroup$ I've been trying to find a contradiction, but with no luck thus far. $\endgroup$ – justin Nov 26 '13 at 3:46
  • 3
    $\begingroup$ How about the parallelogram rule and/or polarization identity? $\endgroup$ – Omnomnomnom Nov 26 '13 at 3:49
  • 3
    $\begingroup$ That is, you can write $\langle x, y\rangle$ in terms of the norm only. $\endgroup$ – user99914 Nov 26 '13 at 3:50
  • $\begingroup$ An strong intuitive argument can be given by noting that a unit sphere in an inner product space contains no affine line segments, while the unit sphere under this norm is a cube, which contains many affine line segments. It wouldn't be difficult to make that idea rigorous, but there's probably a much simpler algebraic proof. $\endgroup$ – Dustan Levenstein Nov 26 '13 at 3:57
  • $\begingroup$ Omnomnomnom, I took your suggestion and think I came up with a proof. Look at the newest edited version of my question. $\endgroup$ – justin Nov 26 '13 at 4:11
4
$\begingroup$

First recall the identity $\|x\|^2 = \langle x,x \rangle$. Now by the Parallelogram law, $\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2.$ Rewriting the Parallelogram law in terms of inner products, we see that $\langle x+y, x+y \rangle + \langle x-y,x-y \rangle = 2\langle x,x \rangle + 2\langle y,y \rangle.$

Assume that there exists an inner product satisfying the specified conditions. Then let $x,y \in \mathbb{R}^n$ be $x = (1,0, \ldots, 0)$ and $y = (0, \ldots, 0, 1)$, so that $x+y = (1,0,\ldots,0,1)$ and $x-y = (1,0,\ldots, 0, -1)$. Then by assumption, $\langle x,x \rangle = 1, \langle y,y \rangle = 1, \langle x+y,x+y \rangle = 1,$ and $\langle x-y,x-y \rangle = 1.$

But plugging these values into the Parallelogram law gives us $1+1=2+2,$ a contradiction. QED

$\endgroup$
1
$\begingroup$

Show that if a norm is índuced by an inner product, then most convex combinations of two vectors of norm 1 have norm different from 1, and then show that this is not so for the maximum norm.

$\endgroup$

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.