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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.

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  • $\begingroup$ I've been trying to find a contradiction, but with no luck thus far. $\endgroup$
    – justin
    Commented Nov 26, 2013 at 3:46
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    $\begingroup$ How about the parallelogram rule and/or polarization identity? $\endgroup$ Commented Nov 26, 2013 at 3:49
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    $\begingroup$ That is, you can write $\langle x, y\rangle$ in terms of the norm only. $\endgroup$
    – user99914
    Commented Nov 26, 2013 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. $\endgroup$ Commented Nov 26, 2013 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
    Commented Nov 26, 2013 at 4:11

2 Answers 2

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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

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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.

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