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Let $Q$ be the cube $[-1,1]^{3}$ and $\pi$ be a plane in $\mathbb{R}^{3}$ that contains the origin but doesn't contain any vertex of $Q$. Suppose that $A$ is an invertible linear transformation from $\mathbb{R}^{3}$ to $\mathbb{R}^{3}$ and $A\left(Q\cap\pi\right)$ is contained in $Q$. Do there always exist vectors $v_1\in\mathbb{R}^{3}\setminus\pi$ and $v_2\in\mathbb{R}^{3}$ such that

$$\left\Vert v+tv_1\right\Vert_{\infty}\geq\left\Vert Av+tv_2\right\Vert_{\infty}.$$

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The equations are not very readable, perhaps someone can fix? –  Igor Rivin Dec 12 '10 at 23:20
    
@Igor Rivin: fixed. @OP: re-tagged. –  Willie Wong Dec 12 '10 at 23:25
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1 Answer 1

You are essentially asking whether a non-expanding linear map $A:(\pi,\|\cdot\|_\infty)\to(\mathbb R^3,\|\cdot\|_\infty)$ can be extended to a non-expanding linear map $B:(\mathbb R^3,\|\cdot\|_\infty)\to(\mathbb R^3,\|\cdot\|_\infty)$.

(A linear map $f:X\to Y$ between normed spaces is non-expanding if $\|f(x)\|_Y\le \|x\|_X$ for all $x\in X$. Given $v_1$ and $v_2$ as in the question, the map $B$ sending every vector of the form $v+tv_1$, where $v\in\pi$, to $Av+tv_2$, is non-expanding.)

The answer is yes, and the key is that the norm on the target space is of $\ell_\infty$ type.

Let $P_i:\mathbb R^3\to\mathbb R$ denote the $i$-th coordinate projection, $i=1,2,3$. Consider linear functions $A_i:\pi\to\mathbb R$ given by $A_i(x)=P_i(Ax)$, $x\in\pi$. Since $A(Q\cap\pi)\subset Q$, we have $|A_i(x)|\le\|x\|_\infty$ for all $x\in\pi$. By Hahn-Banach, there exists an extension $B_i:\mathbb R^3\to\mathbb R$ such that $|B_i(x)|\le\|x\|_\infty$ for all $x\in\mathbb R^3$. Let $B:\mathbb R^3\to\mathbb R^3$ be the linear map whose coordinate functions are $B_1,B_2,B_3$. Then $\|Bx\|_\infty\le\|x\|_\infty$ for all $x\in\mathbb R^3$ and $B=A$ on $\pi$.

Now let $v_1$ be any vector from $\mathbb R^3\setminus \pi$, define $v_2=Bv_1$, and we are done.

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