If a $C^1$ map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is expanding ($\|f(x)-f(y)\|\geq\|x-y\|$), which of these must be true?

Question

Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be a continuously differentiable map satisfying $\|f(x)-f(y)\|\geq\|x-y\|$ for all $x,y\in\mathbb{R}^{n}.$ Then

$A.$ $f$ is onto.

$B.$ $f(\mathbb{R}^{n})$ is a closed subset of $\mathbb{R}^{n}.$

$C.$ $f(\mathbb{R}^{n})$ is a open subset of $\mathbb{R}^{n}.$

$D.$ $f(0)=0.$

According to me $f$ is one one and onto so image is closed and open both. Please suggest me. Thanks in advance.

zhw. · Answer 1 · 2017-12-19T06:46:04.470

D. This fails because if $x_0\ne (0,\dots, 0),$ then the map $x\to x + x_0$ satisfies the condition.

C. Suppose $Df(x_0)$ is singular at some $x_0.$ Then $Df(x_0)(v) = 0$ for some nonzero $v\in \mathbb R^n.$ Thus as $t\to 0$ in $\mathbb R,$

$$f(x_0 + tv)- f(x_0)= Df(x_0)(tv) + o(|tv|) = 0 + o(|tv|) .$$

This violates the given condition on $f.$ Therefore $Df$ is nonsingular everywhere. Hence by the inverse function theorem, $f$ is an open map. Thus $f(\mathbb R^n)$ is open.

B. Suppose $y_k \in f(\mathbb R^n)$ and $y_k \to y_0.$ Each $y_k = f(x_k)$ for some $x_k\in \mathbb R^n.$ The condition on $f$ shows that if $\{x_k\}$ is unbounded, then so is $\{f(x_k)\}.$ That is a contradiction, and therefore $\{x_k\}$ is bounded. Thus there exists a subsequence $x_{k_j}$ converging to some $x_0\in \mathbb R^n.$ It follows by continuity that $f(x_{k_j}) \to f(x_0).$ Since $f(x_{k_j}) \to y_0,$ we have $y_0= f(x_0).$ Thus $y_0 \in f(\mathbb R^n),$ proving $f(\mathbb R^n)$ is closed.

A. We have shown $f(\mathbb R^n)$ is both open and closed in $\mathbb R^n,$ a connectd metric space. Since $f(\mathbb R^n)$ is nonempty, it must be everything. Thus $f$ is onto.

Jack D'Aurizio · Answer 2 · 2017-12-19T05:46:23.160

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$(D)$ is clearly false by considering $f(\overline{x})=2\overline{x}+\overline{x_0}$.
$(C)$ is clearly true, since differentiable maps with a non-vanishing Jacobian are open maps.

All the partial derivatives of $f$ are $\geq 1$ in absolute value, and since $f$ is continuously differentiable the partial derivatives have constant sign. Without loss of generality we may assume they all are $\geq 1$. In particular the image of the ball with radius $\rho$ centered at the origin (with respect to the euclidean norm or the supremum norm) contains the ball with radius $\rho$ centered at $f(0)$. In particular $f$ is surjective and $(A),(B)$ are true.

edited Dec 19 '17 at 05:46

answered Dec 19 '17 at 05:42

Jack D'Aurizio

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Differentiable maps are always open ?? how – neelkanth Dec 19 '17 at 05:44
@neelkanth: https://math.stackexchange.com/questions/645657/proving-that-fu-to-mathbbrn-differentiable-is-an-open-map-when-det-oper – Jack D'Aurizio Dec 19 '17 at 05:45
Yes but its when Jocabian matrix has non zero det... – neelkanth Dec 19 '17 at 05:46
@ Jack D'aurizio in this case no role of J is given... – neelkanth Dec 19 '17 at 05:49
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@neelkanth: which is clearly the case here, since any vector in $\text{ker}, J$ gives a direction in which we cannot have an expansion of the distances. Differentiable expansions are locally invertible. – Jack D'Aurizio Dec 19 '17 at 05:49
I got every point execpt the open map ....please give me more details if possible.... – neelkanth Dec 19 '17 at 05:50
@neelkanth: simply, the Jacobian cannot vanish under the condition $|f(x)-f(y)|\geq |x-y|$! – Jack D'Aurizio Dec 19 '17 at 05:51
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@neelkanth: in layman's terms: consider the skeleton of a polyhedron. The action of $f$ has to increase the length of every edge, hence the enclosed volume. In particular, $f$ cannot shrink non-zero volumes to zero volumes, hence $\det J\neq 0$. – Jack D'Aurizio Dec 19 '17 at 05:54

If a $C^1$ map $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is expanding ($\|f(x)-f(y)\|\geq\|x-y\|$), which of these must be true?

2 Answers2