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Let $f: \mathbb { R^n } \rightarrow \mathbb { R } ^ { n }$ be continuously differentiable, Satisfying $$ \| f ( x ) - f ( y ) \| \geqslant \| x - y \| , \forall x , y \in \mathbb { R } ^ { n } $$ then how to prove that $f$ is onto.

My work: Consider $f:\mathbb R^n \rightarrow Range(f),$ Then clearly $f$ is bijective from $\mathbb { R } ^ { n }$ onto $Range( f )$ also since $$\left\| f ^ { - 1 } ( x ) -f ^ { - l } ( y ) \right\| \leq \| x - y \|$$ $f ^ { - 1 }: \operatorname { Range } (f) \rightarrow \operatorname {\mathbb R^n }$ Continuous. Thus $Range(f) \subseteq \mathbb { R } ^ { n }$ is homeomorphic to $\mathbb { R } ^ { n }$, is the way is correct? I am not getting idea further.

cmk
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Riaz
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  • How exactly did you conclude that the range of $f$ is $\mathbb R^n$ from the fact that $f\simeq\mathbb R^n$? – kcjkgs Oct 31 '19 at 16:16
  • It cannot always be true. For example consider $\operatorname{ln}:(0,\infty)\rightarrow \mathbb R$. It is continuous and has continuous inverse $\operatorname{exp}$ so that $(0,\infty)\simeq\mathbb R$ but they are not equal. – kcjkgs Oct 31 '19 at 16:19
  • If $f$ weren't surjective, there'd be a $z \in \overline{f(\mathbb{R}^n)}\setminus f(\mathbb{R}^n)$. Try to obtain a contradiction from that. – Daniel Fischer Oct 31 '19 at 16:25
  • if the statement to be proved is true I would try a proof by contradiction – Masacroso Oct 31 '19 at 17:01

2 Answers2

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For suppose $z\in\overline{f(\mathbb{R}^{n})}-f(\mathbb{R}^{n})$, there is a sequence $(w_{n})$ such that $f(w_{n})\rightarrow z$. But then $\|f(w_{n})-f(w_{m})\|\geq\|w_{n}-w_{m}\|$, so $(w_{n})$ is Cauchy and hence convergent, say, $w_{n}\rightarrow w$. Continuity gives $f(w_{n})\rightarrow f(w)$ and hence $z=f(w)\in f(\mathbb{R}^{n})$, a contradiction.

This shows that $f(\mathbb{R}^{n})$ is closed.

We claim that $f(\mathbb{R}^{n})$ is open, for then $f(\mathbb{R}^{n})=\mathbb{R}^{n}$ by the connectedness of $\mathbb{R}^{n}$.

The Jacobian is not singular, see here. Hence Inverse Function Theorem gives you that $f$ is an open map, so $f(\mathbb{R}^{n})$ is open.

user284331
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  • This shows that $f(\mathbb{R}^n)$ is closed. I think you need more to deduce that $f(\mathbb{R}^n)=\mathbb{R}^n$, e.g. the fact that it is open as well. – Thorgott Oct 31 '19 at 16:46
  • Thanks for reminding, I think now it works. – user284331 Oct 31 '19 at 17:04
  • If $g:f(\mathbb{R}^{n})$ is continuous, then $f$ is $\mathbb{R}^{n}$-$f(\mathbb{R}^{n})$ open, this is relatively. But we don't know yet $f(\mathbb{R}^{n})$ is itself open or not. – user284331 Oct 31 '19 at 19:52
  • How can you say that Jacobian is non- singular? You got the sum of squares of absolute values of derivatives $\geq 1$, from that? – Riaz Nov 01 '19 at 02:36
  • If the Jacobian is the zero matrix, then each of its component is zero, so all the partials are zeros, then it is not possible for the sum of any of the row of those partials to be greater than one – user284331 Nov 01 '19 at 02:41
  • Ya......its clear that Jacobian is non-zero, but i am not getting how it is non- singular... – Riaz Nov 01 '19 at 03:02
  • I see your point now, I messed up with nonzero and nonsingular, sorry, I will try to repair my proof. – user284331 Nov 01 '19 at 03:04
  • I found a useful link to show that why it is non-singular, sorry for messed up. – user284331 Nov 01 '19 at 03:39
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We show the image is open and closed. Since it's also nonempty, by connectedness of ${\mathbb R}^n$ it's all of ${\mathbb R}^n$:

Openness: This will follow from the inverse function theorem if we can show the matrix $Df(x_0)$ is nonsingular at each $x_0$; the range of $f$ is the union of the open sets that are the images of the open sets provided by the inverse function theorem.

Note that we have $$f(x) = f(x_0) + Df(x_0)(x-x_0) + o(x- x_0)$$ If $v$ is any vector of magnitude $1$, if we let $x = x_0 + tv$ for $0 < t < 1$, the above gives $$f(x_0 + tv) - f(x_0) = tDf(x_0)v + o(t)$$ This implies $${|f(x_0 + tv) - f(x_0)| \over t} = Df(x_0)v + o(1)$$ Since by assumption $|f(x_0 + tv) - f(x_0)| > t$, letting $t$ go to zero we get that $Df(x_0)v $ is nonzero for any vector $v$. Thus the matrix $Df(x_0)$ is nonsingular for all $x_0$ and the image of $f$ is open.

Closedness: Suppose $y_n$ are in the range of $f$ and $\lim_n y_n = y_0$ for some $y_0$. Let $x_n$ be such that $f(x_n) = y_n$. Since the $y_n$ converge, the $y_n$ are a Cauchy sequence. By assumption, $|x_n - x_m| \leq |f(x_n) - f(x_m)| = |y_n - y_m|$. Hence the $x_n$ are a Cauchy sequence too and converge to some $x_0$ in ${\mathbb R}^n$. By continuity $f(x_0) = y_0$ and hence $y_0$ is in the range of $f$. We conclude the range of $f$ is closed.

Zarrax
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  • Can we use the continuity of the inverse $f^{-1}$ from $f(\mathbb R^n)$ to $\mathbb R^n$ to say that $f(\mathbb R^n)$ is open? – Riaz Dec 06 '19 at 05:19
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    No, that would just say that the image $f(U)$ for any open $U$ is open in the relative topology on $f({\mathbb R}^n)$. In other words, $f(U) \cap f({\mathbb R}^n)$ is open. – Zarrax Dec 06 '19 at 14:44