A continuously differentiable function $f$ Satisfying $ \| f ( x ) - f ( y ) \| \geq \| x - y \| , \forall x , y \in \mathbb { R } ^ { n } $ is onto

Question

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.

Answer 1

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.

Answer 2

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.