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Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ a $\mathcal{C}^1$ function such that there exists $\alpha > 0$: $$ \alpha \|x-y\| \leq \|f(x)-f(y)\|, \forall x,y \in \mathbb{R}^n$$ 1)Show that $f(\mathbb{R}^n)$ is a closed set.

I don't know how to approach this exercise. I tried to take a Cauchy sequence $(y_n) = f(x_n)$, then as $\mathbb R^n$ is a Banach space, I immediately have that $(x_n)$ converges, thus $f(x_n)$ converges in $f(\mathbb{R}^n)$, thus is closed. But I don't know how to show this rigorously, and I am unsure if my idea is correct.

Michael Hardy
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John Mayne
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  • I think you wanted to write $f : \mathbb{R}^n \mapsto \mathbb{R}^{\mathbf{p}}$. Because for $n \ge 2$, such a function $f$ would be a continuous injective function from $\mathbb{R}^n$ to $\mathbb{R}$ (which does not exist, cf. https://math.stackexchange.com/questions/189328/existence-of-an-injective-c1-map-between-mathbb-r2-and-mathbb-r) – charmd Feb 19 '18 at 20:44
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    The inequality implies that $f$ is continuous. The domain $\mathbb R^n$ is topologically a connected space. Suppose we know a theorem that says the image of a connected space under a continuous function is connected. And a theorem that says that the only connected subsets of $\mathbb R$ are intervals. For an interval, the problem of proving that it is closed is only a matter of proving it contains its smallest upper bound and largest lower bound if it has such bounds. $\qquad$ – Michael Hardy Feb 19 '18 at 21:05

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Let $f(x_n)$ be a Cauchy sequence of $f(\mathbb{R}^n)$, for every $c>0$ there exists $N$ such that for $n,m>N$ implies that $|f(x_n)-f(x_m)|<c$ this implies that $\|x_n-x_m\|<{1\over\alpha}|f(x_n)-f(x_m)|<{c\over\alpha}$ we deduce that $x_n$ is a Cauchy sequence so converges towards $x$ since $f$ is continuous, $f(x)=\lim_nf(x_n)$ and the limit of the Cauchy sequence $f(x_n)\in f(\mathbb{R}^n)$, we deduce that $f(\mathbb{R}^n)$ is closed.

Michael Hardy
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Tsemo Aristide
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    Just a question — completely for extending understanding — but wondering if there's anyway to do this without using Cauchy sequences and just using topology? E.G. something like building an open ball around a limit point of the image of $f$, showing that there must be then an open ball in the domain of $f$ with diameter determined by the constraint and then using continuity of $f$ to show that as the ball in the domain is squeezed, some element in that domain must approach the limit point chosen above showing that the image is closed? – EDZ Feb 19 '18 at 21:08