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Let $n \in \mathbb{N}$ and $\mathbb{R}^{n}$ be a space with the 1 -norm: $$ \|x\|_{1}=\sum_{i=1}^{n}\left|x_{i}\right|, \quad x=\left(x_{1}, \ldots, x_{n}\right) $$ Consider the map $$ f:\left(\mathbb{R}^{n},\|\cdot\|_{1}\right) \longrightarrow\left(\mathbb{R}^{n},\|\cdot\|_{1}\right), \quad f(x):=\left\{\begin{aligned} \frac{x}{\|x\|_{1}}, & x \neq 0 \\ 0, & x=0 \end{aligned}\right. $$ Is $f$ continuous at any point $x\neq0 \in \mathbb{R}^{n}$？(Induced metric)

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Consider a sequence of points $x_n \in \mathbb{R}^N \setminus \lbrace 0 \rbrace$ converging to $x\neq 0$. Because $\|\cdot \|_{1}$ is a norm, it is Lipschitz and in particular continuous. Therefore, $\displaystyle{\lim_{n\rightarrow \infty}} \frac{x_n}{\|x_n|\ }= \frac{x}{\|x\|}$.

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$f:\mathbb{R}^n \rightarrow \mathbb{R}^n$

$f(x)=(f_1(x), ..., f_n(x))$

$f$ is continuous if and only if every component $f_k$ is continuous (it is a property of the box topology).

In this case $f_k(x)=\frac{x_k}{||x||}$ is the product in $\mathbb{R}$ of two continuous functions, the projection on the $k$ axis $\pi_k(x)=x_k$, and $h(x)=(||x||)^{-1}$ wich is continuous on $\mathbb{R}^n/\{0\}$ as the composition of two continous functions the norm and the inverse function. The continuity of $f$ follows.

Moreover, to show the norm is continuous: every metric $d$ that induces the underling topology is continuous as a function $d:X\times X \rightarrow \mathbb{R}$ in the product topology, hence the norm $N(x)=d(x, 0)$ is the composition of two continuous functions, the map $x \rightarrow (x, 0)$ and the metric.

$||x||_1$, as the euclidean norm $||x||$, induces the usual topology on $\mathbb{R}^n$.

Giovanni Barbarani
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