(Uniform) continuity in $\mathbb{R^n}$

Question

I have the following function $$ \mathbb{R^n} \setminus \{0\} \to \mathbb{R^n} : f(x) = \frac{x}{|x|^2} $$ equipped with the euclidean norm (p-norm with $p = 2$)

I know I can analyze the continuity of multivariate functions by splitting it up into component functions. So I do know that $x \mapsto x$ continous and that every norm is continuous in a metric space.

Thus the division of two continuous functions is continuous if the denominator doesn't become $0$. But by definition we excluded $\{0\}$ from the domain.

And what about uniform continuity? Can I somehow extend the $\epsilon-\delta$ definition from $\mathbb{R}$ to $\mathbb{R^n}$?

Answer 1

You can do the same thing: A function $F=(f_1,\dots,f_n):\mathbb{R}^n\to \mathbb{R}^n$ is uniformly continuous if its component functions are.

Suppose $f_1,\dots,f_n$ are uniformly continuous functions from $\mathbb{R}^n\to \mathbb{R}$. Fix $\epsilon>0$ and let $\delta=\min_{1\leq i\leq n}\delta_i$ be the minimum of the $\delta_i$ meeting the $\sqrt{\epsilon/n}$ challenge for each $f_i$ and any points $x,y\in \mathbb{R}^n$.

Then, for this fixed $\epsilon>0$ and any $x,y\in \mathbb{R}^n$, we have $|x-y|\leq \delta$ implies $$ |f_i(x)-f_i(y)|<\epsilon, 1\leq i\leq n. $$ So, $$ ||F(x)-F(y)||^2=\sum_{i=1}^n |f_i(x)-f_i(y)|^2< \sum_{i=1}^n \epsilon/n=\epsilon $$

It's worth internalizing the mantra, vector valued functions inherit the properties of their component functions, (within reason).

Details on this particular example: If we take $\epsilon=1$, we may choose for any given $\delta>0$, the points $x=(\delta,0,\dots,0), y=(\delta/2,0,\dots,0)$. Then, $||x-y||=\delta/2<\delta$ but $$ ||f_1(x)-f_1(y)||=\left|\frac{x_1}{|x|^2}-\frac{y_1}{|y|^2} \right|=\frac{1}{\delta}\geq 1 $$ for $\delta\leq 1$. I leave $\delta>1$ to you.

Answer 2

Restricted to $\mathbb{R} \times \{0\}^{n-1}$, $f$ si just the usual inverse function $f(x) = \frac{1}{x}$, which is not uniformly continuous. So $f$ cannot be.

Uniform continuity is a metric statement : you just replace $|.|$ by a distance ($d(x,y) < \delta$) and in that case by the norm distance $\|x-y \|$.

Answer 3

If $D\subset\Bbb R$ and $f:D\to \Bbb R$ and if $f$ is unbounded on some bounded $E\subset D$ then $f$ is not uniformly continuous. In the proof of this, just replace $\Bbb R$ with $\Bbb R^n$ and replace $|u|$ with $\|u\|$ for any $u\in \Bbb R^n.$

In your Q let $E=\{x\in \Bbb R^n: 0\ne\|x\|<1\}.$