# Is this piecewise function continuous?

Is the above function continuous? Intuitively, I think the answer is no because at points where the norm is equal to 1 and points where the norm is slightly larger than 1, it feels like there is a "jump discontinuity" but I'm not sure how to show that.

• I would have thought the answer was yes since $\frac{\mathbf x}{||\mathbf x||}=\mathbf x$ when $||\mathbf x||=1$ and the two pieces are continuous Apr 11 '21 at 10:46
• Think of the 1D case. Where would the jump be ?
– user65203
Apr 11 '21 at 12:55

Yes, it is continuous. It is the product of two continuous functions: the identity function and the function$$\begin{array}{rccc}a\colon&\Bbb R^n&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}1&\text{ if }\|x\|<1\\\frac1{\|x\|}&\text{ otherwise.}\end{cases}&\end{array}$$The function $$a$$ is continuous because, for each $$x\in\Bbb R^n\setminus\{0\}$$,$$a(x)=\min\left\{1,\frac1{\|x\|}\right\}=\frac{1+\frac1{\|x\|}-\left|1-\frac1{\|x\|}\right|}2.$$

• ok i get that both piecewise functions are continuous in their domain, but the part that I'm having a problem with are the points where they almost meet.
– Bill
Apr 11 '21 at 10:49
• What I explained is that the function $f$ is continuous, and what this means is that it is continuous everywhere. Apr 11 '21 at 10:52

You can use the pasting lemma for the two closed sets $$A=\{x\mid \|x\| \le 1\}$$ and $$B = \{x\mid \|x\|\ge 1\}$$ where $$f$$ is the identity on $$A \cap B$$. $$f\restriction_A$$ is just the identity, hence continous and $$f\restriction_B$$ is just $$\frac{x}{\|x\|}$$ which is continuous as the quotient of two continuous functions, the second of which vanishes nowhere on $$B$$.

• may I ask why function B was redefined to also include the elements with norm equal to 1? The original function only defined them to be greater than norm 1. I thought about the gluing lemma, but I thought we can't use it because the two sets did not share any domain in common.
– Bill
Apr 12 '21 at 1:25
• @William because then both sets are closed which is a condition in this version of the pasting lemma. Apr 12 '21 at 4:48

Let $$A_1=\{x: \|x\| \leq 1\}$$ and $$A_2=\{x: \|x\| \geq 1\}$$, then $$A_1,A_2$$ are closed in $$\mathbb{R}^n$$ and $$\mathbb{R}^n=A_1 \cup A_2$$. Define $$f_1: A_1 \rightarrow \mathbb{R}^n$$ and $$f_2: A_2 \rightarrow \mathbb{R}^n$$ by $$f_1(x)=x$$ and $$f_2(x)=\frac{x}{\|x\|}$$. Then $$f(x)=f_1(x)$$ for $$x \in A_1$$ and $$f(x)=f_2(x)$$ for $$x \in A_2$$. We will show that for any closed set $$C$$ in $$\mathbb{R}^n$$, $$f^{-1}(C)$$ is closed in $$\mathbb{R}^n$$.

$$x \in f^{-1}(C) \cap A_i \Leftrightarrow f(x) \in C~~\textit{and}~~x \in A_i \Leftrightarrow f_i(x) \in C~~\textit{and}~~x \in A_i \Leftrightarrow x \in f_i^{-1}(C) \cap A_i$$. Thus $$f^{-1}(C) \cap A_i=f_i^{-1}(C) \cap A_i$$. Since $$f_i$$'s are continuous, $$f_i^{-1}(C) \cap A_i$$ is closed in $$A_i$$ hence in $$\mathbb{R}^n$$. So $$f^{-1}(C)=f^{-1}(C) \cap \mathbb{R}^n=f^{-1}(C) \cap (\cup_i A_i)=\cup_i(f^{-1}(C) \cap A_i)$$ is closed in $$\mathbb{R}^n$$.

• Impressive how one can hide the important information (namely that the pieces are continuous and coincide along their border) in indigestible technicalities ! Compare to the answer by José.
– user65203
Apr 11 '21 at 13:21