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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.

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  • $\begingroup$ 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 $\endgroup$
    – Henry
    Apr 11 '21 at 10:46
  • $\begingroup$ Think of the 1D case. Where would the jump be ? $\endgroup$
    – user65203
    Apr 11 '21 at 12:55
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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 $x\in\Bbb R^n\setminus\{0\}$,$$a(x)=\min\left\{1,\frac1{\|x\|}\right\}=\frac{1+\frac1{\|x\|}-\left|1-\frac1{\|x\|}\right|}2.$$

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  • $\begingroup$ 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. $\endgroup$
    – Bill
    Apr 11 '21 at 10:49
  • $\begingroup$ What I explained is that the function $f$ is continuous, and what this means is that it is continuous everywhere. $\endgroup$ Apr 11 '21 at 10:52
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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 $B$.

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  • $\begingroup$ 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. $\endgroup$
    – Bill
    Apr 12 '21 at 1:25
  • $\begingroup$ @William because then both sets are closed which is a condition in this version of the pasting lemma. $\endgroup$ Apr 12 '21 at 4:48
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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 $\mathbb{R}^n$.

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  • $\begingroup$ 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. $\endgroup$
    – user65203
    Apr 11 '21 at 13:21

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