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Let the function $y=f(x)=(1+x^2)sgn(x)$. Prove that $y^{-1}$ is a continuous function.

I know that the $y$ is not continuous, but don't know how to treat the problem at first. Thank you for your help.

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What is $sgn$??? Is it the sign function? – Babak S. Jan 30 '13 at 7:15
Yes, Babak. its – Basil R Jan 30 '13 at 7:19
up vote 5 down vote accepted

Your notation is a bit confusing. I think you mean that $f^{-1}$ (i.e. the inverse function, not the reciprocal) is a continuous function. Thus you're writing $x = f^{-1}(y)$.

There are three cases to consider: $x > 0$, $x < 0$ and $x = 0$. When $x > 0$, $y = 1 + x^2 > 1$ and $x = \sqrt{y-1}$. This is a continuous function on $(1,\infty)$. Similarly, when $x < 0$, $y = -1 - x^2 < -1$ and $x = -\sqrt{-1 - y}$. This is a continuous function on $(-\infty, -1)$. Finally, when $x = 0$, $y = 0$. Thus $$ f^{-1}(y) = \cases{\sqrt{y-1} & for $x > 1$\cr 0 & for $x = 0$\cr -\sqrt{-1-y} & for $x < -1$\cr \text{undefined} & otherwise\cr}$$ Since the three sets $(-\infty, -1)$, $\{0\}$ and $(1,\infty)$ are separated, $f^{-1}$ is continuous on its domain which is their union.

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+1. I think you want $y$'s instead of $x$'s in your expression for $f^{-1}$ (e.g. $\text{ for } y\geq 1$ instead of $\text{ for } x\geq 1$). – Stefan Hansen Jan 30 '13 at 7:30
Thank you robert. – Basil R Jan 30 '13 at 7:30
And you want strict inequalities: $f^{-1}$ is undefined at $\pm1$. – Marc van Leeuwen Jan 30 '13 at 7:35
@MarcvanLeeuwen: yes, of course. I edited it. – Robert Israel Jan 30 '13 at 18:33
I encountered same problem in my book.I read this post but i am confused,in this question co-domain of $(1+x^2)sgn(x)$ is not given,its range we found out to be $(-\infty,1)\cup\left\{0\right\}\cup(1,\infty)$then how can we say that this function is bijective and its inverse exists.If codomain is equal to range then function is surjective.@RobertIsrael – diya Dec 19 '15 at 2:51

The function $f$ is discontinuous (only) at $0$, and since $1+x^2\geq1$ for all $x\in\mathbf R$, the range of $f$ is contained in $(-\infty,-1)\cup\{0\}\cup(1,+\infty)$. As $f$ is strictly increasing and tends to $\pm\infty$ as $x\to\pm\infty$, the range of $f$ is in fact equal to that set, and $f$ is a bijection from $\mathbf R$ to that range.

The inverse of $f$ is therefore a well defined function $ (-\infty,-1)\cup\{0\}\cup(1,+\infty)\to\mathbf R$; continuity is a local property, and therefore determined separately on each of these three connected components of the domain of $f^{-1}$. The component $\{0\}$ is discrete (it is an isolated point), so continuity there is vacuously satisfied. For the other two components, we are dealing with the inverse of a restriction of $f$ to $(-\infty,0)$ respectively to $(0,+\infty)$, and these restrictions are continuous and strictly increasing. You should know how to prove that the inverse of such a function is continuous, and this settles the continuity of $f^{-1}$ everywhere on its domain.

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Let $D \subset \mathbb{R}$ and let $x \in D$. Then continuity at $x$ is vacuous -- i.e., every $f: D \rightarrow \mathbb{R}$ is continuous at $x$ -- iff $x$ is an isolated point of $D$. This holds in the case at hand...but $D$ having empty interior is not enough. – Pete L. Clark Jan 30 '13 at 7:41
@PeteL.Clark: Thanks, you're right. I'll correct. – Marc van Leeuwen Jan 30 '13 at 7:46

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