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Given a discontinuous function: $$ f(x) = e^{|x|}\text{sign}(x) $$ Prove that its inverse function is continuous.

Note: sign is a Signum function.

I've started with the following. Obviously, signum function does not have an inverse, so I decided to split the domain of the function into several parts and treat them separately. Define $f(x)$ as follows: $$ f(x) = \begin{cases} e^x,\ x> 0\\ 0, \ x = 0\\ {-e^{-x}},\ x < 0 \end{cases} $$

Starting with the case $x > 0$ it's pretty obvious that the range (call it $E(f)$) of the function is: $$ E(f) = (1, +\infty) $$ While for the case $x < 0$ the range is: $$ E(f) = (-\infty, -1) $$ At $x = 0$ the value of the function is simply $0$. We may now find the inverse for that different intervals: $$ y = e^x \stackrel{x>0}{\implies} x = \ln y\\ y = -e^{-x} \iff -y = {1\over e^x} \iff e^x = -{1\over y} \stackrel{x<0}{\implies} x = \ln \left(-{1\over y}\right) $$

So for $x \ne 0$ the inverse becomes: $$ f^{-1}(y) = \begin{cases} \ln y,\ y > 1\\ \ln\left(-{1\over y}\right),\ y < -1 \end{cases} $$

The problem here is when $x = 0$. It feels natural that $f^{-1}(y)$ should equal to $0$ for $y \in [-1, 1]$, and then whole definition of inverse function would become: $$ f^{-1}(y) = \begin{cases} \ln y,\ y > 1\\ 0,\ y\in[-1, 1] \\ \ln\left(-{1\over y}\right),\ y < -1 \end{cases} $$

Which is indeed continuous. But how do I justify that? Also if the approach above is invalid then could you please guide me through the proof?

Thank you!

Here is how the problem appears in the book (In russian): enter image description here

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    $\begingroup$ Your problem is ill-posed. It makes no sense as stated. $\endgroup$ – Will M. Jul 1 '19 at 17:21
  • $\begingroup$ While sign does not have a total inverse, you can restrict the domain to $0$ so that it has a partial inverse. Thus $$\text{sgn}^{-1}(x)=\begin{cases}\text{undefined}&x<0\\0&x=0\\ \text{undefined}&x>0\end{cases}$$ $\endgroup$ – R. Burton Jul 1 '19 at 17:26
  • $\begingroup$ @WillM. I've just double checked two editions of the book I've taken the problem from. Both of them list the problem as stated in the OP. Might be author's mistake. $\endgroup$ – roman Jul 1 '19 at 17:29
  • $\begingroup$ Not sure how to deal with ill-posed problems, should I delete the post? $\endgroup$ – roman Jul 1 '19 at 17:30
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    $\begingroup$ @roman Does the author state the domain of the function? If $f$ is defined on $(-\infty,0)\cup(0,\infty)$ then $f^{-1}$ is continuous on $(-\infty,-1)\cup(1,\infty)$. $\endgroup$ – Jam Jul 1 '19 at 17:55
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The range of $x\mapsto sign(x)e^{-|x|}$ is $(-1,0)\cup(1,\infty)$. Thus, th inverse is defined on $(-1,0)\cup(1,\infty)$ by $-\log(-y)$ if $(-1,0)$ and $\log(y)$ of $y>1$. On each part of its domain, the inverse is continuous. The point here is that the domain consist of two disconnected pieces.

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