everyone. I have a question of proving the continuity of a piecewise function.
This question is from Patrick M.Fitzpatrick, <Advanced Calculus, 2nd edition>
Problem
Exercise 4 of the exercises for section 3.6 Images and Inverses, monotone functions, Chapter 3 Continuous functions:
Define
$ f(x)= \begin{cases} x-1, & \text{if x < 0} \\ x+1, & \text{if x >= 0} \end{cases} $
Show that $f: \mathbb{R}\rightarrow\mathbb{R}$ is strictly increasing and that $f^{-1}: f(\mathbb{R})\rightarrow\mathbb{R}$ is continuous at $x=1$.
What I know and My solution
It is simple to prove that $f: \mathbb{R}\rightarrow\mathbb{R}$ is strictly increasing, thus I omit this step here. To show the inverse function $f^{-1}: f(\mathbb{R})\rightarrow\mathbb{R}$ is continuous at $x=1$, I apply Theorem 3.29:
Theorem 3.29: Let I be an interval and suppose that the function $f: I\rightarrow \mathbb{R}$ is strictly monotone. Then the inverse function $f^{-1}: f(I) \rightarrow \mathbb{R}$ is continuous.
Proof
Obviously, the function $f: \mathbb{R}\rightarrow\mathbb{R}$ is defined on $\mathbb{R}$, which is an interval, and it is strictly increasing. Theorem 3.29 asserts that the inverse function$f^{-1}: f(\mathbb{R})\rightarrow\mathbb{R}$ is continuous. Hence, the inverse function is continuous at $x = 1$. $\Box$
My Question
The inverse function is:
$ f^{-1}(x)= \begin{cases} x+1, & \text{if x < -1} \\ x-1, & \text{if x >= 1} \end{cases} $
How to show the inverse function is continuous at $x=1$ by the definition of continuity or by $\epsilon-\delta$ criterion rather than using Theorem 3.29?
The definition of continuity at a point: The function $f: D \rightarrow \mathbb{R}$ is continuous at a point $x_0$ in the domain D provided that for a sequence $\{x_n\}$ in D, $\lim\limits_{x\rightarrow\infty}f(x_n)=f(x_0)$, if $\lim\limits_{x\rightarrow\infty}x_n=x_0$.
The $\epsilon-\delta$ criterion at a point: A function $f: D \rightarrow \mathbb{R}$ is said to satisfy the $\epsilon-\delta$ criterion at a point $x_0$ in the domain D provided that for each positive number $\epsilon$, there is a positive number $\delta$ such that for $x$ in D, $|f(x)-f(x_0)|<\epsilon$, if $|x-x_0|<\delta$.
Thank you very much!