I've been interested in figuring out, if $f^2(x) = f(x) \cdot f(x)$ is continuous, does that confirm that $f$ is continuous? We know that the domain between the two functions are the same, and so for all $P \in E$ where $E$ is the domain of $f^2,$ $\lim_{x \rightarrow P} f^2(x) = f^2(P) = f(P)f(P).$ However, I am wondering if this provides any implications on $f(x).$ Any recommendations on what to look for in this problem?
Asked
Active
Viewed 32 times
1 Answers
2
Let $f$ be defined as $$f(x)=\begin{cases}1 & x \ge 0 \\ -1 & x < 0 \end{cases}$$
Then $f^2\equiv 1$ is continuous, but $f$ is not.
angryavian
- 89,882
-
To complement this answer, we can note, there are even uglier counter-examples. For instance, $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=1$ if $x$ is rational and $f(x)=-1$ if $x$ is irrational. Here $f^2\equiv 1$ as well, but $f$ isn't continuous at any point in its domain. – Nate River Jan 15 '16 at 03:29