If $f^2$ monotonically increasing in $R$ then f monotonic Function $f$ is continuous in $R$.
Prove: If $f^2$ monotonically increasing in $R$ then $f$ monotonic in $R$.
Intuitively its seems pretty clear, But I don't have any idea how to "start" the prove.
Any ideas?
Thanks.
 A: If there were two points $x_1$, $x_2\in{\mathbb R}$ with $f(x_1)f(x_2)<0$ there would be a point $\xi$ between $x_1$ and $x_2$ with $f(\xi)=0$. This would prevent $f^2$ from being monotone between $x_1$ and $x_2$. Therefore one has $f(x)\geq0$ or $f(x)\leq0$ throughout, and this implies that one of the following is true:
$$f(x)=\sqrt{f^2(x)}\quad \forall x\in{\mathbb R}\qquad{\rm resp.}\qquad f(x)=-\sqrt{f^2(x)}\quad \forall x\in{\mathbb R}\ .$$
In both cases $f$ is monotone.
A: Discuss the question as three cases.


*

*If $f=0, \forall x\in\mathbb{R}$, then the conclusion holds.

*Suppose $\exists x_0, f(x_0)<0$, then claim $f(x)\leq 0, \forall x\in\mathbb{R}$. 


Suppose not, $\exists y, f(y)>0$, then if $y<x_0$, then since $f$ is continuous, by intermediate value theorem, $\exists y_0\in (y,x_0), f(y_0)=0$, then $f^2(y)>f^2(y_0)=0$, so $f^2$ is not increasing, contradiction. If $y>x_0$, then since $f$ is continuous, by intermediate value theorem, $\exists y_0\in (x_0,y), f(y_0)=0$, then $f^2(x_0)>f^2(y_0)=0$, so $f^2$ is not increasing, contradiction.
Let $x\leq y$, by assumption, $f^2(x)\leq f^2(y)\iff (f(x)-f(y))(f(x)+f(y))\leq 0$
If $f(x)+f(y)=0$, by previous argument, $f(x)=f(y)=0$. Otherwise $f(x)+f(y)<0$, then $f(x)\geq f(y)$. In both cases, we have $f(x)\geq f(y)$, hence $f$ is decreasing on $\mathbb{R}$.


*Suppose $\exists x_0, f(x_0)>0$, then similarly, argue $f$ is increasing on $\mathbb{R}$. 

