Strictly increasing function If $f(x)$ is a continuous function on $\mathbb R$, and $|f(-x)|< |f(x)|$ for all $x>0$. Does it imply that $|f(x)|$ is strictly increasing on $(0,\infty)$?
I tried to use the definition: let $a,b \in (0,\infty)$ with $a<b$, we need to show that $|f(a)|<|f(b)|$. We have 
$|f(-a)|< |f(a)|$ and $|f(-b)|< |f(b)|$, and I don't know how to proceed!

No, consider 
$$
f(x) = \begin{cases}
1&,x\in[1,\infty)
\\
x&,x\in [0,1)
\\
\frac12x&,x\in [-1,0]
\\
-\frac12&,x\in(-\infty,-1)
\end{cases}
$$
Just to clarify: $|f(1)| = |f(2)|$ thus function is not strictly increasing.
