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!
Best Answer
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.