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!
 A: 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.
A: While this question is long answered, all answers use some piecewise function. Therefore I want to add one additional counterexample which isn't piecewise:
$$f(x) = \frac{\mathrm e^x}{1+\mathrm e^{x^2}}$$
Clearly for $x\in(0,\infty)$ we have $\lvert f(-x)\rvert < \lvert f(x)\rvert$: Since the function is positive, the absolute value bars have no effect, and since $x^2=(-x)^2$, the condition reduces to $\mathrm e^{-x}<\mathrm e^x$ which is true for $x>0$ because the exponential function is strictly increasing. There's also no question that the function is continuous. On the other hand, $f(0)=1/2$, but $\lim_{x\to\infty}f(x)=0$, so it's also obvious that it is not strictly increasing.
Here's a graph of the function:

A: Define $f$ piecewise: $f=0$ on $(-\infty,0]$, an increasing straight line on $(0,1)$, a decreasing straight line on $[1,2]$, and then a positive constant until $+\infty$. It is a positive function, and it is somewhere increasing and somewhere decreasing. Since $f(-x)=0$ for any $x >0$, the condition $f(-x)<f(x)$ is satisfied.
It is boring to write a formula, but you can construct a function that has $y=0$ as a left asymptote, then it slowly increases on $(-\infty,0)$, then it goes up and down and yet $f(-x)<f(x)$ whenever $x>0$. It probably suffices to be careful, so that $f(0)<\inf_{x>0} f(x)$.
A: How about the following counter-example:
$$
f(x) = \begin{cases}
2-x&,x\in[1,\infty)
\\
x&,x\in [0,1)
\\
\frac12x&,x\in [-1,0)
\\
-1-x&,x\in(-\infty,-1)
\end{cases}
$$
$|f(x)|$ is clearly neither increasing nor decreasing in $(0,\infty)$. But $|f(-x)| < |f(x)|$ holds $\forall x \in (0,\infty)$.
