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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, 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!

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    There is something I cannot understand. If $f=0$ on $(-\infty,0]$ and $f$ is positive on $(0,+\infty)$ with a "bump" somewhere, and then it decays to zero, it will probably satisfy the condition $0=f(-x) for $x>0$.2012-08-27
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    You're right, but I should mention that $f$ is nonzero on such big interval, a trivial case!2012-08-27
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    I believe this is not a problem. Actually, I believe that the given condition has almost no influence on the monotonicity of $f$ or $|f|$...2012-08-27

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