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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 $a<b$, 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!

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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 $0=f(-x)<f(x)$ for $x>0$. –  Siminore Aug 27 '12 at 14:14
    
You're right, but I should mention that $f$ is nonzero on such big interval, a trivial case! –  Monca.L Aug 27 '12 at 14:17
    
I believe this is not a problem. Actually, I believe that the given condition has almost no influence on the monotonicity of $f$ or $|f|$... –  Siminore Aug 27 '12 at 14:20

4 Answers 4

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.

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what about $f(-1)$? is is defined? –  Monca.L Aug 27 '12 at 14:09
    
@Monca.L: I didn't decide where to include $x = -1$, so that was a typo :) –  Ilya Aug 27 '12 at 14:09
    
Can we say that under the stated conditions then we must have $|f(x)|$ is nondecreasing (instead of strictly increasing)? –  Monca.L Aug 27 '12 at 14:13
    
I think that you can assume $f$ is positive on the positive axis... –  Siminore Aug 27 '12 at 14:16
    
@Siminore: Yes! –  Monca.L Aug 27 '12 at 14:18

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:

Graph of the function

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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 $f(0)<\inf_{x>0} f(x)$.

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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 $\forall x \in (0,\infty)$.

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I think there are infinitely many counterexamples! –  Siminore Aug 27 '12 at 14:30
    
Right. This is just one simple example which came to my mind. –  Kartik Audhkhasi Aug 27 '12 at 14:31
    
That function is not continuous at $x=-1$. –  celtschk Sep 19 '12 at 17:40

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