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Function $f$ is continuous in $R$.

Prove: If $f^2$ monotonically increasing in $R$ then $f$ monotonic in $R$.

Intuitively its seems pretty clear, But I don't have any idea how to "start" the prove.

Any ideas?

Thanks.

TonyK
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JaVaPG
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    Start by proving that either $f(x) \leq 0$ for all $x \in \mathbb{R}$ or $f(x) \geq 0$ for all $x$. – jxnh Nov 13 '14 at 04:27
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    Note that $|f(x)|$ is monotonic. – user164587 Nov 13 '14 at 04:34
  • @JHance I still don't seem to find a way to prove it, I know that the derivative of $f(x)^2 \leq 0$ or $f(x)^2 \geq 0$, since it monotonically increasing, but what could I say about $f(x)$? – JaVaPG Nov 13 '14 at 04:37
  • Forget about computing derivatives. Prove monotonicity directly: i.e. if $x > y$ then $f(x) > f(y)$ (or the other way around) – jxnh Nov 13 '14 at 04:44
  • @JHance If $x>y$ I can't say that $f(x)>f(y)$, I could say that for $f(x)^2$, but how could I conclude that if $x>y \rightarrow f(x)>f(y)$ – JaVaPG Nov 13 '14 at 04:54
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    At this point I refer you to the first two comments. – jxnh Nov 13 '14 at 07:28

2 Answers2

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If there were two points $x_1$, $x_2\in{\mathbb R}$ with $f(x_1)f(x_2)<0$ there would be a point $\xi$ between $x_1$ and $x_2$ with $f(\xi)=0$. This would prevent $f^2$ from being monotone between $x_1$ and $x_2$. Therefore one has $f(x)\geq0$ or $f(x)\leq0$ throughout, and this implies that one of the following is true: $$f(x)=\sqrt{f^2(x)}\quad \forall x\in{\mathbb R}\qquad{\rm resp.}\qquad f(x)=-\sqrt{f^2(x)}\quad \forall x\in{\mathbb R}\ .$$ In both cases $f$ is monotone.

Christian Blatter
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Discuss the question as three cases.

  1. If $f=0, \forall x\in\mathbb{R}$, then the conclusion holds.
  2. Suppose $\exists x_0, f(x_0)<0$, then claim $f(x)\leq 0, \forall x\in\mathbb{R}$.

Suppose not, $\exists y, f(y)>0$, then if $y<x_0$, then since $f$ is continuous, by intermediate value theorem, $\exists y_0\in (y,x_0), f(y_0)=0$, then $f^2(y)>f^2(y_0)=0$, so $f^2$ is not increasing, contradiction. If $y>x_0$, then since $f$ is continuous, by intermediate value theorem, $\exists y_0\in (x_0,y), f(y_0)=0$, then $f^2(x_0)>f^2(y_0)=0$, so $f^2$ is not increasing, contradiction.

Let $x\leq y$, by assumption, $f^2(x)\leq f^2(y)\iff (f(x)-f(y))(f(x)+f(y))\leq 0$

If $f(x)+f(y)=0$, by previous argument, $f(x)=f(y)=0$. Otherwise $f(x)+f(y)<0$, then $f(x)\geq f(y)$. In both cases, we have $f(x)\geq f(y)$, hence $f$ is decreasing on $\mathbb{R}$.

  1. Suppose $\exists x_0, f(x_0)>0$, then similarly, argue $f$ is increasing on $\mathbb{R}$.
John
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  • Lets take $f(x)=x$ is clear that there is $y_0 \in(-1,1) \rightarrow f(y_0)=0$, (of course $y_0=0$) however, $f^2(-1)\ngtr f^2(0)$ Because $f(f(-1)=-1$. – JaVaPG Nov 13 '14 at 23:44
  • @JaVaPG If I understand your question clearly, you are confusing about the notation. $f^2$ should be multiplication $f(x)f(x)$ instead of composition. – John Nov 14 '14 at 00:36
  • @JaVaPG Hence with your example, $f^2(-1)=1$ – John Nov 14 '14 at 00:37
  • Oh, I feel so stupid, Everything understtod, Thank you:)! – JaVaPG Nov 14 '14 at 01:00