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Let $f:\mathbb{R} _{+}\rightarrow \mathbb{R} _{+}$ be a real function.

Find all conditions on $f$ under which

$f$ is convex on $\left( a;b\right) \subset \mathbb{R} _{+}$$\Leftrightarrow$ $\dfrac {1} {f}$ is convex on $\left( a;b\right) \subset \mathbb{R} _{+}$

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    Do you want us to solve the problem, or to help you with its solution?2012-08-28
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    By $\mathbb{R}_{+}$ you mean $(0,\infty)$?2012-08-28
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    @ Nirakar Neo : Yes.2012-08-28
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    @ Ilya: to solve the problem, if possible.2012-08-28
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    As far as I see, assuming that both $f$ and $1/f$ are convex lead to no contradiction, or any further restriction on $f$.2012-08-28
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    I don't think so. E.g. $\sin x$ is concave in $\left( 0;\dfrac {\pi } {2}\right)$, but $\dfrac {1} {\sin x}$ is convex!2012-08-28
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    I assumed that $f$ is defined on the whole of $(0,\infty)$; after the edit, the question seems more interesting.2012-08-28
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    The question remains a bit unclear to me. Is the clause to be for a fixed interval $(a,b)$, i.e. so that the clause is not met if $f$ is convex in part of the interval and concave in another, or for any interval? Note that if $f$ is concave in any interval, $1/f$ must be convex in that interval.2012-08-28
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    You are not right. Take for example $f\left( x\right) =\dfrac {1-x^{5}} {\left( 2-x\right) ^{5}-1}$. Both $f$ and $1/f$ are convex on $\left( 0;1\right)$ ( Plot it with Wolfram!)2012-08-28
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    ...Or a simpler one; $f\left( x\right) =\dfrac {1-x^{2}} {\left( 2-x\right) ^{2}-1}$...2012-08-28
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    Or $f(x)=\frac{1+x}{1-x}$ on $(0,1)$. (Do not plot it with W|A, think!)2012-08-28
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    Maria: You should direct your replies to comments by adding @user at the start of the comment. In any case, I still find the question unclear. E.g. if the condition should hold for all intervals, you're just asking for the family of functions so that both $f$ and $1/f$ are convex, i.e. twice differentiable with $0.2012-08-30
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    Correction: You can have neither $f$ nor $1/f$ convex if $f$ is highly irregular, e.g. have intervals where it is nowhere differentiable. Is that the type of conditions you are looking for?2012-08-30

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