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I am looking for a function $f$ having the following characteristics:

  • $f$ defined on $[0,1]$
  • $f(0)=0$
  • $f(1)=1$
  • $ \forall x \in ]0,1[, x <f(x) < 1$

  • $f$ differentiable on $]0,1]$

  • $f'>0$
  • $f'(1)=1$
  • $\lim\limits_{x\to0} f'(x)=+\infty$

Finally, I will also need an analytical expression of the inverse function $f^{-1}$.

Do you know such function?

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is that last requirement really $xf(x) \rightarrow 1 $ for $x\rightarrow 0$? Are you requiring continuity of $F$ at $x=0$? – user20266 Dec 31 '11 at 19:57
@Thomas yes, my last requirement is what you have written. And, f has to be continuous on [0,1] – julien Dec 31 '11 at 22:39
Your requirements are incompatible. Suppose $f$ is continuous on $[0,1]$, and $f(0)=0$. Then as $x$ approaches $0$, $f(x)$ approaches $0$. Thus $xf(x)$ cannot approach $1$. But that's what your last condition requires. Maybe modify the requirements, possibly by removing the last one. Or did you mean that the ratio $x/f(x)$ must approach $1$? – André Nicolas Dec 31 '11 at 23:23
oops, I did a mistake in my last requirement: it is $x f'(x)$ and not $x f(x)$. Sorry. – julien Jan 1 '12 at 14:31
@julien: The altered condition that $\lim_{x\to 0} xf'(x)=1$ is still incompatible with the others, though more work is needed to show the incompatibility. You are asking for the derivative to behave like $1/x$ near $0$. That's too high a rate of blowup. – André Nicolas Jan 1 '12 at 16:20
up vote 3 down vote accepted

A solution is the function $f:[0,1]\to[0,1]$ defined by $$ f(x)=\tfrac12(1+x)\sqrt{x}, $$ whose inverse function $g$ is defined by $$ g(y)=\left(\sqrt{y^2+\tfrac1{27}}+y\right)^{2/3}+\left(\sqrt{y^2+\tfrac1{27}}-y\right)^{2/3}-\tfrac23 $$

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Thanks Didier. Are you sure the inverse function is correct? – julien Jan 4 '12 at 16:53
Yes. $ $ $ $ $ $ – Did Jan 4 '12 at 17:05
Jewel-like example! – André Nicolas Jan 7 '12 at 18:22

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