# Prove the concavity of the transformation from a concave function to another

Let's say we have $f_1$ and $f_2$, both strictly increasing and strictly concave on $[0,+\infty)$. $f_1(0)=f_2(0)=0$ and the difference $f_1-f_2$ is strictly positive and strictly increasing. That is, $f_1(x)>f_2(x)$ for $x>0$ and $f^\prime_1(x)>f^\prime_2(x)$ for $x>0$.

Can we prove the following intuitive result:

There exist $\phi$, strictly positive, strictly increasing and strictly concave, such that $f_2=\phi(f_1)$. We would have $\phi'<1$.

Thanks a lot !

• I'd be interested to see a partial result, if this could be applied to $$f_1(x) = ax \quad f_2(x) = bx$$ for $a>b$. – Omnomnomnom Jun 25 '15 at 15:55
• I guess $\phi(x)=\frac{b}{a}x$ would do the trick... – Seneleh Jun 25 '15 at 16:00
• But is this function "strictly concave"? – Omnomnomnom Jun 25 '15 at 16:02
• Sure not. But I don't think the result holds in this case (if it holds at all). – Seneleh Jun 25 '15 at 16:06
• In fact, it's clear when we take $f_1(x) = x$ that we must have $\phi(x) = f_2(x)$. So, it is not possible to generally have a strictly concave $\phi$, nor would we have $\phi' < 1$. – Omnomnomnom Jun 25 '15 at 16:12

When you refer to concave function, I suppose the definition is concave upward function. Now under the condition in your post, if we suppose $f_2=\phi(f_1)$, then $\phi$ is not only existed, but also unique: $f_2=\phi(f_1)\Rightarrow \phi=f_2\circ f_1^{-1},$ since both $f_1$ and $f_2$ are bijection from $[0,+\infty)$ to $Imf_1, Imf_2$, respectively, therefore they are invertible.
Now we study the property of $\phi$, one can easily check:
(1) $\phi$ is a bijection from $Imf_1$ to $Imf_2$.
(2) for $y\in Imf_1,$ and $x=f_1^{-1}(y)$, $$\phi'(y)=f_2'(f_1^{-1}(y))\cdot f_1^{-1}(y)=\frac{f_2'(x)}{ f_1(x)}$$
Since $f^\prime_1(x)>f^\prime_2(x)>0,$ one has $0<\phi'(y)<1.$
(3) The convexity for $\phi$ is uncertain, since $f_1^{-1}$ convex and $f_2$ concave, and convexity of their composition is uncertain.