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Find the following limit for a smooth convex function $f:[0,\infty)\to\mathbb{R}^{\geq0}$ such that $f(1)=1$ and $f'(0)=f(0)=0$ $$\lim_{x\to0^{+}}\frac{f'(f^{-1}(x)a)}{f'(f^{-1}(x))}=L.$$ Is $L$ equal to $\frac{f(a)}{a}?$

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How would the information $f(1) = 1$ have any effect on the final result? – Anon Feb 1 '13 at 8:38
Unless the function is flat, I guess the result is (via L'Hopital) $a^n$ where $n$ is the first natural number where the $n$th derivative of $f$ is non-zero. I don't see how convexity or $f(1) = 1$ will be relevant here... – Anon Feb 1 '13 at 8:54
I have a example here of function satisfying all hypothesis that Op asked, but I can't evaluate the limit. If anyone has some skills with mathematica or any other program try to evaluate the limit please: $\text{Exp}\left[-1\left/\left(1-\left(\left(\frac{1}{3} \left(3-3^{3/4}\right)\right)*x-1\right){}^{\wedge}2\right)\right.\right]/\left(‌​e^{-\frac{1}{1-\left(1+\frac{1}{3} \left(-3+3^{3/4}\right)\right)^2}}\right)$ – Tomás Feb 1 '13 at 10:51
Since $f^{-1}(x)\to0^+$ when $x\to0^+$, the requested limit is equal to $\lim_{w\to0^+}f^\prime(aw)/f^\prime(w)$, so don't get distracted by that nasty $f^{-1}\ldots$ – Matemáticos Chibchas Feb 2 '13 at 18:06
I guess $f(1)=1$ is there to rule out $f(x)=0$. – Maesumi Feb 3 '13 at 16:20
up vote 3 down vote accepted

Counterexample. For $p\in (0,1)$ the function $$ f_p:[0,\infty)\to [0,\infty),\ f_p(x)=(1-p)x^2+px^3 $$ satisfies all the conditions, however for every $a>0$ we have $$ \lim_{x \to0}\frac{f_p'(ax)}{f_p'(x)}=\lim_{x \to 0}\frac{2a(1-p)+3a^2px^2}{2(1-p)x+3px^2}=a\ne (1-p)a+pa^2=\frac{f_p(a)}{a}. $$ Claim: Let $f:[0,\infty) \to [0,\infty)$ be a non-zero smooth function such that $$ f(0)=f'(0)=\ldots=f^{(n-1)}(0)=0\ne f^{(n)}(0). $$ Then $$ \lim_{x \to 0+}\frac{f'(ax)}{f'(x)}=a^{n-1}, $$

Proof. For $\varepsilon>0$ small enough, we have $$ f(x)=\frac{f^{(n)}(0)}{n!}x^n+o(x^n)=:c_nx^n+o(x^n) \quad \forall x \in (0,\varepsilon]. $$ Therefore, for every $x \in (0,\varepsilon]$ we have $$ \frac{f'(ax)}{f'(x)}=\frac{nc_na^{n-1}x^{n-1}+o(x^{n-1})}{nc_nx^{n-1}+o(x^{n-1})}=\frac{a^{n-1}+o(1)}{1+o(1)}. $$ Hence $$ \lim_{x \to 0+}\frac{f'(ax)}{f'(x)}=a^{n-1}. $$ Remark: If $f(x)=x^n$, with $n\ge 2$ an integer, then in fact $$ \frac{f(a)}{a}=a^{n-1} \quad \forall a>0. $$

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