a) Consider\[f\text{:}\left[ \text{0,}\infty \right)\to \left[ \text{0,}\infty \right)\] a differentiable and convex function .Show that $f\left( x \right)\le x$, for every $x\ge 0$, than ${f}'\left( x \right)\le 1$ ,for every $x\ge 0$ b) Determine \[f\text{:}\left[ \text{0,}\infty \right)\to \left[ \text{0,}\infty \right)\] differentiable and convex functions which have the property that $f\left( 0 \right)=0\,$, and ${f}'\left( x \right)f\left( f\left( x \right) \right)=x$, for every $x\ge 0$
Problem
Source:
Tags: function, inequalities, real analysis, real analysis unsolved
dr_Civot
04.08.2013 17:59
a) For sake of contradiction assume that there exist $a\geq 0$ such that $f'(a)>1$. Since $f$ is convex we have $f'(x)\geq f'(a)$ for all $x\in [a,\infty)$. If $g(x)=f(x)-x$ then $g'(x)=f'(x)-1\geq f'(a)-1$ for all $x\in [a,\infty)$. By Lagrange theorem we have $\frac{g(x)-g(a)}{x-a}=g'(t)$ for some $t\in (a,x)$ therefore: \[g(x)\geq (x-a)g'(t)+g(a)\geq (x-a)(f'(a)-1)+g(a)>0\] for $x$ big enoug since $f'(a)-1>0$. Now we will have $g(x)>0$ or $f(x)>x$ which is contradiction. b) Note that $f'(x)>0$ for $x>0$ hence $f$ is increasing. Since $f$ is convex we have $f''(x)\geq 0$ or: \[f''(x)=(f'(x))'=\left(\frac{x}{f(f(x))}\right)'=\frac{f(f(x))-xf'(x)f'(f(x))}{\left(f(f(x))\right)^2}\geq 0\] But we know that $f'(x)=\frac{x}{f(f(x))}$ and $f'(f(x))=\frac{f(x)}{f(f(f(x)))}$ hence: \[f(f(x))\geq\frac{x^2f(x)}{f(f(x))f(f(f(x)))}\Leftrightarrow f(f(x))^2f(f(f(x)))\geq x^2f(x)\] If there exist $a$ with $f(a)<a$ then $f(f(a))<a^2$ and $f(f(f(x)))<f(a)$ which contradicts with above inequality, hence $f(x)\geq x$. Now we have $f'(x)=\frac{x}{f(f(x))}\leq 1$. Again by Lagrange theorem we have $\frac{f(x)-f(0)}{x}=f'(t)\leq 1$ for some $t\in (0,x)$ which means that $f(x)\leq x$ therefore $f(x)=x$ is only solution.