Solve the functional inequality $f(x+1)>(f(x))^2-f(x)+1$

Question

Suppose $f(x):[0,\infty)\longrightarrow [0,\infty)$ satisfies $$f(x+1)>(f(x))^2-f(x)+1$$ for sufficiently large values of $x$.

Can we solve this functional inequality or get some good information about it? We can put some conditions like $f$ is continuous if necessary.

Answer 1

Here is a family of solutions: $f_{a,b}(x)=a^{b^x}$, where $a>1$ and $b\geq 2$.

Proof: if $x \geq 0$, then $$ f_{a,b}(x+1)=a^{b^{x+1}}=a^{bb^x}\geq a^{2b^x}=(a^{b^x})^2>(a^{b^x})^2-a^{b^x}+1 =(f_{a,b}(x))^2-f_{a,b}(x)+1.\quad\square $$

Answer 2

The inequality can be rewritten as $$f(x+1)-f(x)>[f(x)-1]^2$$ Let $g(x)=f(x)-1.$ Then $$g(x+1)-g(x)>g(x)^2$$ For $x-y\notin\mathbb{Z}$ there is no relation between the values $f(x+n)$ and $f(y+m)$ if we do not impose the continuity of $f.$

Fix $0\le x<1$ and set $a_n=g(x+n).$ Then $$a_{n+1}-a_n> a_n^2\quad (*)$$

There are two possible cases $$\lim_na_n=\infty\quad {\rm or}\quad \lim_na_n=0$$

The first case can be described pretty precisely.

As $\lim a_n=\infty,$ we get $a_{n_0}>1$ for some $n_0.$ Then $a_n=e^{b_n}$ for $b_n>0.$ We obtain $$e^{b_{n+1}-b_n}>e^{b_n}+1$$ Hence $b_{n+1}>2b_n,$ i.e. $b_n>2^{n-n_0}b_{n_0}.$ Finally $$a_n> e^{2^nc},\quad n\ge n_0,\ c>0\quad (**)$$ On the other hand the sequence $a_n=e^{2^nc}-1$ satisfies the inequality $(*),$ which shows that the estimate in $(**)$ is sharp.

Concerning the case $\lim_na_n=0$ i.e. $f(x+n)<1$ an example follows. For $f(x)=1-{a\over x+b}$ we get $$f(x+1)-f(x)={a\over (x+b)(x+b+1)}\\ [f(x)-1]^2={a^2\over (x+b)^2}$$ The desired inequality holds for $b>0$ and $0<a<{b\over b+1}.$

Remark Assume $f(x)$ satisfies the inequality. Then $f(x)\neq 1.$ Assume $f$ is continuous. Then $f(x)<1$ for all $x$ or $f(x)>$ for all $x,$ by the IMV property.