# Find all functions f with the following two properties

Let $f(x): [0, +\infty)\mapsto \mathbb{R}$ be a function such that for one $k\in [0, +\infty)$: $$f^2(x)=k^2+x\cdot f(x+k) \quad \forall x\in \{\;[0, +\infty) : x\geq k\;\}\qquad (1)$$ and $$\frac{x+k}{2}\leq f(x) \leq 2\cdot (x+k) \quad \forall x\in \{\;[0, +\infty) : x\geq k\;\}\qquad (2)$$

Find all such functions $f.$

I have tried the following:

From (2), with substitution $x=-k$ we get $0\leq f(-k) \leq 0$, so $f(-k)=0$. Then, from (1) with substitution $x=-k$ we get $0=k^2-k\cdot f(0)$, so $f(0)=k$, which is easily verifiable even when $k=0$.

I also see that $f(x)=x+k$ is a valid function. Any hint how to find ALL such functions? (I believe the above is the only function)

Edit 1: As Hagen Von Eitzen noted, I cannot use the substitution $x=-k$, because both x and k are non-negative numbers...

• You can't use $x=-k$ because $x$ and $k$ must both be $\ge 0$. Also, it seems that $f$ can be arbitrary on $[0,k)$ – Hagen von Eitzen Aug 4 '15 at 8:21
• $f(x)=x+k$ is NOT a solution of the problem. – mathcounterexamples.net Aug 4 '15 at 8:24
• Sorry I'm wrong. – mathcounterexamples.net Aug 4 '15 at 8:41

If we require that $(1)$ and $(2)$ hold for all $x\ge 0$, then $f(x)=x+k$ is the only solution. (It is a solution in the first place because clearly $(x+k)^2=k^2+x(x+2k)$ and $\frac{x+k}{2}\le x+k\le 2(x+k)$).
First consider the case $k=0$. Then $(1)$ translates to $f(x)^2=xf(x)$, so $f(x)=x$ or $f(x)=0$; and $(2)$ becomes $\frac x2\le f(x)\le 2x$ so that indeed $f(x)=x$ for all $x\ge 0$.
Assume $k>0$. For $x\ge 0$ let $c(x)=\frac{f(x)-(x+k)}{k}$. Then $$\tag3-\frac{x+k}{2k}\le c(x)\le \frac{x+k}{k}$$ For $x>0$ we have \begin{align}f(x+k)&=\frac{f(x)^2-k^2}{x}\\&=\frac{(x+(c(x)+1)k)^2-k^2}{x}\\&=\frac{x^2+2(c(x)+1)kx_0+((c(x)+1)^2-1)k^2}{x}\\ &=x+2k\,+\,2c(x)k+\frac{c(x)(c(x)+2)k^2}{x}\end{align} i.e., $$c(x+k)=2c(x)+\frac{c(x)(c(x)+2)k}{x}$$ If for some $x>0$ we have $c(x)>0$ this implies $c(x+k)>2c(x)$ and by induction $c(x+nk)>2^nc(x)\to\infty$. As exponentials grow faster than polynomials, we obtain a contradiction with $(3)$. We conclude $c(x)\le 0$ (that is: $f(x)\le x+k$) for $x>0$.