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...

  • 2
    $\begingroup$ You can't use $x=-k$ because $x$ and $k$ must both be $\ge 0$. Also, it seems that $f$ can be arbitrary on $[0,k)$ $\endgroup$ – Hagen von Eitzen Aug 4 '15 at 8:21
  • $\begingroup$ $f(x)=x+k$ is NOT a solution of the problem. $\endgroup$ – mathcounterexamples.net Aug 4 '15 at 8:24
  • $\begingroup$ Sorry I'm wrong. $\endgroup$ – 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 $x>0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.