Maximum based recursive function definition Does a function other than 0 that satisfies the following definition exist? 
$$
f(x) = \max_{0<\xi<x}\left\{ \xi\;f(x-\xi) \right\}
$$
If so can it be expressed using elementary functions?

Since we cannot be sure if the $\max$ exists, let us consider $f\colon I\to\mathbb R$ with
$$\tag1f(x)=\sup_{0<\xi<x}\xi f(x-\xi)$$
instead, where $I$ is an interval of the form $I=(0,a)$ or $I=(0,a]$ with $a>0$.
If $x_0>0$ then $f(x)\ge (x-x_0)f(x_0)$ for $x>x_0$ and $f(x)\le\frac{f(x_0)}{x_0-x}$ for $x<x_0$.
We can conclude $f(x)\ge0$ for all $x>0$: Select $x_0\in(0,x)$. Note that $f(x_0)$ may be negative. Let $\epsilon>0$. For $0<h<x-x_0$ we have $f(x_0+h)\ge h f(x_0)$ and $f(x)\ge (x-x_0-h)f(x_0+h)\ge h(x-x_0-h)f(x_0)$. If $h<\frac{\epsilon}{(x_1-x_0)|f(x-0)|}$, this shows $f(x)\ge-\epsilon$. Since $\epsilon$ was arbitrary, we conclude $f(x)\ge0$.
Assume $f(x_0)>0$. Then for any $0<\epsilon<1$ there is $x_1<x_0$ with $(x_0-x_1)f(x_1)>(1-\epsilon)f(x_0)$ and especially $f(x_1)>0$. In fact, for a sequence $(\epsilon_n)_n$ with $0<\epsilon_n<1$ and 
$$\prod_n (1-\epsilon_n)=:c>0$$
(which is readily constructed) we find a sequence $x_0>x_1>x_2>\ldots$ such that $(x_n-x_{n+1})f(x_{n+1})>(1-\epsilon_n)f(x_n)$, hence
$$\prod_{k=1}^{n} (x_{k}-x_{k+1})\cdot f(x_{n+1})>\prod_{k=1}^{n-1}(1-\epsilon_k)\cdot f(x_1)>c f(x_1). $$
By the arithmetic-geometric inequality, $${\prod_{k=1}^n (x_{k}-x_{k+1})}\le \left(\frac {x_1-x_n}n\right)^n<\left(\frac {x_1}n\right)^n$$
and 
$$f(x_{n+1})>c f(x_1)\cdot \left(\frac n{x_1}\right)^n$$
The last factor is unbounded. 
Therefore,
$f(x_0)\ge (x_0-x_n)f(x_{n+1})\ge (x_0-x_1) f(x_{n+1})$ gives us a contradiction.
Therefore $f$ is identically zero.
