Does a function other than 0 that satisfies the following definition exist?
$ f(x) = \max_{0<\xi
If so can it be expressed using elementary functions?
Does a function other than 0 that satisfies the following definition exist?
$ f(x) = \max_{0<\xi
If so can it be expressed using elementary functions?
Ok, let us put some considerations together. If we assume that $f$ is a continous function over $[0,+\infty)$ for which: $ \forall x\in\mathbb{R}^+,\quad f(x)=\max_{\xi\in[0,x]}\left(\xi\,f(x-\xi)\right), $ we clearly have $f(x)\geq 0$. Take an $\eta\in(0,1)$ and assume that $f$ reaches a positive maximum over $[0,\eta]$ in $\xi_\eta\leq\eta$. Then: $ f(\xi_\eta) = \max_{\xi\in[0,\xi_\eta]} \xi\;f(\xi-\xi_\eta)\leq \xi_\eta\,\max_{\xi\in[0,\xi_\eta]}f(\xi) \leq \eta\;f(\xi_\eta) $ a contradiction. This gives $f(x)\equiv 0$ over $(0,1]$. Take an $\eta\in(0,1)$ and assume that $f$ reaches a positive maximum over $[1,1+\eta]$ in $\xi_\eta\leq 1+\eta$. Since $ f(x) = \max_{\xi\in[0,x]}\left((x-\xi)\;f(\xi)\right),$ we still have $f(\xi_\eta) \leq \eta\; f(\xi_\eta),$ then $f$ is identically $0$ over $[1,2]$, too, and by induction we have that $f$ is identically zero over the whole $\mathbb{R}^+$.
Update: yes, we don't really need to have $f$ continous, it is sufficient that, for every compact set $K=[0,r]$, there is an $x_K\in K$ such that $f(x_K)=\max_{x\in K}f(x)$.
Claim. There is no function $f:\ ]0,a[\ \to{\mathbb R}_{\geq0}$ satisfying $f(x)=\max_{0<\xi
Proof. Assume the function $f$ has property $(*)$ for some $a>0$ and is not identically $0$. This $f$ has to be nondecreasing: For $h>0$ one has $\eqalign{f(x+h)&=\max_{0<\xi
As a curiosity one might add that the function $f(x):=e^{x/e}\qquad(x>0)$ satisfies the given condition for all $x>e$.
Proof. Fix an $x>e$ and consider the function $g(t):=t f(x-t)= t\ e^{-t/e} f(x)\qquad(0
Since we cannot be sure if the $\max$ exists, let us consider $f\colon I\to\mathbb R$ with $\tag1f(x)=\sup_{0<\xi
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
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
Assume $f(x_0)>0$. Then for any $0<\epsilon<1$ there is $x_1
Therefore $f$ is identically zero.