# 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?

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You should specify the range of $f$, since the values of $f$ on $\mathbb{R}^-$ are irrelevant. Some initial considerations are the following: if there is some $\alpha\in\mathbb{R}^+$ for which $f(\alpha)>0$, for any $\beta>\alpha$ you have $f(\beta)>0$. Moreover, if $f$ is a continous function over some interval of the form $[0,r>0]$ and $f(0)>0$, you have $f(x)\geq f(0)\,x$ for any $x>0$. In any case, you have that $f(2y)>y\,f(y)$ holds for any $y>0$. –  Jack D'Aurizio Nov 4 '12 at 11:30
On the other way, since $ab\leq\frac{a^2+b^2}{2}$, you have $f(x)\leq \frac{x^2}{2}+\frac{1}{2}\left(\max_{0<\xi<x}|f(\xi)|\right)^2$ for any $x>0$. –  Jack D'Aurizio Nov 4 '12 at 11:44
And a consequence of the last inequality is that, for any $y\in(0,1)$, $(M_y-1)^2\geq 1-y^2$ holds, where $M_y=\max_{0<\xi<y}|f(\xi)|$. –  Jack D'Aurizio Nov 4 '12 at 12:15

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 $f(x_K)=\max_{x\in K}f(x)$.

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@Jack D'Aurizio : you don't really need continuity. From your proof, if $f$ is bounded on an interval it must be identically zero there. But on any interval $[0,a]$, $f$ is bounded from above by $\frac{f(2a)}{a}$ (since $f(2a) \geq (2a-t) f(t)$ for any $t<2a$). –  pgassiat Nov 6 '12 at 14:49

Claim. There is no function $f:\ ]0,a[\ \to{\mathbb R}_{\geq0}$ satisfying $$f(x)=\max_{0<\xi<x}\bigl\{\xi f(x-\xi)\bigr\}\qquad(0<x<a)\qquad(*)$$ other than $f(x)\equiv0$.

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<x+h}\bigl\{\xi f(x+h-\xi)\bigr\}\geq \max_{h<\xi<x+h}\bigl\{\xi f(x+h-\xi)\bigr\}\cr &=\max_{0<\xi'<x}\bigl\{(h+\xi') f(x-\xi')\bigr\}\geq \max_{0<\xi'<x}\bigl\{\xi' f(x-\xi')\bigr\}\cr &=f(x)\ .\cr} Let $x_0:=\inf\{x\ |\ f(x)>0\}\geq 0$. There is an $h\in\ ]0,{1\over2}]$ with $f(x_0+h)=:c>0$. There are no positive values of $f$ to the left of $x_0$. Therefore $$c=f(x_0+h)=\max_{0\leq t<h}\bigl\{(h-t)f(x_0+t)\bigr\}\leq h\, c\leq{c\over2}\ ,$$ which is incompatible with $c>0$.$\qquad\square$

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 $$g(t):=t f(x-t)= t\ e^{-t/e} f(x)\qquad(0<t<x)\ .$$ One has $$g'(t)=\Bigl(1-{t\over e}\Bigr) e^{-t/e} f(x)\ .$$ Therefore $g$ is increasing for $0<t\leq e$ and decreasing for $e\leq t <x$. It follows that $g$ takes its maximal value at $t:=e$, and this value is $$g(e)=e \ e^{-e/e}\ f(x)=f(x)\ .\qquad\square$$

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But for $x<e$, $g$ takes its maximal value at $x$ and $g(x)<f(x)$... –  pgassiat Nov 6 '12 at 14:29
@pgassiat: You are right. –  Christian Blatter Nov 6 '12 at 15:40

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.

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