Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

3 Answers 3

up vote 4 down vote accepted
+50

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

share|improve this answer
1  
@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 $$g(e)=e \ e^{-e/e}\ f(x)=f(x)\ .\qquad\square$$

share|improve this answer
1  
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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