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Solve the equation $$f(x+y)=\max(f(x),y)+\min(x,f(y))$$

My work so far:

1) $y=0 $

$f(x)=\max(f(x),0)+\min (x,f(0))$

2) $y=-x$

$f(0)=\max(f(x),-x)+\min(x,f(-x))$

nonuser
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Roman83
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2 Answers2

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If you put $y=x$ you get

$$f(2x) = \max (f(x),x) + \min (f(x),x) = f(x)+x$$ because if one of them is the maximum the other one is the minimum.

Let $g(x)=f(x)-x$ then $g(2x)+2x = g(x)+x+x$ hence $g(2x)=g(x)$ and so (assuming that $f$ is continuous) $g(x)$ is a constant, say $g(x)=c$ for some $c\in\mathbb{R}$.

It follows that $f(x)=x+c$. Now we check for which $c$ we have a solution, we need that

$f(x+y) = x+y+c = \max (x+c,y) + \min(x,y+c)$

clearly we can find $x,y$ such that $\max (x+c,y)=x+c$ and $\min(x,y+c)=y+c$ so we get that

$$x+y+c = x+c+y+c$$ it follows that $c=0$. We conclude that $f(x)=x$ is the only (continuous) solution.

Edit: Without the continuity we can argue as follows: Observe that $$f(x+y) = \max (f(x),y) + \min (x,f(y))$$ $$f(y+x) = \max (f(y),x) + \min (y,f(x))$$ If we add these equations we get:

$$2f(x+y) = f(x)+f(y)+x+y$$

Now let $g(x)=f(x)-x$ we have $$2g(x+y)+2x+2y = g(x)+x+g(y)+y+x+y$$ and so $$2g(x+y) = g(x)+g(y)$$

Let $y=0$ we have that $2g(x) = g(x)+g(0)$ therefore $g(x)=g(0)$, we conclude that $g(x)$ is a constant.

Yanko
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    $g(2x)=g(x)$ does not imply that $g$ is a constant. This is just a perodicity condition and there are lots of non-constant functions satisfying this unless it is given that $f$ is continuous. – Kavi Rama Murthy Dec 17 '18 at 10:23
  • This is not valid; proving that $g(2x)=g(x)$ does not prove that $g(1)=g(3)$, and certainly does not prove $g(x)$ constant. – Benedict Randall Shaw Dec 17 '18 at 10:24
  • @Yanko, as an example, $g(x)=\begin{cases}{\log_2|x|},&x\ne0\0,&x=0\end{cases}, {\cdot}$ is the fractional part – Shubham Johri Dec 17 '18 at 10:37
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    Of course you are all right. I edited the question to include the non-continuous case as well. The only answer is $f(x)=x$ continuous or not. – Yanko Dec 17 '18 at 11:23
  • To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $ – Mohsen Shahriari Dec 18 '18 at 21:53
  • @MohsenShahriari Right corrected. Thanks. – Yanko Dec 19 '18 at 17:04
  • Well, now the problem is $ x - | c | \le y \le x + | c | $ doesn't give $ \max ( x + c , y ) = x + c $ or $ \min ( x , y + c ) = y + c $. – Mohsen Shahriari Dec 20 '18 at 19:30
  • Are you sure? It means that either $x-c\leq y \leq x+c$ or that $x+c\leq y \leq x-c$ and in both cases the next line is correct. Anyway I'll change it a little to make it clear. – Yanko Dec 21 '18 at 10:59
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Since $$\max \{a,b\} = {a+b+|a-b|\over 2}\;\;\;\;{\rm and }\;\;\;\;\min \{a,b\} = {a+b-|a-b|\over 2}$$ we get $$ \boxed{2f(x+y) = f(x)+y+|f(x)-y|+x+f(y)-|x-f(y)|}$$ Let $c=f(0)$. For $y=0, x=t$:

$$ f(t) = |f(t)|+t+c-|t|$$

and for $x=0,y=t$:

$$f(t) =c+t+|c-t| -|f(t)|$$

Adding last two we get: $f(t) = t+c$, for all $t$.

Plugging this in boxed equation we get $$|x-y+c| = |x-y-c|$$ which is valid for all $x,y$, so also for $x=c$ and $y=0$ and so $c=0$. Thus the only solution is $f(x)=x$.

nonuser
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