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












4












$begingroup$



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) $y=-x$



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






functional-equations







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$endgroup$












  • $begingroup$
    I can guess a solution, $f(x)=x$.
    $endgroup$
    – Yanko
    Dec 17 '18 at 9:59










  • $begingroup$
    I did considered cases $x=y=0$ and $f(0)<>=0$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:04






  • 1




    $begingroup$
    I post an answer. It turns out that $f(x)=x$ is the only solution.
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:05










  • $begingroup$
    And if $x=y$ then $f(2x)=max(f(x),x)+min(f(x),x)$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:07










  • $begingroup$
    Yes but $max(f(x),x) + min(f(x),x) = x+f(x)$ (check for $xleq f(x)$ and for $f(x)leq x$).
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:08
















4












$begingroup$



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) $y=-x$



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






functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    I can guess a solution, $f(x)=x$.
    $endgroup$
    – Yanko
    Dec 17 '18 at 9:59










  • $begingroup$
    I did considered cases $x=y=0$ and $f(0)<>=0$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:04






  • 1




    $begingroup$
    I post an answer. It turns out that $f(x)=x$ is the only solution.
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:05










  • $begingroup$
    And if $x=y$ then $f(2x)=max(f(x),x)+min(f(x),x)$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:07










  • $begingroup$
    Yes but $max(f(x),x) + min(f(x),x) = x+f(x)$ (check for $xleq f(x)$ and for $f(x)leq x$).
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:08














4












4








4


2



$begingroup$



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) $y=-x$



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






functional-equations







share|cite|improve this question











$endgroup$





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) $y=-x$



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






functional-equations




functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 18:03









greedoid

39.2k114797




39.2k114797










asked Dec 17 '18 at 9:56









Roman83Roman83

14.4k31856




14.4k31856












  • $begingroup$
    I can guess a solution, $f(x)=x$.
    $endgroup$
    – Yanko
    Dec 17 '18 at 9:59










  • $begingroup$
    I did considered cases $x=y=0$ and $f(0)<>=0$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:04






  • 1




    $begingroup$
    I post an answer. It turns out that $f(x)=x$ is the only solution.
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:05










  • $begingroup$
    And if $x=y$ then $f(2x)=max(f(x),x)+min(f(x),x)$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:07










  • $begingroup$
    Yes but $max(f(x),x) + min(f(x),x) = x+f(x)$ (check for $xleq f(x)$ and for $f(x)leq x$).
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:08


















  • $begingroup$
    I can guess a solution, $f(x)=x$.
    $endgroup$
    – Yanko
    Dec 17 '18 at 9:59










  • $begingroup$
    I did considered cases $x=y=0$ and $f(0)<>=0$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:04






  • 1




    $begingroup$
    I post an answer. It turns out that $f(x)=x$ is the only solution.
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:05










  • $begingroup$
    And if $x=y$ then $f(2x)=max(f(x),x)+min(f(x),x)$
    $endgroup$
    – Roman83
    Dec 17 '18 at 10:07










  • $begingroup$
    Yes but $max(f(x),x) + min(f(x),x) = x+f(x)$ (check for $xleq f(x)$ and for $f(x)leq x$).
    $endgroup$
    – Yanko
    Dec 17 '18 at 10:08
















$begingroup$
I can guess a solution, $f(x)=x$.
$endgroup$
– Yanko
Dec 17 '18 at 9:59




$begingroup$
I can guess a solution, $f(x)=x$.
$endgroup$
– Yanko
Dec 17 '18 at 9:59












$begingroup$
I did considered cases $x=y=0$ and $f(0)<>=0$
$endgroup$
– Roman83
Dec 17 '18 at 10:04




$begingroup$
I did considered cases $x=y=0$ and $f(0)<>=0$
$endgroup$
– Roman83
Dec 17 '18 at 10:04




1




1




$begingroup$
I post an answer. It turns out that $f(x)=x$ is the only solution.
$endgroup$
– Yanko
Dec 17 '18 at 10:05




$begingroup$
I post an answer. It turns out that $f(x)=x$ is the only solution.
$endgroup$
– Yanko
Dec 17 '18 at 10:05












$begingroup$
And if $x=y$ then $f(2x)=max(f(x),x)+min(f(x),x)$
$endgroup$
– Roman83
Dec 17 '18 at 10:07




$begingroup$
And if $x=y$ then $f(2x)=max(f(x),x)+min(f(x),x)$
$endgroup$
– Roman83
Dec 17 '18 at 10:07












$begingroup$
Yes but $max(f(x),x) + min(f(x),x) = x+f(x)$ (check for $xleq f(x)$ and for $f(x)leq x$).
$endgroup$
– Yanko
Dec 17 '18 at 10:08




$begingroup$
Yes but $max(f(x),x) + min(f(x),x) = x+f(x)$ (check for $xleq f(x)$ and for $f(x)leq x$).
$endgroup$
– Yanko
Dec 17 '18 at 10:08










2 Answers
2






active

oldest

votes


















7












$begingroup$

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 $cinmathbb{R}$.



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.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    $g(2x)=g(x)$ 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.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 10:23












  • $begingroup$
    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.
    $endgroup$
    – Benedict Randall Shaw
    Dec 17 '18 at 10:24










  • $begingroup$
    @Yanko, as an example, $g(x)=begin{cases}{log_2|x|},&xne0\0,&x=0end{cases}, {cdot}$ is the fractional part
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 10:37








  • 3




    $begingroup$
    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.
    $endgroup$
    – Yanko
    Dec 17 '18 at 11:23










  • $begingroup$
    To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $
    $endgroup$
    – Mohsen Shahriari
    Dec 18 '18 at 21:53



















3












$begingroup$

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 $f(x)=x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $ x = y = 0 $ doesn't give you $ f ( 0 ) = 0 $.
    $endgroup$
    – Mohsen Shahriari
    Dec 20 '18 at 19:05










  • $begingroup$
    Fixed @MohsenShahriari
    $endgroup$
    – greedoid
    Dec 20 '18 at 19:53










  • $begingroup$
    @MohsenShahriari Do you think this solution of Hagen is correct: math.stackexchange.com/questions/2982120/…
    $endgroup$
    – greedoid
    Dec 20 '18 at 20:02











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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 $cinmathbb{R}$.



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.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    $g(2x)=g(x)$ 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.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 10:23












  • $begingroup$
    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.
    $endgroup$
    – Benedict Randall Shaw
    Dec 17 '18 at 10:24










  • $begingroup$
    @Yanko, as an example, $g(x)=begin{cases}{log_2|x|},&xne0\0,&x=0end{cases}, {cdot}$ is the fractional part
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 10:37








  • 3




    $begingroup$
    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.
    $endgroup$
    – Yanko
    Dec 17 '18 at 11:23










  • $begingroup$
    To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $
    $endgroup$
    – Mohsen Shahriari
    Dec 18 '18 at 21:53
















7












$begingroup$

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 $cinmathbb{R}$.



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.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    $g(2x)=g(x)$ 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.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 10:23












  • $begingroup$
    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.
    $endgroup$
    – Benedict Randall Shaw
    Dec 17 '18 at 10:24










  • $begingroup$
    @Yanko, as an example, $g(x)=begin{cases}{log_2|x|},&xne0\0,&x=0end{cases}, {cdot}$ is the fractional part
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 10:37








  • 3




    $begingroup$
    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.
    $endgroup$
    – Yanko
    Dec 17 '18 at 11:23










  • $begingroup$
    To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $
    $endgroup$
    – Mohsen Shahriari
    Dec 18 '18 at 21:53














7












7








7





$begingroup$

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 $cinmathbb{R}$.



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.






share|cite|improve this answer











$endgroup$



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 $cinmathbb{R}$.



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 11:01

























answered Dec 17 '18 at 10:02









YankoYanko

6,3831528




6,3831528








  • 2




    $begingroup$
    $g(2x)=g(x)$ 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.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 10:23












  • $begingroup$
    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.
    $endgroup$
    – Benedict Randall Shaw
    Dec 17 '18 at 10:24










  • $begingroup$
    @Yanko, as an example, $g(x)=begin{cases}{log_2|x|},&xne0\0,&x=0end{cases}, {cdot}$ is the fractional part
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 10:37








  • 3




    $begingroup$
    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.
    $endgroup$
    – Yanko
    Dec 17 '18 at 11:23










  • $begingroup$
    To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $
    $endgroup$
    – Mohsen Shahriari
    Dec 18 '18 at 21:53














  • 2




    $begingroup$
    $g(2x)=g(x)$ 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.
    $endgroup$
    – Kavi Rama Murthy
    Dec 17 '18 at 10:23












  • $begingroup$
    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.
    $endgroup$
    – Benedict Randall Shaw
    Dec 17 '18 at 10:24










  • $begingroup$
    @Yanko, as an example, $g(x)=begin{cases}{log_2|x|},&xne0\0,&x=0end{cases}, {cdot}$ is the fractional part
    $endgroup$
    – Shubham Johri
    Dec 17 '18 at 10:37








  • 3




    $begingroup$
    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.
    $endgroup$
    – Yanko
    Dec 17 '18 at 11:23










  • $begingroup$
    To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $
    $endgroup$
    – Mohsen Shahriari
    Dec 18 '18 at 21:53








2




2




$begingroup$
$g(2x)=g(x)$ 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.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 10:23






$begingroup$
$g(2x)=g(x)$ 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.
$endgroup$
– Kavi Rama Murthy
Dec 17 '18 at 10:23














$begingroup$
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.
$endgroup$
– Benedict Randall Shaw
Dec 17 '18 at 10:24




$begingroup$
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.
$endgroup$
– Benedict Randall Shaw
Dec 17 '18 at 10:24












$begingroup$
@Yanko, as an example, $g(x)=begin{cases}{log_2|x|},&xne0\0,&x=0end{cases}, {cdot}$ is the fractional part
$endgroup$
– Shubham Johri
Dec 17 '18 at 10:37






$begingroup$
@Yanko, as an example, $g(x)=begin{cases}{log_2|x|},&xne0\0,&x=0end{cases}, {cdot}$ is the fractional part
$endgroup$
– Shubham Johri
Dec 17 '18 at 10:37






3




3




$begingroup$
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.
$endgroup$
– Yanko
Dec 17 '18 at 11:23




$begingroup$
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.
$endgroup$
– Yanko
Dec 17 '18 at 11:23












$begingroup$
To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $
$endgroup$
– Mohsen Shahriari
Dec 18 '18 at 21:53




$begingroup$
To find $ x $ and $ y $ such that $ x - c < y < x + c $, you should have $ c > 0 $
$endgroup$
– Mohsen Shahriari
Dec 18 '18 at 21:53











3












$begingroup$

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 $f(x)=x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $ x = y = 0 $ doesn't give you $ f ( 0 ) = 0 $.
    $endgroup$
    – Mohsen Shahriari
    Dec 20 '18 at 19:05










  • $begingroup$
    Fixed @MohsenShahriari
    $endgroup$
    – greedoid
    Dec 20 '18 at 19:53










  • $begingroup$
    @MohsenShahriari Do you think this solution of Hagen is correct: math.stackexchange.com/questions/2982120/…
    $endgroup$
    – greedoid
    Dec 20 '18 at 20:02
















3












$begingroup$

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 $f(x)=x$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    $ x = y = 0 $ doesn't give you $ f ( 0 ) = 0 $.
    $endgroup$
    – Mohsen Shahriari
    Dec 20 '18 at 19:05










  • $begingroup$
    Fixed @MohsenShahriari
    $endgroup$
    – greedoid
    Dec 20 '18 at 19:53










  • $begingroup$
    @MohsenShahriari Do you think this solution of Hagen is correct: math.stackexchange.com/questions/2982120/…
    $endgroup$
    – greedoid
    Dec 20 '18 at 20:02














3












3








3





$begingroup$

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 $f(x)=x$.






share|cite|improve this answer











$endgroup$



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 $f(x)=x$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 19:58

























answered Dec 19 '18 at 17:18









greedoidgreedoid

39.2k114797




39.2k114797












  • $begingroup$
    $ x = y = 0 $ doesn't give you $ f ( 0 ) = 0 $.
    $endgroup$
    – Mohsen Shahriari
    Dec 20 '18 at 19:05










  • $begingroup$
    Fixed @MohsenShahriari
    $endgroup$
    – greedoid
    Dec 20 '18 at 19:53










  • $begingroup$
    @MohsenShahriari Do you think this solution of Hagen is correct: math.stackexchange.com/questions/2982120/…
    $endgroup$
    – greedoid
    Dec 20 '18 at 20:02


















  • $begingroup$
    $ x = y = 0 $ doesn't give you $ f ( 0 ) = 0 $.
    $endgroup$
    – Mohsen Shahriari
    Dec 20 '18 at 19:05










  • $begingroup$
    Fixed @MohsenShahriari
    $endgroup$
    – greedoid
    Dec 20 '18 at 19:53










  • $begingroup$
    @MohsenShahriari Do you think this solution of Hagen is correct: math.stackexchange.com/questions/2982120/…
    $endgroup$
    – greedoid
    Dec 20 '18 at 20:02
















$begingroup$
$ x = y = 0 $ doesn't give you $ f ( 0 ) = 0 $.
$endgroup$
– Mohsen Shahriari
Dec 20 '18 at 19:05




$begingroup$
$ x = y = 0 $ doesn't give you $ f ( 0 ) = 0 $.
$endgroup$
– Mohsen Shahriari
Dec 20 '18 at 19:05












$begingroup$
Fixed @MohsenShahriari
$endgroup$
– greedoid
Dec 20 '18 at 19:53




$begingroup$
Fixed @MohsenShahriari
$endgroup$
– greedoid
Dec 20 '18 at 19:53












$begingroup$
@MohsenShahriari Do you think this solution of Hagen is correct: math.stackexchange.com/questions/2982120/…
$endgroup$
– greedoid
Dec 20 '18 at 20:02




$begingroup$
@MohsenShahriari Do you think this solution of Hagen is correct: math.stackexchange.com/questions/2982120/…
$endgroup$
– greedoid
Dec 20 '18 at 20:02


















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