Zhero wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = \max(f(x),y) + \min(f(y),x)$. Let $P(x,y)$ be the assertion $f(x+y)=\max(f(x),y)+\min(f(y),x)$ Let $a=f(0)$ Let $x\le a$ and $x\ne 0$ : $P(x,0)$ $\implies$ $f(x)=\max(f(x),0)+x$ and so $f(x)<0$ and $f(x)=x$ So $a\le 0$ and $f(x)=x$ $\forall x\le a$ and $x\ne 0$ If $f(x)<0$ $\forall x>0$, then, choosing $x,y>0$, $P(x,y)$ implies $f(x+y)=y+f(y)=x+f(x)$ and so $f(x)=u-x$ and $f(x+y)=u=u-(x+y)$, impossible. So $\exists b>0$ such that $f(b)\ge 0$ and then $P(b,0)$ $\implies$ $f(b)=f(b)+a$ and so $a=0$ Let $x\ge 0$ : $P(x,0)$ $\implies$ $f(x)=\max(f(x),0)$ and so $f(x)\ge 0$ So $f(0)=0$ and $f(x)=x$ $\forall x\le 0$ and $f(x)\ge 0$ $\forall x\ge 0$ Let then $x\ge 0$ : $P(x,-x)$ $\implies$ $0=f(x)-x$ And so $\boxed{f(x)=x}$ $\forall x$, which indeed is a solution.

My solution is similar to pro's. Let $a=f(0),x\neq 0$. We just consider these two lines; $f(x)=max(a,x)+min(f(x),0)$---(1), $f(x)=min(f(x),0)+max(a,x)$---(2) 1. $f(x)\ge 0,x\ge a$ follows $f(x)=x,f(0)=0$ 2. $f(x)\ge 0, x\le a$ follows $x=0$, contradiction. 3. $f(x)\le 0, x\ge a$ follows $x=0$ again a contradiction. 4. $f(x)\le 0, x\le a$ follows $f(x)=x,f(0)=0$. Then We get $f(x)=x$ for all $x\neq 0$. Then $a=max(x,-x)+min(-x,x)=0$ follows, thus $\boxed{f(x)=x \forall x}$.

Let $P(x,y)$ be the preposition $f(x+y) = \max(f(x),y) + \min(f(y),x)$. Recall that $\max(a,b) + \min(a,b) = a + b$ and $\max(a,b) - min(a,b) = |a-b|$. $P(x,0) + P(0,x)$: $f(x) = x + f(0)$ $P(x,0) - P(0,x)$: $|f(x)| = |f(0) - x|$ Substituting the first into the second and squaring, we get $4xf(0) = 0$. Taking nonzero $x$ gives $f(0) = 0$ and hence $f(x) = x$ for all $x$.

Swapping $x$ and $y$, we have \[\max(f(x),y) + \min(f(y),x) = \max(f(y),x) + \min(f(x), y)\]so \[\max(f(x),y) - \min(f(x),y) = \max(f(y), x) - \min(f(y), x)\]that is, \[|f(x) - y| = |f(y) - x|\]for all $x, y$. Taking $y=f(x)$, we get $f(f(x)) = x$, so $f$ is bijective. Taking $y = 0$, we get $|f(x)| = |f(0) - x|$, so for each $x$, either $f(x) = f(0) - x$ or $f(x) = x - f(0)$. Suppose that $f(a) = f(0) - a$ for some $a$. Taking $y=x=a$ in the original equation, we have $f(2a) = f(a) + a = f(0)$. But $f(2a) \in \{f(0) - 2a, 2a - f(0)\}$, so either $f(0) = f(0) - 2a \implies a=0$, or $f(0) = 2a - f(0) \implies a = f(0)$. Thus, for all $a \not \in \{0, f(0)\}$, we have $f(a) = a - f(0)$. Specifically, if $f(0) \neq 0$, then taking $a=2f(0)$, we have $f(2f(0)) = f(0)$. Since $f$ is injective, we have $2f(0) = 0$, contradicting $f(0) \neq 0$. Thus $f(0) = 0$. It is now easy to conclude $f(x) = x$ for all $x$.

Let $P(x,y)$ be the given assertion. Note that for $a,b\in \mathbb R$, we have $\max(a,b)+\min(a,b)=a+b$. Now\begin{align*}P(x,0)+P(0,x)\implies &2f(x)&&=\left(\max(f(x),0)+\min(f(x),0)\right)+\left(\max(f(0)+x)+\min(f(0)+x)\right)\\ & &&=(f(x)+0)+(f(0)+x)\\ \implies & f(x) &&=f(0)+x.\end{align*}Let's say $f(0)=c$, then $f(x)=x+c\quad \forall x\in\mathbb R$. Now $P(2c,0)$ says $2c+0+c=\max(3c,0)+\min(c,2c)$. If $c>0$, then this becomes $3c=3c+c$, which is bad; if $c<0$, then this gives $3c=0+2c$, which is not good either. So $c=0$ and we have $\boxed{f(x)=x\; \forall x\in\mathbb R}$; which is easily seen to work. $\blacksquare$

This is an old problem posted on AoPS on 2011: https://artofproblemsolving.com/community/c6h406771

This problem is pretty funny. We claim the only solution is $f(x)=x$, and it is easy to see that this works. Suppose $f$ is a solution, and let $P(x,y)$ be the given FE. Note that $P(x,y)+P(y,x)$ implies $2f(x+y)=f(x)+f(y)+x+y$. Setting $y=0$, we see that $f(x)=x+f(0)$. We see $P(f(0),0)$ implies $2f(0)=\max(2f(0),0)+\min(f(0),f(0))$, so $f(0)=\max(2f(0),0)$, so $f(0)=0$. Thus, $f(x)=x$, as desired.

We claim only $f(x)=x$ works. It clearly works. Adding $P(x,y)$ and $P(y,x)$ gives \[ 2f(x+y) = f(x)+f(y)+x+y. \]Let $g(x)=f(x)-x$. Then \[ 2g(x+y)=g(x)+g(y). \]Plugging in $y=0$ gives $2g(x)=g(x)+g(0)$, so $g(x)=c$ for some constant $c$. So $f(x)=x+c$. Plugging back into the original, \[ x+y+c = \max(x+c,y) +\min(y+c,x).\]This must hold for all $x,y$. If $x\ge y+c$, then the RHS is $(x+c)+(y+c) = x+y+2c$, and the LHS is $x+y+c$. So $c=0$. Hence, $f(x)=x$ is the only solution.

The LHS of the problem's equation but the RHS is not, so we can exploit this bump in symmetry to get $\max(f(x), y)+\min(f(y), x)=\max(f(y), x)+\min(f(x), y)$, or equivalently, $\max(f(x), y)-\min(f(x), y)=\max(f(y), x)-\min(f(y), x)$. This is equivalent to $|f(x)-y|=|f(y)-x|$ for all $x, y$. Now, setting $y=0$ we have $|f(x)|=|f(0)-x|$ for all $x$, for each individual $x$, $f(x)$ is $f(0)-x$ or $x-f(0)$. To avoid pointwise, now suppose $f(a)=f(0)-a$, $f(b)=b-f(0)$ for $a, b\ne 0$; then, we have $|f(a)-b|=|f(b)-a|$, so $|f(0)-a-b|=|b-f(0)-a|$, meaning $f(0)-a-b=\pm(b-f(0)-a)$. However in both cases we get $a=0$ or $b=0$, which is a contradiction; thus, either $f(x)=x+c$ for all $x\ne 0$, or $f(x)=-x+c$ for all $x\ne 0$ (where $c=\pm f(0)$). Suppose $f(x)=x+c$ for all $x\ne 0$; then, plugging $(y+|c|, y)$ into the original equation, we get $2y+|c|=2y+c+|c|$ so $c=0$. We can do the same if $f(x)=-x+c$, so therefore $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$.

My solution:

My first ELMO A Solution. The only working function is $f\equiv \text{Id}$. Now we show this is the only one. Let $P(x,y)$ denote the assertion. Now $P(x,x)$ implies $f(2x)=f(x)+x~(\spadesuit)$ We proceed with the following claim as our base. Claim. We have $f(0)=0$ Proof. Assume otherwise. Note that from $P(x,0)$ we have $$f(x)=\max\{f(x),0\}+\min\{f(0),x\}$$If $f(x)\geqslant 0$ then $\min\{f(0),x\}=0$. It follows that $x=0$. Therefore, $f(x)<0~\forall x-\{0\}$. Call this $(\clubsuit)$. Take $x>0$ and $y>0$ then $P$ yields $$f(x+y)=\max\{f(x),y\}+\min\{f(y),x\}\overset{(\clubsuit)}{=}y+f(y)\overset{(\spadesuit)}{=}f(2y)$$Plugging $x=3y$ in last equation gives $$f(y)+3y\overset{(\spadesuit)}{=}f(2y)+2y\overset{(\spadesuit)}{=}f(4y)=f(2y)\overset{(\spadesuit)}{=}f(y)+y$$It follows that $y=0$ which is a contradiction. We conclude $f(0)=0$ as desired. $\square$ Consequently $P(x,0)$ with $x>0$ yields $f(x)=\max\{f(x),0\}$. If $f(x)<0$ then $f(x)=0,$ a contradiction, we conclude $f>0$ for $x>0$. Whereas, $x<0$ in $P(x,0)$ yields $$f(x)-x=\max\{f(x),0\}$$If $f(x)\geqslant 0$ then $x=0,$ contradiction since $x<0$. It follows that $f(x)-x=0$ implying $f(x)=x$ for $x<0$. To finish note by $P(0,y)$ where $y>0$ the following: \begin{align*} f(y)&=\max\{0,y\}+\min\{f(y),0\}\\ &=y+0=y\\ \end{align*}We conclude $f(x)=x$ for all $x>0$. Consequently, it follows $f\equiv\text{Id}$ as desired. $\blacksquare$

Remark: swapping the variable, then sum up the two equation, you get $f(x+y)=f(x)+f(y)+x+y$, then change the function $f(t)-t=g(t)$, etc.

Claim: $f(x)=x$ is the only solution. This works since $x+y=f(x+y)=\max(x,y)+\min(x,y)=x+y$. Let $f(0)=a$. Let $P(x,y)$ be the assertion. Let $x \le a$ and $x \neq 0$. $$P(x,0) \implies f(x)=\max(f(x),0)+\min(a,x)=\max(f(x),0)+x$$If $f(x) \ge 0$, $f(x)=f(x)+x \implies x=0$, contradiction. Now if $f(x)<0$, $f(x)=0+x=x$. This means $f(x)=x$ for $x \le a$. Now let $x > a$ and $x \neq 0$. $$P(x,0) \implies f(x)=\max(f(x),0)+\min(a,x)=\max(f(x),0)+a$$If $f(x) < 0$, $f(x)=0+a=a$. If $f(x) \ge 0$, $f(x)=f(x)+a \implies 0=a$. Assume $a \neq 0$, so $f(x)=a$ for $x>a$ and $f(x)=x$ for $x \le a$. Let $x>a$ and $x>0$. $$P(x,x) \implies f(2x)=f(x)+x \implies \implies a=a+x \implies x=0$$Contradiction. This means $f(0)=0$. This means $f(x)=x$ for $x \le 0$. Let $x>0$. $$P(x,-x) \implies 0=f(0)=\max(f(x),-x)+\min(f(-x),x)=\max(f(x),-x)+\min(-x,x)=\max(f(x),-x)-x$$If $f(x) \le -x$, $0=-x-x=-2x \implies x=0$, contradiction. If $f(x)>0$, $0=f(x)-x \implies f(x)=x$. This means $f(x)=x$ for $x\le 0$ ad $f(x)=x$ for $x>0$, so $f(x)=x$ for all $x$.

Zhero wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = \max(f(x),y) + \min(f(y),x)$. George Xing. Easy for A3 My Different Solution: Claim 1.$f(0)=0$ Proof: $P(f(0),0)\implies f(f(0))=\max(f(f(0)),0)+\max(f(0),f(0))=\max(f(f(0)),0)+f(0)$ $P(0,f(0))\implies f(f(0))=\max(f(0),f(0))+\min(0,f(f(0)))=\min(f(f(0)),0)+f(0)$ Then $\max(f(f(0)),0)=\min(f(f(0)),0)\implies f(f(0))=0$ $P(f(0),0)\implies f(0)=0$ Claim 2.$f(2x)=2f(x)$ Proof: $P(x,x)\implies f(2x)=f(x)+x$ $f(x+y) = \max(f(x),y) + \min(f(y),x)= \max(f(y),x) + \min(f(x),y)$ Then $2f(x+y) = \max(f(x),y) + \min(f(y),x)+ \max(f(y),x) + \min(f(x),y)=f(x)+f(y)+x+y$ $2f(x+y)=f(x)+x+f(y)+y=f(2x)+f(2y)$ $y=0\implies 2f(x)=f(2x)$ Finish: $P(x,x)\implies f(2x)=f(x)+x$ Claim 1$\to f(2x)=2f(x)$ Then $f(x)+x=2f(x)\implies f(x)=x$ Answer: $\boxed{f(x)=x}$ $\forall x$

Zhero wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = \max(f(x),y) + \min(f(y),x)$. George Xing. $P(x, 0) + P(0, x) \implies f(x) = x + f(0)$ and $P(x, -f(0)) \implies f(x - f(0)) $ $= x = $ $\mathrm{max}(x + f(0), -f(0))$ $ +$ $ \mathrm{min}(f(-f(0)), x)$, we choose sufficiently large enough $x$ and get that $x = x + f(0) + f(-f(0)) \implies f(0) = -f(-f(0)) = 0$ and so $f(0) = 0$ implying that $\boxed{f(x)=x}$ for all reals $x$

Let $P(x,y)$ be the assertion $f(x+y)=\max\{f(x),y\}+\min\{f(y),x\}$. $P(x,f(x))\Rightarrow f(x+f(x))=f(x)+\min\{f(f(x)),x\}$ $P(f(x),x)\Rightarrow f(x+f(x))=\max\{f(f(x)),x\}+f(x)$ So $f(f(x))=x$ and $f$ is bijective. Then $P(x,f(x))$ becomes $f(x+f(x))=x+f(x)$, while $P(x,x)\Rightarrow f(2x)=f(x)+x$, so $f(2x)=f(x+f(x))$, by injectivity, $\boxed{f(x)=x}$, which works.

The answer is $f(x)=x$ only, which obviously works. Let $P(x,y)$ be the assertion. Comparing $P(x,y)$ with $P(y,x)$, we obtain $$\max(f(x),y)-\min(f(x),y)=\max(f(y),x)-\min(f(y),x).$$Putting in $y=f(x)$ and noting $\max(a,b)-\min(a,b)=|a-b|$, we have $$|f(f(x))-x|=0 \implies f(f(x))=x.$$Thus let $f(0)=a$, so $f(a)=0$. Then from $P(a,0)$ we have $$f(a)=\max(0,0)+\min(a,a)=a,$$hence $a=0$ and $f(0)=0$. Now pick some $x>0$. From $P(0,x)$ we obtain $$f(x)=\max(0,x)+\min(f(x),0)=x+\min(f(x),0)$$If $f(x)<0$, then $\min(f(x),0)=0$ and we obtain $f(x)=x+f(x) \implies x=0$, which is absurd. Hence $f(x)>0$, so $f(x)=x$. Similarly, for $x<0$, we can consider $P(x,0)$ and obtain the same result. This gives $f(x)=x$ for all $x$, as desired. $\blacksquare$

Dumb solution with dumb casewhack but I think it still works Let $P(x,y)$ be the assertion that $f(x+y)= max (f(x),y) + min(f(y),x)$ $$P(x,x) : f(2x)= max (f(x),x) + min (f(x),x)$$$$f(2x)=f(x)+x $$ $$P(x,0) : f(x) = max(f(x),0)+ min (f(0),x)$$ $ f(x) = f(x) + f(0) -(1)$ or $f(x)= f(x) +x -(2)$ or $f(x)=f(0)+0 - (3)$ or $f(x)= x+ 0 -(4)$ Clearly, $(2)$ fails, and $(3)$ fails upon inspection $(4)$ gives us that $f(x)=x$, which upon checking works. Hence, let us now only consider $(1)$ For $(1)$, not that we just get $f(0)=0$ and that since we chose $f(x)$ over $0$, $$f(x)>0$$if $x>0$ $P(x,-x): f(0)= max (f(x), -x) + min (f(-x), x)$ $0= f(x)+ f(-x) -(1)$ or $0= f(x)+x -(2)$ or $0= -x + f(-x) -(3)$ or $0= -x+x -(4)$ Obviously, $(4), (3), (2),$ fail after inspection due to what we have established earlier. So, we are left with $$f(x)=-f(-x)$$$P(x,-2x): f(-x)= max( f(x),-2x) + min (f(-2x),x) = f(x) + (-f(x))-x$ ( this is assuming )$x>0$ So we get that $f(-x)=-x$ $$f(x)=x$$ Hence, the only solution is $f(x)=x$ for all $x$.

Zhero wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that $f(x+y) = \max(f(x),y) + \min(f(y),x)$. George Xing. Cute

Neat. The answer is $\boxed{f(x)=x}$. It is easy to see that this holds. Claim: $f$ is linear with slope $1$ Proof: Substitute $(x,0)$ and $(0,x)$ into the equation and add. We get that $f(x) = x + f(0)$. To finish, I claim that $f(0) = 0$. Suppose the contrary, and set $f(0) = c$, where $c \neq 0$. Pick an arbitrary $n$ such that $c > 2n > -c$, and then substitute $(n,-n)$. We get $c = \max{(f(n),-n)} + \min{(f(-n),n)} = f(n) + f(-n) = 2c$. Hence $c=0$, contradiction. $\blacksquare$

srisainandan6 wrote: $f(2x)-f(x) = x$. Hence $f(x) = x + f(0)$ Why?

uwu this is amazing As usual let $P(x,y)$ denote the given assertion. $P(x,f(x)) - P(f(x),x) \implies$ \begin{align*} 0&= \max \{f(x) , f(x)\} + \min \{f(f(x)) , x\} - \max \{f(f(x)) , x\} - \min \{f(x) , f(x)\}\\ &= \min \{f(f(x)) , x\} - \max \{f(f(x)) , x\} \end{align*}So $f(f(x)) = x$. $P(x,0) + P(0,x) \implies$ \begin{align*} 2f(x) &= \max\{f(x),0\} + \min\{f(0),x\} + \max\{f(0),x\} + \min\{f(x),0\} \\ &= f(x) + 0 + f(0) + x \end{align*}So $f(x) = x + f(0)$ and using $f(f(x)) = x$ we get that $f(x)=x$ which is the only solution.

The only solution is $\boxed{f(x)=x}$, which clearly works. Let $P(x,y)$ denote the assertion that $f(x+y) = \max(f(x),y) + \min(f(y),x)$. $P(x,f(x)): f(x+f(x))=f(x)+\min(f(f(x)),x)$ $P(f(x),x): f(x+f(x))=\max(f(f(x)),x)+f(x)$. Subtracting the two equations gives that $f(f(x))=x$. Let $f(0)=k$. So $f(k)=0$. Claim: $k=0$. $P(k,-k): k=\max(0,-k)+\min(f(-k),k)$. $P(-k,k): k=\max(f(-k),k)+\min(0,-k)$. Suppose $k\ge0$. Then $k=\min(f(-k),k)=\max(f(-k),k)-k$, so $f(-k)=2k$. $P(-k,0): 2k=2k+\min(k,-k)$, so either $-k$ or $k$ are zero, so $k=0$. Suppose $k\le0$. Then $k=-k+\min(f(-k),k)=\max(f(-k),k)$, so $\min(f(-k),k)=2k$, but if $k\ne2k$, then this is not possible. So $k=0$. $P(x,0): f(x)=\max(f(x),0)+\min(0,x)$. $P(0,x): f(x)=\max(0,x)+\min(f(x),0)$. Case 1: $x>0$. So $f(x)=\max(f(x),0)=x+\min(f(x),0)$. So $\max(f(x),0)=f(x)$, which implies $f(x)\ge0$, so $f(x)=x$. Case 2: $x<0$. So $f(x)=\max(f(x),0)+x=\min(f(x),0)$. Thus, $f(x)\le0$, which implies $f(x)=x$. Case 3: $x=0$. We know that $f(0)=0$. Thus, we have shown that $f(x)=x$ is the only solution.

Let $P(x,y)$ denote the given assertion. By combining $P(x,0)$ and $P(0,x)$ we have $f(x)=x+f(0).$ Then $P(f(0),0)$ implies $f(0)=0.$ The identity function clearly works.

A bit different solution than all the other ones: Denote $P(x, y)$ as usual. Comparing $P(x, f(x))$ and $P(f(x), x)$, we obtain $max(x, f(f(x)))$ = $min(x, f(f(x)))$, so $f(f(x)) = x$. Knowing this, $P(x, f(x))$ gives $f(x+f(x))=x+f(x))$. However, by $P(x, x)$, $f(2x) = x + f(x)$ ($max(a, b) + min(a, b) = a+b$), so: $x+f(x)=f(x+f(x))=f(f(2x))=2x$, and therefore $f(x)=x$. This clearly satisfies the equation.

MathGuy1729 wrote: A bit different solution than all the other ones: Really? jasperE3 wrote: Let $P(x,y)$ be the assertion $f(x+y)=\max\{f(x),y\}+\min\{f(y),x\}$. $P(x,f(x))\Rightarrow f(x+f(x))=f(x)+\min\{f(f(x)),x\}$ $P(f(x),x)\Rightarrow f(x+f(x))=\max\{f(f(x)),x\}+f(x)$ So $f(f(x))=x$ and $f$ is bijective. Then $P(x,f(x))$ becomes $f(x+f(x))=x+f(x)$, while $P(x,x)\Rightarrow f(2x)=f(x)+x$, so $f(2x)=f(x+f(x))$, by injectivity, $\boxed{f(x)=x}$, which works.

Time to complete the problem which I didn't solve at because I was too scared. Here is a long solution? We refer $(1)$ as the original Functional Equation. Solution: We will show that $f \equiv x$ is the only possible solution. It is easy to check it works. Let $P(x,y)$ denote the assertion to the Functional Equation. We now have the following claim. Claim: $f(0) = 0$. Proof: Just put $y = 0$. One would get \[f(x) = \max(f(x), 0) + \min(f(0),x).\]If there exists some $x$ such that $f(x)>0$ then we can easily conclude $f(0) = 0$. But if $f(x) \le 0$ always, then we get $f(x) = \min(f(0), x)$. Fixing $y$ in $(1)$ and varying $x$ to be large enough such that $x+y < f(0)$ would show that this is not a solution. $\square$ Let $\alpha$ be some negative real. Again from $P(\alpha,0)$ we have that \[f(\alpha) = \max(f(\alpha), 0) + \alpha\]If $f(\alpha)<0$, then we get $f(x) = x$ for all negative $x$. If not, we would get a contradiction. To conclude the problem, set $x>0$ and $y<0$ in $(1)$ such that $x+y<0$. One would eventually get that \[\max(f(x), y) = x>0\]As $y<0$, clearly $\max(f(x),y) = f(x) = x$ and the solution is complete. $\blacksquare$

The answer is $f(x)=x$. It is easy to check that this works. Plugging in $x=a$ and $y=0$ gives \[f(a+0) = \max (f(a),0) + \min (f(0),a). \]Plugging in $x=0$ and $y=a$ gives \[f(0+a) = \max (f(0),a) + \min (f(a),0).\]Summing these two equations gives \begin{align*} f(a+0)+f(0+a) &= \left( \max (f(a),0) + \min (f(0),a) \right) + \left( \max (f(0),a) + \min (f(a),0) \right) \\ 2f(a) &= \left( \max(f(a),0) + \min(f(a),0) \right) + \left( \max(f(0),a) + \min(f(0),a) \right) \\ 2f(a) &= f(a)+f(0) + a \\ f(a) &= a+f(0). \end{align*}Now, let $x=1$ and $y=2$. We have \begin{align*} f(1+2) &= \max (f(2), 1) + \min (f(1),2) \\ 3+f(0) &= \max (2 + f(0), 1) + \min (1 + f(0), 2). \end{align*}Solving this equation for $f(0)$ gives $f(0) = 0$. Therefore, the only possible solution is $f(x) = x$.

Let $P(x,y)$ be the assertion. Add $P(x,y), P(y,x)$ to get $$2f(x+y)=f(x)+f(y)+x+y,$$and plug in $y=0$ to get $f(x)=x+f(0)$. Plugging in $x+c$ into the original FE, we have \[ x+y+c = \max(x+c,y) +\min(y+c,x).\]Since this must hold for all $x,y$, take the case where $x\geq y+c$, which gives $c=0$. The only solution is $f(x)=x$.

We claim that the answer is $f(x)=x$ which clearly works. Since we have, $f(x + y) = \max(f(x), y) + \min(f(y), x),$ we can swap the variables to get $$f(x+y)=\max(f(y), x) + \min(f(x), y)$$Adding these, we have $$2f(x+y)=\min(f(y),x)+\max(f(y),x)+\min(f(x),y)+\max(f(x),y)$$$$2f(x+y)=x+y+f(x)+f(y).$$Letting $y=0$ yields $f(x)=x+f(0),$ so $f(x)=x+c$ for some constant $c$. However, if we look back at $$x+y+c = \max(x+c, y) + \min(y+c, x),$$for sufficiently large $x$ we have $$x+y+c=x+c+y+c,$$contradiction for $c\neq 0$, so $c=0$ are we are done.

We have that $f(x+f(x))=f(f(x)+x)=f(x)+\min{(f(f(x)),x}=f(x))+\max{(f(f(x)),x)}$. From this we conclude that $f(f(x))=x$ and $f(x+f(x))=x+f(x)$. Plugging in $0$ for $x$, we can also conclude that $f(0)=0$. Therefore $f(x+0)=f(0+x)=\max{(0,f(x))}+\min{(0,x)}=\max{(0,x)}+\min{(0,f(x))}$. Taking this in cases(on whether $f(x)\geq{}0$ or not, etc), we get two final solutions of $f(x)=x$ or $f(x)=-x$. Since $f(x+f(x))=x+f(x)$(from earlier), we know that $f(x)=-x$ is an extraneous solution, so $f(x)=x$ is our only working solution. Plugging this back into the original equation shows that this does indeed work, and we are done.

By symmetry we may conclude $$\max(f(x), y) + \min(f(y), x) = f(x+y) = \min(f(x), y) + \max(f(y), x).$$As a result, $$f(x) + f(y) + x+y =2 f(x+y).$$Setting $y=0$ here yields $f(x) + x + c = 2f(x)$, so $f$ is linear. Checking, $f(x) = x$ is the only solution.

We claim the answer is $f(x)=x$ which works. Let $P(x,y)$ denote the given assertion. Notice $P(x,x)$ gives $f(2x)=f(x)+x$. Swapping $x$ and $y$ and adding this to our original equation, we find \[2f(x+y)=f(x)+x+f(y)+y=f(2x)+f(2y)\]so $f(x)$ satisfies Jensen's functional equation so $g(x)=f(x)-f(0)$ satisfies Cauchy's functional equation. From $P(x,0)$, we see \[f(x)=\max(f(x),0)+\min(f(0),x)\]so for $x>0$ we have $f(x)\ge 0+\min(f(0),0)$ so $f$ is bounded below on $(0,\infty)$. Therefore, $g$ is also bounded on this interval so $g$ is linear. Thus, $f(x)=kx+c$ and plugging into $f(2x)=f(x)+x$ we see $k=1$ and $c=0$. $\square$

Solution: We have four cases: $f(x) \le y, f(y) \ge x$: $f(x+y) = x+y \Rightarrow \boxed{f(x)=x}$ seems like a solution, which works for all reals. $f(x) \le y, f(y) \le x$: $f(x+y) = y + f(y)$, which means that $f$ must be constant. So, $f \equiv 0$, but this doesn't satisfy the FE for all reals. $f(x) \ge y, f(y) \ge x$: $f(x+y) = x + f(x)$, which means that $f$ must be constant. So, $f \equiv 0$, but this doesn't satisfy the FE for all reals. $f(x) \ge y, f(y) \le x$: $f(x+y) = f(x)+f(y)$. According to our initial condition, $f$ can be shown to be bounded above $f \equiv y_0$ over $(y_0, y_0 + 1)$, which means $f \equiv cx$, for some real number $c$. To check this, we can plug in $(x, x)$ into our original FE to get $f(2x) = f(x) + x \Rightarrow 2cx = cx + x = (c+1)x \Rightarrow 2c=c+1 \Rightarrow c=1$, so the only form of $cx$ that works is $\boxed{f(x) = x}$, which we already established as a solution that works. In conclusion, the only solution to the FE is $\boxed{f(x)=x}$.

First note 0,0 yields f(0)=2f(0) or 0, which either way yields f(0)=0. Using the symmetry, we can swap x and y in that equation and sum to get $$f(x)+f(y)+x+y=2f(x+y)\implies^{y=0}f(x)+f(0)+x=2f(x)\implies f(x)=x,$$as desired. $\blacksquare$

huashiliao2020 wrote: First note 0,0 yields f(0)=2f(0) or 0, which either way yields f(0)=0. No. $0,0$ yields $f(0)=\max(f(0),0)+\min(f(0),0)=f(0)+0=f(0)$. Nothing more.

pco wrote: huashiliao2020 wrote: First note 0,0 yields f(0)=2f(0) or 0, which either way yields f(0)=0. No. $0,0$ yields $f(0)=\max(f(0),0)+\min(f(0),0)=f(0)+0=f(0)$. Nothing more. Sorry, my mistake. It's easily fixed with f(x)=x+f(0) and checking all solutions of the form x+c finishes. (For example, comparing "c"s implies that we either only have f(x)>y and f(y)>x or y>f(x) and x>f(y). WLOG x>y; if we let c be positive, y>f(x)=x+c is impossible, so in this case we must ALWAYS have the first condition satisfied if c is nonzero, upon which choosing sufficiently large x>y+c implies f(y)=y+c>x, a contradiction. The analogous case c being negative follows.)

Claim: $f(x)$ is linear. Proof: Note that $f(x+y)=min(y,f(x))+max(x,f(y)),$ by symmetry. So, $2f(x+y)=max(y,f(x))+min(x,f(y))+min(y,f(x))+max(x,f(y))=f(x)+f(y)+x+y,$ now take the assertion $P(x,y)=P(x,0),$ then $f(x)+x+f(0)=2f(x) \implies f(x)=x+f(0).$ If we check the linear functions, $\boxed{f(x)=x},$ is the only solution $\blacksquare$

Plugging in $(x,y)$ and $(y,x)$ and comparing gives \[\max(f(x),y) + \min(f(y),x) = \max(f(y),x) + \min(f(x), y)\]\[\max(f(x),y) - \min(f(x),y) = \max(f(y), x) - \min(f(y), x)\] Hence, \[|f(x) - y| = |f(y) - x|\] Setting $y=f(x)$ gives us $f(f(x))=x$, so $f$ is bijective. Setting $y=0$ gives us $|f(x)|=|f(0)-f(x)|$ so for all $x$, we have either $f(x) = f(0) - x$ or $f(x) = x - f(0)$. Let $K$ be the set of all values $x$ such that $f(x) = f(0) - x$. Select a value $k \in K$ and set $(x,y)=(k,k)$ in the original expression. This gives us $f(2k)=f(k)+k=f(0)$. This means we have \[f(0) = f(0) - 2k \text{ or } f(0) = 2k - f(0)\] so $k=0, f(0)$ and these are the only elements in $K$. Thus, for all $a \notin K$, we have $f(a) = a - f(0)$. Specifically, if $f(0) \neq 0$, then setting $a=2f(0)$, we have $f(2f(0)) = f(0)$. Since $f$ is bijective, we must have $2f(0) = 0$, contradicting $f(0) \neq 0$. This means $f(0)=0$, making it easy to conclude $\boxed{f(x) = x}$ for all $x$.

Exploiting the asymmetry of this equation, we can get \begin{align*} 2f(x+y) &= f(x+y) + f(y+x) \\ &= \left(\max(f(x),y) + \min(f(x), y)\right) + \left(\max(f(y),x) + \min(f(y),x)\right) \\ &= f(x)+f(y)+x+y \end{align*} Substituting $g(x)=f(x)-x$, we can rewrite this equation as \[2g(x+y)=g(x)+g(y).\] Note if we substitute $(k,1)$, we find $g(k)=g(0)$ for all $k$. Thus $g(x)$ is a constant function, so $f(x)=x+c$. Testing, we see that we can only have $\boxed{f(x)=x}$. $\blacksquare$

$f(x)=x$ works. Firstly note that $\max \left\{a,b\right\} + \min \left\{a,b\right\} = a+b$. $P(x,0) \implies f(x) = \max \left\{f(x),0\right\} + \min \left\{f(0),x\right\}$. $P(0,x) \implies f(x) = \max \left\{f(0),x\right\}+\min \left\{f(x),0\right\}$. Adding both together, we get $f(x) = x+f(0) = x+c$. Now we have, \begin{align*} &f(x+y) = \max \left\{f(x),y\right\}+\min \left\{f(y),x\right\} \\ \implies &x+y+c = \max \left\{x+c,y\right\}+\min \left\{y+c,x\right\} \\ \implies &x+y=\max \left\{x+c,y\right\}+[\min \left\{y+c,x\right\}-c]\\ \implies &x+y = \max \left\{x+c,y\right\} + \min \left\{y,x-c\right\} .\end{align*} Now if $c > 0$, pick $y\in (x-c,x+c)$, $y\neq x$. Then $x+y = x+c + x-c = 2x \implies y=x$, contradiction. If $c<0$, pick $y\in (x+c,x-c)$, $y\neq x$. Then $x+y = 2y \implies y=x$, contradiction. Therefore, we must have that $c=0$ and we are done.

We claim that $f(x) = x$ which clearly works. Swap $x$ and $y$ to yield \[\max(f(x), y) + \min(f(y), x) = \min(f(x), y) + \max(f(y), x)\]which rearranges to \[|f(x) - y| = |f(y) - x|\]using the identity \[\max(a, b) - \min(a, b) = |a - b|\text{.}\]From this we can substitute $y = f(x)$ to yield \[f(f(x)) = x\text{.}\]In particular, this implies that $f$ is injective. Now simply take $y = f(x)$ and $y = x$ into the original equation, which yields \[f(f(x) + x) = f(x) + x\]and \[f(2x) = f(x) + x\text{.}\]Thus, \begin{align*} f(f(x) + x) &= f(2x) \\ f(x) + x &= 2x \\ f(x) &= x \end{align*}as desired.

Let $P(x,y)$ denote the assertion. $P(x,0)$ gives us $f(x)=\min(f(0),x)+\max(f(x), 0)$. If $x\geq f(0)$, then we must have $\max(f(x),0)=f(x)-f(0)$, so we either have $0\leq f(x)=f(x)-f(0)$ or $f(x)\leq 0 = f(x)-f(0)$. In the first case, we get $f(0)=0$ and in the second case, we get $f(0)\leq 0$. Either way, we have $f(0)\leq 0$. If $x\leq f(0)$, then we must have $\max(f(x),0)=f(x)-x$, so we either have $0\leq f(x)=f(x)-x$ or $f(x)\leq 0=f(x)-x$. In the first case, we get $x=0$, so $f(0)\geq 0$, and in the second case, we get $f(x)=x$, so $f(0)=0$. In either case, we get $f(0)\geq 0$. Therefore, since $f(0)\leq 0$ as well as $f(0)\geq 0$, we have $f(0)=0$. So, $P(x,0)$ becomes $f(x)=\min(0,x)+\max(f(x),0)$. This tells us that if $x>0$, then we must have $f(x)\geq 0$, and otherwise having $x<0$ implies $f(x)=x$. Now, using $P(x,-x)$, we get $0=f(x+(-x))=\max(f(x),-x)+\min(f(-x),x)$. For $x>0$, since $f(x)\geq 0$ and $f(-x)=-x$, we have $0=f(x)+(-x)$, giving $f(x)=x$. Therefore, the answer is $f(x)=x$ for all $x$, and it is easy to check that this works.