$f(x)=x+c(c\in R)$

A proof would be nice. Don't just post answers without proofs.

It's so long: if $\forall x\in R,f(x)=x$,then $f(x)=x$,it's obvious that this one satisfied. or there is $a\in R,f(a)\neq a,f(a)-a=c\neq0$ WLOG,suppose $c>0$,since $f(x)+g(x)=2x,g(a)=a-c,f(a)=a+c$ $f(a-c)=a,g(a+c)=a$. $g(a-c)=2(a-c)-a=a-2c,f(a+c)=a+2c$ By induction:$g(a+nc)=a+(n-a)c,f(a+nc)=a+(n+1)c,n\in Z$ If there exists $b\in R,f(b)-b=d\neq c$,then there exists$n\in Z,a+nc<b<a+(n+1)c$. Similar,$f(b+md)=b+(m+1)d$,

Since $f(x)$ is strictly increasing. $f(a+nc)<f(b)<f(a+(n+1)c),a+(n+1)c<f(b)<a+(n+2)c$ So $d=f(b)-b\in(0,2c)$. If $d<c$,there exists $m\in Z$ such that $\{\frac{b+md-a}{c}\}<1-\frac{d}{c}$ $c\{\frac{b+md-a}{c}\}<c-d,l=[\frac{b+md-a}{c}]$ $a+lc\leq b+md<b+(m+1)d<a+(l+1)c$ $a+lc\leq b+md$ but $f(b+md)<f(a+nc)$,A contradiction. If $d>c$,Similar there exist $l,m\in Z$,$b+md\leq a+lc<a+(l+1)c<b+(m+1)d$,a contradiction. So for all $b\in R$,$f(b)-b=c$,$f(x)=x+c(c\in R)$.

Now we prove there exists $m\in Z,\{\frac{b+md-a}{c}\}<1-\frac{d}{c}(1)$ Just let $t=\{\frac{b-a}{c}\},s=\frac{d}{c},(1)\Leftrightarrow\{t+ms\}<1-s(0\leq t<1,0<s<1)$ For $\{t+(m+1)s\}-\{t+ms\}=s \mbox{or} 1-s$ If $s\leq\frac{1}{2}$, (1)$t<1-s$,then let $m=0$; (2)$t\geq1-s$,let m be the smallest one such that $t+ms\geq1$, then $t+(m-1)s<1,\{t+ms\}=t+ms-1<s<1-s$. If $s>\frac{1}{2}$ $\{t+ms\}=\{t-m(1-s)\}$ Just the same with $s\geq\frac{1}{2}$,replace $s$ by $1-s$.

See exercice 35,But its in french!http://www.animath.fr/cours/eqfonc.pdf

A start...

The Zuton Force wrote: A start... (is it provable from the given information that $y$ is differentiable?) You cannot assume that! Your proof doesn't work.

I remember writing this argument (or a very similar one) on the forum before, but since I made the effort of finding it again, I'm posting it . First we prove that $f$ must be continuous: Suppose that for some real $x$ we have $\lim_{x+}f>f(x)\ (*)$. For every $t>0$ we have $f(x+t)+f^{-1}(x+t)=2x+2t$; taking limits in this expression as $t\to 0$ and using $(*)$ and the fact that $f(x)+f^{-1}(x)=2x$, we get $\lim_{x+}f^{-1}<f^{-1}(x)$, contradicting the fact that $f^{-1}$ is increasing. In the same way we show that the left limit of $f$ in $x$ is $f(x)$, and we're done with the continuity. Let $g(x)=f(x)-x$. From the functional equation we find $g(x+mg(x))=g(x),\ \forall m\in\mathbb Z,\ \forall x\in\mathbb R$. Clearly, we can always find $x\ne y$ with $g(x)=g(y)$ (if $g(x)=0,\ \forall x$ there's nothing to prove, and otherwise take $x$ with $g(x)\ne 0$ and $y=x+g(x)$). Suppose WLOG that $x<y$. The function $u: \left(-\infty,\frac{x+y}2\right)\times\left(\frac{x+y}2,\infty\right)$ defined by $u(x,y)=\frac{g(x)-g(y)}{x-y}$ is continuous, and takes the value $0$. If $g$ weren't constant, we could find $(x,y)$ in the domain of $u$ with $\left|u(x,y)\right|=\frac1n$ for some positive integer $n$. We may assume WLOG that $u(x,y)=\frac1n$. We then have $x-ng(x)=y-ng(y)$. Apply $g$ to this to get $g(x)=g(y)$, which implies $x=y$, and this contradicts the fact that $(x,y)$ belongs to the domain of $u$, which contains no elements of the form $(x,x)$. The contradiction above shows that $g$ must be constant, and this is what we wanted to prove.

shobber wrote: Determine all functions $ f$ from the reals to the reals for which (1) $ f(x)$ is strictly increasing and (2) $ f(x) + g(x) = 2x$ for all real $ x$, where $ g(x)$ is the composition inverse function to $ f(x)$. (Note: $ f$ and $ g$ are said to be composition inverses if $ f(g(x)) = x$ and $ g(f(x)) = x$ for all real $ x$.) The Putnam problem book suggests trying linear recursion for this problem. However, I can't seem to incorporate the first condition! What's wrong in the following?

Ok, for some constant $x$ we have A constant but this doesn't solve the problem

shobber wrote: Determine all functions $f$ from the reals to the reals for which (1) $f(x)$ is strictly increasing and (2) $f(x) + g(x) = 2x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$. (Note: $f$ and $g$ are said to be composition inverses if $f(g(x)) = x$ and $g(f(x)) = x$ for all real $x$.) Denote $f_0(x)=x$ and $f_n(x)=f\big(f_{n-1}(x)\big),$ the same for $g_n(x).$ Replacing $x$ by $f(x)$ in $(2),$ we get \[f_2(x)=2\cdot f(x)-x,\quad \forall x \in \mathbb R\] and so, by induction, we can prove that \[f_n(x)=2\cdot f_{n-1} (x) -f_{n-2}(x),\quad \forall n \ge 2.\] It follows that \[f_n(x)-f_{n-1}(x)=f_{n-1}(x)-f_{n-2}(x)=\cdots = f(x)-x,\] from which we can easily deduce that \[f_n(x)=x+n\big[f(x)-x\big] ,\quad \forall n \in \mathbb N.\quad (3)\] Since $f$ and $g$ has the same property, we can also prove that \[g_n(x)=x+n\big[g(x)-x\big]=x-n\big[f(x)-x\big],\quad \forall n \in \mathbb N.\quad (4)\] Since $f$ is an inceasing bijective, it is clear that $f_n(x)$ and $g_n(x)$ are also increasing. For $x >y ,$ we have $f_n(x) >f_n(y)$ and $g_n(x)>g_n(y).$ Therefore, \[x+n\big[f(x)-x\big] >y+n\big[ f(y)-y\big]\Leftrightarrow n\Big\{ \big[f(x)-x\big] -\big[f(y)-y\big]\Big\}>y-x,\quad (5)\] and \[x-n\big[ f(x)-x\big] >y-n\big[f(y)-y\big]\Leftrightarrow n\Big\{\big[f(x)-x\big]-\big[f(y)-y\big]\Big\}<x-y.\quad (6)\] If $f(x)-x>f(y)-y,$ we may take $n \to +\infty$ in $(6)$ to get a contradiction. If $f(x)-x<f(y)-y,$ we may take $n \to +\infty$ in $(5)$ to get a contradiction. Therefore, we must have $f(x)-x=f(y)-y,$ or $f(x)=x+c,\,\forall x \in \mathbb R.$

Claim: $f(x)=x+c \forall x,c\in \mathbb{R},$ which fits. Proof: Assume the set $S_c$ is non empty and let $\forall x \in S_c: f(x)=x+c.$ Now, $f(x)=x+c\implies g(x+c)=x \implies f(x+c)=x+2c \implies x+c\in S_c.$ Assume exists a set $S_{c'}: c'<c$. We see both the sets are infinite. If $x\in S_c$ and $x\leq y \leq x+(c-c'),$ then $y\notin S_{c'}.$ Assume $y\in S_{c'}$. From the monotonicity of $f$ we have $y+c'=f(y) \geq f(x)=x+c$ which contradicts our assumption. By induction, $\forall y:x+k(c-c') \leq y \leq x(k+1)(c-c') \notin S_{c'}.$ However this contradicts the properties of the two sets. Similarly, if $c'>c$, switching $c$ and $c'$ gives a contradiction. So we conclude $S_{c'}=\{\varnothing\}$ and $S_c=\mathbb{R}.$ The claim follows. $\blacksquare$