Also this is my soln: 1. Proof of bijectivity: Equation equivalent to f(c) = c or f(c) = ff(c-f(y))+y for all real y. If always first case then obviously bijective. If second case then by varying y it is clearly surjective. Also if f(a) = f(b) then ff(c-f(a))+a=ff(c-f(b))+b, so a=b, which proves injectivity. 2. Case where no fixed points (so second equation always holds): ff(y)=ff(0)+y. Let t = ff(0) for convenience plugging in f(y) for y gives: f(x+y+t) = x+f(y) +t Setting a as x+t: f(y+a) = f(y)+a for any a. Then clearly follows f(x) = x + constant. 3. Case where at least one fixed point and f(f(0)) is not zero: i. Proof that if c not fixed point but k fixed point then c-k fixed point: Assume k is a fixed point but c-k and c are not fixed points. Then, plugging in y = f-1(c-k) gives: f(c)=f-1(c-k) + k Also f(c) = ff(c-f(k))+k = ff(c-k)+k, so: f-1(c-k) = ff(c-k) so f^6(c-k) = c-k However, we know that for non fixed points c-k, ff(c-k) = c-k + ff(0), meaning f^6(c-k) = c-k + 3ff(0), contradiction However, given x,y which are fixed points, we can see that f(x+y) = x+y, so f(c-k+k) = c-k+k, contradicting the fact that c is not a fixed point, so f(x) = x for every x. 4. Case where ff(0)=0: Then ff(x)=x for every x So f(x+0) = x or x+f(0) for every x So f(f(0)) = f(0) or 2f(0) either way, f(0)=0, so f(x) = x for every x
Also this is my soln:
1. Proof of bijectivity:
Equation equivalent to f(c) = c or f(c) = ff(c-f(y))+y for all real y. If always first case then obviously bijective. If second case then by varying y it is clearly surjective. Also if f(a) = f(b) then ff(c-f(a))+a=ff(c-f(b))+b, so a=b, which proves injectivity.
2. Case where no fixed points (so second equation always holds):
ff(y)=ff(0)+y. Let t = ff(0) for convenience
plugging in f(y) for y gives:
f(x+y+t) = x+f(y) +t
Setting a as x+t:
f(y+a) = f(y)+a for any a. Then clearly follows f(x) = x + constant.
3. Case where at least one fixed point and f(f(0)) is not zero:
i. Proof that if c not fixed point but k fixed point then c-k fixed point:
Assume k is a fixed point but c-k and c are not fixed points. Then, plugging in y = f-1(c-k) gives:
f(c)=f-1(c-k) + k
Also f(c) = ff(c-f(k))+k = ff(c-k)+k, so:
f-1(c-k) = ff(c-k)
so f^6(c-k) = c-k
However, we know that for non fixed points c-k, ff(c-k) = c-k + ff(0), meaning f^6(c-k) = c-k + 3ff(0), contradiction
However, given x,y which are fixed points, we can see that f(x+y) = x+y, so f(c-k+k) = c-k+k, contradicting the fact that c is not a fixed point, so f(x) = x for every x.
4. Case where ff(0)=0:
Then ff(x)=x for every x
So f(x+0) = x or x+f(0) for every x
So f(f(0)) = f(0) or 2f(0)
either way, f(0)=0, so f(x) = x for every x
Thanks for this cool solution, I like it. Especially your start with the bijectivity is more elegant than mine.
Flashbacks to aquesulian
Relatable :D