Midnight fun(ctional equations)
by shiningsunnyday, Dec 12, 2016, 4:10 PM
Functional Equations - David Arthur - Canada 2014 wrote:
Find all functions
such that
for all 



Solution
Let
be given assertion. If we fix
and vary
we easily see
is surjective. Sweet. Perhaps we can prove it's injective as well? Let
Taking advantage of symmetry, we have
Therefore,
is bijective
In addition, there exists
such that
Drop in
To finish, put
Checking, we indeed have
and thus the only solution is 





![\[P(a,b) = P(b, a) \Rightarrow a + f(b) = b + f(a) \Rightarrow a =b.\]](http://latex.artofproblemsolving.com/d/1/f/d1f81ef628cccba7e4ca32df12bfbbff55f1480a.png)








Introduction to Functional Equations - Evan Chen / Taiwan IMO Training wrote:
Find all continuous functions
such that for any real
and
![\[f(f(x+y))=f(x)+f(y).\]](//latex.artofproblemsolving.com/7/a/1/7a1c6497ace6ada9c6f036e13cc79db68db305aa.png)



![\[f(f(x+y))=f(x)+f(y).\]](http://latex.artofproblemsolving.com/7/a/1/7a1c6497ace6ada9c6f036e13cc79db68db305aa.png)
Solution
Notice the resemblance to Cauchy's functional equation. Since we're given continuity, we automatically have 
Let
We then have
Plugging this back into the original equation,
We seem ALMOST there, but we would like
as opposed to
Using the idea of "shifting" in Jensen's functional equation, define
and so we have
To finish, we plug in to the original equation: ![\[f(f(x+y)) = f(k(x+y)+c) = k^2(x+y)+ck + c = kx+c+ky+c \Rightarrow k^2(x+y) + ck = k(x+y) + c\]](//latex.artofproblemsolving.com/0/7/e/07e90d22878c88f986dd1b13db5bb2f6700f399a.png)
If
, we may easily check all
work. Otherwise,
and since
can vary,
and we may check only
work in this case. Therefore, the only solutions are 

Let

![\[f(f(x)) = f(x) + c \Rightarrow f(f(x+y)) = f(x+y) + c.\]](http://latex.artofproblemsolving.com/f/e/a/feaacdf9b9bbbe2154772c28390d48212f96f764.png)
![\[f(f(x+y)) = f(x+y) + c = f(x) + f(y).\]](http://latex.artofproblemsolving.com/2/c/7/2c72e743114b6db8825300ad2cc235944787d1b8.png)



![\[f(x+y) + c = f(x) + f(y) \Rightarrow g(x+y) = g(x) + g(y) \Rightarrow g(x) = kx, f(x) = kx+c.\]](http://latex.artofproblemsolving.com/f/0/4/f040f0b0fccf407961a333c292ee0c20b2d444cf.png)
![\[f(f(x+y)) = f(k(x+y)+c) = k^2(x+y)+ck + c = kx+c+ky+c \Rightarrow k^2(x+y) + ck = k(x+y) + c\]](http://latex.artofproblemsolving.com/0/7/e/07e90d22878c88f986dd1b13db5bb2f6700f399a.png)
![\[(x+y)(k^2-k)+c(k-1)=0, (k-1)(k(x+y)+c)=0.\]](http://latex.artofproblemsolving.com/8/1/a/81a7eb5ffdee92cdeb2969f40b16f8852e9254c2.png)







Tidbit
I didn't even solve the second one completely myself. I swear, this fe thing is even harder than a certain part of my body when I see my crush.