Real Mode

If f(a+b) = f(a) + f(b) + 2ab, what is f?

December 1, 2006 iconjack Leave a comment

OK, I think I have a proof of the following:
If $f(a+b) = f(a) + f(b) + 2ab$ , then $f(x) = x^2 + mx$ for some real m.

Proof:
Setting $a=0$ and $b=0$ , we have $f(0+0) = f(0) + f(0) + 0$ ;
in other words $f(0) = 0$ .
Next, set $a=x$ and $b=-x$ , which yields $f(x-x) = f(x) + f(-x) - 2x^2$ .
Since $f(x-x) = f(0) = 0$ , we have $f(x) + f(-x) = 2x^2$ , or
$f(x) - x^2 = -[f(-x) - (-x)^2]$ .
Let $g(x) = f(x) - x^2$ (the left side)
Then we see that $f(x) = x^2 + g(x)$ and $g$ is an odd function, i.e. $g(x) = -g(-x)$ .
Now,
$f(a+b) = (a+b)^2 + g(a+b) = a^2 + b^2 + 2ab + g(a+b)$
but also
$f(a+b) = f(a) + f(b) + 2ab$
Setting equal the right sides of the above two equations and simplifying,
we get
$g(a+b) = g(a) + g(b)$ ,
which is the definition of linear function with $g(0) = 0$ . (y-intercept = 0)