The functional equation $f(x + y) = f(x) + f(y) + xy$

Question

Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy the functional equation $$f(x + y) = f(x) + f(y) + xy\text,$$ for all $x,y \in \mathbb{R}$.

I need hints for this problem.

Answer 1

Since $f(0) = 0$ and the "inhomogeneous term" $xy$ looks like a polynomial, I suggest you try the ansatz $f(x) = a_{1}x + a_{2}x^{2}$.

To prove uniqueness, note that if you assume that $f'(0)$ exists, then \begin{align} f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{f(h)}{h} + x = f'(0) + x \end{align} so all differentiable solutions must be of the for $\frac{x^2}{2} + kx$.

Answer 2

Suppose $x=n\in\mathbb{N}$ and $y=1.$ Then $$f(n+1)-f(n)=f(1)+n.$$ Therefore by telescoping $$f(n)-f(1)=(n-1)f(1)+\dfrac{n(n-1)}{2}$$ $$f(n)=nf(1)+\dfrac{n(n-1)}{2}$$ $$f(n)=\dfrac{n^2}{2}+An$$ for some $A\in\mathbb{R}.$

Answer 3

Define $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - \frac { x ^ 2 } 2 $. Then the original functional equation leads to $$ g ( x + y ) = g ( x ) + g ( y ) \text , $$ which is the famous Cauchy's functional equation.