I am interested in classifying solutions $f\,:\,\mathbb R\longrightarrow \mathbb R$ to the functional equation \begin{equation} f(x+y)=f(x)+f(y)+2xy\qquad\qquad(\dagger) \end{equation} and in particular, how to minimize the underlying assumptions of $f$. For instance, one can show that for every $a\in\mathbb R$, there exists a unique solution $f$ of $(\dagger)$ which is differentiable at zero with $f'(0)=a$. Indeed, it is a simple induction to verify that if $f$ is such a solution and $x\in\mathbb R$ then $$f(x)=2^nf(2^{-n}x)+x^2(1-2^{-n})\qquad\qquad(*)$$ for every $n\in\mathbb N$. (Those who are curious might like to derive this formula for themselves.) Since $2^nf(2^{-n})\to f'(0)=a$ as $n\to\infty$, it follows that $$f(x)=ax+x^2$$ which proves uniqueness, and existence follows simply by observing that $f$ as defined above is indeed a solution.

What if we drop the assumption that $f$ is differentiable at zero? It is simple to derive some basic properties of the solution, such as $f(0)=0$ and $f(x)+f(-x)=2x^2$. However, I cannot seem to get much more without referring to formula $(*)$ above, which is not that useful if we do not know a priori that $f'(0)$ exists.

Does anyone have any insights as to how to weaken the assumptions? Perhaps only continuity at zero, or even continuity everywhere?


Define $g$ such that $g(x)=f(x)-x^2$ for every $x$. So by the original equation we have $g(x+y)+(x+y)^2=g(x)+g(y)+x^2+2xy+y^2$. Hence $g(x+y)=g(x)+g(y)$. This is the well-known Cauchy's functional equation. You can find some useful properties of its solutions here.


This is a moderately simple comment.

$f(x+y)=f(x)+f(y)+2xy $

$f(x+h)=f(x)+f(h)+2xh $ or $\frac{f(x+h)-f(x)}{h} =2x+\frac{f(h)}{h} =2x+\frac{f(h)-f(0)}{h} $, so if $f$ is differentiable anywhere, it is differentiable everywhere.

  • $\begingroup$ Good point - in that case, we have that the unique solution with $f$ differentiable at $x_0$ and $f'(x_0)=a$ is $f(x)=(a-2x_0)x+x^2$. $\endgroup$ – Jason Sep 30 '15 at 3:59

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