A double differentiable function satisfies $$f(x^2+y^2)=f(x^2-y^2)+2xy \quad \forall\; x,y\in\Bbb R$$
Given that $f(0)=0$ and $f''(0)=2$, determine $f(x)$.
I tried to convert the equation into some form of Cauchy's additive function by observing that $f(x)$ is an even function. However, I am unable to make any significant progress. Any hints will be appreciated.
Edit: My apologies for posting the incorrect question. Correct question:
$$f(x^2+y^2)=f(x^2-y^2)+f(2xy) \quad \forall\; x,y\in\Bbb R$$
Given that $f(0)=0$ and $f''(0)=2$, determine $f(x)$.
This has already been answered here Find all functions $f:\mathbb{R} \to [0, \infty)$such that $f(x^2 + y^2)=f(x^2 – y^2)+ f(2xy)$.
Best Answer
There is no such function. Putting $x=y$ we get $f(2x^{2})=2x^{2}$ which implies that $f(t)=t$ for all $t \geq 0$. Since $f$ is also even (as seen by putting $x=0$) we get $f(t)=|t|$ which is not even differentiable.