Find all function $f : \mathbb{R} \to \mathbb{R}$ s.t. $\lim_{x \to 0} \frac{f(x)}{x}=1$ & $f(x+y)=f(x)+f(y)+2xy$.

Question

Find all function $f : \mathbb{R} \to \mathbb{R}$ such that : $\lim_{x \to 0} \frac{f(x)}{x}=1$ and for all reals $x,y$ : $$f(x+y)=f(x)+f(y)+2xy$$

I tried to solve it and I got the following :

If $x=y=0$ we have $f(0)=0$

I wanted to use the limit but I couldn't : $$\text{We know that :} f(x+y)=f(x)+f(y)+2xy$$ Thus : $$f(x)=f(x+y)-f(y)-2xy$$ $$\therefore \frac{f(x+y)-f(y)-2xy}{x}=\frac{f(x+y)-f(y)}{x}-2y$$ I think that I need to substitute that $f(x+y)$ but I don't even know if this is true or no !

Answer 1

I will edit what you have done, the functional equation can be written as : $$f(x+y)-f(y)=f(x)+2xy$$ Thus : $$\forall x,y\in \mathbb{R}^*\times \mathbb{R}: \ \ \ \frac{f(x+y)-f(y)}{x}=\frac{f(x)}{x}+2y$$ If $x\to 0$ then : $$\lim_{x\to 0}\frac{f(x+y)-f(y)}{x}=1+2y$$ Thus $f$ is differentiable and $f'(y)=1+2y$, note that $f(0)=0$ .

Finally : $$\int f'(y)\ dy=\int 1+2y \ dy=y+y^2+C$$ Therefore your function is : $f(x)=x+x^2$

Answer 2

Hint: $$ \lim_{x\to 0} \frac{f(x)}{x}= \left( \lim_{x\to 0} \frac{f(x+y)-f(x)}{x} \right)-2y, \forall y\in \mathbb{R}$$

Hence you have $f'(y)=2y+1, \forall y \in \mathbb R$

$$\therefore f(x)=x^2+x+c , \forall x \in \mathbb R $$ where $c$ is a constant .

Now from the functional equation we can show that $c=0$