# Find all functions such that $f(x^2+y)=xf(x)+f(y)$

Find all functions (over real numbers) such that $$f(x^2+y)=xf(x)+f(y).$$

My idea: Put $x=1$. Therefore the given equation becomes $f(y+1)=f(y)+f(1)$. It is in the Cauchy's first form, so we get $f(x)=cx$. It also satisfies that given functional equation.

• That's not in Cauchy's form. – Saad Jun 16 '18 at 6:21
• @Alex Francisco $f(x+y)=f(x)+f(y)$ is a Cauchy form! And its solution is $f(x)=cx$ – Sufaid Saleel Jun 16 '18 at 6:23
• What you derived is $f(y+1)=f(y)+f(1)$. – Saad Jun 16 '18 at 6:24
• No, you are not right. If $g$ is a function of period $1$ with $g(0)=0,$ then $f(x)=g(x) +cx$ satisfies $f(x+1)=f(x)+f(1).$ – Thomas Andrews Jun 16 '18 at 6:24
• Is there any statement of continuity? – DanielV Jun 16 '18 at 6:25

Here is a proof that $f$ does satisfy Cauchy's functional equation. For now, I do not have a complete answer, unless some sort of continuity conditions is given.

Let $P(x,y)$ denote the condition that $f\left(x^2+y\right)=x\,f(x)+f(y)$. Then, $P(1,0)$ implies that $f(0)=0$. Now, $P(x,0)$ implies that $f\left(x^2\right)=x\,f(x)$ for all $x\in\mathbb{R}$. This shows that $f\left(x^2+y\right)=f\left(x^2\right)+f(y)$ for all $x,y\in\mathbb{R}$. Next, $P\left(x,-x^2\right)$ leads to $f\left(-x^2\right)=-f\left(x^2\right)$, and consequently, $f(-x)=-f(x)$ for every $x\in\mathbb{R}$. From this, it can be easily shown that $f$ satisfies Cauchy's functional equation.

• $$( x + 1 ) f ( x ) + ( x + 1 ) f ( 1 ) = ( x + 1 ) f ( x + 1 ) = f \left( ( x + 1 ) ^ 2 \right) \\ = f \left( x ^ 2 + 2 x + 1 \right) = f \left( x ^ 2 \right) + f ( 2 x ) + f ( 1 ) = ( x + 2 ) f ( x ) + f ( 1 )$$ and thus $f ( x ) = f ( 1 ) x$. – Mohsen Shahriari Nov 9 at 19:35

Assuming $f \in C$ and arranging as

$$\frac{f(y+x^2)-f(y)}{x^2}=\frac{f(x)}{x} = \phi(x)$$

So

$$\lim_{x\to 0}\frac{f(y+x^2)-f(y)}{x^2}=\phi(0)$$

which is independent of $y$ hence $f(x) = C_0 x$

• It's enough to assume $f$ is continuous. Then your argument shows that the derivative exists and is constant. – Wojowu Jun 16 '18 at 8:26
• Yes. I left an exponent in excess at $C$. Thanks. – Cesareo Jun 16 '18 at 9:01
• Why do you know that $\lim_{x\rightarrow 0} \phi(x)$ exists? – Severin Schraven Jun 16 '18 at 11:46
• Because this implies on that the first limit does not exists for any $y$. – Cesareo Jun 16 '18 at 11:48