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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.

Is my answer correct?? If not then what is the complete solution?? Please help me!

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

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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 $f(x) = C_0 x$

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

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