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$$f(x+y^2) = f(x) + f(y)^2$$ Hello everybody. I am doing this exercise, I need to find all functions real valued $f: \mathbb{R} \to \mathbb{R}$ that satisfy the condition above. I have already found some type of functions with a little of guess and check, but I would like to go beyond.

I thought that would be a good idea to derivate both sides, but I am a little confused how to do it.

As it was said, it is a function from $\mathbb{R}$ to $\mathbb{R}$, not from $\mathbb{R}^2$ to $\mathbb{R}$, so it makes me thought that would be a simple derivative, but how to do it?

Can I say that $\frac{df(x+y^2)}{d(x+y^2)} = \frac{df(u)}{du} = \frac{df(x)}{dx}$ ?

So that it would lead us to: $f' = f' + 2f(x)f'$

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    $\begingroup$ $f$ is not supposed being differentiable, but if it were differentiable, you can differentiate with respect to $x$ or with respect to $y$. In the first case you would get $f'(x+y^2)=f'(x)$ and in the second case you would have $2yf'(x+y^2)=2f'(y)f(y)$. $\endgroup$
    – Tuvasbien
    Dec 19, 2020 at 3:30
  • $\begingroup$ @Tuvasbien So when you say differentiate with respect to you mean partial derivative, right? But it does not require that the function has two variable? $\endgroup$ Dec 19, 2020 at 3:33
  • $\begingroup$ I think i need to say that x and y is not necessarily variables, but can be just any real numbers $\endgroup$ Dec 19, 2020 at 3:35
  • $\begingroup$ If you want to differentiate the left hand side, you will need to use partial derivatives. In general, if $g$ is a function of $x$ and $y$ then $$dg = \frac{\partial g}{\partial x}dx + \frac{\partial g}{\partial y}dy$$ where $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial y}$ are called partial derivatives of $g$ with respect to $x$ and $y$. You derive each of them (say, $\frac{\partial g}{\partial x}$) by assuming the other variable (in this case, $y$) as a constant. $\endgroup$
    – Saeed
    Dec 19, 2020 at 3:49
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    $\begingroup$ You can see the equality $f(x+y^2)=f(x)+f(y)^2$ as an equality of functions of $x$ when $y$ is fixed, or as an equality of functions of $y$ when $x$ is fixed. In the first case, differentiating the equality means differeniating with respect to $x$ and in the second case, differentiating the equality means differentiating with respect to $y$. $\endgroup$
    – Tuvasbien
    Dec 19, 2020 at 3:55

1 Answer 1

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Yon cannot make any assumptions of continuity, differentiability etc. Since $(f(y))^{2} \geq 0$ we see that $f(x+y^{2}) \geq f(x)$. So $f(x+t) \geq f(x)$ for all $t \geq 0$. Thus $f$ is an increasing function.

Now $x=y=0$ gives $f(0)=0$ and $x=0$ gives $f(y^{2})=(f(y))^{2}$. Now the equation beconme s $f(x+y^{2})=f(x)+f(y^{2})$ so $f(x+t)=f(x)+f(t)$ for $t \geq 0$. Putting $x=-y^{2}$ in the given equation prove that $f$ is an odd function. We now have $f(x+y)=f(x)+f(y)$ for all $x,y$ It is well known that the only increasing functions with this property are of the form $f(x)=cx$ for some constant $c \geq 0$.

EDIT: As pointed out by user bof we must have $c=0$ or $c=1$ so $f\equiv 0$ or $f(x)=x$ for all $x$.

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  • $\begingroup$ @bof Thank you for the comment. $\endgroup$ Dec 19, 2020 at 5:06

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