$$f(x+y^2) = f(x) + f(y)^2$$ 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: $$f' = f' + 2f(x)f'$$

• $f$ 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 $2yf'(x+y^2)=2f'(y)f(y)$. Dec 19 '20 at 3:30
• @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? Dec 19 '20 at 3:33
• I think i need to say that x and y is not necessarily variables, but can be just any real numbers Dec 19 '20 at 3:35
• 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. Dec 19 '20 at 3:49
• 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 $y$. Dec 19 '20 at 3:55

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 $$x$$.