# Strategies to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy a given functional equation

My question is as follows: What methods can be used to find the set of functions $f:\mathbb{R}\to\mathbb{R}$ satisfying a certain functional equation. An example of a case where this applies is the following:

Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following equation: $$f(x^{3})+f(y^{3})=(x+y)(f(x^{2})+f(y^{2})-f(xy)):\forall x, y\in\mathbb{R}$$

I'm curious as to whether there are general methods (or strategies) for solving this type of question, or whether questions like these should just be handled on a case-by-case basis.

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I believe that a case-by-case analysis is necessary. For example, it took a long time to find all solutions to the Cauchy Equation. I dont see how other cases would be simpler. en.wikipedia.org/wiki/Cauchy's_functional_equation – Ravi Jun 5 '12 at 15:35
@FortuonPaendrag Ahh, okay. I was just curious as to see whether there were general methods which I was not aware of that I could employ strictly for functions $f:\mathbb{R}\to\mathbb{R}$, it appears not. Thanks! – Shaktal Jun 5 '12 at 15:37
Also, since you need a form of the axiom choice to find many non-trivial solutions of Cauchy's equation, it makes it near-impossible to describe the solutions. Right now, I feel like you might have similar problems in your case. – Ravi Jun 5 '12 at 15:40
There are general strategies that are used by people looking at problems of this type in contests. Start by seeing what happens if you let one or both of $x$ and $y$ equal to $0$. Continue by exploring things like $y=-x$. Doing that in this question tells you that $f(-u)=-f(u)$, which is a nice start! – André Nicolas Jun 5 '12 at 15:45
Are you sure that you are not looking for continuous solutions? My approach would be trying to solve this for rational numbers and then use continuity. – Simon Markett Jun 5 '12 at 15:48

Here is a typical start. Put $x=y=0$. The right side is $0$, so $f(0)=0$.

Now set $y=0$, and let $x$ roam freely. Since $f(0)=0$, we get $f(x^3)=xf(x^2)$.

Set $y=-x$. We get $f(x^3)+f(-x^3)=0$. Since everything is a cube, we have $f(-u)=-f(u)$ for all $u$.

Now explore $x=y$. Can we learn anything from setting $x$ and/or $y$ qual to $1$?

A little playing has gotten us a lot of information, enough that we should be able to complete things.

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Ahh; this is basically what I was looking for. Thank you. Are there any other common strategies which may also be applied to cases like this (not just this one specifically)? – Shaktal Jun 5 '12 at 16:02

Using the equation, we can see that $f(0) = 0$ is necessary. Let $y = -x$, then \begin{gather} f(x^3) + f(-x^3) = 0 \end{gather} This implies that $f$ must be an odd function. Also letting $y = x$ we see that $$f(x^3) = x f(x^2)$$ Plugging this relationship into the LHS gives \begin{align} xf(x^2) + yf(y^2) & = (x + y)(f(x^2) + f(y^2) - f(xy)) \\ 0 & = x f(y^2) + y f(x^2) - (x+y)f(xy) \tag{*} \end{align} If we let $y \to -y$ then by the symmetry of the function we have \begin{align} 0 = xf(y^2) - yf(x^2) + (x-y)f(xy) \tag{**} \end{align} Adding $(^*)$ and $(^{**})$ we see that $$x f(y^2) = y f(xy)$$ If we let $y = 1$ we obtain $$f(x) = x f(1)$$

EDIT: At this point we should check whether this function indeed satisfies the above relation. Letting $f(1) = c$, and plugging in we have $$c(x^3 + y^3) = c(x+y)(x^2 + y^2 - xy) = c(x^3 + xy^2 -x^2 y + x^2 y + y^3 - xy^2) = c(x^3 + y^3)$$

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Nicely done, except you must check if $f(x)=cx$ indeed works, for any constant $c\in\Bbb R$. You've only found a necessary condition, but not necessarily sufficient. – user26486 Jun 22 '15 at 0:56
@user26486 good point. Though is not straight forward enough to show that it is sufficient? – Gregory Jun 22 '15 at 1:02
Indeed, it's very straightforward. But you still should've mentioned it. It may happen it doesn't suffice. In an olympiad you'd probably lose a point for not checking it. – user26486 Jun 22 '15 at 1:04
Fair point. I will edit it. Thanks for the comment. – Gregory Jun 22 '15 at 1:47
The question was reasked in June '15, and merged with this one. Therefore this independent answer has a later date than the others. Please take that into account when voting. – Jyrki Lahtonen Jun 22 '15 at 11:24

If it is of any help the only continuous solutions are of the form $f(x)=xf(1)$. Indeed we can use the property $f(x^3)=xf(x^2)$ to show $$f(x)=x^\frac{1}{3}f(x^\frac{2}{3})=x^\frac{1}{3}x^\frac{2}{9}f(x^\frac{4}{9})=\cdots=x^{\frac13\sum_i(\frac23)^i}f(1)=xf(1).$$ In fact we only need continuity at $1$.

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