# Finding functional equations that a given function satisfies

Suppose we're given a function, for example a function $f:\mathbb{C}\rightarrow \mathbb{C}$ such that $f(x)=ax+b$ with $a,b \in \mathbb{C}$. I would like to know which functional equations are satisfied by the given function; but not any functional equation will do as I want to impose some conditions on the equations I'm looking for.

For example, let's consider the equation $\left(x-y\right)f(x+y)=xf(x)-yf(y)$, which the function $f(x)=ax+b$ satisfies. One can see that constants do not appear in the equation, and that it generally gives a relation between $f(x)$, $f(y)$ and $f(x+y)$ and nothing more, as $f$ does not appear in any other form (at least generally). I, for instance, would like to know if there are other equations such that the above function satisfies, that it gives a relation between $f(x)$, $f(y)$ and $f(x+y)$ and nothing more, and that it has no constants appearing.

I would like to know if there are general methods to achieve the above (both specifically as in the example and generally), aside from gaining experience in solving functional equations and dealing with each case individually.

• The words "functional equation" do not have a precise definition. For this reason, usually people would start with a class of functional equations and ask what kind of functions can satisfy them. – Alexandre Eremenko May 3 '16 at 18:34
• I know, but my problem is to go the other way around. I do agree that clearer definitions are needed, as even I felt somewhat uncomfortable asking the question because of it. – Hellbound May 4 '16 at 3:29

Your $f$'s form a two-dimensional vector space.

A homogeneous linear equation $u(x,y) f(x+y) + v(x,y) f(x) + w(x,y) f(y) = 0$ is satisfied by $f(x) = a x + b$ for all $a, b$ iff it is satisfied by $f(x) = x$ and $f(x)=1$, i.e. \eqalign{ (x+y) u + x v + y w &= 0\cr u + v + w &= 0\cr} Solving this as a system of linear equations in $u,v,w$, \eqalign{u &= \dfrac{x-y}{y} w\cr v &= -\dfrac{x}{y} w\cr}

EDIT: Similarly, you can obtain all polynomial equations that are homogeneous of a given degree $d$. Write such an equation as $$\sum_{i+j+k = d} c_{ijk}(x,y) f(x)^i f(y)^j f(x+y)^k = 0$$ With the given form of $f$, the left side is a polynomial in $a$ and $b$ homogeneous of degree $d$, and the coefficients of each $a^m b^{d-m}$ give you $d+1$ linear equations in the $c_{ijk}$. However, in this case I doubt that you'll get anything really new: the left sides of the polynomial equations form an ideal in the ring of polynomials in $x, y, f(x), f(y), f(x+y)$ which I suspect is generated by $(x-y) f(x+y) - x f(x) + y f(y)$.

• This is correct, but what about other (nonlinear) relations? – Hellbound May 4 '16 at 7:31
• Regarding the edit: That's interesting. I have to give it more thought. I guess these are evidence to point to the fact that there aren't any general methods to doing these things, yet. – Hellbound May 4 '16 at 21:48