find functions $f$ such that $f(x)+f(y) = f(g(x,y))$, $g$ is given and symmetric

Question

I want to find solutions $f$ of the following functional equation given a function $g(x,y)$, which is symmetric ($g(x,y)= g(y,x)$) and strictly monotonic $\forall x,y \in $ Reals:

$f(x)+f(y) = f(g(x,y))$

An observation I have been able to make is that $f(g(x,y))$ cannot contain terms that couple $x$ and $y$, e.g. $f(g(x,y)) \neq x*y $. I.e. if we rewrite the functional equation like $f(x) = f(g(x,y))-f(y)$, then with a coupled term (e.g. $f(g(x,y)) = x*y)$, one could change the right hand side by varying $y$ without changing the left hand side.

What can be said about $f$? Any insights would be helpful.

Answer 1

Taking kimchi lover's suggestion, and assuming everything in sight is sufficiently differentiable, we have: \begin{align*} \partial_x[f(x)+f(y)&=f(g(x,y))]\\ f'(x)&=\frac{df(g(x,y))}{dg}\,g_x(x,y)\\ \partial_y\bigg[f'(x)&=\frac{df(g(x,y))}{dg}\,g_x(x,y)\bigg]\\ 0&=\frac{d^2f(g(x,y))}{dg^2}\cdot g_y(x,y)\cdot g_x(x,y)+\frac{df(g(x,y))}{dg}\cdot g_{yx}(x,y). \end{align*} You can let $h(g)=df/dg$ and use the first-order linear formula, or separate out and integrate:

\begin{align*} 0&=\frac{dh}{dg}\,g_y\,g_x+h\,g_{yx}\\ \frac{dh}{dg}&=-h\,\frac{g_{yx}}{g_y g_x}\\ h&=C_1\exp\left(-\frac{g_{yx}}{g_yg_x}\,g\right). \end{align*}

Special case: $g_{yx}=0.$ Here we have $$\frac{dh}{dg}\,g_y\,g_x=0, $$ with different possibilities depending on which factor is zero. Suppose $g_x=0.$ Then $g(x,y)=g(y).$ But because $g(y,x)=g(x,y),$ we must have $g(x,y)=g(x).$ The only way this could happen is if $g$ is a constant. It would follow, then, that $f(x)+f(y)=f(\text{const}),$ the only solution being a constant, namely, $f(x)=0.$

On the other hand, if $dh/dg=0,$ with neither of $g_x$ or $g_y$ zero, then $h$ is a constant, and hence $f=C_1g+C_2.$

So, back to $g_{yx}\not=0:$ integrating with respect to $g$ yields $$f(g)=-\frac{C_1 g_yg_x}{g_{yx}}\,\exp\left(-\frac{g_{yx}}{g_yg_x}\,g\right)+C_2.$$ You can absorb the overall minus sign into $C_1$ if you like.

Putting it all together: $$f(g)=\begin{cases}0,\;& g_x=0\;\text{or}\;g_y=0 \\ C_1g+C_2, &g_{yx}=0,\; g_x\not=0,\;\text{and}\;g_y\not=0 \\ \frac{C_1 g_yg_x}{g_{yx}}\,\exp\left(-\frac{g_{yx}}{g_yg_x}\,g\right)+C_2, &g_{xy}\not=0\end{cases}.$$

Answer 2

My idea is the following:

1) Let's assume $f$ is injective.

2) It is easy to check that $f(x)+f(y)+f(z) = f(g(g(x,y),z))$.

3) Since we can permute $x,y$, and $z$, and $g$ is symmetric, it follows that $f(g(g(x,y),z))=f(g(x,g(y,z)))$. Using the injectivity of $f$ we obtain $$g(g(x,y),z)=g(x,g(y,z)).$$

4) So the $g:\mathbb{R}\times\mathbb{R}\to \mathbb{R}$ is an associative symmetric operation, and $f:(\mathbb{R},g)\to (\mathbb{R},+)$ is an injective morphism of associative symmetric operations in $\mathbb{R}$.

5) Then we can identify $(\mathbb{R},g)$ with its image, let's say $A\subseteq\mathbb{R}$, which is closed under the sum $+$;

6) We conclude that the pair of funcions $(f,g)$ satisfying these hypothesis are in bijections with the subsets $A\subseteq\mathbb{R}$, which are closed under $+$, and have the same cardinality of $\mathbb{R}$ (there is also a choosing of bijection between $\mathbb{R}$ and $A$).

7) For example, we can take $A=(0,+\infty)$ and take any bijection $f:\mathbb{R}\to A$.

8) It would be good if there were a characterization of the subsets of $\mathbb{R}$ closed under $+$. A special case would be the characterization of the subgroups of $(\mathbb{R},+)$.