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.

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    $\begingroup$ If all the functions in sight are smooth, the mixed partial derivative $\partial^2/\partial x\partial y$ will annihilate $f(g(x,y))$, leading to a 2nd order differential equation in $f$. $\endgroup$ – kimchi lover Jun 7 at 16:04
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    $\begingroup$ $f(x) = f(g(x,y)) - f(y)$ is fine. It means that no matter the choice of $y$, the expression $f(g(x,y)) - f(y)$ will always give you the same result as $f(x)$. If you think of $f(g(x,y)) - f(y)$ as a function over the $y$ variable, it is the constant function! $\endgroup$ – frabala Jun 7 at 16:09
  • $\begingroup$ When you say that $g$ is given, do you actually have a formula for it? $\endgroup$ – Adrian Keister Jun 7 at 16:48
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    $\begingroup$ Please read about the logramithm of a commutative formal group law in Wikipedia. $\endgroup$ – Somos Jun 7 at 18:39
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    $\begingroup$ @Somos: I think this is an excellent suggestion. I tried using the logarithm of formal group laws to solve this problem for $g(x,y) = \frac{1}{1+e^{-x}}\frac{1}{1+e^{-y}}$, but it doesn't seem to work since this choice of $g$ isn't a formal group, right? $\endgroup$ – MRT Jun 11 at 15:22

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

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    $\begingroup$ Good! I was daunted by the mass of formulas, & am glad you took it this far. $\endgroup$ – kimchi lover Jun 7 at 17:18
  • $\begingroup$ @AdrianKeister, actually I am not sure the latter half of your calculation is correct, i.e. after you make the notational change $h(g) = \frac{\partial f}{\partial g}$. Consider $g(x,y) = x+y$. An $f$ that satisfies the equation is $f(x) = x$. But note that in your solution $f(g)$, the $g_{xy}$ in the denominator blows up, so this can't be correct. However, if you plug $g(x,y)=x+y$ into the differential equation you wrote down, one recovers $f(g) = C_1 g + C_2$, which is correct. I think the mistake is that one can't assume that the terms $g$, $g_x$ etc can be integrated independently. $\endgroup$ – MRT Jun 7 at 17:37
  • $\begingroup$ @Ammar: Yep, just saw that myself. Working on it, as it's a special case. $\endgroup$ – Adrian Keister Jun 7 at 17:38
  • $\begingroup$ Think I've fixed it. See what you think. $\endgroup$ – Adrian Keister Jun 7 at 17:47
  • $\begingroup$ Your discussion of the special case $g_{xy}=0$ now seems correct. For the case $g_{xy} \neq 0$, I don't see how there is a $g$ multiplying $C_2$, is that a typo? Thanks for writing all this down btw. $\endgroup$ – MRT Jun 7 at 17:59

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 $(\mathbb{R},+)$.


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