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

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.

• 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 $f$. – kimchi lover Jun 7 at 16:04
• $f(x) = f(g(x,y)) - f(y)$ 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! – frabala Jun 7 at 16:09
• When you say that $g$ is given, do you actually have a formula for it? – Adrian Keister Jun 7 at 16:48
• Please read about the logramithm of a commutative formal group law in Wikipedia. – Somos Jun 7 at 18:39
• @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? – 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: $$f(g)=\begin{cases}0,\;& g_x=0\;\text{or}\;g_y=0 \\ C_1g+C_2, &g_{yx}=0,\; g_x\not=0,\;\text{and}\;g_y\not=0 \\ \frac{C_1 g_yg_x}{g_{yx}}\,\exp\left(-\frac{g_{yx}}{g_yg_x}\,g\right)+C_2, &g_{xy}\not=0\end{cases}.$$

• Good! I was daunted by the mass of formulas, & am glad you took it this far. – kimchi lover Jun 7 at 17:18
• @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. – MRT Jun 7 at 17:37
• @Ammar: Yep, just saw that myself. Working on it, as it's a special case. – Adrian Keister Jun 7 at 17:38
• Think I've fixed it. See what you think. – Adrian Keister Jun 7 at 17:47
• 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. – 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 $$(\mathbb{R},+)$$.