# A functional equation $\log f(x,y) = f(\log x, \log y)$

Solve the functional equation $$\log f(x,y) = f(\log x, \log y)$$ The motivation comes from generalizing the arithmetic and geometric means, but I have no idea how to find a simpler expression for the function $$f(x,y)$$ with $$f(x,y) = f(y,x)$$

What I have tried:

1. $$f(ax,y) = e^{f(\log x +\log a, y)}$$
2. $$f(1,x) = e^{f(0,\log x)}$$
3. $$f(x,y) \equiv \infty$$ solves the equation, so I am looking for non-trivial solutions.
4. Letting $$f(1+x,1+y) = \sum_{m,n=0}^\infty \frac {a_{m,n}}{x^my^n} = a_{0,0} + \frac {a_{1,1}}{xy} + \frac {a_{2,0}}{x^2}+ \frac {a_{0,2}}{y^2}+\cdots$$ Using the functional equation $$f(1+x,1+y) = e^{f(\log (1+x), \log (1+y))}$$ we have $$a_{0,0} + \frac {a_{2,0}}{x^2}+ \frac {a_{0,2}}{y^2}+\cdots = 1+f\left(x-\dfrac{x^2}2 + \dfrac{x^3}3 - \cdots, y-\dfrac{y^2}2 + \dfrac{y^3}3 - \cdots\right) + \cdots$$ which does not give anything pleasant.
5. If I consider a recursive definition $$f_{n+1} (x,y) = e^{f_n (\log x, \log y)}$$, then modulo convergence issues, $$f_n$$ converges to $$f$$. Motivated by AM and GM, I consider $$f_0 (a,b) = \frac{a+b}2$$. Then $$f_1 (a,b) = \operatorname{GM}(a,b)$$.
6. Test: $$f_0 = \dfrac{x+y}2, f_1 = \sqrt {xy}, f_2 = e^{\operatorname{GM}(\log x,\log y)} = \exp\sqrt{(\log x)(\log y)}$$, $$f_3 = \exp(\exp\sqrt{(\log\log x)(\log\log y)})$$. I guess the pattern is $$\exp(\exp\cdots(\exp\sqrt{(\log\log\cdots \log x)(\log\log\cdots \log y)})$$
7. Note that by AM$$\geq$$ GM, $$f_n (x,y) \leq \underbrace{\exp(\exp\cdots(\exp}_n\dfrac{\underbrace{\log\log\cdots \log}_n x+\underbrace{\log\log\cdots \log}_n y}2)$$ = $$\underbrace{\exp(\exp\cdots(\exp}_{n-1}\sqrt{(\underbrace{\log\log\cdots \log}_{n-1} \,x)(\underbrace{\log\log\cdots \log}_{n-1}\,y)}) = f_{n-1}(x,y)$$.
8. I have tested out some explicit values: for $$x = 2000, y= 8000$$, $$f_0 = 6000, f_1 = 4000, f_2 = 3832.6491..., f_3 = 3762.9980...$$ by Wolfram Alpha. Some larger $$n$$'s give complex numbers. If $$f_n$$ is real, it is a priori apparent that $$f_n\geq 0$$.

But I have no idea how to go on further.

• Did you see this thread that I found on SearchOnMath? Mar 17, 2022 at 12:21
• Oh, you ask the same question! I think I have provided a (partial) answer by infinite interations of exp and log Mar 17, 2022 at 12:29

Take any function $$g(x,y):A\to \Bbb R$$ where $$A = \{(x,y) \mid x \le 0 \text{ or } y \le 0\}$$.
Now define $$f(x,y)$$ as follows: $$f(x,y) = \begin{cases} g(x,y) &\text{if } (x,y)\in A\\ \exp(f(\log x, \log y))& \text{otherwise} \end{cases}$$
• I wonder what happens when one logarithm is zero and the other one is positive (I think infinities appear) . Also, I think you meant $f(\log x, \log y)$. Mar 17, 2022 at 12:51
• I think it would be better if you write $f_{n+1}$ on the left hand side and $f_n$ on the right for easier reading. Mar 17, 2022 at 12:54