Exploring the Solution Space of a Complex Functional Equation: fxy fx e^y fy e^x

The functional equation given is of a complex nature, involving exponential terms and multiple functions. Specifically, we have:

fxy fx e^y fy e^x

To approach this problem, we will break it down into different cases, considering whether the function f is differentiable or not. This will help us understand the various solution methods and the implications of differentiability on the functional equation.

Differentiable Case: Exploring Differentiability

Assuming that the function f is differentiable, we start by differentiating the given equation with respect to x and performing the necessary manipulations:

1. Start with the given equation:

fxy fx e^y fy e^x

2. Differentiate with respect to x:

fxy fx e^y fy e^x fx e^y fxy - fx e^y

3. Define g(x) f'(x) - f(x). Then:

gxy g(x)e^y

4. Set x 0, which implies:

gy ke^y

5. Thus, we have:

f'y - fy ke^y

fy ce^y

6. Substituting this back into the original equation to ensure no extraneous solutions:

ce^x e^y ce^y ce^x ce^y ce^y 2ce^x e^y

This implies:

c 0

Thus, the solution is:

f(x) kxe^x [1]

Non-Differentiable Case: Exploring Pathological Solutions

For the case where f is not differentiable, we take a different approach. We define a new function h(x) fx e^{-x}. This step is inspired by the above solution and assumes the domain to be the real numbers:

1. Starting with the given equation:

hxy e^{xy} hxe^x e^y hxe^y e^x

2. Simplifying, we get:

hxy hx hy

This is the Cauchy functional equation. With sensible regularity conditions (e.g., f is continuous at one point, f is monotonic on any interval), we get:

hx kx

Thus, we recover solution [1]. Without such conditions, there are infinitely many pathological solutions.

Functional Equation Simplification

Another approach to the problem involves analyzing specific simplifications of the functional equation:

1. Given the functional equation:

fxy fx e^y fy e^x

2. From this, we can simplify by setting specific values for x and y:

When x 0:

fy f0 e^{-y} f1 e^{y}

This implies:

fy (f1 / e) f-1 e^1

Therefore, understanding the solution space and exploring differentiability properties are crucial in solving complex functional equations. The methods and conditions discussed above provide a comprehensive approach to addressing such equations.