Unique Functional Solutions for the Equation ( f(x,y) f(x)f(y) - 2f(x,y) )

This article explores the unique solutions to the functional equation ( f(x,y) f(x)f(y) - 2f(x,y) ). We will discuss two distinct approaches to solving this equation, both based on differentiability assumptions and a calculus-free method. The primary goal is to demonstrate that the only function satisfying the given equation for all ( x, y ) is the zero function. This exploration not only enriches our understanding of functional equations but also provides insights into their unique solutions.

Method 1: Calculus-Based Approach

First, we differentiate both sides of the functional equation with respect to ( x ) while keeping ( y ) constant. This leads to:

[ f'_{xy} f'_{x} - 2yf'_{xy} ]

By letting ( x 0 ), we obtain:

[ f'_y f'_0 - 2yf'_0 ]

For convenience, we can swap ( y ) for ( x ) and denote ( f'_0 A ). This yields:

[ f'_x -2AxA ]

We can then integrate this expression to get:

[ f(x) -Ax^2 AxB ]

where ( B ) is a constant. Substituting this expression into the original functional equation:

[ 2Ax^2y^2 - B 0 ]

For this equation to hold for all ( x ) and ( y ), we must have ( A B 0 ). Therefore, ( f'(x) 0 ), implying that ( f(x) ) is a constant function.

Method 2: Calculus-Free Approach

Without using differentiability assumptions, we can still find a solution. Let's define ( f ) such that:

[ f_{xy} f_x f_y - 2f_{xy} ]

Step 0: Letting ( x y 1 ) yields:

[ f_2 2f_1 - 2f_1 0 ]

Step 1: For arbitrary ( x ) and ( y 1 ), we get:

[ f_{x1} f_x f_1 - 2f_x f_1 - f_x ]

Step 2: Substituting ( x ) by ( x_1 ) in the expression above:

[ f_{x2} f_1 - f_{x1} ]

Step 3: Combining Step 1 and Step 2 gives:

[ f_{x2} f_x ]

Step 4: Using the original relation ( f_{xy} f_x f_y - 2f_{xy} ), take ( y 2 ) and combine with Step 0 and Step 3:

[ f_{2x} f_{2} - 2f_{x2} Rightarrow f_{2x} 0 ]

Since ( f_{2x} 0 ) holds for all ( x ), the proof is complete. This shows that the only function satisfying the given functional equation is the zero function.

Conclusion and Special Cases

The exploration proves that the only function satisfying the equation ( f(x,y) f(x)f(y) - 2f(x,y) ) for all ( x, y ) is the zero function. This conclusion holds under both the calculus-based and calculus-free approaches. Moreover, these methods can be extended to different domains such as the set of integers ( Z ) and the set of real numbers ( R ). For the domain ( R ) or any set closed by division by 2, it is necessary and sufficient that the function is the zero function.

Through this analysis, we have enriched our understanding of the nature of functional equations and the importance of domain considerations in finding unique solutions. The zero function represents a trivial yet crucial solution in these contexts.