Analysis of the Functional Equation fxy fx * fy - 2fxy

Analysis of the Functional Equation fxy fx * fy - 2fxy

The functional equation you provided is:

fxy fx * fy - 2fxy

To analyze this equation, we can explore specific forms of solutions. A common approach is to test simple functions such as linear functions or polynomials.

Testing for Linear Solutions

Assume that fx ax b for some constants a and b. We will compute each side of the equation and match them.

Left-hand side

fxy axy b

Right-hand side

fx * fy - 2fxy (ax b)(ay b) - 2axy

Simplifying this, we get:

(ax b)(ay b) - 2axy a^2xy abx aby b^2 - 2axy a^2xy abx aby b^2 - 2axy a^2xy abx aby b^2 - 2axy

This simplifies to:

a^2xy abx aby b^2 - 2axy

Matching the left-hand side and the right-hand side:

axy b a^2xy abx aby b^2 - 2axy

Step-by-Step Analysis

For the equation to hold true for all x and y, the constant terms and the coefficients of xy must match.

Constant term:

b b^2

This implies that b 0.

Coefficient of xy term:

1a a^2 - 2a

Therefore, we have:

0 a^2 - 3a

This simplifies to:

a(a - 3) 0

Hence, a 0 or a 3.

Considering a 0:

When a 0, the function becomes:

fx 0

This is a valid solution.

Considering a 3:

If a 3, the equation becomes:

3x 0 0

This implies:

3x 0

This is not a valid function for all x.

Therefore, the only valid solution is:

fx 0

Calculus-free Approach

Let's present a calculus-free solution to show that the only function satisfying the functional relation is fequiv 0.

Step 0: Introductory Step

Letting x y 1 yields:

f2 2f1 - 2f1 0

Step 1: Substitution with Arbitrary x

For x arbitrary, taking y 1 yields:

fx1 fx * f1 - 2fx f1 - fx

Step 2: Substitution with x1

In the expression above, substitute x by x1, reaching:

fx2 f1 - fx1

Step 3: Combining Steps 1 and 2

Combining Step 1 and Step 2 gives:

fx2 fx

Step 4: Applying the Original Relation

In the original relation namely fxy fx * fy - 2fxy, take y 2 and combine with Steps 0 and 3:

fx2 fx * f2 - 2f2x

This simplifies to:

f2x 0

Since f2x 0 holds for all x, the proof is complete.

Thus, the only solution to the functional equation is:

fequiv 0