Functional Equation: Solving For F(x^2 - Y^2)

Hey guys! Today, we're diving deep into the fascinating world of functional equations. Specifically, we're tackling a problem where we need to determine the set of all functions f that satisfy a given equation. This type of problem often appears in mathematical competitions and can seem daunting at first, but with a systematic approach, we can break it down and find the solution. Let's get started!

Understanding the Functional Equation

Our main goal here is to find all the functions ${ f : \mathbb{R} \to \mathbb{R} }$ that comply with the following equation:

${ f(x^2 - y^2) = (x - y)(f(x) + f(y)) }$ for every ${ (x, y) \in \mathbb{R}^2 }$.

This equation essentially links the function's value at ${ x^2 - y^2 }$ to the values at ${ x }$ and ${ y }$. The task at hand is to figure out what these functions ${ f }$ could be. Think of it like a puzzle where we're trying to discover the hidden rules that the function must follow. To crack this, we'll try out some clever substitutions for ${ x }$ and ${ y }$ and observe what happens. This is a classic technique in solving functional equations. We're essentially poking and prodding the equation to see how it behaves under different conditions. So, let's roll up our sleeves and get into the nitty-gritty!

Initial Substitutions and Observations

The initial approach to solving functional equations often involves substituting specific values for the variables to simplify the equation and reveal underlying properties of the function. Let's begin with some simple substitutions that can provide valuable insights.

1. Setting x = y

Let's substitute ${ x = y }$ into the equation. This gives us:

${ f(x^2 - x^2) = (x - x)(f(x) + f(x)) }$

Simplifying, we get:

${ f(0) = 0 }$

This is a significant piece of information. We've discovered that the function f must equal zero when its input is zero. This is like finding the first piece of a jigsaw puzzle; it gives us a fixed point to work from.

2. Setting x = 0

Next, let's try setting ${ x = 0 }$. The equation becomes:

${ f(0^2 - y^2) = (0 - y)(f(0) + f(y)) }$

Since we know ${ f(0) = 0 }$, this simplifies to:

${ f(-y^2) = -y \, f(y) }$

This equation is interesting because it relates the function's value at ${ -y^2 }$ to its value at ${ y }$. It suggests a symmetry or antisymmetry property that the function might possess. This is another crucial clue that we'll keep in mind as we continue.

3. Setting y = 0

Now, let's set ${ y = 0 }$. The original equation transforms into:

${ f(x^2 - 0^2) = (x - 0)(f(x) + f(0)) }$

Again, using ${ f(0) = 0 }$, we simplify to:

${ f(x^2) = x \, f(x) }$

This equation is similar to the previous one but gives us information about ${ f(x^2) }$. It tells us how the function behaves for square inputs. These two equations, ${ f(-y^2) = -y \, f(y) }$ and ${ f(x^2) = x \, f(x) }$, are like two sides of the same coin. They provide complementary information about the function's behavior.

Combining the Results

By combining the results from setting ${ x = 0 }$ and ${ y = 0 }$, we can gain even more insight. We have:

${ f(-y^2) = -y \, f(y) }$

${ f(x^2) = x \, f(x) }$

Let's replace ${ y }$ with ${ x }$ in the first equation:

${ f(-x^2) = -x \, f(x) }$

Now, we have expressions for ${ f(x^2) }$ and ${ f(-x^2) }$. This is a significant step forward. It's like fitting more pieces into our puzzle, allowing us to see the bigger picture more clearly. These initial substitutions have given us a solid foundation to build upon. We've uncovered key properties of the function, and we're now better equipped to explore further substitutions and manipulations.

Exploring Symmetry and Antisymmetry

Based on the relationships we've derived, we can start to explore the function's symmetry properties. Remember, we found that:

${ f(-x^2) = -x f(x) }$ and ${ f(x^2) = x f(x) }$

These equations suggest a possible connection between the function's values at ${ x^2 }$ and ${ -x^2 }$. Let’s delve deeper into this.

Deriving f(-x) in terms of f(x)

To understand the function’s symmetry, we aim to express ${ f(-x) }$ in terms of ${ f(x) }$. This will help us determine if the function is even (${ f(-x) = f(x) }$), odd (${ f(-x) = -f(x) }$), or neither.

Let's return to our original equation:

${ f(x^2 - y^2) = (x - y)(f(x) + f(y)) }$

Now, substitute ${ x = 0 }$:

${ f(-y^2) = -y f(y) }$

Next, substitute ${ y = 0 }$:

${ f(x^2) = x f(x) }$

To find ${ f(-x) }$, we need to make a strategic substitution in the original equation. Let's try setting ${ y = -x }$:

${ f(x^2 - (-x)^2) = (x - (-x))(f(x) + f(-x)) }$

This simplifies to:

${ f(0) = 2x(f(x) + f(-x)) }$

Since we know ${ f(0) = 0 }$, we have:

${ 0 = 2x(f(x) + f(-x)) }$

For ${ x \neq 0 }$, we can divide both sides by ${ 2x }$:

${ 0 = f(x) + f(-x) }$

Thus, we get:

${ f(-x) = -f(x) }$

This is a crucial finding! It tells us that the function f is an odd function. This means it exhibits symmetry about the origin. Knowing this significantly narrows down the possibilities for the function's form.

Implications of Odd Function Property

The fact that ${ f(-x) = -f(x) }$ has several implications. It means that if we know the function's behavior for positive values of ${ x }$, we automatically know its behavior for negative values. This simplifies our task considerably.

Moreover, since ${ f(-x) = -f(x) }$, we can substitute ${ -x }$ for ${ x }$ in some of our previous equations and see if we gain any new insights. For example, we know:

${ f(x^2) = x f(x) }$

Replacing ${ x }$ with ${ -x }$, we get:

${ f((-x)^2) = -x f(-x) }$

${ f(x^2) = -x (-f(x)) }$

${ f(x^2) = x f(x) }$

This confirms our earlier result and reinforces the consistency of our findings. We've successfully established that f is an odd function, a key piece of the puzzle that will guide our next steps in solving the functional equation.

Determining the Form of f(x)

Now that we know ${ f }$ is an odd function, we're closer to determining its general form. We'll use this information along with our previously derived equations to deduce the structure of ${ f(x) }$.

Revisiting the Original Equation

Let's go back to the original functional equation:

${ f(x^2 - y^2) = (x - y)(f(x) + f(y)) }$

We'll try a different set of substitutions to see if we can extract more information about ${ f }$.

Strategic Substitutions

1. Swapping x and y

Swap ${ x }$ and ${ y }$ in the original equation:

${ f(y^2 - x^2) = (y - x)(f(y) + f(x)) }$

Notice that ${ y^2 - x^2 = -(x^2 - y^2) }$. So, we can write:

${ f(-(x^2 - y^2)) = (y - x)(f(x) + f(y)) }$

Since ${ f }$ is an odd function, we have ${ f(-z) = -f(z) }$. Applying this to the left side:

${ -f(x^2 - y^2) = (y - x)(f(x) + f(y)) }$

Multiply both sides by -1:

${ f(x^2 - y^2) = (x - y)(f(x) + f(y)) }$

This is the same as our original equation, so swapping ${ x }$ and ${ y }$ didn't immediately give us new information, but it confirms the consistency of our approach.

2. A Key Substitution: y = -x

We already used this substitution to show that f is odd. Let's consider another one.

3. Trying x = 2y

Let’s substitute ${ x = 2y }$ into the original equation:

${ f((2y)^2 - y^2) = (2y - y)(f(2y) + f(y)) }$

${ f(4y^2 - y^2) = y(f(2y) + f(y)) }$

${ f(3y^2) = y(f(2y) + f(y)) }$

This equation relates ${ f(3y^2) }$ to ${ f(2y) }$ and ${ f(y) }$. It might be useful in establishing a pattern or a relationship between the function's values at different points.

The Linear Solution

At this point, let's make a crucial observation. A simple function that satisfies the odd function property is a linear function of the form:

${ f(x) = ax }$

where ${ a }$ is a constant. Let's test if this form satisfies the original equation:

${ f(x^2 - y^2) = a(x^2 - y^2) }$

And,

${ (x - y)(f(x) + f(y)) = (x - y)(ax + ay) }$

${ = a(x - y)(x + y) }$

${ = a(x^2 - y^2) }$

So, ${ f(x) = ax }$ indeed satisfies the equation. This is a major breakthrough! We've found a family of solutions. But are these the only solutions? To be sure, we need to rigorously prove that no other functions satisfy the given conditions.

Proving the General Solution

We've discovered that linear functions of the form ${ f(x) = ax }$ satisfy the functional equation. Now, we need to demonstrate that these are the only solutions. This requires a more rigorous approach.

Utilizing Existing Equations

We have a few key equations at our disposal:

1. ${ f(x^2 - y^2) = (x - y)(f(x) + f(y)) }$ (Original equation)
2. ${ f(-x) = -f(x) }$ (Odd function property)
3. ${ f(x^2) = x f(x) }$

We'll use these equations to build a case that any function satisfying them must be linear.

Substituting y = 0 in the Original Equation

We already did this, but it’s a crucial step, so let’s reiterate. Substituting ${ y = 0 }$ into the original equation gives us:

${ f(x^2) = x f(x) }$

Substituting x = 0 in the Original Equation

Similarly, substituting ${ x = 0 }$ gives us:

${ f(-y^2) = -y f(y) }$

Combining These Results

We know that ${ f(x^2) = x f(x) }$. Let's try to express ${ f(x) }$ in a more useful form. Since we suspect ${ f(x) = ax }$, let's try to show that ${ f(x)/x }$ is constant for all ${ x \neq 0 }$.

Dividing by x

From ${ f(x^2) = x f(x) }$, if we let ${ x \neq 0 }$, we can write:

${ \frac{f(x^2)}{x^2} = \frac{x f(x)}{x^2} = \frac{f(x)}{x} }$

Let's define a new function ${ g(x) = \frac{f(x)}{x} }$ for ${ x \neq 0 }$. Then, the above equation tells us:

${ g(x^2) = g(x) }$

This is an interesting property. It says that the function ${ g }$ has the same value at ${ x }$ and ${ x^2 }$. This suggests that ${ g }$ might be constant, but we need to prove it rigorously.

Proving g(x) is Constant

To show that ${ g(x) }$ is constant, we need to show that ${ g(x) = g(y) }$ for any ${ x, y \neq 0 }$. Let's go back to the original equation and try a different approach.

Consider the original equation:

${ f(x^2 - y^2) = (x - y)(f(x) + f(y)) }$

Divide both sides by ${ x^2 - y^2 }$ (assuming ${ x^2 \neq y^2 }$):

${ \frac{f(x^2 - y^2)}{x^2 - y^2} = \frac{(x - y)(f(x) + f(y))}{x^2 - y^2} }$

${ \frac{f(x^2 - y^2)}{x^2 - y^2} = \frac{(x - y)(f(x) + f(y))}{(x - y)(x + y)} }$

${ \frac{f(x^2 - y^2)}{x^2 - y^2} = \frac{f(x) + f(y)}{x + y} }$

Now, let's rewrite ${ f(x) }$ as ${ x g(x) }$ and ${ f(y) }$ as ${ y g(y) }$:

${ \frac{f(x^2 - y^2)}{x^2 - y^2} = \frac{x g(x) + y g(y)}{x + y} }$

We want to show that ${ g(x) = g(y) }$. Let's consider the case when ${ x^2 - y^2 = z }$, where ${ z }$ is some real number. Then, the left side becomes ${ \frac{f(z)}{z} = g(z) }$. If we can show that ${ g(z) }$ is independent of ${ x }$ and ${ y }$, we'll be closer to proving that ${ g }$ is constant.

This is where things get a bit tricky. We need to manipulate the equation further to isolate ${ g(x) }$ and ${ g(y) }$ and show that they are equal. This often involves clever substitutions and algebraic manipulations.

A Different Approach to Proving Linearity

Instead of directly proving ${ g(x) }$ is constant, let's try a different tack. We suspect ${ f(x) = ax }$, so let's define a function ${ h(x) = f(x) - ax }$ for some constant ${ a }$. If we can show that ${ h(x) = 0 }$ for all ${ x }$, then we've proven that ${ f(x) = ax }$.

Let's choose ${ a = \frac{f(1)}{1} = f(1) }$. Now, we have ${ h(x) = f(x) - f(1)x }$. We want to show that ${ h(x) = 0 }$ for all ${ x }$.

First, notice that ${ h(0) = f(0) - f(1) \cdot 0 = 0 }$.

Also, ${ h(1) = f(1) - f(1) \cdot 1 = 0 }$.

Now, let's substitute ${ f(x) = h(x) + f(1)x }$ into the original equation:

${ h(x^2 - y^2) + f(1)(x^2 - y^2) = (x - y)(h(x) + f(1)x + h(y) + f(1)y) }$

${ h(x^2 - y^2) + f(1)(x^2 - y^2) = (x - y)(h(x) + h(y) + f(1)(x + y)) }$

${ h(x^2 - y^2) + f(1)(x^2 - y^2) = (x - y)(h(x) + h(y)) + f(1)(x^2 - y^2) }$

Subtracting ${ f(1)(x^2 - y^2) }$ from both sides:

${ h(x^2 - y^2) = (x - y)(h(x) + h(y)) }$

This equation looks similar to our original functional equation, but now for the function ${ h }$. If we can show that ${ h(x) = 0 }$ is the only solution to this equation, we'll have proven our result.

Now, let's substitute ${ y = 0 }$:

${ h(x^2) = x h(x) }$

And substitute ${ x = 0 }$:

${ h(-y^2) = -y h(y) }$

Also, ${ h(-x) = -h(x) }$ because ${ f }$ is odd and ${ x \mapsto f(1)x }$ is odd.

Now, let's try setting ${ y = -x }$ in the equation for ${ h }$:

${ h(0) = 2x(h(x) + h(-x)) }$

Since ${ h(0) = 0 }$, we have ${ 0 = 2x(h(x) - h(x)) }$, which is always true.

This approach is promising, but we still need to show that ${ h(x) = 0 }$ for all ${ x }$. This often involves a more advanced technique, such as using the density of rational numbers or a continuity argument.

Finalizing the Proof (Hint: Advanced Techniques Needed)

The final step to rigorously prove that ${ f(x) = ax }$ are the only solutions often requires concepts beyond basic algebra. Here’s a brief overview of the common approaches:

1. Continuity Argument: If we assume that ${ f }$ is continuous, then we can leverage the density of rational numbers to extend the solution from rational to real numbers. This involves showing that if ${ f(x) = ax }$ for rational ${ x }$, it must also hold for real ${ x }$.

2. Density of Rationals: Use the fact that rational numbers are dense in the reals. This means any real number can be approximated by a sequence of rational numbers. If we can show that ${ f(x) = ax }$ for all rational ${ x }$, then, with some additional arguments (like continuity), we can extend this to all real numbers.

The full rigorous proof is quite involved and often seen in advanced mathematical texts or problem-solving courses. However, we've successfully identified the form of the solution and outlined the key steps for a complete proof.

Conclusion

In summary, we've embarked on a journey to solve the functional equation:

${ f(x^2 - y^2) = (x - y)(f(x) + f(y)) }$

We started with initial substitutions, uncovered the odd function property, and deduced that linear functions of the form ${ f(x) = ax }$ are solutions. We then outlined the steps to rigorously prove that these are the only solutions, which often involves advanced techniques like continuity arguments or the density of rational numbers.

Solving functional equations is a challenging but rewarding endeavor. It requires a mix of algebraic manipulation, strategic thinking, and a deep understanding of function properties. Guys, I hope this walkthrough has been insightful and has equipped you with some useful techniques for tackling similar problems. Keep practicing, and you'll become functional equation masters in no time! Remember, the journey of a thousand miles begins with a single step – or, in this case, a clever substitution! Keep exploring, keep questioning, and most importantly, keep having fun with math!