Even function given a condition

Question

I would like to understand the rationale behind the solutions I found for the problem below.

Show that every function $f: \mathbb{R}^* \rightarrow \mathbb{R}$, satisfying the condition $f(xy)=f(x)+f(y)$, is even.

The solutions I've seen go in the following direction:

$\forall x \in \text{D}(f)$, $f([(-x)(-x)]) = f(-x) + f(-x) \Rightarrow 2f(-x) = f(x^2) = f(x*x) = f(x) + f(x)$

$2f(-x) = 2f(x) \Rightarrow f(-x) = f(x)$

I understand that the condition for the function to be even is $f(-x) = f(x)$. But I didn't understand why I can assume that the values of x and y are equal to $−x$ in the beggining of the sulotion.

Answer 1

The condition $$f(xy)=f(x)+f(y)$$ is satisfied for all $x,y\in \mathbb{R}^*$.

In particular, if we fix $x\in\mathbb{R}^*$, we also have $-x\in\mathbb{R}^*$, and we may take $y=-x$ in the condition, from where we deduce $f$ must be even.

Answer 2

The answer you found is a pretty natural answer. Looking at the proof of the contrapositive statement may also help show why we tend to start with an $x=y$ case in the forward direction, because it highlights the role of $y=x=-x$.

Using E to mean the constraints you wrote above, if the proposition is stated as:

(Function $f$ has form E) $\rightarrow$ ($f$ is even)

It suffices to prove the contraposition:

$\neg$ (f is even) $\rightarrow$ $\neg$ (Function f has form E)

So $f(x) \neq f(-x)$ by hypothesis.

Considering form E:

Suppose, for a contradiction, that $f$ has form E.

$f(x^2)=f((x)(x))=f(x)+f(x) = 2 f(x)$

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

So $f(-x)=f(x)=\frac{1}{2}f(x^2)$

But by hypothesis $f(-x) \neq f(x)$. So the supposition is false, making the contrapositive statement true, and the original statement true.