Find all functions s.t.$f(x^2-y^2)=xf(x)-yf(y)$

Question

Find all functions such that $\forall x,y\in \mathbb R$ $$f(x^2-y^2)=xf(x)-yf(y)$$ It’s obvious that $f(0)=0$, and by setting $x=0$ then $y=0$, we get $$\cases{f(x^2)=xf(x) \\ f(-y^2)=-yf(x)=-f(y^2)}$$ Hence $$f(x^2-y^2)=f(x^2)-f(y^2)$$ My guess is that $f(x)=x$, but I don’t know if that’s the only solution.

Answer 1

I'm not sure about the general case, but here's a solution that assumes $f$ is continuous at $1$. Fix some $a > 0$. Your first functional equation, $f(x^2) = xf(x)$, implies that $$f(a) = \sqrt{a} f(\sqrt{a}).$$ Applying this functional equation again yields $$f(a) = a^{3/4} f(a^{1/4}).$$ Iterated applications of the functional equation in this way then yields, in general, $$f(a) = a^k f(a^{\frac{1}{2^n}}),$$ where $k = \sum_{i=1}^n 2^{-i}$. Thus, $$\frac{f(a)}{a^k} = f(a^{\frac{1}{2^n}})$$ for all $k$. As $n \to \infty$, $k \to 1$, so the LHS tends to $f(a)/a$, while the right-hand side tends to $f(1)$ (since $f$ is continuous at $1$). Thus, $$f(a) = ca$$ for some constant $c$ for all $a > 0$. Since you have already shown that $f(0) = 0$, your second functional equation $f(-y^2) = -f(y^2)$ then extends this relation for all $a \in \mathbb{R}$.

Addendum: Your final equation, $f(x^2 - y^2) = f(x^2) - f(y^2)$, is equivalent to Cauchy's functional equation $f(x + y) = f(x) + f(y)$. Therefore, any of the following additional assumptions on $f$ would allow you to conclude that $f$ is linear (according to Wikipedia):

1. $f$ is continuous at some point.
2. $f$ is bounded on any interval.
3. $f$ is monotonic on any interval.
4. $f$ is Lebesgue measurable.

Without some additional assumption, you cannot conclude that $f$ is linear, and in fact nonlinear solutions to this functional equation exist. The question is whether or not your first equation $f(x^2 - y^2) = xf(x) - yf(y)$ is a stronger condition than $f(x + y) = f(x) + f(y)$, which I assume it must be (because otherwise, you probably wouldn't be asking this question). However, I'm not entirely sure how you would go about proving that.

Answer 2

Starting from what you found, we can set $a = x^2$ and $b = y^2$. Now, we have $$f(a - b) = f(a) - f(b)$$

With $a = 0$, we have $f(-b) = -f(b)$, so $f(b)$ is odd. Which permits to write

$$f(a - (-b)) = f(a+b) = f(a) - f(-b) = f(a) + f(b)$$

Also, notice that, if $a = b$, $f(a - b) = f(a - a) = f(0) = f(a) - f(a) = 0$, so we also proved that $f(0) = 0$.

Knowing the two pieces of information we just proved, some people would directly state that $f(a) = ma$, $m \in \mathbb{R}$, but let's prove that these two statements are equivalent. Manipulating $f(a + b) = f(a) + f(b)$, we can write the following:

$$f(a + b) - f(a) = f(b) - f(0) \Leftrightarrow \frac{f(a + b) - f(a)}{b} = \frac{f(0 + b) - f(0)}{b}$$

Up here, we can add $f(0)$ because $f(0) = 0$, and then we divide by $b \neq 0$. The cool fact here, is that we ended up to the definition of the derivative, when $\lim_{b\rightarrow 0}$. So, we can write,

$$\lim_{b\rightarrow 0}\ \frac{f(a + b) - f(a)}{b} = \lim_{b\rightarrow 0} \ \frac{f(0 + b) - f(0)}{b} = f'(a) = f'(0)$$

What this line tells us, is that the derivative of $f(a)$ is a constant: $f'(0)$, which means that $f(a)$ is equal to $ma + c$, but we also know that $f(0) = 0 = m\times 0 + c \Leftrightarrow c = 0$, so $f(a) = ma$.

By putting back the variable $x$, we have $f(x^2) = mx^2 \Leftrightarrow f(x) = mx$ with $m \in \mathbb{R}$

So, as you said, $f(x) = x$ is a valid solution, like $f(x) = \pi x$ is.

Answer 3

Another way of doing it is using calculus. For any fixed $y$, if you derivate the expression $f(x^2-y^2)=xf(x)-yf(y)$ with respect to $x$, you get $$ 2xf'(x^2-y^2) = f(x) + xf'(x).$$

Since this holds for every $y$, in particular, taking $y=x^2$ (and using that $f(0)=0)$, this yields $$ f(x) = -xf'(x) = -\frac{xdf}{dx}$$

Combining with the fact that $f(0)=0$, we have the EDO $$ -\frac{dx}{x} = \frac{df}{f}, \quad f'(0)=0.$$ Now solving this EDO yields the general solution $f(x)=cx+b$, but the initial condition $f'(0)=0$ implies that $b=0$, and therefore you must get the same $f(x)=cx$.