Let $\mathbb{R}$ be the set of Real numbers. Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$f(x^2-y^2)=xf(x)-yf(y)$$ for all pairs of real numbers $x$ and $y$.
This is a problem from USAMO 2002. The solution on the page I linked, is a bit tedious (maybe for good reason). I tried it this way. Is it valid?
Putting $x=y$ yields $f(0)=0$.
Putting $x=-y$ gives $$0=f(0)=xf(x)+xf(-x)=x(f(x)+f(-x))$$ which is true for all $x\in\mathbb{R}$ and hence, $f(-x)=-f(x)$ and the function is odd. Since the function is odd, I'll consider only $x\in\mathbb{R}_+$
Now, this is the part I'm not sure about.
$$f(x^2)=xf(x)=x\sqrt{x}f(\sqrt{x})=\dots=\lim_{n\to\infty}x^{1+\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2n}}f(x^\frac{1}{2n})=x^2\lim_{n\to\infty}f(x^\frac{1}{2n})$$
For non-zero $x$, $$\lim_{n\to\infty}x^\frac{1}{2n}=1$$ (How do I prove this? I've seen it work on a calculator, and I know that for $x\in(0,1)$, $x^{\frac{1}{n}}>x$ and for $x\in(1,\infty)$, $x^{\frac{1}{n}}<x$ and that $1$ is a fixed point. Is that sufficient?)
So, $$f(x^2)=x^2f(1)=cx^2$$ and with $x^2=t$ we arrive at $$f(t)=ct$$ This also satisfies the odd criterion.
Is the proof above correct? Am I assuming somethings without realising? I know these are risky waters, so I'd rather be careful.
EDIT: I realise that I'm assuming continuity when I take the limit inside: $$\lim_{n\to\infty}f(x^\frac{1}{2n})=f(\lim_{n\to\infty}x^\frac{1}{2n})$$ So, the naturally next question is how can I prove continuity?
As the solution on the page you linked notes, all solutions must be additive: $$ f(x+y) = f(x) + f(y)$$ Now there are discontinuous additive functions. But in this case we have additional structure. In fact, taking $x = s+1$ and $y = s-1$ in your original equation, you can simplify the result (using additivity) to get $$ f(s) = s f(1)$$