Let $f$ be a function ​satisfying such that $f(x+y)=f(x)+f(y)$ Let $f$ be a function ​satisfying such that $f(x+y)=f(x)+f(y)$ and $f(x)=x^2g(x)$ for all $x$ and $y$, $g(x)$ is continuous function.then find $f''(x)$.
As we can see that $f(ax)=af(x)$, so can we take $f(x)=ax$? But if do so, then there is no use of $g(x)$. What am I missing here?

We immediately get $f(0)=0$, $f(-x)=-f(x)$ and by induction $f(nx)=nf(x)$ for $n\in\Bbb N$. In fact, as $f$ is odd, $f(nx)=nf(x)$ for all $n\in\Bbb Z$ and hence $qf(\frac pqx)=f(px)=pf(x)$ for $p,q\in \Bbb Z$, $q\ne 0$, i.e., $f(cx)=cf(x)$ at least for $c\in\Bbb Q$. In particular, $f(x)=xf(1)$ for $x\in\Bbb Q$. As $g$ is continuous, so is $f$, an we conclude that $f(x)=xf(1)$ for all $x\in \Bbb R$. This already shows that $f'(x)=f(1)$ is constant, hence $f''(x)=0$.
But in fact we already have $f(x)=0$ for otherwise there is no continuous $g$ with $cx=x^2g(x)$ for all $x$.
