# prove $f$ is bounded given $f(t^2+u)=tf(t)+f(u)$

I have been trying to solve the functional equation $$f:\Bbb R \to \Bbb R$$ $$f(t^2+u)=tf(t)+f(u)$$. So far i have managed to show that $$f$$ is additive i.e. $$f(a+b)=f(a)+f(b)$$ which means that the condition can be simplified to $$f(t^2)=tf(t)$$. To show that $$f$$ is linear i need to show that it is bounded above or below on some interval. I have been trying to do this for some time but have gotten nowhere. Could someone please give me a hint?

• If you managed to show that $f$ is additive, then $f$ is a Cauchy functional equation, hence, it has only one family of solutions, namely $f(x)=cx$. Check en.wikipedia.org/wiki/Cauchy%27s_functional_equation Jan 7, 2019 at 0:00
• @EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity. Jan 7, 2019 at 0:03
• @KaviRamaMurthy True. Forgive my absent-mindedness. Jan 7, 2019 at 0:06
• @EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation Jan 7, 2019 at 0:10

$$af(a)+bf(b)+ bf(a) + af(b) = (a+b)(f(a)+f(b)) = (a+b)f(a+b) = f((a+b)^2)$$
$$=f(a^2+b^2+2ab)= f(a^2)+f(b^2)+f(2ab) = af(a)+bf(b)+f(2ab)$$.
Thus $$bf(a)+af(b)=f(2ab)$$. For $$a=1$$ this yields $$bf(1)+f(b)=f(2b)=f(b+b)=f(b)+f(b)$$. Thus $$f(b)=bf(1)$$, so $$f$$ is linear.