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?

  • $\begingroup$ 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 $\endgroup$
    – EuxhenH
    Jan 7, 2019 at 0:00
  • $\begingroup$ @EuxhenH It is not true that every solution of Cauchy equation has this form. OP is not assuming continuity. $\endgroup$ Jan 7, 2019 at 0:03
  • $\begingroup$ @KaviRamaMurthy True. Forgive my absent-mindedness. $\endgroup$
    – EuxhenH
    Jan 7, 2019 at 0:06
  • $\begingroup$ @EuxhenH My issue is trying to prove one of the conditions that allows me to use the solution to that Cauchy equation $\endgroup$
    – Ben Martin
    Jan 7, 2019 at 0:10

1 Answer 1


$af(a)+bf(b)+ bf(a) + af(b) = (a+b)(f(a)+f(b)) = (a+b)f(a+b) = f((a+b)^2)$

$=f(a^2+b^2+2ab)= f(a^2)+f(b^2)+f(2ab) = af(a)+bf(b)+f(2ab)$.

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.

  • $\begingroup$ Thank you for your answer! $\endgroup$
    – Ben Martin
    Jan 7, 2019 at 0:25
  • $\begingroup$ You are welcome! $\endgroup$ Jan 7, 2019 at 0:40

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