KAIST POW 2012-14 : Equation with integration

Equation with integration

 Determine all continuous functions $f:(0, \infty) \to (0, \infty)$ such that $\int_{t}^{t^3} f(x) dx =2 \int_{1}^{t} f(x) dx$ for all $t > 0$.

 
 Proof. It's easy to check that $f(t)=k/t$ satisfies the equation for any constant $k>0$. The purpose is to show that these are all solutions of the equation. Since $f$ is continuous, by Fundamental Theorem of Calculus, we can take derivatives to both sides and get $3f(t^3 )t^2 - f(t) = 2f(t)$. That is, $f(t^3)t^2 = f(t)$. We define a continuous function $g$ by $g(t)=tf(t)$ for $t>0$. Then we get $g(t^3)=g(t)$. Let $a=g(1) > 0$.

  Claim : $g(t)=a$ for all $t\in (0, \infty)$.

  When $t\in (0, \infty )$ is given, we define a sequence $\{t_n\}$ by $t_1 =t$ and $t_{n}={t_{n-1}}^{1/3} = t^{1/{3^n}}$. Then $g(t_n)$ is a constant sequence. $\{t_n\} \to 1$ since $\log t_n = {(\log t )}/ {3^n} \to 0$. Since $g$ is continuous, $g(t)=\lim_{n\to\infty} g(t_n)=g(1)=a$. Therefore, $g(t)=a$ for all $t\in(0,\infty)$.

This shows that if $f$ satisfies given equation, $f(t)=a/t$ for some positive $a=g(1)=f(1)$. Therefore, a continuous $f$ satisfies the given equation if and only if $f(t)=k/t$ for some constant $k>0$.