Get ready Part 27

Consider the function $\psi(x)= f^2(x)+f'^2(x) .$ Differentiating both sides, we get $\psi'(x)= 2f'(x)\big(f(x)+f''(x)\big)=-2xg(x)f'(x)^2,$ where, in the last step, we have used the given condition. Hence, we see that $\psi'(0)=0,$ and since $g(x) \geq 0$ , we find that the function $\psi(x)$ attends its (finite) maximum value $\psi(0)=C$ (say) at $x=0$ . Hence $\psi(x) \leq C, \forall x$ . Thus, $0\leq |f(x)| \leq \sqrt{C}, \forall x \in \mathbb{R}.$ Hence, $|f(x)|$ is bounded.