$\int_0^\infty |f|^2<\infty, \int_0^\infty |f''|^2<\infty$, then $\int_0^\infty |f'|^2<\infty$. If $f$ is twice continuously differentiable on $[0,+\infty)$, and $\int_0^\infty |f|^2<\infty, \int_0^\infty |f''|^2<\infty$, then $ \int_0^\infty |f'|^2<\infty$? 
My attempt: $\int |f'|^2
=\int f'd f
=ff'|_0^\infty-\int_0^\infty f'' f $. The last term is easy By Schwarz. But what about $ff'$ at $0$ and $\infty$?

This is a variant of Landau-Kolmogorov and it follows essentially the way you wrote it, noting that $\int_{0}^{X} |f'(x)|^2dx \to \infty$ as $X \to \infty$ if and only if $(ff')(X) \to \infty$ (precisely as in your argument and noting that LHS is positive as it the integral of a square, so $ff'$ is increasing plus a bounded term, in particular it would be eventually positive) but then $2\int_{0}^{X} (ff'(x))dx \to \infty$ and that is not possible since the previous integral is obviously $f^2(X)-f^2(0)$ ( $f \in L^2$ means $f^2(x)$ cannot go to infinity as $x \to \infty$)
