# $\int f^2$ and $\int f''^2$ is convergent then so is $\int f'^2$

$$f$$ is second order differentiable in $$[0,+\infty)$$. And $$\int_0^\infty f^2$$ and $$\int_0^\infty f''^2$$ is convergent. Prove that $$\int_0^\infty f'^2$$ is convergent.

I can prove the case that $$f$$ and $$f''$$ is monotonic. In this case $$f \rightarrow 0$$ and $$f'' \rightarrow 0$$ when $$x \rightarrow +\infty$$. Therefore $$\int f'^2 \mathrm{d}x = \int f' \mathrm{d}f = f'f|_0^\infty - \int ff''\mathrm{d}x$$

and

$$\int ff'' \le (\int f^2 )^{\frac{1}{2}} (\int f''^2)^{\frac{1}{2}}$$ in convergent, so is $$\int f'^2$$.

But I don't know how to do in the general case.

$$|ff^{\prime\prime}|\leq f^2+{f^{\prime\prime}}^2$$ so that $$ff^{\prime\prime}$$ is integrable. But an integration by parts gives $$\int_a^b{f^\prime}^2=\left[ff^\prime\right]_a^b-\int_a^bff^{\prime\prime}$$ Thus, $${f^\prime}^2$$ is integrable iff $$ff^\prime$$ has finite limits at $$\pm\infty$$.
If $${f^\prime}^2$$ is not integrable on $$\mathbb R_+$$, then $$\int_{0}^{+\infty}{f^\prime}^2=+\infty$$ so that $$ff^\prime\to_{+\infty}+\infty$$. Then, $$ff^\prime(x)\geq 1$$ for $$x\geq x_0$$, so $$\frac12(f^2(x)-f^2(x_0))\geq x-x_0$$ which contradicts the fact that $$f^2$$ is integrable. So, $${f^\prime}$$ is integrable on $$\mathbb R_+$$. The same argument show that it is also integrable on $$\mathbb R_-$$, so, it is integrable on $$\mathbb R$$.