$f$ 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 $$\int f'^2 \mathrm{d}x = \int f' \mathrm{d}f = f'f|_0^\infty - \int ff''\mathrm{d}x$$


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

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


1 Answer 1


$|ff^{\prime\prime}|\leq f^2+{f^{\prime\prime}}^2$ 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 $\mathbb R$.


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