If $f$ and $f''$ are square integrable is $f'$?

Question

Let $f:[0,\infty)\rightarrow \mathbb{R}$ be twice differentiable. Such that $\int_{0}^{\infty}f(x)^{2}dx<\infty$ and $\int_{0}^{\infty}f''(x)^{2}dx<\infty $, show that

$$\int_{0}^{\infty}f'(x)^{2}dx<\infty$$.

I'm pretty stumped on this problem. I've tried integration by parts but didn't really get anywhere. A short hint would be very appreciated. Thanks.

Answer 1

Consider $$\int_0^N f^\prime(x)^2 d x.$$ Integrate it by parts, to get: $$f(N) f^\prime(N) - f(0) f^\prime(0) - \int_0^N f(x) f^{\prime \prime}(x) dx.$$ Now note that the last integral can be bounded by Cauchy-Schwarz, so the only thing you need to bound is $f(N) f^\prime(N).$ Can you?

Answer 2

If $f$ and $f''$ are square integrable, then, by the Cauchy-Schwarz inequality, $ff''$ is integrable on $\mathbb{R}$. So, the limit of the following exists as $x\rightarrow\infty$: $$ \int_{0}^{x}f(t)f''(t)\,dt = f(x)f'(x)-f(0)f'(0)-\int_{0}^{x}f'(t)^{2}dt. $$ If $f'$ is not square-integrable, then $\lim_{x\rightarrow\infty}f(x)f'(x)=+\infty$ because the left side of the above is bounded and $\lim_{x\rightarrow\infty}\int_{0}^{x}f'(t)^{2}dt=\infty$. That means that, for any $M > 0$, there exists $R > 0$ such that $f(x)f'(x) > M$ whenever $x > R$. But that means $(f^{2})' > M/2$ for all $x > R$, which clearly contradicts the fact that $f$ is square-integrable. So $f'$ must be square-integrable.

Answer 3

A heavy handed answer: Let the Fourier transform of $f$ be $g,$ then the Fourier transform of $f^\prime$ is $s g,$ and the Fourier transform of $f^{\prime\prime}$ is $s^2 g.$ A function is in $L^2$ if and only if its Fourier transform is. But note that the Fourier transform of $f^\prime$ is dominated at $\pm \infty$ y that of $f^{\prime \prime}$ and at $0$ by that of $f,$ so must be in $L^2$ (since the other two are).