Function $f$ such that $f''\in L^2(\mathbb{R})$

Question

Let us assume that $f\in L^2(\mathbb{R})$ and $f''\in L^2(\mathbb{R})$ ($f'$ - first derivative,$f''$ - second derivative), i.e. $f$ is square-integrable, $f$ is differentiable, its first deriviative is absolutely continuous and second derivative of $f$ (which is defined almost everywhere) is square-integrable.

What might be said about the limits:

$$ \lim_{x\to\pm\infty} f(x)$$

and

$$ \lim_{x\to\pm\infty} f'(x).$$

Is it true that $f'\in L^2(\mathbb{R})$?

Note that if $f\in L^2(\mathbb{R})$ and $f'\in L^2(\mathbb{R})$ then $(|f|^2)'=\overline{f'}f+\overline{f}f'\in L(\mathbb{R})$ by Schwarz inequality and $\lim_{x\to\pm\infty} f(x) = 0$.

Answer 1

Hints: If $f,f''\in L^2(\Bbb R)$ does it follow that $f'\in L^2$? Look at the Fourier transform.

About limits at infinity: If $f$ is absolutely continuous, $f'\in L^2$ and $a<b$ then $$|f(b)-f(a)|=\left|\int_a^bf'(t)\,dt\right|\le(b-a)^{1/2}||f'||_2,$$so $f$ is uniformly continuous.

Answer 2

Suppose that $f,f'$ are absolutely continuous on $\mathbb{R}$ with $f,f''\in L^2$. Then $ff''\in L^1$, which means that the following limits exist $$ \int_{0}^{\pm\infty}ff''dt=-f(0)f'(0)+\lim_{x\rightarrow\pm\infty}f(x)f'(x). $$ If either of these limits is non-zero, then the following diverges in magnitude to $\infty$: $$ \int_{0}^{\infty}ff'dt= \left.\frac{f^2}{2}\right|_{0}^{\infty}. $$ But that is impossible because $f\in L^2$. Therefore $\lim_{x\rightarrow\pm\infty}f(x)f'(x)=0$. Knowing that gives $f'\in L^2$ because the right side of the following must converge: $$ \int_{-\infty}^{\infty}|f'|^2dt=f\overline{f}'|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}f\overline{f}''dt $$