TITLE: Does existence of the second weak derivative of $f\in L^2$ imply existence of the first?
QUESTION [9 upvotes]: Let's consider a function $f\in L^2(\mathbb{R})$ for which the second weak derivative exists and lie in $L^2(\mathbb{R})$,
i.e. there exists $f''\in L^2(\mathbb{R})$ such that for all $\varphi\in C_0^\infty(\mathbb{R})$ the following integral equation stands:
$$
\int\limits_\mathbb{R}f(x)\varphi''(x)dx=\int\limits_\mathbb{R}f''(x)\varphi(x)dx.
$$
My question is, having this can we assume that weak $f'$ also exists in $L^2(\mathbb{R})$?

Suppose we found a normal (not generalized) function $g:\mathbb{R}\to\mathbb{C}$ such that for all $\varphi\in C_0^\infty(\mathbb{R})$
$$
\int\limits_\mathbb{R}f(x)\varphi'(x)dx=-\int\limits_\mathbb{R}g(x)\varphi(x)dx.
$$
Then
$$
\langle -f'', f \rangle_{L^2}=-\int\limits_\mathbb{R}f''(x)f(x)dx=\int\limits_\mathbb{R}g(x)g(x)dx=\|g\|_{L^2}^2
$$
which means, that $g$ is in $L^2(\mathbb{R})$. But what guarantees us the existence of such $g$?

REPLY [4 votes]: This is clear if you look at the Fourier transform. If $f,f''\in L^2$ then  $$\int|\hat f(\xi)|^2<\infty$$ and $$\int|\xi|^4|\hat f(\xi)|^2<\infty,$$and hence $$\int|\xi|^2|\hat f(\xi)|^2<\infty,$$because $|\xi|^2\le\max(1,|\xi|^4)$.