Does existence of the second weak derivative of $f\in L^2$ imply existence of the first?

Question

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 $L^2(\mathbb{R})$?

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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 $g$?

Answer 1

Define $ f'(x) = \int_0^x f''(t)d t $, then $|f'(y)|\leq (\int _0^y |f''|^2 )^{1/2}\sqrt{y}\leq ||f''||_2 \sqrt{y}$.

Therefore $$\int\limits_{\mathbb{R}} f''(x) \phi (x) d x = -\int\limits_{\mathbb{R}} f'(x) \phi '(x) d x $$

And so $$\int\limits_{\mathbb{R}} f(x) \phi''(x) d x=-\int\limits_{\mathbb{R}} f'(x) \phi '(x) d x $$

Answer 2

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 $|\xi|^2\le\max(1,|\xi|^4)$.