Distributional derivative of a square integrable function and dual of Sobolev space

Question

This is a simple question: Let $f\in L^2(\mathbb{R})$, $f'$ be its distributional derivative, then is $f'$an element of $H^{-1}(\mathbb{R})$, the dual of Sobolev space $H^1(\mathbb{R})$? Also, if I take an element of $H^{-1}(\mathbb{R})$, say $q$, then is it true that there exists an element $Q\in L^2(\mathbb{R})$ such that $Q'=q$ in distributional sense? Why?

Answer 1

I am gonna work with the fact that $H^{-1}(\mathbb{R})=(H^1(\mathbb{R}))^\star$ and $H^1(\mathbb{R})=H_0^1(\mathbb{R})$. By definition $$\langle f',\varphi\rangle=-\int_\mathbb{R} f\varphi',\ \forall\ \varphi\in C_c^\infty(\mathbb{R})$$

Because $C_c^\infty$ is dense in $H_0^1$ we can conclude by the last integral and the Lebesgue theorem that for $v\in H_0^1$, $\langle f',v\rangle$ is well defined by the formula $$\langle f',v\rangle=\lim\langle f',\varphi_n\rangle$$ where $\varphi_n\in C_c^\infty$ converges to $v$ in $H_0^1$. Thereofore, $f'$ defines a bounded linear functional, i.e. $f'\in H^{-1}$.

For your second question, take a look in Brezis proposition 8.14.