Prove that $f : \mathbb{R} \rightarrow \mathbb{R}, f(x) = |x|$ belongs to $D'(\mathbb{R})$ and find its first and second distributional derivatives, $f', f''$.
To prove its linearity I used the linearity of the integral so: $\langle f, \alpha\phi+\beta\psi \rangle = \int_\mathbb{R}{|x|[\alpha\phi(x) + \beta\psi(x)]dx} = \alpha \langle f, \phi \rangle +\beta \langle f, \psi \rangle$
Then I tried to proved its continuity showing that $|\int_\mathbb{R}{|x|\phi(x)dx}| <M||\phi||_{D(\mathbb{R})}$ : $$\left|\int_\mathbb{R}{|x|\phi(x)dx}\right| \leq \left|\int_\mathbb{R}{|x| |\phi(x)|dx}\right| = \left|\int_{\operatorname{supp}(\phi)}{|x| |\phi(x)|dx}\right| \leq ||\phi||\int_{\operatorname{supp}(\phi)}{|x|dx}$$ but $\int_{\operatorname{supp}(\phi)}|x|dx$ doesn't converge if $\operatorname{supp}(\phi) = \mathbb{R}$.
So I tried to prove continuity by showing that it's continuous in zero.
$\langle f, 0 \rangle = 0$
$\langle f, \phi_k \rangle - \langle f, \phi \rangle = \int_\mathbb{R}|x|(\phi_k-\phi)dx \leq \sup|\phi_k-\phi|\int_\mathbb{R}|x|dx \rightarrow 0$ for $\phi_k \rightarrow\ \phi$
Is this correct? Do I have to make some observations to justify it? Is generally more convenient or easier to prove continuity in the latter way than in the former?