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?
In order to prove continuity, we have to go back to the definition, namely, we have to show that if $\varphi_n\to 0$ in $\mathcal D(\mathbb R)$, then $\langle f,\varphi_n\rangle\to 0$. Convergence to $0$ in $\mathcal D(\mathbb R)$ means that the supports of the functions $\varphi_n$ is contained in a common compact and if $K$ is compact, then $\sup_{x\in K}|\varphi^{(k)}_n(x)|\to 0$ for each integer $k$.
In the particular case of $f(x)=|x|$ (or any locally integrable function), we have $$\left|\int_{\mathbb R} |x|\varphi_n(x)\mathrm dx\right|=\left|\int_{-R}^R|x|\varphi_n(x)\mathrm dx\right|\leqslant2R\sup_{x\in [-R,R]}|\varphi_n(x)|,$$ where $R$ is such that $\mathrm{supp}(\varphi_n)\subset [-R,R]$ for each $n$.