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?

  • 3
    $\begingroup$ The support of $\phi \in \mathcal{D}$ can't be unbounded... Moreover, the space of test functions isn't a normed space, so you can't write $\|\phi\|_{\mathcal{D}}$. You should use seminorms... $\endgroup$
    – Siminore
    Nov 5, 2014 at 12:21

1 Answer 1


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 $n$.

  • $\begingroup$ Thank you, I understand it now. $\endgroup$ Nov 5, 2014 at 13:05

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