Let $T : \mathcal{D}(\mathbb{R}) \to \mathbb{R}$ be given by $T(\phi) = |\phi(0)|$. Show that $T$ is not a distribution.

Question

As the title states, I wish to show that $T(\phi) = |\phi(0)|$ is not a distribution. I assume I need to show that the bound $|T(\phi)| \leq C \sum_{|\alpha| \leq n} ||D^{\alpha}\phi||_{L^{\infty}}$ fails to hold.

I am confused since the dirac distribution is defined as $\delta_0(\phi) = \phi(0)$ which means that $T(\phi) = |\phi(0)| = |\delta_0(\phi)|$, right?

What am I missing?

Answer 1

You have only to check that $\langle T, \varphi_1 + \varphi_2\rangle=\langle T, \varphi_1 \rangle +\langle T, \varphi_2\rangle$ is false in general with the definition of $\langle T,\varphi\rangle=|\varphi(0)|$.

Briefly said: the absolute value blocks the linearity.

Your last question deals implicitely with the meaning of $|\delta|$ ; but we have precisely shown that this definition of $|T|$ is faulty...