Show that $T(\theta)= | \theta (0) | $ is not a distribution

Question

I have the following to show:

Exercise: Let $ $ $T:\mathcal{D}(\mathbb{R})\rightarrow\mathbb{R},$ $ \theta \in \mathcal{D}(\mathbb{R}),$ $ if $ $ T(\theta)=|\theta(0)| $, then T is not a distribution.

I have the following defition and theorem, the theorem is not needed, I added it for the context:

Definition: Let $\Omega \subset \mathbb{R}^{n} $, the dual space $\mathcal{D}'(\Omega)$ of the space $ \mathcal{D}(\Omega)$ of test function is called space of distributions, elements of $\mathcal{D}'(\Omega)$ are continuous linear functionals $T:\mathcal{D}(\Omega)\rightarrow \mathbb{C} $. Where $ $ $\mathcal{D} $ is the space of rapidly decreasing functions.

Theorem: If $ $ $T:\mathcal{D}(\Omega)\rightarrow \mathbb{C} $ is linear, then the following are equivalent

1. $T \in \mathcal{D}'(\Omega)$;
2. $\forall $ $ K \subset \Omega$ compact $ $ $\exists $ $ m\in \mathbb{N}$ and $M \ge0 $ $ $ s.t: $ \forall $ $ \psi \in \mathcal{D}_{K}(\Omega): |T(\psi)|\le M \cdot p_{m}(\psi)$.

Where $ $ $p_{m}:=sup\{ |\sigma^{\alpha} \psi|: \alpha \in \mathbb{N}^{n}_{0}, |\alpha| \le m \}$

My attempt: Let $ $ $\psi, \theta \in \mathcal{D}(\mathbb{R})$, then $ $ $T(\psi + \omega)=|(\psi + \omega)(0)|=|\psi(0)+\omega(0)| \le |\psi(0)|+ |\omega(0)|=T(\omega)+T(\psi)$, so $T$ is not linear and so not a distribution.

I am not sure how to achive a strict inequality, if it is needed, and if not I can't understand if this is sufficient. Any suggestion will be really appreciated, thank you in advance.

Answer 1

Your idea is correct. Take $\omega=-\theta$. Then $T(\omega+\theta)=T(0)=0$. But $|\theta(0)|+|\omega (0)|=2|\theta(0)|$ and you can find $\theta$ such that $\theta(0)\neq 0$.

Answer 2

Take some $\theta \in\mathcal D(\mathbb R)$ with $\theta(0) = 1$. Then, you have : $$T(\theta) = 1$$ If $T$ was linear, this would imply $T(-\theta) = -1$, whereas we have : $$T(-\theta) = 1$$ $T$ is not linear, so it cannot be a distribution.