$\text{supp }u =\{0\}$ implies $u=c\delta_0$ in distributional sense?

Question

Given $\text{supp } (u) =\{0\}$ where $u\in D(X)^\prime$ is a distribution and $X\subset \mathbb{R}$. Does this already imply that $u=c \delta_0$? for some constant $c$.

Answer 1

No, for example $\operatorname{supp} \delta'$ Is also $\{0\}$. More generally, linear combinations of derivatives of $\delta$ are supported by a single point.

Answer 2

In the case of the single-element-Supplement (here: 0) you have simply $u=c (rect(x))$ where $rect(x)$ is the rectangle function which is 1 in $[- \epsilon, \epsilon]$ for $\epsilon \rightarrow 0$ and Zero elsewhere. To match this with Distribution properties, you have to require normalizing; to obtain a normalized Distribution you have to choose $rect(x) \rightarrow \frac{1}{2 \epsilon}rect(x)=\delta(x)$. Therefore, a single-element-Supplement would imply $u=c \delta_0$.