How can I prove $\delta_{\lambda} \in \mathcal{E}'(\mathbb{R})$?

Question

How can I prove, formally, by its definition, that $\delta_{\lambda}=\delta(x-\lambda)$ is a distribution in $\mathcal{E}'(\mathbb{R})$ for every $\lambda$ where I choose to center my distribution?

Answer 1

Hopefully, it's clear that $\delta_\lambda$ is a distribution, through the same argument as $\delta_0$ ($=\delta$). Also, $\text{supp}\ \delta_\lambda=\{\lambda\}$ (in the same way as $\text{supp}\ \delta_0=\{0\}$), which is compact. So, $\delta_\lambda\in\mathcal{E}'$.

Answer 2

Alternatively, one can show it is a continuous linear functional on the space $\mathscr{E}(\mathbb{R})$. The topology of this space can be defined by the collection of seminorms $$ ||f||_{k,m}=\max_{0\le j\le k}\ \max_{x\in [-m,m]}|f^{(j)}(x)| $$ for $k,m$ nonnegative integers. The linear map $\delta_{\lambda}:f\mapsto f(\lambda)$ satisfies $$ |\delta_{\lambda}(f)|\le ||f||_{0,m} $$ for fixed $m$ chosen such that $|\lambda|\le m$. So $\delta_{\lambda}$ is continuous and belongs to the topological dual $\mathscr{E}'(\mathbb{R})$.