I want to show that the space $C^{\infty}(\Omega)'$ of continuous linear functionals on $C^{\infty}(\Omega)$ can be identified with the subspace $\mathcal{E}'(\Omega)$ (distributions with compact support) of $\mathcal{D}'(\Omega)$ in such a way that when $\Lambda \in C^{\infty}(\Omega)'$ is identified with $\Lambda_1 \in \mathcal{D}'(\Omega)$, then $$\Lambda(\phi) = \left\langle \Lambda_1,\phi \right\rangle .$$

What exactly must be shown here?


First, suppose that $u$ is a continuous linear functional on $C^\infty(\Omega)$. Since $\mathcal{D}(\Omega)$ is a subspace of $C^\infty(\Omega)$, obviously $u$ is a linear functional on $\mathcal{D}(\Omega)$. But does it act continuously with respect to the topology of $\mathcal{D}(\Omega)$?

Second, suppose that $u\in \mathcal{D}'(\Omega)$ and that the support of $u$ is a compact set, say $K\subset\Omega$. Now since $C^\infty(\Omega)$ is a superset of $\mathcal{D}(\Omega)$, you need some way to extend the action of $u$ from $\mathcal{D}(\Omega)$ to $C^\infty(\Omega)$, with the additional requirement that this action must be continuous with respect to the topology of $C^\infty(\Omega)$. Can it be done? Would this boil down to showing that $u:\mathcal{D}(\Omega)\to\mathbb{R}$ is continuous with respect to the topology on $\mathcal{D}(\Omega)$ that is induced by the embedding $\mathcal{D}(\Omega)\subset C^\infty(\Omega)$? Is the extension unique?


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