> [!definition] > > Let $U, V \subset \real^d$ be [[Open Set|open]], $\cc \subset \cd'(U)$ be a subspace, and $T: \cc \to \cd'(V)$ be a [[Linear Transformation|linear map]]. If there exists a [[Continuity|continuous]] linear map $T^*: \cd(V) \to \cd(U)$ on the [[Space of Test Functions|test functions]] such that > $ > \angles{TF, \phi} = \angles{F, T^*\phi} \quad \forall F \in \cc, \phi \in \cd(U) > $ > then there is a unique extension $\ol T$ of $T$ to $\cd'(U)$ into a continuous linear map > 1. $\langle \ol TF, \phi \rangle = \angles{F, T^*\phi}$ for all $F \in \cf'(V), \phi \in \cd(U)$. > 2. $\ol T: \cd'(U) \to \cd'(V)$ is continuous with respect to the [[Weak Topology|weak-* topology]]. > > This extension is commonly directly denoted as $T$. > > *Proof*. Take the [[Weak Adjoint|adjoint]] of $T^*$ to recover the desired map.