[proofplan] We extend the [Fourier transform](/page/Fourier%20Transform) from $\mathcal{S}(\mathbb{R}^n)$ to $\mathcal{S}'(\mathbb{R}^n)$ via the transposition formula $\hat{u}(\phi) := u(\hat{\phi})$. Well-definedness follows from the [continuity](/page/Continuity) of $\mathcal{F}$ on $\mathcal{S}$ (Theorem 228). Bijectivity follows by constructing the inverse via $(\mathcal{F}^{-1}u)(\phi) := u(\mathcal{F}^{-1}\phi)$ and verifying the composition identities. Continuity on $\mathcal{S}'$ (with the weak-* [topology](/page/Topology)) is immediate from the definition. [/proofplan] [step:Define $\hat{u}$ and verify it belongs to $\mathcal{S}'(\mathbb{R}^n)$] Let $u \in \mathcal{S}'(\mathbb{R}^n)$. Define $\hat{u}: \mathcal{S}(\mathbb{R}^n) \to \mathbb{C}$ by $\hat{u}(\phi) := u(\hat{\phi})$ for every $\phi \in \mathcal{S}(\mathbb{R}^n)$. Since $\mathcal{F}: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ is a continuous [linear map](/page/Linear%20Map) by the [Fourier Transform as Automorphism of Schwartz Space](/theorems/228), the map $\phi \mapsto \hat{\phi}$ is continuous on $\mathcal{S}(\mathbb{R}^n)$. The composition $\hat{u} = u \circ \mathcal{F}$ is therefore a continuous linear functional on $\mathcal{S}(\mathbb{R}^n)$, so $\hat{u} \in \mathcal{S}'(\mathbb{R}^n)$. [/step] [step:Verify linearity of $\mathcal{F}$ on $\mathcal{S}'(\mathbb{R}^n)$] For $u, v \in \mathcal{S}'(\mathbb{R}^n)$, $\lambda \in \mathbb{C}$, and $\phi \in \mathcal{S}(\mathbb{R}^n)$: \begin{align*} \widehat{u + \lambda v}(\phi) &= (u + \lambda v)(\hat\phi) = u(\hat\phi) + \lambda v(\hat\phi) = \hat{u}(\phi) + \lambda\hat{v}(\phi), \end{align*} so $\mathcal{F}: \mathcal{S}'(\mathbb{R}^n) \to \mathcal{S}'(\mathbb{R}^n)$ is linear. [/step] [step:Establish bijectivity by constructing the two-sided inverse] Define $\mathcal{F}^{-1}$ on $\mathcal{S}'(\mathbb{R}^n)$ by $(\mathcal{F}^{-1}u)(\phi) := u(\mathcal{F}^{-1}\phi)$. Since $\mathcal{F}^{-1}: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)$ is continuous (by the [Fourier Transform as Automorphism of Schwartz Space](/theorems/228)), $\mathcal{F}^{-1}u \in \mathcal{S}'(\mathbb{R}^n)$. For every $\phi \in \mathcal{S}(\mathbb{R}^n)$: \begin{align*} (\mathcal{F}^{-1}(\mathcal{F}u))(\phi) &= (\mathcal{F}u)(\mathcal{F}^{-1}\phi) = u(\mathcal{F}(\mathcal{F}^{-1}\phi)) = u(\phi), \end{align*} using $\mathcal{F} \circ \mathcal{F}^{-1} = \mathrm{Id}$ on $\mathcal{S}$. Similarly $\mathcal{F}(\mathcal{F}^{-1}u) = u$. Hence $\mathcal{F}$ is a bijection on $\mathcal{S}'(\mathbb{R}^n)$. [/step] [step:Prove continuity with respect to the weak-* topology on $\mathcal{S}'(\mathbb{R}^n)$] If $u_k \to 0$ in $\mathcal{S}'(\mathbb{R}^n)$ (meaning $u_k(\phi) \to 0$ for every $\phi \in \mathcal{S}(\mathbb{R}^n)$), then for every $\phi \in \mathcal{S}(\mathbb{R}^n)$: \begin{align*} \hat{u}_k(\phi) = u_k(\hat\phi) \to 0, \end{align*} since $\hat\phi \in \mathcal{S}(\mathbb{R}^n)$. Hence $\hat{u}_k \to 0$ in $\mathcal{S}'(\mathbb{R}^n)$, proving continuity of $\mathcal{F}$. The continuity of $\mathcal{F}^{-1}$ follows by the same argument. [/step]