Rigged Hilbert space

Remarks

Injectivity of $H \to \Phi^\times$ comes from condition 2: if $\ip{h}{\varphi} = 0$ for all $\varphi$ in a dense subspace, then $h = 0$. Many authors use the continuous dual $\Phi'$ of linear forms instead of $\Phi^\times$; then the map $H \to \Phi'$ is conjugate-linear.

Example: $\Phi = \mathcal{S}(\R)$, the Schwartz space with its Fréchet topology, $H = L^2(\R)$, and $\Phi^\times$ the conjugate-linear tempered distributions. The inclusion is continuous because $\norm{\varphi}_{L^2} \le \sqrt{\pi}\, \sup_x \abs{(1 + x^2)^{1/2}\varphi(x)}$, and $\mathcal{S}(\R)$ is dense in $L^2(\R)$. The delta $\varphi \mapsto \overline{\varphi(a)}$ lies in $\Phi^\times$ but not in $H$.

Non-example: let $\Phi$ be the Schwartz functions that vanish on $[0,1]$. Its closure in $L^2(\R)$ consists of functions that vanish almost everywhere on $[0,1]$, so condition 2 fails. Then every $h \in L^2(\R)$ supported in $[0,1]$ gives the zero functional on $\Phi$, and $H \to \Phi^\times$ is not injective.