If you have a linear map $h\mapsto T(h)$ from $H_1$ a real separable space, to Hilbert space $H_2$, it seem that this maps provides an isometry of $H_1$ onto a closed subspace of $H_2$. I try to understood this result. The closed supspace can be $\overline{\mathrm{Im}\,\, T}$. And \begin{align*} \|T(h)\|_{H_1}^2=\langle T(h), T(h) \rangle_{H_1} \end{align*} I don't see where is the isometry.
Thanks for reading and helps.