An isomorphism between the dual space of a Hilbert space and a larger space containing the same Hilbert space Let $G \subset H \subset F$ be three Hilbert spaces such that the smaller ones are continuously and densely embedded into the larger ones. Furthermore, assume
$$|\langle h, g\rangle_H|\le \|h\|_{F}\|g\|_G,$$
for all $h\in H$ and $v\in G$.
Define a mapping from $F$ to $G^*$ (the continuous dual space of $G$),
$$(Jf) (g):= \lim_{n\to \infty} \langle h_n, g \rangle_H,$$
where $h_n \to f$ in $\|\cdot\|_F$.
I believe that $J$ is an isometric isomorphim between $F$ and $G^*$ (i.e. a bijective mapping preserving norm).
It is easy to show that $J$ is well-defined (not dependent on the converging sequence) and $\|Jf\|_{G^*} \le \|f\|_F$. How to proceed further?
 A: Consider $G=H=F$ and the embedding map being for $i_1: G\to H$ multiplication with $\frac12$ and for $i_2:H\to F$ just the identity. Then your condition is 
$$|\langle h, i_1(g)\rangle| =\frac12 |\langle h, g\rangle|≤ \|h\|\,\|g\|,$$
which is satisfied.
What is $J$? Note that $Jf(g)=\frac12 \langle f,g\rangle$, so $J$ is not an isometry. But $J$ is topologically an isomorphism, in fact this is always true.
Consider a general scenario as sketched in your question:
Note that $J$ is injective, for if $Jf(g)=0$ for all $g$ then $\langle h_n, g\rangle_H\to0$ for all $g$ and and $\|h_n\|_H\to 0$ since $G$ has dense image in $H$, then by continuity of the embedding $H\to F$ you get $\|h_n\|_F\to0$ so $f=0$.
Next suppose there is a $g\in G$ so that $Jf(g)=0$ for all $f\in F$. In particular you get that $\langle h, g\rangle_H =0$ for all $h\in H$ which implies that $g=0$. Thus the orthogonal complement of the image of $J$ is zero, and $J$ is surjective.
By the open mapping theorem $J$ is a topological isomorphism.
