If $Y$ is a dense subspace of a Banach space $(X,\|\cdot\|_1)$ and $(Y,\|\cdot\|_2)$ is a Banach space such that the inclusion from $(Y,\|\cdot\|_2)$ into $(X,\|\cdot\|_1)$ is continuous, then it is well defined, linear, injective, and continuous in the dual norm topology the map:
$$j:X'\to Y', f\mapsto f|_{Y},$$
where $X'$ is the topological dual of $(X,\|\cdot\|_1)$ and $Y'$ is the topological dual of $(Y,\|\cdot\|_2)$.
So, we can identify $X'$ as a subset of $Y'$.
Is it true that $X'$ is dense in $Y'$ in the norm topology? If not, is true that $X'$ is dense in $Y'$ at least in the weak* topology?
Edit: in this question it is addressed the case where $(Y,\|\cdot\|_2)$ is reflexive, obtaining that in this case (thanks to Hahn-Banach theorem) $X'$ is dense in the norm topology of $Y'$. In the answer to this question it is shown that counterexamples to density in norm topology exist if the reflexivity of $(Y,\|\cdot\|_2)$ is not assumed, e.g. by taking $(X,\|\cdot\|_1):=(l^2,\|\cdot\|_{l^2})$ and $(Y,\|\cdot\|_1):=(l^1,\|\cdot\|_{l^1})$. However, in this counterexample $X'$ is still dense in the weak* topology of $Y'$.
So, it remains to answer only the following part of the original question:
Is true that $X'$ is dense in $Y'$ in the weak* topology?
Best Answer
With this $G$ is dense in $X^*$ in weak* sense if and only if $G$ is total set fact, the proof is rather easy.
Let me denote the embedding of $Y$ into $X$ by $i$. Then $i(Y)$ is dense in $X$. And the question is, whether $i^*(X')$ is weak-star dense in $Y'$.
Due to the result above, we have this density if and only if for all $y\in Y$ $$ (i^*f)(y) =0 \quad \forall f\in X' \Rightarrow y=0. $$ Now let $y\in Y$ be given such that $(i^*f)(y)=f(iy)=0$ for all $f\in X$. This implies $iy=0$, and by injectivity of $i$, $y=0$. So $i^*X'$ is total and hence dense in $Y'$.