TITLE: If $Y\subset X$ are Banach spaces such that $Y$ is dense in $X$, is it true that $X'$ is dense in $Y'$?
QUESTION [8 upvotes]: 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?

REPLY [2 votes]: Example.  Sending each sequence to itself, we get
$$
l^1 \longrightarrow c_0
$$
injective with dense range.  Taking adjoint, we get
$$
l^\infty \longleftarrow l^1
$$
injective but non-dense range.  
Dense range of a bounded linear transformation by itself implies that the adjoint is injective.