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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?

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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 $Y'$.

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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.

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