Suppose $X$ and $Y$ are Banach spaces, and $X \stackrel{\Phi}{\hookrightarrow} Y$ is an embedding (continuous linear injection). Let $\Phi' : Y' \rightarrow X'$ be the induced map on the dual spaces.
If $\Phi(X)$ is closed, then by a Hahn-Banach argument, $\Phi'$ is surjective, i.e. a bounded linear functional on $\Phi(X)$ can be extended to $Y$.
Question: If $\Phi(X)$ is not closed, the Hahn-Banach argument breaks down, correct? If so, counter-example where $\Phi$ is not surjective?
For example, let $X$ and $Y$ be $\ell^2$, the map $\Phi$ being $(x_n)_{n=1}^\infty\mapsto (x_n/n)_{n=1}^\infty$.
The induced $\Phi'$ is defined by the same formula (essentially the same map), and it's not surjective as, for example, $(1/n)_{n=1}^\infty$ is not in its range.