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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?

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    Note that "Hahn" is not spelled "Hanh". – 2017-01-13

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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.

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    That's a compact self adjoint operator, with spectrum containing 0...is it an embedding? – 2017-01-13
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    @Michael, given that you only want a continuous injection, it certainly meets these criteria. – 2017-01-13
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    @TomekKania Right, 0 not in the point spectrum. – 2017-01-13