Let $ X $ and $ Y $ be Banach spaces and let $ T : X \to Y $ be a bounded linear operator. Suppose that $ T $ is surjective, and thus by the Open Mapping Theorem, $ T $ is open.

I have come across a paper that then $ (\ker T)^{**} $ and $ \ker (T^{**}) $ are isomorphic. I can't seem to see why. Perhaps I am missing a theorem?

Any help would be appreciated.

Here's my thoughts so far. Since $ T $ is surjective:

$$ X/\ker (T) \cong Y $$

Is it true that $ T^{**} $ is also surjective? If so, then:

$$ X^{**}/\ker(T^{**}) \cong Y^{**} $$


$$ \left ( X / \ker (T) \right )^{**} \cong X^{**} / \ker (T^{**}) $$

Is this enough to conclude that $ (\ker (T))^{**} \cong \ker (T^{**}) $? And if so, I do not see where the openness of $ T $ is used.

  • $\begingroup$ Maybe the closed range theorem helps. $\endgroup$ – gerw Apr 10 at 7:20

I suggest to write your problem in terms of (topologically) short exact sequences (SES) in the category of Banach spaces: For $K=$ker$(T)$ you then have a SES $$ 0\to K \stackrel{i}{\to} X\stackrel{T}{\to} Y\to 0. $$ The dual sequence $$ 0\to Y^* \stackrel{T^*}{\to} X^*\stackrel{i^*}{\to} K^*\to 0 $$ is again exact (the exactness at $X^*$ requires ker$(i^*)=$im$(T^*)$ and follows from the the fact that $T$ is open and the exactness at $K^*$ is the Hahn-Banach theorem, that $T^*$ and $i^*$ are open onto their images follows from the open mapping theorem). Going to the second duals we get the SES $$ 0\to K^{**} \stackrel{i^{**}}{\to} X^{**}\stackrel{T^{**}}{\to} Y^{**}\to 0. $$ Hence, $K^{**}$ is isomorphic to ker$(T^{**})$. Moreover, they are not just isomorphic (by some isomorphism constructed in a fancy way) but canonically isomorphic (the bitransposed of the inclusion $i$ is an isomorphism).


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