3
$\begingroup$

Suppose we have a surjective bounded linear operator acting between Banach spaces. By the Open Mapping Theorem it maps open sets in the domain to open sets in the codomain. Must the image of a closed linear subspace of the domain be closed then?

$\endgroup$
5
$\begingroup$

A construction very similar to that seen in the question Do there exist closed subspaces $X$, $Y$ of Banach space, such that $X+Y$ is not closed? applies here.

If $f:E\to F$ is any bounded linear operator such that $f(E)$ is not closed, then $g:E\oplus F\to F$ defined by $$g(x+y)=f(x)+y$$ is surjective but maps $E\oplus 0$ to the non-closed subspace $f(E)$.

$\endgroup$
3
$\begingroup$

No not necessarily: Consider the map $\Phi$ sending $l^2$ to $l^2$, where $\Phi(x)_n=x_{2n}$. That is: the map just pulls out the even coordinates. Clearly this map is surjective and bounded.

Now consider the subspace $Y$ of $\ell^2$ consisting of points where $x_2=x_1/2$, $x_4=x_3/4$, $x_6=x_5/6$ etc. This is a closed subspace (it's the intersection of the kernels of a countable collection of bounded linear functionals).

The claim is that $\Phi(Y)$ is not closed. Why not? You can see that $\Phi(Y)$ contains all points with finitely many non-zero terms. This means that the closure of $\Phi(Y)$ is all of $\ell^2$. But there are points of $\ell^2$ not in $\Phi(Y)$.

I'll leave you to find one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.