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?
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2 Answers
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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 $f(E)$.
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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.
