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Are there natural conditions that ensure that a continuous linear map $\phi:V\to W$ between TVS maps any closed subspace $L\subset V$ to a closed subspace in $W$?

It is obviously satisfied if $W$ is finite-dimensional.

Do you know any reasonable generality that ensures it?

Thanks a lot

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If $V$ and $W$ are Banach spaces, the complete answer goes as follows. (1) If $T(V)$ is finite-dimensional then every subspace of $T(V)$ is closed, so $T$ does take all closed subspaces to closed subspaces. (2) If ${\rm ker}(T)$ is finite-dimensional, factor $T$ through a bounded linear map $\tilde{T}: V/{\rm ker}(T) \to W$; then $T$ will take closed subspaces to closed subspaces if and only if $\tilde{T}$ is bounded below, i.e., there exists $c > 0$ such that $\|\tilde{T}(x)\| \geq c\cdot \|x\|$ for all $x \in V/{\rm ker}(T)$. (3) If neither $T(V)$ nor ${\rm ker}(T)$ is finite-dimensional, then there is always a closed subspace of $V$ whose image under $T$ is not closed.

Proof of (2): Suppose $\tilde{T}$ is bounded below and let $V_0$ be any closed subspace of $V$. Since ${\rm ker}(T)$ is finite-dimensional, $V_0 + {\rm ker}(T)$ is closed and so the projection $\pi: V \to V/{\rm ker}(T)$ takes $V_0$ to a closed subspace of the quotient, and since $\tilde{T}$ is bounded below it also takes closed subspaces to closed subspaces, so $T = \tilde{T}\circ \pi$ must take $V_0$ to a closed subspace. Conversely, if $\tilde{T}$ is not bounded below then there exists a sequence $(x_n)$ in $V/{\rm ker}(T)$ with $\|x_n\| = 1$ for all $n$ and $\|\tilde{T}(x_n)\| \to 0$. Letting $V_0$ be the closed span of this sequence, $\tilde{T}$ takes $V_0$ bijectively onto its image, so by the Banach isomorphism theorem its image cannot be closed. QED

Proof of (3): according to the answer to this question, if neither $T(V)$ nor ${\rm ker}(T)$ is finite-dimensional, then there exists a closed subspace $V_0$ of $V$ such that $V_0 + {\rm ker}(T)$ is not closed. Then $T^{-1}(T(V_0)) = V_0 + {\rm ker}(T)$ is not closed, which shows that $T(V_0)$ cannot be closed. QED

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  • $\begingroup$ At least in the case of Hilbert spaces, these conditions (i.e. either dim kerT or dim ran T being finite) also seems necessary. $\endgroup$
    – Pietro Majer
    Commented Dec 21, 2022 at 3:45
  • $\begingroup$ Yes ... I think I can prove this if ker(T) is complemented, which would imply the Hilbert case. $\endgroup$
    – Nik Weaver
    Commented Dec 21, 2022 at 3:55
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    $\begingroup$ You shouldn't have deleted your comment, it was a really clever construction! $\endgroup$
    – Nik Weaver
    Commented Dec 21, 2022 at 3:55
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    $\begingroup$ Here it is again :) In the particular case where $T$ is the map $T: X\times Y\to Y\times Z$, $(x,y)\mapsto (y,0)$ the last condition reads either X or Y are finite dimensional. In this case it is necessary, because otherwise there would be bounded operators $L:X\to Y$ with non-closed range, and T(graph(L))=L(X). And for Hilbert spaces one can reduce to this situation. $\endgroup$
    – Pietro Majer
    Commented Dec 21, 2022 at 4:10
  • $\begingroup$ Thanks a lot. Does the situation improves if V,W are nuclear $\endgroup$
    – Rami
    Commented Jan 3, 2023 at 8:40

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