2
$\begingroup$

Let $A$ be a bounded linear operator ($A$ is not assumed to be an open mapping!) from a subspace $X_1$ of a Banach space to a subspace $Y_1$ of another Banach space, and that $\mathrm{codim}X_1+\mathrm{codim}Y_1<\infty$. I know that if $A$ maps one open subset $X$ of a Banach space onto one open subset $Y$ of another Banach space, since an open subset of a Banach space is second category, $A$ is surjective and an open map.

If $A$ maps one open subset $X$ of $X_1$ onto one open subset $Y$ of $Y_1$, is it true that $A$ maps $X_1$ onto $Y_1$, i.e. $A$ is an open map?

$\endgroup$
  • 1
    $\begingroup$ Open subset of the subspace or (open subset of the space)$\subset$ subspace?. The second case is impossible because nhoods of $0$ are absorbent. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 11 '17 at 10:55
  • $\begingroup$ I noticed that you edited your post, adding comments about the Baire category theorem. Surely Baire category is used to prove that surjective implies open? But isn't your question about showing that open implies surjective? $\endgroup$ – Kenny Wong Apr 11 '17 at 11:48
1
$\begingroup$

Edited in response to clarification:

I believe the following statement is true: If $f : X_1 \to Y_1$ is a linear mapping between normed spaces such that there exists a non-empty open set $U \subset X_1 $ such that $f(U)$ is open in $Y_1$, then $f$ is surjective.

(The fact that $X_1$ and $Y_1$ are finite-codimension subspaces of Banach spaces seems irrelevant.)

To prove this, pick any $x \in U$, and let $\widetilde U = -x + U$. Then $\widetilde U$ is an open set in $X_1$ containing the origin of $X_1$, and $f(\widetilde U) = -f(x) + f(U)$ is an open set in $Y_1$ containing the origin of $Y_1$. Since $f(\widetilde U)$ is open, there exists a $\delta$ such that $B_{Y_1}(\delta) \subset f(\widetilde U)$. But then, the union of the images of $n \widetilde U$ for all $n \in \mathbb N$ contains the union of $B_{Y_1}(n\delta)$ for all $n \in \mathbb N$, hence contains the whole of $Y_1$.

$\endgroup$
  • $\begingroup$ Thank you. There is ambiguity in my question and I have edited it. I do not assume $A$ to be an open map between $X_1$ and $Y_1$, but $A$ only maps an given open subset $X$ of $X_1$ onto an given open subset $Y$ of $Y_1$. $\endgroup$ – SimonChan Apr 11 '17 at 11:53
  • $\begingroup$ @SimonChan Okay, I edited the answer. I do hope this is correct - I didn't sleep much last night so my brain isn't working today! $\endgroup$ – Kenny Wong Apr 11 '17 at 13:12
0
$\begingroup$

Translate $A(\text{open set})$ converting the translated set in a nhood of $0$. Nhoods of $0$ are absorbent.

$\endgroup$

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.