# Surjectivity in open mapping theorem

The open mapping theorem states that a surjective, bounded, linear operator $T$ between Banach spaces $E,F$ is open.

Let $B_k$ denote the open disk with radius $k$ around $0$ and $\overline\cdot$ the closure operation. Then one does something like

$$F=T(E)=\bigcup_{n\in\mathbb N} T\left(\overline{B_n}\right)=\bigcup_{n\in\mathbb N} \overline{T\left(B_n\right)}.$$

The first equality is directly surjectivity and the last one implicitly uses surjectivity as we use that we can't get larger then the whole space.

Now we have a decomposition of the space into closed subsets and can invoke Baire.

My question is: Can we assume something weaker than surjectivity? Clearly closed image would suffice but that is not a real weakening since the image is then a Banach space on its own. Clearly we need that the image has an inner point. Is it possible/interesting to require something weaker than closed image such that we may pull the closures out in the way we did it above without dropping out the image?

• It suffices to demand that $T(E)$ is non-meagre/of the second category. It then follows that $T$ is surjective, of course. – Daniel Fischer Feb 26 '16 at 14:26
• I don't know if this is what you asked, but this theorem is actually an equivalence: $T$ is open if, and only if, it is surjective. – Gustavo Dec 7 '16 at 3:35

## 1 Answer

It is enough to require that $T$ is almost open, i.e., $\overline{T(B_1)}$ has interior points. This implies that $0$ is an interior point, so that $B_1\subseteq \overline{T(B_k)}$ for some $k\in \mathbb N$.

Compared to surjectivity (where you have to solve all equations $T(x)=y$ exactly) you only have to produce approximate solutions $\|T(x)-y\| <\varepsilon$ but with an additional bound $\|x\|\le k \|y\|$.

Almost openness can be nicely characterized using the theorem of bipolars by the condition $\left(T^*\right)^{-1}(B^\circ) \subseteq k B^\circ$ (preimages of equicontinuous sets under the transposed operator are equicontinuous).