Open mapping theorem (functional analysis)

Suppose A : X → Y is a surjective continuous linear operator. In order to prove that A is an open map, it is sufficient to show that A maps the open unit ball in X to a neighborhood of the origin of Y.

Let ${\displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0).}$ Then

${\displaystyle X=\bigcup _{k\in \mathbb {N} }kU.}$

Since A is surjective:

${\displaystyle Y=A(X)=A\left(\bigcup _{k\in \mathbb {N} }kU\right)=\bigcup _{k\in \mathbb {N} }A(kU).}$

But Y is Banach so by Baire's category theorem

${\displaystyle \exists k\in \mathbb {N} :\qquad \left({\overline {A(kU)}}\right)^{\circ }\neq \varnothing }$.

That is, we have c ∈ Y and r > 0 such that

${\displaystyle B_{r}(c)\subseteq \left({\overline {A(kU)}}\right)^{\circ }\subseteq {\overline {A(kU)}}}$.

Let v ∈ V, then

${\displaystyle c,c+rv\in B_{r}(c)\subseteq {\overline {A(kU)}}.}$

By continuity of addition and linearity, the difference rv satisfies

${\displaystyle rv\in {\overline {A(kU)}}+{\overline {A(kU)}}\subseteq {\overline {A(kU)+A(kU)}}\subseteq {\overline {A(2kU)}},}$

and by linearity again,

${\displaystyle V\subseteq {\overline {A\left(LU\right)}}}$

where we have set L=2k/r. It follows that for all y ∈ Y and all