Recall that a mapping , where and are topological spaces, is called open if the image under of each open set of is open in .
Open Mapping Theorem
A bounded linear operator between Banach spaces is open if and only if it is surjective.
Proof: Let and be Banach spaces and let be a bounded linear operator. Let and denote the open unit balls in and , respectively.
First, suppose that is surjective. The balls , , cover . Since is surjective, their images cover . This remains true, if we take closures: . Hence, we have written the space , which is assumed to have a complete norm, as the union of countably many closed sets. It follows form the Baire Category Theorem that has nonempty interior for some . Thus there are and such that . Choose a with . It is a well known fact, that in a normed space the translation by a vector and the multiplication with a nonzero scalar are homeomorphisms and thus compatible with taking the closure. We conclude . Since is a bounded set, it is contained in a ball for some . Thus, . With we obtain .
Clearly, every is the limit of a sequence , where . However, the sequence may not converge! We show that it is possible to find a convergent sequence in such that . To construct , we recursively define a sequence with for . The sequence starts with . Given , one has . By the definition of closure, there exists a such that lies in the open -ball about . This means that . Now define as the -th partial sum of the series . The series converges, because it converges absolutely (Here we use the completeness of ). The latter is true because . This also shows that each and the limit lie in . The auxiliary sequence converges to by construction. Therefore, in the limit
as desired. It follows from the continuity of that , thus .
In the preceding paragraph it was proven that . Hence, where . To show that is open, consider any open set . If lies in , there exists a such that . Since is open, there is an such that . Applying , we find . Combine with to see . Hence, is open. This shows that is indeed an open mapping.
Conversely, suppose that is open. TODO
For a bijective mapping between topological spaces, to say that it is open, is equivalent to saying that its inverse is continuous. The inverse of a bijective linear map between normed spaces is automatically linear and thus continuous if and only if it is bounded. As a corollary to the Open Mapping Theorem we obtain the following:
Bounded Inverse Theorem
If a bounded linear operator between Banach spaces is bijective, then its inverse is bounded.
Also known as Inverse Mapping Theorem.