In functional analysis, the open mapping theorem, also known as the Banach–Schauder theorem or the Banach theorem (named after Stefan Banach and Juliusz Schauder), is a fundamental result that states that if a bounded or continuous linear operator between Banach spaces is surjective then it is an open map. A special case is also called the bounded inverse theorem (also called inverse mapping theorem or Banach isomorphism theorem), which states that a bijective bounded linear operator T {\displaystyle T} from one Banach space to another has bounded inverse T − 1 {\displaystyle T^{-1}} .
Statement and proof
The proof here uses the Baire category theorem, and completeness of both E {\displaystyle E} and F {\displaystyle F} is essential to the theorem. The statement of the theorem is no longer true if either space is assumed to be only a normed vector space; see § Counterexample. The proof is based on the following lemmas, which are also somewhat of independent interest. A linear map f : E → F {\displaystyle f:E\to F} between topological vector spaces is said to be nearly open if, for each neighborhood U {\displaystyle U} of zero, the closure f ( U ) ¯ {\displaystyle {\overline {f(U)}}} contains a neighborhood of zero. The next lemma may be thought of as a weak version of the open mapping theorem.
Proof: Shrinking U {\displaystyle U} , we can assume U {\displaystyle U} is an open ball centered at zero. We have f ( E ) = f ( ⋃ n ∈ N n U ) = ⋃ n ∈ N f ( n U ) {\displaystyle f(E)=f\left(\bigcup _{n\in \mathbb {N} }nU\right)=\bigcup _{n\in \mathbb {N} }f(nU)} . Thus, some f ( n U ) ¯ {\displaystyle {\overline {f(nU)}}} contains an interior point y {\displaystyle y} ; that is, for some radius r > 0 {\displaystyle r>0} ,
B ( y , r ) ⊂ f ( n U ) ¯ . {\displaystyle B(y,r)\subset {\overline {f(nU)}}.}
Then for any v {\displaystyle v} in F {\displaystyle F} with ‖ v ‖ < r {\displaystyle \|v\|<r} , by linearity, convexity and ( − 1 ) U ⊂ U {\displaystyle (-1)U\subset U} ,
v = v − y + y ∈ f ( − n U ) ¯ + f ( n U ) ¯ ⊂ f ( 2 n U ) ¯ {\displaystyle v=v-y+y\in {\overline {f(-nU)}}+{\overline {f(nU)}}\subset {\overline {f(2nU)}}} , which proves the lemma by dividing by 2 n {\displaystyle 2n} . ◻ {\displaystyle \square } (The same proof works if E , F {\displaystyle E,F} are pre-Fréchet spaces.) The completeness on the domain then allows to upgrade nearly open to open.
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