Closed range theorem explained

In the mathematical theory of Banach spaces, the closed range theorem gives necessary and sufficient conditions for a closed densely defined operator to have closed range.

The theorem was proved by Stefan Banach in his 1932 Théorie des opérations linéaires.

Statement

Let

X and Y be Banach spaces, T : D(T) \to Y a closed linear operator whose domain D(T) is dense in X, and T' the transpose of T . The theorem asserts that the following conditions are equivalent:R(T), the range of T, is closed in Y. R(T'), the range of T', is closed in X', the dual of X. R(T) = N(T') ^{\perp} = \left\{ y \in Y : \langle x ^{*,y} \rangle = 0 { forall }\quad x^* \in N(T') \right\}.R(T') = N(T) ^{\perp} = \left\{x ^{*} \in X' : \langle x ^{*,y} \rangle = 0 { forall }\quad y \in N(T) \right\}.

Where

N(T) and N(T') are the null space of T and T' , respectively.

Note that there is always an inclusion

R(T)\subseteq N(T') ^{\perp} , because if y = Tx and x ^{*\in} N(T') , then \langle x ^{*,y\rangle} = \langle T'x ^{*,x\rangle} = 0 . Likewise, there is an inclusion R(T')\subseteq N(T) ^{\perp} . So the non-trivial part of the above theorem is the opposite inclusion in the final two bullets.

Corollaries

Several corollaries are immediate from the theorem. For instance, a densely defined closed operator

T as above has R(T) = Y if and only if the transpose T' has a continuous inverse. Similarly, R(T') = X' if and only if T has a continuous inverse.

Sketch of proof

Since the graph of T is closed, the proof reduces to the case when

T : X \to Y is a bounded operator between Banach spaces. Now, T factors as X \overset{p}\to X/\operatorname{ker}T \overset{T _{0}\to} \operatorname{im}T \overset{i}\hookrightarrow Y . Dually, T' isY' \to (\operatorname{im}T)' \overset{T _{0'}\to} (X/\operatorname{ker}T)' \to X'. Now, if \operatorname{im}T is closed, then it is Banach and so by the open mapping theorem, T _{0} is a topological isomorphism. It follows that T _{0'} is an isomorphism and then \operatorname{im}(T') = \operatorname{ker}(T) ^{\bot} . (More work is needed for the other implications.) \square

References

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