Closed range theorem

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title: "Closed range theorem" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["banach-spaces", "theorems-in-functional-analysis"] description: "Mathematical theorem about Banach spaces" topic_path: "general/banach-spaces" source: "https://en.wikipedia.org/wiki/Closed_range_theorem" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

::summary Mathematical theorem about Banach spaces ::

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 \quad {\text{for all}}\quad x^ \in N(T') \right}.
• R(T') = N(T)^\perp = \left{x^* \in X' : \langle x^*,y \rangle = 0 \quad {\text{for all}}\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' is :Y' \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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