Closed range theorem

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.) $$\square$$