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 ${\displaystyle X}$ and ${\displaystyle Y}$ be Banach spaces, ${\displaystyle T:D(T)\to Y}$ a closed linear operator whose domain ${\displaystyle D(T)}$ is dense in ${\displaystyle X,}$ and ${\displaystyle T'}$ the transpose of ${\displaystyle T}$. The theorem asserts that the following conditions are equivalent:

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

Where ${\displaystyle N(T)}$ and ${\displaystyle N(T')}$ are the null space of ${\displaystyle T}$ and ${\displaystyle T'}$, respectively.

Note that there is always an inclusion ${\displaystyle R(T)\subseteq N(T')^{\perp }}$, because if ${\displaystyle y=Tx}$ and ${\displaystyle x^{*}\in N(T')}$, then ${\displaystyle \langle x^{*},y\rangle =\langle T'x^{*},x\rangle =0}$. Likewise, there is an inclusion ${\displaystyle R(T')\subseteq N(T)^{\perp }}$. 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 ${\displaystyle T}$ as above has ${\displaystyle R(T)=Y}$ if and only if the transpose ${\displaystyle T'}$ has a continuous inverse. Similarly, ${\displaystyle R(T')=X'}$ if and only if ${\displaystyle T}$ has a continuous inverse.

Sketch of proof

Since the graph of T is closed, the proof reduces to the case when ${\displaystyle T:X\to Y}$ is a bounded operator between Banach spaces. Now, ${\displaystyle T}$ factors as ${\displaystyle X{\overset {p}{\to }}X/\operatorname {ker} T{\overset {T_{0}}{\to }}\operatorname {im} T{\overset {i}{\hookrightarrow }}Y}$. Dually, ${\displaystyle T'}$ is

${\displaystyle Y'\to (\operatorname {im} T)'{\overset {T_{0}'}{\to }}(X/\operatorname {ker} T)'\to X'.}$

Now, if ${\displaystyle \operatorname {im} T}$ is closed, then it is Banach and so by the open mapping theorem, ${\displaystyle T_{0}}$ is a topological isomorphism. It follows that ${\displaystyle T_{0}'}$ is an isomorphism and then ${\displaystyle \operatorname {im} (T')=\operatorname {ker} (T)^{\bot }}$. (More work is needed for the other implications.) ${\displaystyle \square }$