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 $T^{\prime }$ the transpose of $T$. The theorem asserts that the following conditions are equivalent:

• $R(T),$ the range of $T,$ is closed in $Y.$
• $R(T^{\prime }),$ the range of $T^{\prime },$ is closed in $X^{\prime },$ the dual of $X.$
• $R(T)=N(T^{\prime }{)}^{\perp }=\{y\in Y:⟨x^{*},y⟩=0\text{for all}{x}^{*}\in N(T^{\prime })\}.$
• $R(T^{\prime })=N(T{)}^{\perp }=\{x^{*}\in X^{\prime }:⟨x^{*},y⟩=0\text{for all}y\in N(T)\}.$

Where $N(T)$ and $N(T^{\prime })$ are the null space of $T$ and $T^{\prime }$, respectively.

Note that there is always an inclusion $R(T)\subseteq N(T^{\prime }{)}^{\perp }$, because if $y=Tx$ and $x^{*}\in N(T^{\prime })$, then $⟨x^{*},y⟩=⟨T^{\prime }{x}^{*},x⟩=0$. Likewise, there is an inclusion $R(T^{\prime })\subseteq N(T{)}^{\perp }$. 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 $T^{\prime }$ has a continuous inverse. Similarly, $R(T^{\prime })=X^{\prime }$ 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 $X\stackrel{p}{\to }X/\ker T\stackrel{T_0}{\to }\mathrm{im}T\stackrel{i}{\hookrightarrow }Y$. Dually, $T^{\prime }$ is

$Y^{\prime }\to (\mathrm{im}T{)}^{\prime }\stackrel{{T_0}^{\prime }}{\to }(X/\ker T{)}^{\prime }\to X^{\prime }.$

Now, if $\mathrm{im}T$ is closed, then it is Banach and so by the open mapping theorem, $T_0$ is a topological isomorphism. It follows that ${T_0}^{\prime }$ is an isomorphism and then $\mathrm{im}(T^{\prime })=\ker (T{)}^{\mathrm{\perp }}$. (More work is needed for the other implications.) $◻$

References

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