2

I need a hand with the proof of this result:

If we have an operator between Banach spaces $T:X\to Y$, with closed range, then the adjoint operator $T^*:Y^*\to X^*$ has also closed range.

Thanks in advance for any help.

hbghlyj
  • 5,361
Mark_Hoffman
  • 1,579

2 Answers2

8

This is how it's proved in Dunford and Schwartz:

Using the Open Mapping Theorem, you can prove the following:

Lemma: If $T:X\rightarrow Y$ is a bounded linear operator between the Banach spaces $X$ and $Y$ with closed range, there is a constant $K$ so that to each $y\in TX$ there is an $x\in X$ with $Tx=y$ and $\Vert x\Vert\le K\Vert y\Vert$.

Using the Lemma, you can prove the:

Theorem: For $T$ as in the Lemma, the range of $T^*$ is the set of $x^*\in X^*$ such that $Tx=0$ implies $x^* x=0$.

This in particular implies $T^*$ has closed range.

To prove the non-trivial inclusion of the Theorem:

Suppose $x^*$ satisfies the given condition. Then one uniquely defines $y_0^*$ on $TX$ via $y_0^*(Tx)=x^* x$. Use the Lemma to show $y_0^*$ is continuous. Then extend by Hahn-Banach to a $y^*\in Y^*$ with $T^*y^*=x^*$.

David Mitra
  • 76,313
6

If you factor $T$,

$$X \xrightarrow{\pi} X/\ker T \xrightarrow{\tilde{T}} \mathcal{R}(T) \xrightarrow{\iota} Y,$$

and then take the dual of that sequence,

$$Y^\ast \xrightarrow{\iota^\ast} \mathcal{R}(T)^\ast \xrightarrow{\tilde{T}^\ast} (X/\ker T)^\ast \xrightarrow{\pi^\ast} X^\ast,$$

can you see how that quickly leads to the result?

Daniel Fischer
  • 211,575
  • I don't get this...could you detail it a bit? How this leads to $T^*$ having closed range? – Mark_Hoffman Feb 27 '14 at 15:33
  • You explicitly know (or can straighforwardly determine, using Hahn-Banach) what $\iota^\ast$ and $\pi^\ast$ are, and what their ranges are. So it all hinges on $\tilde{T}^\ast$. Now, $\tilde{T}$ is an isomorphism (of Banach spaces, in particular its inverse is also continuous), so what does that tell you about $\tilde{T}^\ast$? – Daniel Fischer Feb 27 '14 at 15:36