Linear operator, its dual, proving surjectivity

  • #1
Linux
7
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Let \(\displaystyle T: X \rightarrow Y\) be a continuous linear operator between Banach spaces.

Prove that $T$ is surjective \(\displaystyle \iff\) \(\displaystyle T^*\) is injective and \(\displaystyle im T^*\) is closed.

I've proven a "similar" statement, with \(\displaystyle imT^*\) replaced with \(\displaystyle imT\).

There I used these facts: $\overline{imT}= ^{\perp}(kerT^*)$ and $\overline{imT^*} \subset (kerT)^{\perp}$

However, I do not know how to prove the equivalence above.

Could you give me some ideas?
 
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  • #2
Linux said:
Let \(\displaystyle T: X \rightarrow Y\) be a continuous linear operator between Banach spaces.

Prove that $T$ is surjective \(\displaystyle \iff\) \(\displaystyle T^*\) is injective and \(\displaystyle im T^*\) is closed.

I've proven a "similar" statement, with \(\displaystyle imT^*\) replaced with \(\displaystyle imT\).

There I used these facts: $\overline{imT}= ^{\perp}(kerT^*)$ and $\overline{imT^*} \subset (kerT)^{\perp}$

However, I do not know how to prove the equivalence above.

Could you give me some ideas?
A classical theorem of Banach (you'll find it somewhere in his book Opérations linéaires – it's a consequence of the open mapping theorem) says that a bounded linear operator between Banach spaces has closed range if and only if its adjoint also has closed range. I think that will be the key ingredient in your problem.
 
  • #3
Thank you. Could you tell me where I can find a proof of this theorem, apart from Stefan Banach's book (which is in French)?
 
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  • #4
(From my phone by Tapatalk)
Try Yosida's Functional Analysis.
 
  • #5
Thanks. I've just found the book.
 
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