MHB Linear operator, its dual, proving surjectivity

Linux
Messages
5
Reaction score
0
Let $$T: X \rightarrow Y$$ be a continuous linear operator between Banach spaces.

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

I've proven a "similar" statement, with $$imT^*$$ replaced with $$imT$$.

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?
 
Physics news on Phys.org
Linux said:
Let $$T: X \rightarrow Y$$ be a continuous linear operator between Banach spaces.

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

I've proven a "similar" statement, with $$imT^*$$ replaced with $$imT$$.

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.
 
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)?
 
Last edited:
(From my phone by Tapatalk)
Try Yosida's Functional Analysis.
 
Thanks. I've just found the book.
 
I posted this question on math-stackexchange but apparently I asked something stupid and I was downvoted. I still don't have an answer to my question so I hope someone in here can help me or at least explain me why I am asking something stupid. I started studying Complex Analysis and came upon the following theorem which is a direct consequence of the Cauchy-Goursat theorem: Let ##f:D\to\mathbb{C}## be an anlytic function over a simply connected region ##D##. If ##a## and ##z## are part of...
Back
Top