Hi there! I am interesting about homological methods in functional analysis, maybe someone else interesting too?

@kp9r4d That's definitely too advanced for me. Maybe somebody else will react - you will see. If not, it could be worth asking here, too (judging by the fact that the word homological is in the name of that chatroom):

@MartinSleziak For me too :3 i wanna to find naturally proof of next fact: "if $T^* : F^* \to E^*$ surjective then $T : E \to F$ injective". Proof of that fact are exist, but it use very difficult techniques. I think naturally proof can be finded if we find some good properties of banach adjoint functor. I tried proof that fact the third week in a row 3:

What are E, F? Banach spaces? Topological vector spaces?

Banach spaces. T - linear bounded operator. * - adjoint.

Is there a proof of moderate difficulty simply from Hahn-Banach?

Sorry. I confused problems. Right problem is that: "if $T^* : F^* \to E^*$ is injective than $T : E \to F$ surjective".

I was thinking along these lines: If $T(x)=T(y)$ for some $x\ne y$ (i.e., if $T$ is not injective), and we take an $f\in E^*$ such that $f(x)\ne f(y)$....

Ok, so we have a new problem.

So for the modified problem if we assume that $T[E]\ne F$ (i.e. $T$ is not surjective) and if we take and $f\in F^*$ such that $f\ne 0$ but the restriction to $T[E]$ is zero.

Then $T^*(f)=0$, i.e. $f\in\ker T$ and $T$ is not injective.

There must be some catch, since you said the problem is difficult.

Is the problem that I do not know whether $T[E]$ is closed in $F$? Or something else?

Just a minute, I think about your proof.

Aw, yes. I think that T(E) may be not closed is problem. I found proof of that fact in Rudin "Functional analysis" theorem 4.15 but he use more powerful methods.

Yes, the same thing is written here:

"The main problem, in those things, is that linear operators in infinite dimensional spaces need not have closed image."

In fact, the answer to this posts claims to give a counterexample: If $T^{*}$ is injective then $T$ is surjective?

To be honest, I am bit confused now.

I am too, wait a few minutes...

This is probably the same example: Example: operator injective, then the adjoint is NOT surjective. Although it seems to be formulated less clearly.

Maybe you want in fact prove this? Prove that $T^*$ is injective iff $ImT$ Is dense

I think it is error in Khelemsky book.

That's the book called "Exercises in functional analysis" or something similar?

amazon.com/…

Which exercise is it?

I will screen your a page that confused me.

I have the book right here.

Just page number is sufficdient.

(Assuming it is the same edition.)

Exercise 7 on Page 160?

Ok, but I read it in russian language, so I need to download english version to reffer page

It is fast, 2 minutes.

(i) An operator between Banach spaces is topologically injective $\Leftrightarrow$ its Banach adjoint operator is surjective (or, equivalently, topologically surjective.

(ii) an operator between Banach spaces that is Banach adjoint to a surjective operator, is topologically injective.

Yes but at bottom there must be remark: "Statement (ii) true in two sides, but proof of that difficult, see <refference on Rudin>" or something like that.

Sorry, problems with internet connection.

They say that: "The required arguments are presented in [72, Theorem 4.15]. (Perhaps you will succeed in finding a simpler proof.)"

Its ok :3 Do you find remark?

But if I look at Theorem 4.15 in Rudin, this is not what it claims.

It basically says: $T$ is injective $\Leftrightarrow$ $T^*$ is injective and $\operatorname{Rng}(T^*)$ is norm-closed in $X^*$.

Yes I also notice that, but I think that I am too stupid to understand that is the same.

Ok. Hence, it is inaccuracy in the Helemskii textbook.

What is the definition of topologically injective in Helemskii? Might that be the crucial difference.

T - topological injective iff T - homeomorphism between dom T and Im T (in topological sense)

$\varphi \colon \Omega \to \Delta$ is topologically injective if it provides a homeomorphism between $\Omega$ and $\operatorname{Im}\varphi$.

Or T is injective and exists some constants c, C > 0 that c ||x|| \leqslant ||Tx|| \leqslant C ||x|| (it is equivalent)

Yes, thats it.

This is condition (b) in Rudin.

So the claim of equivalence is not with $T^*$ injective but with $T^*$ topologically injective (which seems to be the same thing as an embedding).

Oh yes... I knew that injective and topological injective it is not the same, but forget it ):

ok, in fact I should be doing something else now.

But we at least clear some misunderstanding. So we did something useful.

Ok, good luck on your deals, and thanks for your helping! It is really useful for me!

I am willing to take Helemskii's word for the fact the proof is not easy. (Although in Rudin it seems to be only a few sentences long, but in clearly relies on some facts shown earlier in the book.)

Ok, but I like to think about that problem, I always find something new in trying to solve it. :3 So, I will continue to try.

It is always possible to choose nonzero functional $f$ such that $f|_E = 0$ and $E \subset F$ - closed subspace?

@kp9r4d You can define $f$ on the subspace $E\oplus[v]$ first and then extend it using Hahn-Banach theorem.

Proof a corollary on Hahn Banach theorem (Although my impression is that the proof in the answer there is unnecessary complicated.)

You will need that $T[E]$ is not dense in order to use this argument.

Truly...

@kp9r4d What you asked here is formulated as Proposition 2.7 in the book by Fabián, Habala et al. They also give a detailed proof.

Ok, thanks! I think that your first argument: "You can define $f$ on the subspace $E\oplus[v]$ first and then extend it using Hahn-Banach theorem.", - much simplier than argument on your book.

Well, it seems simple because I did not bother proving that such $f$ is continuous (bounded).

That still has to be done.

And they seem to prove this in quite an elegant way.

Anyway, it is probably time for me to go home and get some sleep. See you later!

See ya! Thanks for helping and good night! :3