An example of Banach algebra with a specified property

I asked this question
(Https://math.stackexchange.com/questions/3076735/an-example-of-a-banach-algebra-satisfying-given-conditions)
but unfortunately nobody answered it. Please, help me find an example of Banach algebra (if any) with the following property:

Non commutative noncommutative Banach algebra $ A $ For who $ aa_0 -a_ {0} a $ is in the annihilator of $ A $ for all $ a in A $.

Right here $ a_0 $ is an element of $ A $ not belonging to his center $ Z (A) $.

Could you please suggest me a good reference (on Banach algebras) with examples like this?

Any help is appreciated.

linear algebra – Dual and bidual on a Banach space

Problem: Let $ E $ to be a normed space on $ mathbb {C} $.

  1. Show that we are able to integrate $ E $ in the second double space (bidual) $ E & # 39; $ of $ E $ by linear isometry, that is to say we can consider $ E $ as a subspace of $ E & # 39; $.
  2. For each subset $ A $ of $ E $, let $ A ^ { perp} = {f in E}: f | _A = 0 } $. Show that a Banach space $ E $ is reflexive $ iff $ for any closed subspace $ F $ of $ E $ we have $ F ^ { perp perp} = J (F) $ in which $ J is canonical entrenchment $ E $ in $ E $.
  3. Fix a continuous $ f: left[ a,b right] rightarrow E $ with $ left[ a,b right] subset mathbb {R} $. Consider $ varphi: E rightarrow mathbb {C} $ given by $ varphi (y): = displaystyle int_a ^ b (y circ f) (t) dt, forall y in E $. CA watch $ varphi in E & # 39; $ and if $ E $ is the Banach space then $ varphi in E $.

My attempt:

  1. Let $ E $ to be a normed space. For each $ x in E $, let $ (Jx) (u) = u (x) $ for everyone $ u in E & # 39; $.

We show that $ J: E rightarrow E & # 39; $ is a linear isometry. Clearly, $ Jx is a linear form on $ E $. Since $ (Jx) (u) = | u (x) | le || u || $ $ || x || $, $ Jx is continuous on $ E $and $ || Jx || le || x || $. So, $ Jx in E & # 39; $. It is common to show that $ J is a linear map of $ E $ in $ E & # 39; $. Take everything $ x in E $. There is $ v in E & # 39; $ such as $ || v || = $ 1 and $ v (x) = || x || $. Therefore
$$ || Jx || = sup {| (Jx) (u) |: || u || le 1 } $$
$$ = sup {| u (x) |: u in E & # 39 ;,| u || le 1 } ge v (x) = || x ||. $$
So $ J is a linear isometry.
Anyone can help me solve the issue number $ 2 $ and $ 3 $? Thank you all!

There are many derivations of discontinuous points on the Banach algebra $ ( mathbb {C ^ {(n)}[0,1], | | _n}) $

This is a 6.2.55 exercise in Garth Dales, Introduction to Banach Algebra

Show that there are many derivatives of discontinuous points on the Banach algebra $ ( mathbb {C ^ {(n)}[0,1], | | _n}) $ or
$$ | f | _n = sum_ {k = 0} ^ {n} frac {1} {k!} | f ^ {(k)} | $$ for everyone $ f in mathbb {C ^ {(n)}[0,1]} $

so if you give reasonable clues, I will be very happy. Thank you

Automatic continuous in Banach algebra

I have the following two questions

1: Let $ A $ and $ B $ to be Banach algebra and suppose that $ B $ is semi-simple. Let $ T: A to B $ to be a homomorphism with $ overline {TA} = B. $ is $ T $ automatically continuously?

2: Let $ A $ and $ B $ to be Banach algebra. Let $ T: A to B $ to be a homomorphism with $ overline {TA} = B. $ is $ T $ automatically continuously?

Any help is very appreciated! Thank you!

Functional Analysis – If the closed unit ball of the Banach space has at least one extreme point, should the Banach space be a double space?

Let $ X $ to be a Banach space.
By the theorems of Banach-Alaoglu and Kerin-Milman, it can be shown that if $ X $ is a double space, so $ X $ must have at least one extreme point.

I'm interested in his conversation.
More precisely,

Question: Let $ X $ to be a Banach space.
If the closed unit ball of $ X $ at least one extreme point, must $ X $ to be a double space?

I think the statement above is negative.
However, I could not produce a counterexample.

In fact, the only Banach spaces that are not double spaces are $ c_0 $ and $ C_0 ( mathbb {R}) $ (The last set is the collection of all real-valued continuous functions that disappear to infinity) because the two sets have no extreme point.

Real Analysis – The Neat Banach Space Concept

In this article, the authors used the notion of ordered Banach space.

Definition: Let $ mathcal E $ to be a Banach space with the standard $ left | . right | $, whose positive cone is defined by $ K = {x in mathcal E : : x geq 0 } $. then $ ( mathcal E, left |. right |) $ is now a partially ordered Banach space with the order relationship $ sqsubseteq $ induced by the cone K $.

and they used this result:

Theorem:Let $ ( mathcal E, left |. right |, sqsubseteq) $ to be an orderly space of Banach, whose positive cone K $ Is normal(?). Let $ (u_n) _ {n in mathbb N} $ a monotonous sequence (let's say $ u_n sqsubseteq u_ {n + 1} $), such as $ {u_n} $ has a convergent subsequence. Then the whole sequence $ u_n $ is convergent?

I have two questions here:

$ 1. $ What is it? $ a sqsubseteq u _ {n + 1} $ means, is there an interpretation of that?

$ 2. $ I wonder why the whole sequence $ u_n $ is convergent?

Please share your thoughts and thank you in advance!