Banach Sub-Algebra of C*-Algebra: Proving Completeness

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In summary, according to the conversation, ##\mathcal A_1## is a Banach sub-algebra of ##\mathcal B(\mathcal A)##. It is closed under addition, scalar multiplication and multiplication, but is not complete. The C*-identity holds, but it is not clear that the limit of a convergent sequence in ##\mathcal A_1## is in ##\mathcal A_1##.
  • #1
Fredrik
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Edit: I originally wrote that ##\mathcal A## is a Banach algebra. The assumption that goes into the theorem is stronger. It's a C*-algebra. I am however still mainly interested in the claim that ##\mathcal A_1##, as defined below, is a Banach sub-algebra of ##\mathcal B(\mathcal A)##.

Let ##\mathcal A## be a C*-algebra without identity. For each ##x\in\mathcal A##, define ##L_x:\mathcal A\to A## by ##L_xy=xy## for all ##y\in\mathcal A##. It's trivial to show that ##L_x\in\mathcal B(\mathcal A)## for all ##x\in\mathcal A##. The map ##x\mapsto L_x## with domain ##\mathcal A## will be denoted by L. It's easy to show that L is an isometric algebra homomorphism into ##\mathcal B(\mathcal A)##. Define ##\mathcal A_1=\{L_x+\lambda|x\in\mathcal A,\, \lambda\in\mathbb C\}##. The notation ##\mathcal A_1=L(\mathcal A)+\mathbb C## makes this definition easier to remember. ##\mathcal A_1## is closed under addition, scalar multiplication and multiplication. Supposedly* (this is what I want to prove), it's also complete. Since it's a subset of ##\mathcal B(\mathcal A)##, every Cauchy sequence is convergent. It's just not obvious that the limit of a convergent sequence in ##\mathcal A_1## is in ##\mathcal A_1##. So I want to prove that it is.

*) The claim is made by Conway in "A course in operator theory", in the proof of theorem 1.5, on page 3. http://books.google.com/books?id=gt...way operator&hl=sv&pg=PA3#v=onepage&q&f=false. His notation is slightly different from mine. (In particular, he writes λ where I write L)

Some of my thoughts: Let ##(T_n)_{n=1}^\infty## be an arbitrary convergent sequence in ##\mathcal A_1##. Let ##(x_n)## and ##(\lambda_n)## be sequences in ##\mathcal A## and ##\mathbb C## respectively, such that ##T_n=L_{x_n}+\lambda_n##. It would be nice if we could use that ##(T_n)## is Cauchy, to show that these two are Cauchy, and therefore convergent. Then we can define ##x=\lim_nx_n## and ##\lambda=\lim_n\lambda_n##, and perhaps show that ##L_{x_n}+\lambda_n\to L_x+\lambda##. But I don't see a way to proceed from
$$\varepsilon>\|T_n-T_m\|=\|(L_{x_n}+\lambda_n)-(L_{x_m}+\lambda_m)\| =\|L_{x_n-x_m}+(\lambda_n-\lambda_m)\|.$$ When we're dealing with Hilbert spaces, the usual way to continue a calculation like this would be to square both sides of the inequality and then use the pythagorean theorem (assuming that the terms are orthogonal), but I don't see what to do here.

It looks like (I haven't worked through that part of the proof yet) that we can prove that the C*-identity holds, without proving completeness first, when the involution is defined by ##(L_x+\lambda)^* =L_{x^*}+\bar\lambda##. I was thinking that maybe we can use that somehow, but when I tried, it just made the calculation longer, and I ran into the same issue as above.
 
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  • #2
Can't you use the triangle inequality on the norms and pick n, and m each such that the component terms are within epsilon/2 of each other to show their difference (or sum) are within epsilon of each other?

I vaguely recall that being the trick to show the sum of two Cauchy sequences is Cauchy in the most generic setting.

(Or am I misunderstanding your issue?)
 
  • #3
You mean like this?
$$\|L_{x_n-x_m}+(\lambda_n-\lambda_m)\|\leq\|L_{x_n-x_m}\|+\|\lambda_n-\lambda_m\| =\|x_n-x_m\|+\|\lambda_n-\lambda_m\|<\frac\varepsilon 2+\frac\varepsilon 2=\varepsilon.$$ In that case, the answer is no, because we don't know that ##(x_n)## and ##(\lambda_n)## are Cauchy sequences. In my idea for a proof, the first step is to prove that they are.

What we know is that ##(L_{x_n}+\lambda_n)## is a Cauchy sequence. That's our starting point. But since we have ##\varepsilon>## at the start of the calculation, it only makes sense to use the version of the triangle inequality with a minus sign, i.e. ##\|x+y\|\geq\|x\|-\|y\|##, to get a ≥ instead of a ≤.

$$\varepsilon>\|T_n-T_m\|=\|(L_{x_n}+\lambda_n)-(L_{x_m}+\lambda_m)\| =\|L_{x_n-x_m}+(\lambda_n-\lambda_m)\|\geq\|x_n-x_m\|-\|\lambda_n-\lambda_m\|.$$ This doesn't imply that the norms on the right are small. They could both be huge.
 
  • #4
Fredrik said:
...In that case, the answer is no, because we don't know that ##(x_n)## and ##(\lambda_n)## are Cauchy sequences. In my idea for a proof, the first step is to prove that they are.
Oh... I thought there must be something more to it. But then, or course you won't get there from here because...
...This doesn't imply that the norms on the right are small. They could both be huge.
Clearly a sum or difference of sequences which is itself Cauchy need not imply each sequence is Cauchy. One could easily construct counter examples.

Hmmm... I have to run to class. I'll think on this some more.
 
  • #5
jambaugh said:
Clearly a sum or difference of sequences which is itself Cauchy need not imply each sequence is Cauchy. One could easily construct counter examples.
Right, but this isn't an arbitrary Cauchy sequence that's written as a sum of two sequences in an arbitrary way. Most, if not all, of the terms in the first sequence belong to a different subspace than the terms in the other*. Consider e.g. a convergent sequence in ##\mathbb R^2## that's written as a sum of a sequence ##(x_ne_1)## with all terms on the x-axis and a sequence ##(y_ne_2)## with all terms on the y axis. Both of those sequences will be Cauchy. This is easy to prove using the pythagorean theorem. The key steps are:
\begin{align}
&\varepsilon^2>\|(x_n e_1+y_ne_2)-(x_me_1+y_me_2)\|^2 =\|(x_n-x_m)e_1+(y_n-y_m)e_2\|^2 =\|(x_n-x_m)e_1\|^2+\|(y_n-y_m)e_2\|^2\\
&>\begin{cases}\|x_ne_1-x_ne_1\|^2\\ \|y_ne_2-y_me_2\|^2.\end{cases}
\end{align} The last equality in the first line is the pythagorean theorem, so this proof relies on orthogonality, but the orthogonality isn't essential. It just makes the proof easier. If the subspaces aren't orthogonal, we can use the law of cosines instead of the pythagorean theorem.

*) Can ##L_x## can be a multiple of the identity in ##\mathcal B(\mathcal A)##. I'm not sure. If ##L_x=\mu## for some ##\mu\in\mathbb C##, then ##L_x-\mu## is an identity element of addition in ##\mathcal A_1##. The additive identity is unique, so this would imply that ##L_x-\mu=L_0+0##. But this doesn't seem to imply that ##\mu=0## and ##x=0##. If it did, L would be injective, and I'm not sure it can be. So my guess is that there's no ##x\in\mathcal A## such that ##L_x## is a multiple of the identity.

Edit: I figured out how to prove that ##L_x## can't be a multiple of the identity. Suppose that it is. Let ##\mu\in\mathbb C## be such that ##L_x=\mu##. Since L is linear, this equality implies that ##L_{\frac x \mu}## is the identity. So for all ##y\in\mathcal A##, we have ##y=L_{\frac x \mu}y=\frac x \mu y##. This means that ##\frac x \mu## is a left identity in ##\mathcal A##. But that implies that ##\big(\frac x\mu\big)^*## is a right identity, because for all ##y\in\mathcal A##, we have
$$y\left(\frac x \mu\right)^* =\left(\frac x \mu y^*\right)^* =(y^*)^*=y.$$ Since ##\frac x \mu## is a left identity and its adjoint is a right identity, we have
$$\frac x\mu =\frac x \mu \left(\frac x \mu\right)^* =\left(\frac x \mu\right)^*.$$ This means among other things that ##\frac x \mu## is also a right identity, and therefore an identity. This contradicts the assumption that ##\mathcal A## doesn't have an identity.
 
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  • #6
I found a proof strategy in Sunder. Apparently I'm supposed to look at the quotient vector space ##\mathcal B(\mathcal A)/L(\mathcal A)## and use that the associated projection map is continuous to show that ##L(\mathcal A)+\mathbb C## is a closed set. I will have to think about how to do this.

It's really frustrating to read Conway's books sometimes. I don't know how he can think that this isn't even worthy of a comment.
 
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1. What is a Banach sub-algebra?

A Banach sub-algebra is a subset of a C*-algebra that is closed under the algebraic operations of addition and multiplication, and is also a Banach space with respect to the norm induced by the C*-algebra.

2. What is a C*-algebra?

A C*-algebra is a complex algebra equipped with an involution operation and a norm that satisfies certain properties, such as the norm being sub-multiplicative and the involution being an isometric anti-homomorphism.

3. How is completeness proven in a Banach sub-algebra of a C*-algebra?

Completeness in a Banach sub-algebra of a C*-algebra can be proven by showing that every Cauchy sequence in the sub-algebra converges to a unique element in the sub-algebra. This can be done by using the completeness of the C*-algebra and the fact that the sub-algebra is a closed subset of the C*-algebra.

4. What is the importance of proving completeness in a Banach sub-algebra of a C*-algebra?

Proving completeness in a Banach sub-algebra of a C*-algebra is important because it ensures that all Cauchy sequences in the sub-algebra converge to a unique element in the sub-algebra. This is a fundamental property that allows for further analysis and applications of the sub-algebra.

5. Are there any applications of Banach sub-algebras in real-world problems?

Yes, Banach sub-algebras have various applications in fields such as physics, engineering, and economics. They are used to model and study complex systems and phenomena, and have also been applied in the development of quantum mechanics and financial risk management strategies.

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