Can someone please confirm or deny this assertion about absolute convergence?

[AxiomOfChoice]

EDIT: On pg. 390 of Kreyszig's Functional Analysis text, we have: "If T is a bounded linear operator on a nonempty Banach space, then the series

<br /> \sum_{k=0}^\infty \left( \frac{T}{\lambda} \right)^k<br />

converges absolutely for |\lambda| &gt; 2\| T \|."

The argument presented in Kreyszig for why this is the case is something like, "This happens because the series converges (in the operator norm) for |\lambda| &gt; \| T \|". I don't see how the one follows from the other, though.

Here's my initial, erroneous statement of the problem.

There's a proof in Kreyszig's Functional Analysis text (if you have the text handy, it's the proof at the bottom of page 390 about the spectrum being nonempty) that seems to suggest the following:

Consider a Banach space. If \sum_0^\infty x_k converges in norm for \| x_k \| &lt; 1, then the series converges absolutely for \| x_k \| &lt; 1/2.

I ran through a proof of this using the Cauchy criterion, and it appears to hold, but my proof tells me a couple of things:

(1) The 1/2 doesn't *have* to be there; it could be any number less than one.

(2) The first hypothesis, "If ... converges in norm for ...", is unnecessary.

(1) is not so surprising, but (2) is. This leads me to believe I did the proof wrong.

 

[HallsofIvy]

I don't understand why you think the statement that the series converges in norm is "unecessary"- that is the entire hypothesis. If you drop it, the statement becomes "Every series \{x_n\} converges absolutely for ||x_n||&lt; 1/2" which is certainly NOT true.

 

[AxiomOfChoice]

I don't understand why you think the statement that the series converges in norm is "unecessary"- that is the entire hypothesis. If you drop it, the statement becomes "Every series \{x_n\} converges absolutely for ||x_n||&lt; 1/2" which is certainly NOT true.

Ugh...sorry...I'm pretty sure I misstated the problem. I'm just going to state it as close to what actually shows up in Kreyszig as I can:

"If T is a bounded linear operator on a nonempty Banach space, then the series

<br /> \sum_{k=0}^\infty \left( \frac{T}{\lambda} \right)^k<br />

converges absolutely for |\lambda| &gt; 2\| T \|."

The argument presented in Kreyszig for why this is the case is something like, "This happens because the series converges (in the operator norm) for |\lambda| &gt; \| T \|". I don't see how the one follows from the other, though.

Sorry, again...I was in kind of a hurry when I made my first post. I'll edit the first post accordingly.

 

[AxiomOfChoice]

Okay...let me see if I can explain my issue a little more carefully: If |\lambda| &gt; \| T \|, we know (and I've managed to assimilate this much, at least) that

<br /> R_\lambda = (T-\lambda)^{-1} = \frac{1}{\lambda} \left( \frac{T}{\lambda} - I \right)^{-1} = - \frac{1}{\lambda} \sum_{k=0}^\infty \left( \frac{T}{\lambda} \right)^k.<br />

In other words, the series converges in (the operator) norm and converges to R_\lambda. Taking the norm of both sides and applying the triangle inequality, as well as the fact that \|T^k\| \leq \|T\|^k, we have

<br /> \| R_\lambda \| \leq \frac{1}{|\lambda|} \sum_{k=0}^\infty \left( \frac{\| T \|}{|\lambda|} \right)^k.<br />

But, since we are assuming 1 &gt; \|T\|/|\lambda|, don't we have a geometric series on the RHS, which therefore converges to

<br /> \| R_\lambda \| \leq \frac{1}{|\lambda|} \left( \frac{1}{1 - \| T \|/|\lambda|} \right) = (|\lambda| - \| T \|)^{-1}<br />

So I don't see why it's necessary to make the additional assumption, as Kreyszig does, that |\lambda| &gt; 2\| T \| to get the convergence of the series of norms.

 

[Stephen Tashi]

Just a question from someone who is hazy about the theory of linear operators:

You gave an argument why a series of real numbers converges, but the the original question is about a series of operators. How does Kreysig use the convergence of the series of numbers (norms) to establish the convergence of the series of operators?

 

[AxiomOfChoice]

Just a question from someone who is hazy about the theory of linear operators:

You gave an argument why a series of real numbers converges, but the the original question is about a series of operators. How does Kreysig use the convergence of the series of numbers (norms) to establish the convergence of the series of operators?

I'm not quite sure I understand your question, but I think this answers it at least in part: The original series (which is an infinite sum of operators) converges in the operator norm

<br /> \| T \| = \sup_{\|x\| = 1} \| Tx \|<br />

if it satisfies the Cauchy criterion with respect to that norm: For every \epsilon &gt; 0 there exists an N \in \mathbb N such that for all m,n &gt; N,

<br /> \| \sum_{k = m}^n T^k \| &lt; \epsilon.<br />

Of course, T^k is the operator defined by the k-fold composition of T.

It can thereby be shown that if \| T \| &lt; 1, then (T-I)^{-1} = -\sum_0^\infty T^k. Make the replacement T \mapsto T/\lambda for |\lambda| &gt; \| T \| and you have the same kind of result for (T/\lambda - I)^{-1} (provided, of course, that \lambda is in the resolvent set).

The problem now is to figure out when the sum is absolutely convergent.

 

[Landau]

There's a proof in Kreyszig's Functional Analysis text that seems to suggest the following:

Consider a Banach space. If \sum_0^\infty x_k converges in norm for \| x_k \| &lt; 1, then the series converges absolutely for \| x_k \| &lt; 1/2.

No. He's using the following result (here c and \mu are scalars, A a bounded operator between Banach spaces):

Lemma: If \sum_{k} (\mu A)^k converges in norm, then \sum_{k}(cA)^k converges absolutely for |c|&lt;|\mu|.

Proof: Let M=\sup_k\|(\mu A)^k\|. Note that M is finite:
\|(\mu A)^k\|=\left\|\sum_{j=k}^{k-1}(\mu A)^k \right\|\to 0
by the Cauchy criterion. Hence
\sum_{k}\|(c A)^k\|=\sum_{k}\|(\mu A)^k\|\frac{|c|^k}{|\mu|^k}\leq M\sum_k\left|\frac{c}{\mu}\right|^k,
which is finite by the geometric series, as \left|\frac{c}{\mu}\right|&lt;1 by assumption.


He applies this to \mu,c such that

|c|&lt;\frac{1}{2\|T\|}&lt;|\mu|&lt;\frac{1}{\|T\|}, and c=\frac{1}{\lambda}, knowing that for such \mu the series indeed converges in norm.

(Of course the 1/2 is not special, it's just a concrete choice of something strictly between 0 and 1.)

 

[AxiomOfChoice]

No. He's using the following result (here c and \mu are scalars, A a bounded operator between Banach spaces):

Lemma: If \sum_{k} (\mu A)^k converges in norm, then \sum_{k}(cA)^k converges absolutely for |c|&lt;|\mu|.

Proof: Let M=\sup_k\|(\mu A)^k\|. Note that M is finite:
\|(\mu A)^k\|=\left\|\sum_{j=k}^{k-1}(\mu A)^k \right\|\to 0
by the Cauchy criterion. Hence
\sum_{k}\|(c A)^k\|=\sum_{k}\|(\mu A)^k\|\frac{|c|^k}{|\mu|^k}\leq M\sum_k\left|\frac{c}{\mu}\right|^k,
which is finite by the geometric series, as \left|\frac{c}{\mu}\right|&lt;1 by assumption.He applies this to \mu,c such that

|c|&lt;\frac{1}{2\|T\|}&lt;|\mu|&lt;\frac{1}{\|T\|}, and c=\frac{1}{\lambda}, knowing that for such \mu the series indeed converges in norm.

(Of course the 1/2 is not special, it's just a concrete choice of something strictly between 0 and 1.)

Thanks for the complete and thorough explanation - something I have come to expect from you. But if we assume |\lambda| &gt; \| T \|, then \| T \| / |\lambda| &lt; 1, so don't we have

<br /> \sum_0^\infty \left\| \left( \frac{T}{\lambda} \right)^k \right\| \leq \sum_0^\infty \left\| \frac{T}{\lambda} \right\|^k = \sum_0^\infty \left( \frac{\|T\|}{|\lambda|} \right)^k = \frac{1}{1 - \|T\|/|\lambda|}<br />

In other words, don't we have absolute convergence of the series \sum (T/ \lambda)^k whenever |\lambda| &gt; \|T\|? I just don't see why it's necessary to look beyond a radius *larger* than that to get the absolute convergence...

 

[Landau]

Sorry, I hadn't read your other question.

Yes, I agree with you! I opened my pdf file of Kreyszig and I don't understand why he's doing it either: the series (8) on page 390 already converges absolutely for |\lambda|>||T||. In fact, he's quoting Theorem 7.3.4, in which he's quoting Theorem 7.3.1. But in the first paragraph of the proof of the latter, he's saying himself that the series converges absolutely (and hence in also norm).

So yeah, I understand your confusion, it seems unneccesary to look at a strictly smaller radius.