Bounded operators on Hilbert spaces

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Homework Help Overview

The discussion revolves around proving that the direct sum of two bounded operators on Hilbert spaces, denoted as \( T \oplus S \), is itself bounded. The operators \( T \) and \( S \) are defined on Hilbert spaces \( H \) and \( K \), respectively, and the problem involves understanding the properties of the direct sum space \( H \oplus K \) and its inner product structure.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to establish the boundedness of \( T \oplus S \) by analyzing the norms of the operators and their effects on elements of the direct sum space. Some participants suggest that the properties of the norms in the direct sum could simplify the proof. Others express confusion about the notation and the implications of the definitions involved.

Discussion Status

Participants are actively engaging with the problem, sharing insights about the definitions of bounded operators and the norms involved. There is a mix of attempts to clarify the mathematical reasoning and to derive the necessary inequalities. Some guidance has been provided regarding the structure of the direct sum and the implications of the norms, but no consensus has been reached on the proof itself.

Contextual Notes

Participants are navigating through the definitions of bounded operators and the direct sum of Hilbert spaces. There are discussions about the notation and the assumptions required to prove the boundedness of \( T \oplus S \), with some members questioning the logical flow of the arguments presented.

  • #31
There's nothing special about infinite dimensions here. I suggest trying the following: let H and K be one dimensional spaces, pick T and S to be whatever bounded linear functions you want on them. Then write down what ##H\oplus K## and ##T \oplus S## are.
 
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  • #32
$$||T \oplus S (h,k)||^{2} = \sqrt( \langle ((T(h), S(k))) | (T(h), S(k)) \rangle)^{2}$$
 
  • #33
Office_Shredder said:
There's nothing special about infinite dimensions here. I suggest trying the following: let H and K be one dimensional spaces, pick T and S to be whatever bounded linear functions you want on them. Then write down what ##H\oplus K## and ##T \oplus S## are.
Do you mind showing me, I am a bit stuck right now...
 
  • #34
HeinzBor said:
Do you mind showing me, I am a bit stuck right now...
The linear functions on ##\mathbb R## for example are of the form ##T(x) = ax##, for some constant ##a##.
 
  • Skeptical
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  • #35
PeroK said:
The linear functions on ##\mathbb R## for example are of the form ##T(x) = ax##, for some constant ##a##.
$$|| T \oplus S || = || T, S||$$. $$||T,S||^{2} = ||T||^{2} + ||S||^{2} = ||T \oplus S||^{2}$$And by boundedness of T,S we get

$$||Th, Sk||^{2} = ||Th||^{2} + || Sk ||^{2} \leq C ||h||^{2} + W ||k||^{2}$$... and then I am stuck again.
 
Last edited:
  • #36
HeinzBor said:
Well I am currently in grad school in pure maths coming back from 1 year break from mathematics. I just finished a course in functional analysis and I am now taking a course on operator algebras.

I think you should recap all exercises you got in functional analysis. Your question is rather basic, which in my opinion shows a lack of practice in solving problems.
 
  • #37
HeinzBor said:
... and then I am stuck again.
Maybe you should think from the end. What does it mean that ##T\oplus S## is bounded? Which inequality do you have to end up with?
 
  • #38
fresh_42 said:
Maybe you should think from the end. What does it mean that ##T\oplus S## is bounded? Which inequality do you have to end up with?
It was mentioned earlier in this thread. $$||T \oplus S (h \oplus k)|| \leq C ||h \oplus k||$$ for $$h \oplus k \in H \oplus K$$
 
  • #39
HeinzBor said:
It was mentioned earlier in this thread. $$||T \oplus S (h \oplus k)|| \leq C ||h \oplus k||$$ for $$h \oplus k \in H \oplus K$$
Yes. But what does this mean?

Let us see what you already have. E.g. we actually have
$$0\leq ||T \oplus S (h \oplus k)|| \leq C ||h \oplus k||$$
This is equivalent to $$0\leq ||T \oplus S (h \oplus k)||^2 \leq C^2 ||h \oplus k||^2.$$
You used equality earlier, which is wrong, but the inequality holds because the root function is strictly increasing. With the hint you got in post #8 we have
$$0\leq ||T \oplus S (h \oplus k)||^2=\|T(h)\|^2+\|S(k)\|^2 \leq C^2 ||h \oplus k||^2.$$
This means we have to show:
$$0\leq ||T \oplus S (h \oplus k)||^2=\|T(h)\|^2+\|S(k)\|^2 \leq \ldots\leq \ldots\leq \ldots \leq C^2 ||h \oplus k||^2.$$
In other words: You have to fill in the gaps. And what is ##\|h\oplus k\|^2##?

Note that ##C## is used as a variable, which was your decision in post #38. So whatever you do, I don't want to see something like ##\|T(h)\|\leq C\|h\|##.
 
  • Informative
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  • #40
fresh_42 said:
Yes. But what does this mean?

Let us see what you already have. E.g. we actually have
$$0\leq ||T \oplus S (h \oplus k)|| \leq C ||h \oplus k||$$
This is equivalent to $$0\leq ||T \oplus S (h \oplus k)||^2 \leq C^2 ||h \oplus k||^2.$$
You used equality earlier, which is wrong, but the inequality holds because the root function is strictly increasing. With the hint you got in post #8 we have
$$0\leq ||T \oplus S (h \oplus k)||^2=\|T(h)\|^2+\|S(k)\|^2 \leq C^2 ||h \oplus k||^2.$$
This means we have to show:
$$0\leq ||T \oplus S (h \oplus k)||^2=\|T(h)\|^2+\|S(k)\|^2 \leq \ldots\leq \ldots\leq \ldots \leq C^2 ||h \oplus k||^2.$$
In other words: You have to fill in the gaps. And what is ##\|h\oplus k\|^2##?

Note that ##C## is used as a variable, which was your decision in post #38. So whatever you do, I don't want to see something like ##\|T(h)\|\leq C\|h\|##.
Thanks a lot for your detailed explanation. Okay so what I thought is the following, following your hint:

$$0 \leq ||T \oplus S (h \oplus k)||^{2} = ||T(h)||^{2} + ||S(k)||^{2}\\
\leq || T ||^{2} || h ||^{2} + || S ||^{2} || k ||^{2} \ (using \ boundedness \ of \ T,S)$$
$$\leq max \{ ||T||^{2}, ||S||^{2} \} (||h||^{2}_{H} + ||k||_{K}^{2}) $$
$$= max \{ ||T||^{2}, ||S||^{2} \} ||h \oplus k||^{2}_{H \oplus K}$$
Put $$C:= max \{ ||T||^{2}, ||S||^{2} \} $$, then

$$0 \leq ||T \oplus S (h \oplus k)||^{2} \leq C ||h \oplus k||^{2}_{H \oplus K}$$
 
  • #41
Looks good. Except for a missing +.
 
  • #42
fresh_42 said:
Looks good. Except for a missing +.
I think it should be fixed now
 
  • #43
HeinzBor said:
I think it should be fixed now
The plus sign is still missing (1st line).
 
  • #44
fresh_42 said:
The plus sign is still missing (1st line).
ahh yes of course I see it now thanks a lot!
 

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