Bounded operators on Hilbert spaces

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SUMMARY

The discussion focuses on proving that the direct sum of two bounded operators, \( T \in B(H) \) and \( S \in B(K) \), defined as \( (T \bigoplus S)(\alpha, \gamma) = (T\alpha, S\gamma) \), is itself a bounded operator on the Hilbert space \( H \bigoplus K \). The participants derive that the norm of \( (T \bigoplus S) \) can be expressed in terms of the norms of \( T \) and \( S \), specifically showing that \( ||(Th, Sk)||^2 = ||Th||^2 + ||Sk||^2 \). The conclusion is that since both \( T \) and \( S \) are bounded, \( T \bigoplus S \) is also bounded, satisfying the condition \( ||(T \bigoplus S)(h, k)|| \leq C ||(h, k)|| \) for some constant \( C > 0 \).

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  • #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
Likes HeinzBor
  • #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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