Bounded operators on Hilbert spaces

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The discussion centers on proving that the direct sum of two bounded operators, T and S, on Hilbert spaces H and K, respectively, is also bounded. The formula for the direct sum operator is defined as (T ⊕ S)(α, γ) = (Tα, Sγ). Participants explore the implications of the inner product structure on the direct sum space H ⊕ K and how to compute the norm of the operator T ⊕ S in relation to the norms of T and S. A key point made is that the boundedness of T and S implies that the norm of the direct sum can be derived from the norms of the individual operators. The conversation highlights the importance of understanding the definitions and properties of norms in the context of direct sums in functional analysis.
  • #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
Likes HeinzBor
  • #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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