Bounded operators on Hilbert spaces

In summary: You need to use the definition of the direct sum norm, not the direct product.Here is a summary:In summary, the problem at hand is to show that for two bounded operators on Hilbert spaces ##H,K##, i.e. ##T \in B(H)## and ##S \in B(K)##, the formula ##(T \bigoplus S) (\alpha, \gamma) = (T \alpha, S \gamma)##, defined by the linear map ##T \bigoplus S: H \bigoplus K \rightarrow H \bigoplus K## is bounded. To do this, we must show that ##||(T \bigoplus S) \Gamma|| \
  • #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.
 
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  • #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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