# Bounded operators on Hilbert spaces

• HeinzBor
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|| \
HeinzBor
Homework Statement
Bounded operators on hilbert spaces direct sum.
Relevant Equations
operators
I have to show that for two bounded operators on Hilbert spaces ##H,K##, i.e. ##T \in B(H)## and ##S \in B(K)## that 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. After that I must compute the norm of ##T \bigoplus S## in terms of the norms of ##T## and ##S##.

Furthermore, it is given that the direct sum ##H \bigoplus K## is the vector space ##H \times K## with the standard coordinate vector space structure. with the inner product given by

$$\langle (x,y) | (x',y') \rangle := \langle x | x' \rangle_{H} + \langle y | y' \rangle_{K}\, , \,(x,y), (x',y') \in H \times K$$.

However, I am not sure if that is of any use in this problem.

What I have done so far:

We say that an operator ##A: H \rightarrow K## is bounded if there exists ##C > 0## such that ##||Ah|| \leq C ||h|| \forall h \in H##.So naturally what I must show in the first part is that ##||(T \bigoplus S) \Gamma|| \leq C ||\Gamma|| \ \forall \Gamma \in (T \bigoplus S)##. At this point I am a bit unsure of how this element ##\Gamma## looks like, hence the notation. About the "norm-part" of the question I believe that since both ##T## and ##S## are bounded operators I have to use the fact that

\begin{align*}
\|T\|=\sup \{ \frac{\|Ah\|}{ \|h\|} : h \neq 0 \}.
\end{align*}

The only way of thinking how to solve this problem would be to somehow show that
##\|TS\| \leq |\|T\| \|S\|##, which should be fairly simple with the norm I just mentioned. But to me that does not seem right because in that regard I don't show anything about the direct sum of the two bounded operators. I am really stuck any help is appreciated.

Last edited by a moderator:
You need double dollars or double hashes to delimit Latex.

PeroK said:
You need double dollars or double hashes to delimit Latex.
Still doesn't work...

works

HeinzBor said:
The only way of thinking how to solve this problem would be to somehow show that
$$\||TS|| \leq ||T|| \||S||$$, which should be fairly simple with the norm I just mentioned. But to me that does not seem right because in that regard I don't show anything about the direct sum of the two bounded operators. I am really stuck any help is appreciated.
Isn't the result a natural consequence of the definition of the norms on the direct sum?

Let ##h \in H## and ##k \in K##, then compute the norm of ##(h, k)## and ##(Th, Sk)##, and the result should just fall out naturally.

PeroK said:
Isn't the result a natural consequence of the definition of the norms on the direct sum?

Let ##h \in H## and ##k \in K##, then compute the norm of ##(h, k)## and ##(Th, Sk)##, and the result should just fall out naturally.
Could you show me, I am really stuck on this problem. What I derived earlier is that $$||(x,y)_{H \bigoplus K}|| = \sqrt{\langle (x,y) | (x,y) \rangle_{H \bigoplus K}}, \ \forall (x,y) \in H \oplus K.$$

HeinzBor said:
Could you show me, I am really stuck on this problem. What I derived earlier is that $$||(x,y)_{H \bigoplus K}|| = \sqrt{\langle (x,y) | (x,y) \rangle_{H \bigoplus K}}, \ \forall (x,y) \in H \oplus K.$$
Using simplified notation we have:
$$||(Th, Sk)||^2 = \langle (Th,Sk)|(Th, Sk) \rangle = \langle Th|Th \rangle + \langle Sk|Sk \rangle = ||Th||^2 + ||Sk||^2$$ Does that help?

PeroK said:
Using simplified notation we have:
$$||(Th, Sk)||^2 = \langle (Th,Sk)|(Th, Sk) \rangle = \langle Th|Th \rangle + \langle Sk|Sk \rangle = ||Th||^2 + ||Sk||^2$$ Does that help?
Thanks a lot! I will sit down and try to work out a solution with that hint now!

PeroK said:
Using simplified notation we have:
$$||(Th, Sk)||^2 = \langle (Th,Sk)|(Th, Sk) \rangle = \langle Th|Th \rangle + \langle Sk|Sk \rangle = ||Th||^2 + ||Sk||^2$$ Does that help?
A calculation shows that:
$$||Th, Sk||^{2} = || Th ||^{2} + || Sk ||^{2} \iff$$ $$||Th, Sk|| = || Th || + || Sk ||$$.

By assumption we know that $$T$$ and $$S$$ are bounded and thus,

$$|| Th || \leq C ||h||$$ for $$C, W > 0$$. and likewise $$||Sk|| \leq W ||k||$$ for $$C, W >0$$ and $$h \in T \ and \ k \in S$$.

Thus,
$$||(Th, Sk)|| = || Th || + || Sk || \leq C||h|| + W||k||$$.

Which shows that $$T \oplus S$$ is bounded.

HeinzBor said:
A calculation shows that:
$$||Th, Sk||^{2} = || Th ||^{2} + || Sk ||^{2} \iff$$ $$||Th, Sk|| = || Th || + || Sk ||$$.
That can't be right!

HeinzBor said:
$$||(Th, Sk)|| = || Th || + || Sk || \leq C||h|| + W||k||$$.

Which shows that $$T \oplus S$$ is bounded.
That doesn't follow. Where is the norm of ##(h, k)##?

PeroK said:
That can't be right!That doesn't follow. Where is the norm of ##(h, k)##?
By the same logic as you used to calculate the norm of the operators it should be the case that $$|| (h,k) ||^{2} = ||hk||^{2}$$?

HeinzBor said:
By the same logic as you used to calculate the norm of the operators it should be the case that $$|| (h,k) ||^{2} = ||hk||^{2}$$?
What is ##hk##?

By definition:$$||(h, k)||^2 = \langle (h, k)| (h, k) \rangle = \langle h|h \rangle + \langle k|k \rangle = ||h||^2 + ||k||^2$$

HeinzBor
PeroK said:
What is ##hk##?

By definition:$$||(h, k)||^2 = \langle (h, k)| (h, k) \rangle = \langle h|h \rangle + \langle k|k \rangle = ||h||^2 + ||k||^2$$
Okay so this is what I typed up so far

$$T \oplus S : H \oplus K \rightarrow H \oplus K$$ is bounded if $$|| T \oplus S \Gamma || \leq C || \Gamma ||$$ for $$\Gamma \in H \oplus K \ and \ C> 0$$.
But an element of $$H \oplus K$$ has the form $$(h,k) = \Gamma.$$ so it must be shown that,
$$|| T \oplus S (h,k) || \leq C || (h,k) ||.$$
but
$$|| (h,k) ||^{2} = || h ||^{2} + || k ||^{2}.$$
So it must be shown that
$$|| T \oplus S (h,k) ||^{2} \leq C ||h||^{2} + W||k||^{2}, \ for \ C,W >0.\\ \iff ||Th||^{2} + ||Sk||^{2} \leq C ||h||^{2} + W ||k||^{2}, \ for \ C,W >0.$$hmmm.. I feel like I am getting closer, but a bit confused about if I may assume that
$$|| T \oplus S (h,k) ||^{2} = || (Th, Sk) ||^{2}$$

HeinzBor said:
Okay so this is what I typed up so far

$$T \oplus S : H \oplus K \rightarrow H \oplus K$$ is bounded if $$|| T \oplus S \Gamma || \leq C || \Gamma ||$$ for $$\Gamma \in H \oplus K \ and \ C> 0$$.
But an element of $$H \oplus K$$ has the form $$(h,k) = \Gamma.$$ so it must be shown that,
$$|| T \oplus S (h,k) || \leq C || (h,k) ||.$$
but
$$|| (h,k) ||^{2} = || h ||^{2} + || k ||^{2}.$$
So it must be shown that
$$|| T \oplus S (h,k) ||^{2} \leq C ||h||^{2} + W||k||^{2}, \ for \ C,W >0.\\ \iff ||Th||^{2} + ||Sk||^{2} \leq C ||h||^{2} + W ||k||^{2}, \ for \ C,W >0.$$
This is heavy on formal notation, but lacks much relevant to the actual proof. In addition, your post lacks the logical flow of proceeding from definitions and assumptions.

In particular:

HeinzBor said:
So it must be shown that
$$|| T \oplus S (h,k) ||^{2} \leq C ||h||^{2} + W||k||^{2}, \ for \ C,W >0.\\ \iff ||Th||^{2} + ||Sk||^{2} \leq C ||h||^{2} + W ||k||^{2}, \ for \ C,W >0.$$
This makes no logical sense to me. I'll post separately what you need to prove.

HeinzBor said:
hmmm.. I feel like I am getting closer, but a bit confused about if I may assume that
$$|| T \oplus S (h,k) ||^{2} = || (Th, Sk) ||^{2}$$
That's the definition of ##T \oplus S## we are working with. We can stick with the notation:
$$(T \oplus S)(h \oplus k) = Th \oplus Sk$$if you prefer.

Here is what you must prove. Note that this is simply the definition of what it means for ##T \oplus S## to be bounded:
$$\exists C: \ \forall \ h \in H, \ k \in K, \ ||(T \oplus S)(h \oplus k)|| \le C||h \oplus k||$$

HeinzBor
PeroK said:
Here is what you must prove. Note that this is simply the definition of what it means for ##T \oplus S## to be bounded:
$$\exists C: \ \forall \ h \in H, \ k \in K, \ ||(T \oplus S)(h \oplus k)|| \le C||h \oplus k||$$
But in this setting what exactly is $$||T \oplus S|| and ||h \oplus k||$$? I mean for an ordered pair we define the sum $$(a,b) + (c,d)$$ to be $$(a+c, b+d)$$

HeinzBor said:
But in this setting what exactly is $$||T \oplus S|| and ||h \oplus k||$$? I mean for an ordered pair we define the sum $$(a,b) + (c,d)$$ to be $$(a+c, b+d)$$
The first thing is to understand what is meant by direct sum. So, yes, these are ordered pairs of vectors with an appopriate definition of vector sum, scalar multiplication and inner product (hence norm) - that you quoted in your original post.

You need to sort out your conception of that before you tackle this problem.

It appears, although I may be wrong, that you don't have much experience with pure mathematical proofs? The problems you are having should have been ironed out on an introduction to abstract algebra or analysis course. Bounded operators on Hilbert space is advanced undergraduate mathemathics, so you will need a good grasp of pure mathematical and proof-writing techniques.

PeroK said:
The first thing is to understand what is meant by direct sum. So, yes, these are ordered pairs of vectors with an appopriate definition of vector sum, scalar multiplication and inner product (hence norm) - that you quoted in your original post.

You need to sort out your conception of that before you tackle this problem.

It appears, although I may be wrong, that you don't have much experience with pure mathematical proofs? The problems you are having should have been ironed out on an introduction to abstract algebra or analysis course. Bounded operators on Hilbert space is advanced undergraduate mathemathics, so you will need a good grasp of pure mathematical and proof-writing techniques.
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. But I must admit that I haven't studied algebra for a really long time, so some of the algebraic structures are really not that clear in my memory at this point.

PeroK
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. But I must admit that I haven't studied algebra for a really long time, so some of the algebraic structures are really not that clear in my memory at this point.

PeroK said:
Here is what you must prove. Note that this is simply the definition of what it means for ##T \oplus S## to be bounded:
$$\exists C: \ \forall \ h \in H, \ k \in K, \ ||(T \oplus S)(h \oplus k)|| \le C||h \oplus k||$$
PeroK said:
Using simplified notation we have:
$$||(Th, Sk)||^2 = \langle (Th,Sk)|(Th, Sk) \rangle = \langle Th|Th \rangle + \langle Sk|Sk \rangle = ||Th||^2 + ||Sk||^2$$ Does that help?
PeroK said:
What is ##hk##?

By definition:$$||(h, k)||^2 = \langle (h, k)| (h, k) \rangle = \langle h|h \rangle + \langle k|k \rangle = ||h||^2 + ||k||^2$$

HeinzBor
... now, at this point you might consider using the fact that ##T## and ##S## are bounded!

HeinzBor
PeroK said:
... now, at this point you might consider using the fact that ##T## and ##S## are bounded!
I think it should be $$||(T \oplus S)(h,k) || \leq C ||(h,k)||$$ instead of $$||(T \oplus S)(h \oplus k) || \leq C ||(h \oplus k)||$$ right?

HeinzBor said:
I think it should be $$||(T \oplus S)(h,k) || \leq C ||(h,k)||$$ instead of $$||(T \oplus S)(h \oplus k) || \leq C ||(h \oplus k)||$$ right?
It's just different notation for the same thing: $$(h, k) \equiv h \oplus k$$

Must show
$$||(T \oplus S) (h \oplus k)|| \leq C ||h \oplus k||$$
Which is the same as
$$||(Th, Sk)|| \leq C ||(h,k)||$$
Which I assume can be shown if we show that the following holds

$$||(Th, Sk)||^{2} \leq C ||(h,k)||^{2} \iff ||Th||^{2} + ||Sk||^{2} \leq C(|| h ||^{2} + || k ||^{2})$$. But since $$T,S$$ are both bounded the last inequality holds automatically.

Am I missing something? If so do you mind showing what I was supposed to do, at this point I would really like to know what I am missing/ doing wrong.

HeinzBor said:
Must show
$$||(T \oplus S) (h \oplus k)|| \leq C ||h \oplus k||$$
Which is the same as
$$||(Th, Sk)|| \leq C ||(h,k)||$$
Which I assume can be shown if we show that the following holds

$$||(Th, Sk)||^{2} \leq C ||(h,k)||^{2} \iff ||Th||^{2} + ||Sk||^{2} \leq C(|| h ||^{2} + || k ||^{2})$$. But since $$T,S$$ are both bounded the last inequality holds automatically.

Am I missing something? If so do you mind showing what I was supposed to do, at this point I would really like to know what I am missing/ doing wrong.
What's ##C##?

PeroK said:
What's ##C##?
The constant, which I can pick since T,S are bounded?

HeinzBor said:
The constanr
In general ##T## and ##S## will have different bounds.

If you move on to finding the norm of ##T \oplus S## you might see what you're missing.

PeroK said:
If you move on to finding the norm of ##T \oplus S## you might see what you're missing.
$$||(t,s)||_{T \oplus S} = \sqrt{(\langle t,s \rangle | \langle t,s \rangle )_{T \oplus S}}$$ so before was just for elements in $$H \oplus K$$, but I haven't considered for elements in $$T \oplus S$$ is what you mean?

HeinzBor said:
$$||(t,s)||_{T \oplus S} = \sqrt{(\langle t,s \rangle | \langle t,s \rangle )_{T \oplus S}}$$ so before was just for elements in $$H \oplus K$$, but I haven't considered for elements in $$T \oplus S$$ is what you mean?
I'm not sure I understand what you are saying there. I suggest you try to find the norm of ##T\oplus S##.

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.

$$||T \oplus S (h,k)||^{2} = \sqrt( \langle ((T(h), S(k))) | (T(h), S(k)) \rangle)^{2}$$

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...

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##.

HeinzBor
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.

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