Proving a relation is a partial order

[Oxymoron]

Im considering a complex Hilbert space. Then define a relation $\preceq$ on the set of all self-adjoint, bounded, linear operators (denote by A) by

$$S \preceq T \, \Leftrightarrow \langle Sx\,|\,x \rangle \leq \langle Tx\,|\,x \rangle$$

for all $x \in \mathcal{H}$.

Now I want to prove that $\preceq$ is a partial order on $A$, for all $T \in A$ such that

$$-\|T\|I \leq T \leq \|T\|I$$.


Now to prove that a relation is a partial order then it must satisfy three things:

1. $$T \preceq T$$ for all $$T \in A$$
2. If $$T \preceq S$$ and $$S \preceq T$$, then $$T = S$$ for all $$S,T \in A$$.
3. If $$T \preceq S$$ and $$S \preceq U$$, then $$T \preceq U$$, for all $$S,T,U \in A$$.

This relation says that a self-adjoint, bounded, linear operator is greater than another operator of the same category if and only if the inner product of that operator with an arbitrary element in the complex Hilbert space is larger.

So to prove 1. we have to show that $T \leq T$ if and only if $\langle Tx\,|\,x \rangle \leq \langle Tx\,|\,x \rangle$. Well, quite obviously, $\langle Tx\,|\,x \rangle = \langle Tx\,|\,x \rangle$, for any choice of $x\in\mathcal{H}$ and for any S-A,B,L operator on complex Hilbert space. Therefore (1) is proved.

How does this sound so far?

 

[AKG]

So far so good. Are you asked to prove two things:

1) that $\preceq$ is a partial order

and

2) that $-\|T\|I \preceq T \preceq \|T\|I$ for all T in A?

 

[Oxymoron]

That's right AKG, and thanks for replying. I haven't got to that second part yet.

To prove the second axiom of partial orderedness:

If $T \preceq S$ and $S \preceq T$ then $\langle Tx\,|\,x \rangle \leq \langle Sx\,|\,x \rangle$ and $\langle Sx\,|\,x \rangle \leq \langle Tx\,|\,x \rangle$. Therefore since $\langle \cdot\,|\,\cdot \rangle$ is a number, $\langle Tx\,|\,x \rangle = \langle Sx\,|\,x \rangle$

Therfore $T = S$


And for 3

$T \preceq S \, \Rightarrow \langle Tx\,|\,x \rangle \leq \langle Sx\,|\,x \rangle$ and $S \preceq U \, \Rightarrow \langle Sx\,|\,x \rangle \leq \langle Ux\,|\,x \rangle$.

Therefore

$$\langle Tx\,|\,x \rangle \leq \langle Sx\,|\,x \rangle \leq \langle Ux\,|\,x \rangle$$

And so certainly $\langle Tx\,|\,x \rangle \leq \langle Ux\,|\,x \rangle$ which implies $T \preceq U$


Therefore $(\preceq)$ is a partial order on the set of self-adjoint, bounded, linear operators.

How does this look?

 

[AKG]

Looks good. I wouldn't say, "since <.|.> is a number..." since technically it is just a complex number, and in general, <Tx|x> need not be a real number, and hence saying <Tx|x> < <Sx|x> may not make sense if <Tx|x> are complex (not purely real) numbers. There's no need to talk about what kind of number <Tx|x> is (although, is there a theorem that says it is real for self-adjoint operators?), just use the fact that < is a partial-order (in fact, a total order on the reals but meaningless in general on C) to conclude that if <Tx|x> < <Sx|x> and vice versa, that <Tx|x> = <Sx|x> and then you do as you did, concluding that T = S.

 

[Oxymoron]

Thanks AKG.

But proving the final bit: that for all $T$ (self-adjoint, bounded, linear)...

$$-\|T\|I \leq T \leq \|T\|I$$

...where $I$ is the identity operator on the complex Hilbert space.

Im not sure I understand this statement. Is it saying that $T$ is bounded above and below? That is, let $m=-I$ and $M=I$, then this says that $T$ is an isomorphism no?

Is this a correct understanding of the last statement? I want to try to understand it before I prove it.

 

[Oxymoron]

Just jumping back a second...you said "may not make sense if <Tx|x> are complex (not purely real) numbers.".

Then the statement "...since $\leq$ is a partial order on $\mathbb{R}$ (but meaningless in $\mathbb{C}$) we can conclude that $\langle Tx\,|\,x \rangle = \langle Sx\,|\,x \rangle$..." doesn't work because we are considering a complex Hilbert space so the inner products can be complex..wait...wait...

OMG...

I see what you mean now...aha! Self-adjoint operators only have REAL inner products! and we are considering only self-adjoint, bounded, linear operators. Yes, so $\leq$ is actually a partial order. Great.

 

[Hurkyl]

Self-adjoint operators only have REAL inner products!

At least for those of the form <Tx|x>. (All bets are off for <Tx|y>)

 

[Oxymoron]

All bets are off for <Tx|y>)

If you don't mind, I wouldn't mind a quick explanation of this.

I was wondering, to prove this part (the $-\|T\|I \leq T \leq \|T\|I$ for all S-A,B,L operators bit) will I need to use the following fact:

If you have two normed linear spaces X and Y and you also have a surjective linear map T such that $T:X\rightarrow Y$. Then T is an isomorphism if and only if there exist $m,M > 0$ such that

$$m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X \quad \forall \, x\in X$$

The idea of this may help me prove it. However, I want to fully understand this proposition first before I use it (thats IF it is applicable). Any comments?

 

[Hurkyl]

Let |y> = i |x>. Then, what is <Ix|y>?


I think that proving -||T|| I <= T <= ||T|| I is supposed to be a rather elementary thing, involving little more than the definitions. (At least, given that <Tx|x> is real for any self-adjoint T)

 

[Oxymoron]

Let |y> = i |x>. Then, what is <Ix|y>?

Actually, sorry Hurkyl, I am not at all familiar with bra-ket notation. :(

 

[Hurkyl]

Ah!

| > is just a decoration we like to put on a symbol to indicate that it's supposed to be a vector.

Simlarly, we like to use < | to indicate that the thing inside is a dual vector. (a.k.a. linear functional)

In particular, given the inner product < , >, <x| is the function that maps |y> to <x, y>. But, we like to write the inner product as < | >, so we get the pleasing notation that <x||y> = <x|y>.

|Tx> just means T|x>. So, with a bit of effort, you should be able to work out that <Tx| just means $\langle x\,|T^{*}$.

I find it helps to think of matrices -- |x> is a column vector, and <x| is merely its conjugate transpose.


By |y> = i |x>, I just mean that the vector |y> is the scalar i times the vector |x>.

 

[AKG]

Strictly speaking, what you have doesn't make sense Oxymoron. T < ||T||I? What does it mean for one operator to be less than another? I think you mean:

$$T \preceq \|T\|I$$

Well I'm not sure how you've defined $\|T\|$ but I think the following might work:

$$T \preceq \|T\|I \Leftrightarrow \langle Tx|x\rangle \leq \|T\|\langle x|x\rangle \ \forall x \in \mathcal{H}$$

which is true by definition of $\|T\|$, isn't it?

 

[Oxymoron]

You are right AKG - the question is mistyped I think.

I want to try to prove the given two normed vector spaces, X and Y, and a surjective linear map, T:X->Y, then

T is an isomorphism <=> there exist m,M > 0 such that

[tex]m\|x\| \leq \|Tx\| \leq M\|x\| \quad \forall \, x \in X[/itex]

But all I could come up with is this:

$$(\Rightarrow)$$
Suppose $T$ is an isomorphism. Then $T$ is continuous. Take $\epsilon > 0$ and in particular $\epsilon = 1$. Then we have a $\delta > 0$ such that

$$\|x\| \leq \delta \Rightarrow \|Tx\| \leq 1 \quad \quad \forall \, x \in X$$

Therefore

$$\|T\left(\frac{\delta x}{\|x\|}\right)\| \leq 1 \Rightarrow \frac{\delta}{\|x\|}\|Tx\| \leq 1$$

Which implies

$$\|Tx\| \leq \frac{1}{\delta}\|x\|$$

So let $M = 1/\delta$.

If $M$ satisfies $\|Tx\| \leq M\|x\|$ for all $x\in X$, then $M$ is an upper bound for the set:

$$\|T\| = \sup \{\|Tx\| \leq M \|x\|\,:\, \forall \, x\in X \}$$

and hence

$$\|T\| \leq M$$

...because $\|T\|$ is the least upper bound. Therefore $\|T\|$ is also a lower bound for $S$. That is. for any $x \neq 0$, we have

$$\|\frac{x}{\|x\|}\| = 1 \Rightarrow \frac{1}{\|x\|}\|Tx\| = \|T\left(\frac{x}{\|x\|}\right)\| \leq \|T\|$$

...taking $m=1$, therefore

$$m\|x\| \leq \|Tx\| \leq M\|x\|$$

How does this look so far??

 

[AKG]

Something is definitely wrong, where you said that you take m = 1. Let T(x) = x/2, you'll see you can't take m = 1. Here's one thing that's useful. If there is an m > 0 such that m|x| < |Tx| for all x, then the kernel of T must be 0, so T is injective. Given that it's a surjective linear map, it's an isomorphism. Conversely, assume that T is an isomorphism, hence continuous. I don't know what kind of theorems you have at your disposal and whether you can make use of notions like compactness, but if you can, consider the unit sphere in X, that is {x in X : |x| = 1} = S. The function R : X --> R defined by R(x) = |T(x)| is a composition of continuous functions, hence it's continuous and real-valued. Since S is compact, then a generalized version of the extreme value theorem tells you that R(S) achieves a maximum and minimum value. These values will naturally be positive since |.| is positive, and will be non-zero because the minimum value is non-zero, since if it were, then T(x) = 0 for some non-zero x, but T is an isomorphism, so this wouldn't be the case. Even if you can't use some of these theorems and ideas, it might be a good way to think about it.

Look at the set S. If R were unbounded above, then it would be that as you approach some s on S, R increases. But R must take on some fixed value at s itself, so it can't increase without bound.

 

[Oxymoron]

Just a quick question before I go on. Have I got the $\|Tx\| \leq M\|x\|$ bit right?

 

[Oxymoron]

AKG, so you are saying that I can prove boundedness of a surjective linear operator by considering a compact unit sphere?

So if T is my surjective linear operator and the unit sphere is compact. Then T is continuous and from the triangular inequality the norm will be continuous. That is

T continuous <=> |T| continuous

And therefore |Tx| is a continuous operator on the compact unit sphere. (When you say unit sphere you could mean the unit cube or pyramid, etc... depending on the norm right?).

So what we really have is a continuous, linear operator which takes point in the unit sphere and maps them to a real number?

If so, the image is compact - which implies the image is closed and bounded right?

This means that T is bounded. BUT does this mean:

$$\|Tx\| \leq M\|x\|$$

or could it mean

$$m\|x\| \leq \|Tx\| \leq M\|x\|$$

for all x?

Is all this reasoning correct?

 

[AKG]

AKG, so you are saying that I can prove boundedness of a surjective linear operator by considering a compact unit sphere?

No, I'm saying that you can prove the boundedness of a continuous function by considering a compact domain. In this case, the compact domain happens to be the unit sphere. Note that the function R defined by R(x) = |Tx| is not a linear operator. Use the fact that T is linear to conclude that it is continuous. Use the fact that absolute value is continuous, and that R is absolute value composed with T, to conclude R is continuous is continuous. Use the fact that S is compact to deduce R achieves its minimum and maximum values on S. Use the fact that T is injective (since it's an isomorphism) and the fact that |x| = 0 iff x = 0 to conclude that R(S) does not contain 0. Use the fact that |x| > 0 for all x to conclude that R(S) is contained in the positive reals. Let m be the minimum value of R, and M be the maximum value of R, on S.

When I say unit sphere, I mean {x in V (or whatever your vector space is) | |x| = 1}. It doesn't matter what the norm is, I'm calling it a sphere anyways. It might end up looking like something else, so if you want, you can say that this is the sphere with respect to this norm.

For a function to be bounded, that generally means bounded above and bounded below. However, all you need to know is that R achieves its minimum and maximum values. What you will be able to conclude is that for all x on the sphere:

m < R(x) < M
m|x| < |Tx| < M|w|

Let v be any vector now. v = |v|w, for some w on the unit sphere (w is the unit vector of v, like v-hat). Given the above inequalities, you can say:

m|w| < |Tw| < M|w|

since w is on the unit sphere. Multiply the whole thing by |v| and you're done.

Remember, a linear operator is a linear mapping from V to V. A linear transformation is more generally a linear mapping from V to W, vector spaces (so if in particular W = V, we call the transformation an operator). The function that sends x to |Tx| is not even linear, and certainly not an operator. Linear transformations, I believe, are always continuous, but you may want to check that. At very least, you have that T is an isomorphism, and hopefully that will be good enough to give you continuity.

Now I don't know about T continuous <=> |T| continuous. It may be true, but only because linear functions are always continuous, absolute value is always continuous, and the composistion of continuous functions are always continuous. But you can conceivably have a discontinous function T such that |T| is continuous, e.g. T = 1 for x < 0, T = -1 otherwise.

 

[Oxymoron]

Ok AKG, I have to turn that into a proof.

Firstly, it seems like you are using the extreme value theorem which says that a continuous function defined on some closed interval attains its minimum and maximum.

I want to know how you made that connection. I mean, I was talking about surjective linear maps between normed vector spaces, and somehow you come up with a method which turns |Tx| into a composition of continuous functions (absolute value and T itself) then let R be that function.

Also, am I correct in saying that to use this theorem we need a closed interval, and you have come up with an equivalent compact sphere, which I understand does the same thing.

So all I am saying is, how did you know to convert this problem of using operators into functions and then use the Extreme value property. How did you do it man!?

 

[AKG]

The domain has to be compact (not just closed, and doesn't have to be an interval of any sort). Specifically, if f : X --> R is a continuous function and X is a compact space, then f achieves its minimum and maximum. To prove that our domain, S, is compact, it suffices to show that it is closed and bounded. It is bounded because we defined it to have all its elements with norm 1, and {1} is certainly a bounded set. It is closed since its complement is open. How do we know? Well, pick any element not in S, say x with norm 1 + e, where e > 0. Choose an open ball (where the "ball" may look like a cube, or diamond, etc. based on the metric) around x with radius e/2. Suppose there is some y in this ball with norm 1. Then:

|x| = |(x-y) + y| < |x-y| + |y| < e/2 + 1 < 1 + e, contradiction.

You can show a similar thing for e < 0. So every point not in S is contained in an open ball that doesn't intersect S, so S-complement is open, so S is closed. When our domain is a subset of R, then the compact domains are the closed and bounded subsets of the reals, which are unions of finitely many closed intervals [a,b]. But the theorem I used is the generalized extreme value theorem, which works for any compact domain.

How did I know to do this? I'm not sure I can think of a general reason, since I would use this method only when the problem is specifically like this, i.e. to show that there is some M such that |Tx| < M|x| for all x. One reason is that on the unit sphere, |x| = 1. Since T is linear, you can know what T does to any vector v = |v|w, where w is the unit vector of v, v-hat, just by looking at Tw, and multiplying by |v|.

 

[Hurkyl]

To prove that our domain, S, is compact, it suffices to show that it is closed and bounded.

While true for spaces like C^n, that's not true in general... I don't know if it's true for Banach spaces or not.

 

[Oxymoron]

It's not true for Banach spaces - the unit sphere is not necessarily compact

 

[Oxymoron]

When proving that there exists some $m,M > 0$ such that

$$m\|x\| \leq \|Tx\| \leq M\|x\|$$

for $T$ and linear surjective map between normed spaces, is this saying that $T$ is bounded AND its inverse $T^{-1}$ is bounded?

If so, couldn't I just apply the Inverse Mapping Theorem?

 

[Hurkyl]

When proving that there exists some $m,M > 0$ such that

$$m\|x\| \leq \|Tx\| \leq M\|x\|$$

for $T$ and linear surjective map between normed spaces, is this saying that $T$ is bounded AND its inverse $T^{-1}$ is bounded?

If so, couldn't I just apply the Inverse Mapping Theorem?

No. That inequality cannot possibly say anything about the inverse of T until you are able to show that T is actually invertible. There are plenty of noninvertible operators that satisfy that inequality. For example, take the right-shift operator I mentioned in your other thread: m and M are both 1 for that operator!

 

[Oxymoron]

Ok, I just went through all of what AKG said in detail and I think I like it. If I have anymore problems I'll post later. Thanks guys.

 

[AKG]

Oooh, well I don't like it so much any more, since you don't even know you have a compact domain. Do you have some other reason to believe that our unit sphere in the space we're working with is compact (I know it's closed and bounded, but this doesn't appear to be enough)? Suppose you take the unit sphere in an infinite dimensional vector space. Then take the isomorphic operator T which maps (x₁, x₂, ..., x_n, ...) to (x₁, 2x₂, ..., nx_n, ...). Assuming some reasonable norm which would place (1,0,0,...), (0,1,0,...), (0,0,1,0,0,...) etc. on the unit-sphere, it would place (1,0,0,...), (0,2,0,...), (0,0,3,0,0,...) in the image of the sphere under T, and again assuming this norm is what I'd like it to be, these vectors would have norms of 1, 2, 3, ... etc. i.e. they would form an unbounded sequence, and so there would be no M such that |Tx| < Mx for all x.

 

[Oxymoron]

Yeah, I have just spotted that problem too AKG. I have reason to believe that the compact domain/extreme value theorem is not the right avenue.

Here is the question again:

Prove that $T$ is an isomorphism $\Leftrightarrow$ there exists $m,M > 0$ such that

$$m\|x\| \leq \|Tx\| \leq M\|x\| \quad\quad \forall \, x\in X$$

where $T:X\rightarrow Y$ is a surjective linear map between two normed vector spaces.


Now, unless I am mistaken, $T$ is an isomorphism if and only if (1) $T$ is bounded, (2) $T^{-1}$ exists, and (3) $T^{-1}$ is also bounded.

We want to prove that if $T$ is an isomorphism then it is bounded above and below by positive, non-zero $m$ and $M$. So assume $T$ is an isomorphism, then it is bounded, invertible, and its inverse is bounded.

Since $T$ is bounded, then

$$\|Tx\| \leq M\|x\| \quad\quad \forall\,x \in X$$

Then, since $T$ is an isomorphism it has an inverse: $T^{-1}$ and this is bounded. BUT, how can we be sure that

$$m\|x\| \leq \|Tx\|$$

and that $m > 0$?

Well, I am pretty sure that that is where is surjective bit comes into play(?) If $T$ is a surjective isomorphism we can check where $T$ maps $x=0$, i.e. $Tx = 0$ if and only if $x=0$. I am not too sure about this, but I think we might find that considering the kernel of the operator we can assure that $m > 0$.

Warning! For $T$ to be invertible it needs to be bijective. We already know that it is surjective but not injective. But we know that $\ker(T) = \{0\}$

Therefore (assuming the kernel bit works, I don't know maybe you guys could help me there), $T$ is bounded above and below by positive, nonzero numbers m and M.

Thats just the forward inclusion tho...

 

[Oxymoron]

I think for the reverse inclusion $(\Leftarrow)$ it is trivial. Because if $m\|x\| \leq \|Tx\|$ and $\|Tx\| \leq M\|x\|$ for all $x \in X$ and $m,M > 0$ then $T$ is bounded by $M$ and $T^{-1}$ is bounded by $m$.

Therefore both $T$ and $T^{-1}$ are bounded. But I am not sure if this is enough to prove that $T$ is an isomorphism, let alone whether or not $T^{-1}$ even exists.

But since $T$ is a surjective linear map it is okay(?) - not too sure. So close I think!?

 

[AKG]

Well I've already given you the reverse implication. If there is a positive m such that m|x| < |Tx| for all x, then Tx = 0 iff x = 0, therefore T is injective. You should know this, but if not, observe:

Tx = Ty
iff Tx - Ty = 0
iff T(x-y) = 0 since T is linear
iff x-y = 0 since T is bounded below
iff x = y

hence T is injective. So we have that T is a surjective linear map that is also injective, so its an isomorphism.

The forward implication says that given T is an isomorphism (thus linear, surjective, injective, bijective, has an inverse), show that it is bounded above and below. Now somehow you claim that you know that T will be bounded. I assume you have some theorem that's telling you this? What specifically does it state, and what is specifically meant when it says bounded? I'm used to seeing "bounded" used to refer to real valued functions that are bounded above and below, i.e. there are real (not necessarily positive) m and M such that m < f(x) < M for all x in the domain. I assume by "bounded" you mean that there is some real M such that |Tx| < M|x| for all x in the domain.

The problem with your, "Well, I'm pretty sure that this is where the surjective bit comes into play..." is that, well, for one, it is the injectivity of T (which follows from it being an isomorphism) that Tx = 0 iff x = 0, not from its surjectivity. But the real problem is that this does nothing for you. Tx = 0 iff x = 0, that's true. But consider the unit sphere. T may not map any of it to 0, but it may map it in some weird way so that it spirals towards 0 (but never reaches it). So you would have that |Tx| > 0 for all x on the sphere, but you couldn't pick a specific m such that |Tx| > m for all x on the sphere. I mean, linear functions don't normally do things like map a sphere or a circle into something that spirals towards the origin, but the issue is proving it.

However, assuming you're right about T being bounded (above) and T^-1 being bounded (above), let T(S) be the image of the unit sphere under T. Since T^-1 is supposedly bounded above, then there is an M' > 0 such that for every x in T(s), |T^-1x| < M'|x|. This tells you that T can't send S to something that spirals towards 0, in fact, it can't send it to anything with norm less than 1/M'.