A question on bounded linear operators (Functional Analysis)

[Ricky2357]

Suppose T: X -> Y and S: Y -> Z , X,Y,Z normed spaces , are bounded linear operators. Is there an example where T and S are not the zero operators but SoT (composition) is the zero operator?

 

[matt grime]

Yes. Don't try to make S and T very complicated though. This has nothing to do with functional analysis or boundedness of operators at all, really.

 

[Ricky2357]

Okay, can you think of any such simple example?

 

[Ricky2357]

By the way, it has to do with functional analysis since I am asking for an expamle of BOUNDED linear operators and not just linear operators.

 

[matt grime]

Yes. I can. But what I can think of isn't the point. What are some simple examples of normed spaces and (bounded) linear operators? (Hint: you did these very early on in maths.)

 

[Ricky2357]

R with absolute value is simple enough

 

[Ricky2357]

So can you help me out here ?

 

[DeadWolfe]

Note that in a finite dimensional vector space the boundedness condition is trivial.

Thus you would be asking whether it is possible to have a map S which maps the image to T of zero, even though S is not the zero operator.

In other words, you are asking whether there are matrices other than the zero matrix which do not have full rank.

 

[matt grime]

R with absolute value is simple enough

And what are the linear maps from R to R? Just multiplication by a real number, so that won't do. So what's next?

 

[Ricky2357]

Why is that? A linear operator from R into R of the form Tx=k*x is bounded.
(absolute value as norm)

 

[DeadWolfe]

For god's sake! You are asking whether the ring of operators in a vector space V has zero divisors. In the case V=R that ring of operators is a field, so obviously not. However, in the case where V has dimension greater than one, that ring is the ring of matrices.

If you don't know where the ring of matrices has zero divisors you are definitely not ready to do functional analysis.

 

[Ricky2357]

DeadWolfe you have my respect. Your idea was brilliant! It is enough to find two nonzero matrices such that their product is nonzero. Gongratulations man and thank you !

 

[matt grime]

It is enough to find two nonzero matrices such that their product is zero [note correction].

Yes. That was clear from the outset, since bounded linear maps are after all linear maps.

 

[matt grime]

Why is that? A linear operator from R into R of the form Tx=k*x is bounded.
(absolute value as norm)

If you don't see that the product of two non-zero real numbers is non-zero, then I echo deadwolfe in saying you shouldn't be doing functional analysis, and instead relearn your first course in algebra. And your first course in linear algebra where the answer for this comes from.

A slicker idea comes from projection onto subspaces (maps XxX-->XxX sending (x,y) to (x,0) and (0,y) would also be trivial examples).

 

[Ricky2357]

Okay something else matt grime. Is it true that the range of a bounded linear operator is always a bounded set?

 

[matt grime]

What is the definition of a bounded linear operator? And range?

 

[DeadWolfe]

No, the identity map on R is a counterexample.

Do you make any effort to answer these questions before you pose them to us?

 

[mathwonk]

come on man, this is trivial from the very definition of "bounded".

i.e.a bounded operaTOR IS ONE SUCH that the image of the unit ball is a bounded set. so hellooo...?

of course i could be wrong and hence appear (correctly) very stupid, but i am just saying look at the definition of bounded operator and answer your own question.oops i blew it again, as i assumed you meant that the image of a bounded set is bounded, but you said range, well there you go, i am an idiot.

i have noticed that any time I make fun of someone elses ignorance, it is in fact myself who is out of the loop. good lesson as usual. i will probably forget it quickly anyway.

 

[mathwonk]

now that we have settled this issue, i am thinking maybe you meant "compact" operator, but i forget the definition of those. i am guessing though that they have bounded images. i will look it up on wiki stupedia.wait, how stupid can I be?! any non zero operator has in its image a line which is unbounded!duhhh?! ]\\is this correct?? help me out here.

 

[DeadWolfe]

Yes Roy, the image of any nonzero operator is unbounded.

To the original poster: the questions you have posed in this thread are all trivially answered with examples in finite dimensional vector spaces. You should brush up on your linear algebra before trying your hand at functional analysis.

 

[Ricky2357]

First of all I never said I can teach functional analysis or linear algebra, so take it easy. Second, the image (range) of a bounded linear operator is not always a bounded set. A bounded linear operator is called bounded because it maps bounded sets into bounded sets.

 

[matt grime]

Who mentioned anything about accusing you of wanting to teach functional analysis?

Contrary to the belief that the US university system, for one, propogates, maths is not modularized into unrelated areas. Depth of knowledge is needed, and you can't blithely assume that you'll pick up all you need to know to do functional analysis purely from doing a course in functional analysis. There are certain prerequisites, and a sound understanding of linear algebra is definitely one of them. People have your best interests at heart when they suggest you relearn that material before proceeding, if possible.

You need a thorough understanding of eigenvalues, which will come up when you start studying the spectrum of an operator. If you understand the linear algebra before hand it will make your life a whole lot easier.

 

[DeadWolfe]

Second, the image (range) of a bounded linear operator is not always a bounded set. A bounded linear operator is called bounded because it maps bounded sets into bounded sets.

Yes, it takes bounded sets to bounded sets. But it is not defined on a bounded set, it is defined on the whole space, which is unbounded.

If F is a nonzero operator, then (as mathwonk said) there is a one dimensional subspace V which is mapped into some other one dimensional subspace. Thus the image of the operator contains a line, and is thus unbounded.

I'm sorry if I offended you, but functional analysis extends linear algebra in some complicated ways, and if you don't understand linear algebra to begin with, you won't have any idea what's really going on, even if you can understand and repeat the definitions. (Not that I understand it myself...)

 

[Jimmy Snyder]

It's an example of inconsistent terminology. A bounded linear operator is, in general, not a bounded function.