Welcome to our community

Be a part of something great, join today!

Compact operator definitions

Fermat

Active member
Nov 3, 2013
188
Let T:X->Y be a bounded linear operator. One definition of T being compact is that T maps bounded sequences to sequences with convergent subsequences. An equivalent definition is that T maps sequences of vectors with norms not exceeding 1 to sequences with convergent subsequences.
A couple of questions: What do these definitions require of X and Y. Can they be (incomplete)
normed spaces?

Also for the second definition, does it only make sense when X=Y? Every definition I've seen has assumed that but I've looked at a proof and couldn't see where they had used that assumption.

Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
Re: compact operator definitions

Let T:X->Y be a bounded linear operator. One definition of T being compact is that T maps bounded sequences to sequences with convergent subsequences. An equivalent definition is that T maps sequences of vectors with norms not exceeding 1 to sequences with convergent subsequences.
A couple of questions: What do these definitions require of X and Y. Can they be (incomplete)
normed spaces?

Also for the second definition, does it only make sense when X=Y? Every definition I've seen has assumed that but I've looked at a proof and couldn't see where they had used that assumption.
The definitions do not require X and Y to be complete, but I doubt whether you would ever find compactness used in the context of incomplete spaces. It is also not necessary to have Y = X, but in practice that is usually the case.

Another equivalent definition is this: if $T:X\to Y$ is a bounded linear operator, and $B$ is the unit ball of $X$, then $T$ is compact if the closure of $T(B)$ is a compact subset of $Y$.

A finite-rank operator is always compact. The set of compact operators is closed, so any operator in the closure of the finite-rank operators is compact. For operators on Hilbert space, the converse is true: If $X$ is a Hilbert space then $T:X\to X$ is compact if and only if it is in the closure of the finite-rank operators. On a general Banach space that is no longer true. A Banach space is said to have the approximation property if every compact operator on it can be approximated by finite-rank operators. For a long time it was unknown whether there were any Banach spaces without the approximation property. The first example was constructed about 40 years ago.
 

Fermat

Active member
Nov 3, 2013
188
Re: compact operator definitions

Thanks Opalg. Thinking about, it seems that whether an operator is compact depends on what co-domain we choose. For example, take T:X->X and suppose T is compact. Well if T wasn't surjective so we could restrict the co-domain to some subset Y of X, then Y might not contain all the limit points of the subsequences convergent in X. So then T would not be compact. I don't believe this to be true but where is the flaw?

EDIT: Answering my own question here, but this is probably why the co-domain is normally assumed to be complete.
 
Last edited: