# Compact operator definitions

#### Fermat

##### Active member
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
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
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.

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