A linear map between normed linear spaces is called compact if is compact for every norm-bounded subset of .
It is enough to check for since each norm-bounded subset for some .
Compact operators are bounded operators because compact subsets of a metric space are bounded in the metric.
For , implies that is compact.
If is infinite dimensional, then (which equals ) is non-compact, so is a bounded operator but not a compact operator on .
Recall a metric space is compact if and only if every sequence has a convergent subsequence.
An operator is compact if and only if for every bounded sequence in , the sequence has a convergent subsequence.
Proof. (⇒) compact, bounded implies compact.
(⇐) bounded, a sequence in then get in such that . Since then converges for subsequence , get to converge (to the same limit).
With , , in and , , in ,
The Volterra operator defined by is compact. To see why, first notice and also because (so in particular, ). Let be a bounded sequence in , say for all . Then and . That is to say, you can make this small for all simultaneously by making sufficiently small, and is pointwise (here in fact uniformly) bounded, and equicontinuous. Then Arzela's Theorem says there exists a uniformly convergent subsequence of , say , with for some . So, , which makes in .
For normed linear spaces and , let be the set of compact operators: such that is compact.
If is a Banach space, then is a closed subspace of .
Proof. If and are in and is bounded in , then there exists such that converges, and there is also a further subsequence such that converges. Then we have that is convergent, which shows .
Scalar multiplication is clearly ok. For
is a sequence of compact operators converging in norm to
for all . Using compactness of
…, get successive subsequences
… such that
converges for in
converges for in
be the th item in the th sequence. This makes
eventually a subsequence of
…. It follows that
converges for every .
there exists such that
and this is less than
for sufficiently large and . So the
sequence is Cauchy, and we're done since is complete.
The spectral theory of compact operators has the following special features.
Let be a compact operator on a Banach space . Then
Suppose contrariwise that but . As usual, we may assume . The range of cannot be all of , since if it were, the injective operator would be invertible by the open mapping theorem. Thus is a closed proper subspace of , and is a linear homeomorphism of with . It follows that the subspaces () are all closed, with for each , Lemma I.65 yields a sequence with for every . Take and write . Then . Since , we conclude that for all , which contradicts .
We now have a complete picture of the spectrum of . If is finite-dimensional, then of course can be any subset of of cardinality not exceeding . If is infinite-dimensional, then must belong to because the unit ball of is not compact. The rest of consists of either a finite set of non-zero eigenvalues, or a sequence of non-zero eigenvalues converging to . For every non-zero , (eigenvalue or not) the subspace is finite-dimensional.
Composition of a compact operator with a bounded operator (in either order) gives a compact operator. Thus is a two-sided ideal in which is closed if is a Banach space. is called the Calkin algebra of .
Call finite rank if . Let be the set of finite rank operators in .
is also an ideal in . We have if is complete. Often, perhaps universally, when is a Banach space.
Now, specialize to a Hilbert space .
Let . Then
Proof. and are in .
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