So, the question of this lecture is how do we check compactness for a given operator.

So if I give you, for example, following mapping from small l2 to small l2, and such a sequence

x1, x2 is mapped to x1, x2 divided by 2 squared, x3 divided by 3 squared, and so on.

So the nth entry is divided by n squared. Is this a compact operator or not?

You can try to apply the definition. So you would have to take any bounded sequence of

sequences and you have to show that there exists a subsequence which converges in a

sequence. And you would probably manage to do this in some way, but there are easier

ways and this will be something that the following lemma shows us. So this lemma is about properties

of compact operators. The first property is the following. Every compact operator is bounded.

So this can be used, for example, in the following form. Thus every unbounded operator is not

compact. Well, let's write linear. Linear and compact operator. We are only looking at linear

operators for now. So every unbound operator is not compact. So this is how you can show

that something is not compact by checking that it's unbounded. The second property is

let k1 and k2 be two compact operators from x to y. Compact and linear operators. Then

for all lambda and mu, the real numbers, lambda times k1 plus mu k2 is also compact. So you

can take two compact operators, take any linear combination of them, you will also get another

compact operator. The third one is let k1 from x to y and k2 from y to z be linear and

bounded operators. If either k1 or k2 are compact operators, then

k2 concatenated with k1, which is a mapping from x to y to z, then this is compact. So

we don't require both mappings to be compact. It's enough that k1 or k2 is compact, then

this concatenation of these two operators will be compact. And the fourth, which will

be the most useful for actually proving compactness for a given operator, if kn, so all kn go

from x to y, so this is a compact and linear operator for every n in N. So this is a sequence

of compact operators and k from x to y is an operator such that the operator norm of

kn minus k converges to zero for n to infinity, then k is compact. So why is that useful?

Because it is usually quite easy, if you want to prove contactness for a given operator,

to define some sort of finite rank approximation for this operator. We will see this in practice

later. The trick is to define operators kn, which have finite rank, so finite image, so

to speak, so they map into finite dimensions. This means that they're essentially finite

dimension operators and finite dimension operators, which are bounded, always compact. And if

you can show that those kn converge to k in the operator norm, then the resulting k is

compact. So you can usually start with some k, you want to check whether it's compact.

And if you manage to define an approximation with compact operators, such that this approximation

is converges in the operator norm, then you're done. You have shown that k is compact. So

this may be the most useful part of this lemma. Okay, so let's now show each of these properties.

We start with the first one. The first one is we want to show that every linear and compact

operator is bounded. How do we do that? We assume that k from x to y is compact, linear,

and unbounded. So we assume something which we'll see can't be true. This means that one

of these conditions can't be true. For example, it cannot be compact and unbounded. It has

to be compact and bounded, for example. Or it could be unbounded, but then it's also

not compact. So we have to derive some sort of contradiction here. Under these assumptions,

linear exists by unboundedness a sequence xn in X with the property that it is bounded

in X, but kxn is unbounded in Y. So let's say we can choose by taking a subsequence,

we can assume that this converges to infinity, unbounded in Y. And now we see that something

is wrong here. So we have a sequence which is bounded in X, and we apply a supposedly

compact operator on it, and then it's unbounded. But that is of course a contradiction to the

fact that any compact operator maps bounded sequences to sequences which have a converging

subsequence. So on the other hand, k is compact, so by boundedness of this sequence xn, there

exists a subsequence kxnk which converges in Y. But any subsequence of kxn also is unbounded,

and that is a contradiction between those two points here. These can't be both true.