Hi, we have now exercised ourselves in using finite drag approximation in order to prove

compactness, but this still doesn't help us all the way to proving or to disproving whether

this is a compact operator. So this object here, taking a function, let's say on the

line 0, 1 and taking the primitive of that, I have told you that this is a compact operator,

but so far it's hard to show this. So we would have to find some approximation via finite

rank operators and prove uniform convergence of that. And that's still quite difficult,

but that's a detour that we can make in order to derive some more machinery that we can

use in order to prove compactness of all integral operators of this kind. And this detour will

be talking about Hilbert-Schmidt operators and that they are compact. So let's start

right away with the definition. Definition 2.10. And this works on Hilbert spaces. Let

h1 and h2 be separable Hilbert spaces. And then, well, let's also write E1, sorry, EI1

and EI2. Let them be orthonormal basis on h1 and h2. Orthonormal basis, I will abbreviate

ONB, orthonormal basis. And a linear and bounded operator A from h1 to h2 is called Hilbert-Schmidt

operator. So I will abbreviate that by HS Hilbert-Schmidt operator. If the following

object is finite, A HS squared, so the squared Hilbert-Schmidt norm, so we call this the

Hilbert-Schmidt norm. And this is defined as the sum from 1 to infinity of A EI1 squared.

This of course is the h2 norm. So what we do here is we kind of test the effect that

A has on all the basis vectors of this separable Hilbert space. And then we sum, we kind of

take an L2 norm of the results in the h2 norm. Well, it looks like a complicated object,

but all we're doing is we sum the squared h2 norms of all the effects that A has on

the first Hilbert space's basis. So this is just an object we can compute. So for a specific

operator we can compute this quantity. And if that is finite, then we call this operator

a Hilbert-Schmidt operator. So, there's a, let's make a remark first. We can alternatively

write something else. Well, we start by rewriting x. So any x in h1 can be written as the sum

of the inner product x with the first space's basis vectors and then times E1i. So this

is the basis decomposition of x. So we can write x in this form. And similarly we can

write ax as the sum j from 1 to infinity of ax tested with Ej2 times Ej2. Now we can plug

this in ax, j equal 1 to infinity of a of, now let's plug this in here, so i is 1 to

infinity x Ei1 Ei1 Ej2 Ej2. So you can see that there's inner product inside the inner

product because we're applying a to this vector and so on. And now we can pull this out again.

So we take sum over i and j from 1 to infinity of x Ei1 a Ei1 Ej2 Ej2. So this is applied

to that. This is pulled out here and then we can also pull this real number out of the

inner product and that's what we get. And what can we do with that? More specific, a

applied to a specific basis vector Ei1 will then be, well this double sum collapses to

a single sum. Let's see this here. Let's take Ei but Ek, no to not mix up the indices. So

this is Ek1 tested with Ei1, Ej2, Ej2. So this is an orthonormal basis. So this is 1

if and only if i is equal to k. So it means that the sum over i drops to a single term

which is i equal to k. This becomes sum from j equal 1 to infinity. This is 1, i is k,

so this is a Ek1 Ej2 Ej2. So this is also a nice formula. We will need all those formulas

later. It will be nice to have them later. Now we can plug this in, especially in the

definition of the Hilbert-Schmidt norm. So what's the definition of the Hilbert-Schmidt

norm again? This is the sum over i of the squared Aei1 H2 norms. So we plug this in.

Sum over k equal 1 to infinity of Aek1 H2 squared. This is U2, well this, we're just

plugging this in here. Maybe let's do one more computation before we do that. Put this

down here, we'll need it later. The first thing we have to do is take the norm of Aek1

H2. Well this is the same as, let's write the square here, j equal 1 to infinity of

Aek1 Ej2 Ej2. And you know, a sum, I don't want to write j again because this would clash

with the j here, so let's write i 1 to infinity of Aek1 Ei2 Ei2. Again, writing i and j outside

of the inner product, we have, pull this out, Aek1 Ej2, pull this out of the product, Aek1

Ej2 Ej2, and the remaining thing is Ej2 in the inner product with E12, so Ej2 E1 Ei2,