Hi, it remains to show this lemma here. We want to show that the Hilbert-Schmidt operator

is always compact. And this proof essentially just boils down to using a specific set of

bases for this space and this space and also the space of the Cartesian product of omega

and sigma. So what we do is, so L2 of omega, L2 of sigma are separable Hilbert spaces with

orthonormal bases phi i and psi i for L2 of sigma and L2 of omega respectively. So this

means that a function f of x can be written as a sum ci times phi i of x if f is in L2

of omega and ci can be written down as the integral of f of x times phi i of x dx, where

we integrate over omega. And same thing with L2 of sigma. Okay. And because these orthonormal

bases functions for L2 of sigma and L2 of omega, we can also write down an O and B for

sigma times omega as orthonormal bases. Let's maybe, so what's the best order here? I will

always mix up x, y and omega and sigma here, I'm afraid. So let's say omega, we call the

variable x here and y here. Does it make sense? We will have, we have to plug in L2 of omega.

That's not the right way to do this. Let's write this as y here, x here and then, okay,

does that make sense? k of u of x integral k of x, y, u of y. I think that's now consistent.

Okay, good. So this has orthonormal bases. Again, this is y and x. So orthonormal bases

phi i of x times psi j of y. And now we have to let the indices run from i and j from 1

to infinity. So this countable combination is an orthonormal basis for this space L2

of sigma times omega. Okay, so this will always be a bit difficult to keep apart, but that's

essentially it. And we know that k, so small k, is in L2 of sigma times omega. So that

means there exists ckl in R such that k of x and y is the sum kl from 1 to infinity ckl

phi k of x and psi l of y. And not just that, but because k is in L2, this means that this

here is in L2, which means that the double integral sigma and omega, the sum kl 1 to

infinity of ckl phi k of x, psi l of y, this here is finite, dy dx. Okay, so what does

that mean? That means we still have the integrals sigma and omega. What happens now if we take

the square of this sum? That means we have k and l from 1 to infinity. We also have,

let's call them k prime and l prime going from 1 to infinity. Then we have ckl times

ck prime l prime phi k of x psi l of y phi k prime of x and psi l prime of y dy dx.

Okay, so that looks very complicated, but it reduces fairly quickly because, well let's

look at the integral dy, so integral omega, blah blah blah dy. We have this term here

and this term here. Because psi l is an orthonormal basis, this means that these integrals are

0 if l and l prime are not equal. So it means that, yes, they are two different indices

l, l and l prime, but they have to coincide, otherwise the term will be 0. And the same

for k and k prime, so this actually is just 1 times k and 1 times l. And the integral

then is the orthonormal of phi k and psi k and so on and so on, so this just is this

term here and this has to be finite. We can also write down what ckl is, so ckl is this

integral of kxy phi k of x psi l of y dy dx. That's also a useful way of writing this down.

Okay, I think we have now written almost everything in terms of the basis function that we can

use. And now in order to prove that this operator is a Hilbert-Spitt operator, we have to consider

the object what is k of some psi j. So is that the right object to consider? Let's just

check really quickly. k goes from l2 of omega into something, l2 of omega has those psi

as basis vectors, so that makes sense. Right, okay. So what's k applied to psi j? Well,

this is the integral over omega, so maybe we could write down an x here, integral over

omega k of x y psi j of y and x, this argument is the argument that we plug in later, dy.

Okay, so what can we do now? We can use what we know about k, k is this sum here, so this

is actually sum over, ah I see, k is not that good of an index. So this k is the kernel

function name, this is the next year, and it's maybe a bit unfortunate, let me undo

this mistake, so m and n, c m n, phi m of x, is that the right order? Phi m x psi n

of y, yes, psi n of y, this is k, but now we integrate and multiply by psi j of y dy.

And again, because we're integrating over y, n has to be equal to j and this term will

then be 1, so this integral is nothing else but sum over all m from 1 to infinity, c m,