Hi, we are still in the business of trying to prove compactness for specific operators

because that will allow us to judge whether certain inverse problems are well posed or

ill posed and whether they are difficult due to this.

And we will now employ one tool that we have proven to be working in the last lecture,

which was that, as a reminder, if An is linear and compact and An converges in the operator

norm topology to A, then A is also compact.

This is a very useful result that we will need a few times, and this allows us to prove

the following compactness result.

We write A as being the map from L2 to L2, where x, so this sequence x1, x2, and so on,

is mapped to a different sequence given by x1 divided by 1, x2 divided by 2, x3 divided

by 3, and so on.

So this is similar to the identity map, but not keeping all the entries constant, but

dividing them by a factor proportional to the position, to the dimensionality, so to

speak.

I like to think of operators such as these as being funnels, because they squeeze the

dimensions in the tail down, so the general shape of this operator is, well, in my mind,

it's something like that.

So the first dimension is not touched at all, it's the identity operator, then it's on

half the identity operator in the second dimension, and so on.

So roughly, no matter what the end point, the beginning, the arguments are, the tail

dimensions, you ought to call them, so I call them the tail dimensions, they are squeezed

down to zero, at least roughly, approximately.

And operators such as these are compact, we have learned that the identity operator is

not compact, but this one is, so this is what we want to prove, this is a compact operator,

from L2 to L2, and the proof here is to use this lemma here to construct approximations

which are linear and compact and which uniformly approximate this operator A, in the sense

that the difference in the operator norm, sorry, this mistake here, this goes to zero.

Okay, we define An as another operator with the same domain and range, and such an x is

mapped to x1 divided by 1, x2 divided by 2, and so on, so very similar to what we have

here, xn divided by n, but then the rest is just kept constant at zero.

So first, An is obviously linear, and, and it's compact because it's finite rank, so

the dimensionality of the image is finite dimension, right, so it doesn't matter what

you plug in here, the range of An is always isomorphic to Rn, and because of this finite

dimensionality, we can easily see that this is a compact operator, so what we now have

to show is that the difference of A and An in the operator norm converges to zero for

n to infinity.

Okay, so let's apply the definition of this norm.

This is the supremum, and now I'm using the following characterization of the operator

norm, I'm taking x in L2, where the norm of x, the L2 norm, of course, is bounded by

1, and now I'm plugging this in, A minus A of x, and here, the two norm here.

In order to drop square roots, I will take the square of this, so this, this two means

L2 norm, this means square of the L2 norm.

Now, what was this?

Sorry, there's an n missing here.

That is supremum of x, which are bounded by 1, and well, what's that?

Ax minus Anx, let's look at the difference between those two images.

This one minus this one is zero, this one minus this one is zero, so this difference

vector Ax minus Anx has zero components in the first n slots, but then it's x n plus

1 divided by n plus 1, and so on.

The vector whose norm we take is zero, zero, zero, and then x n plus 1 divided by n plus