Hi, we're going to start with a new chapter now, and this chapter will be slightly more

mathematical than the previous one, and it will be a fundamental chapter where we try

to understand the necessary mathematics in order to understand ill-posed problems in

infinite dimensions better, but we will start with a very practical setting, and it is definitely

true that many inverse problems, and maybe almost all interesting ones, so many function

valued inverse problems are in the form of integral equations.

What does it mean?

The data f is the image of an operator k applied to the parameter u.

Maybe there's an additional error term as well.

It's a function valued inverse problem, so all quantities involved are functions.

The parameter is a parameter function, as in the case of computer tomography, there

the parameter is the density mapping, the density field, so the function that assigns

a density to each point in the patient's body, so that's a function valued inverse problem,

and also the PDE-based inverse problems that I showed at the beginning of this course.

So function valued inverse problems are very common examples in those problems.

And this operator k of u, well it's a function, so we can apply this on x, we can plug some

variable x in, and this is a form, an integral of a sum domain of u of y, but not just u

of y, but it's coupled with some form of integral kernel, so we call this k an integral kernel

dy.

Or some k mapping from sigma times omega to r, and sigma omega domains, for example in

r2 or in r3 or whatever this setting is.

So that's a very broad type of inverse problem, and it looks very specific, but it covers

many of the examples we have seen before.

So for example, integration and differentiation, if the data is the primitive of u, so let's

make this concrete again, f is k of u, where k of u of x is the integral from 0 to x of

u of y dy.

That means that we are measuring the integral of the function, again that's this example

where the inverse mapping is the differentiation operator, which is discontinuous, so this

recurring example that we've seen three times already.

So in this setting, this is a specific example of this framework, we just have to pick the

right integral kernel in order to see this.

So by setting the integral kernel small k of x y as follows, 1 if x is larger or equal

than y, and 0 if x is smaller than y, then we see that the integral over u of y and k

of x y, and the domain here is, domains omega and sigma are the same, and they are the closed

interval 0 1.

So this integral omega is the actual, it's actually the integral from 0 to 1 dy, well

what is that?

That is integral 0 1, u of y k of x y dy, and well let's plug this in, y goes from 0

to 1, and if y is lesser equal than x, something happens, and if it's bigger than x, something

else happens, so we have to decompose this integral 0 to x, u of y times 1 dy, plus integral

from x to 1, u of y times 0 dy, and that is exactly the same thing as integral 0 to x,

u of y dy, and that is k of u evaluated at x.

So you see that this inverse problem is a special case of such an integral equation,

where the integral kernel is of that form that we have just picked.

So that's the first example, second example is the example of image deconvolution, so

omega and sigma are again the same thing, and this is the image, 2D image, that's the

domain of the image, so some rectangle for example.

And if we now pick k of x and y as some, well let's say convolution kernel of x minus y,

for example, so it's not the only choice that we could take, but it's a very common choice,

x minus y squared divided by 2 sigma squared, then we can see that the integral k of x,