Hi, we're starting with a new chapter which is finite dimensional inverse problems via

discretization. In the last few weeks we have talked about continuously defined even infinite

dimensional inverse problems and this allowed us to understand how to define ill-posed inverse

problems but still in the end if you want to analyze them and if you want to implement

them, if you want to work with inverse problems you will have to practically do this on a

computer which means you have to discretize your inverse problem. You have to discretize

your parameters, you have to discretize the forward problem so it means that at some point

you will have to go to your computer and make everything finite dimensional. And this chapter

will be about understanding this process a bit better and how the ill-posedness of the

underlying maybe infinite dimensional inverse problem, how this ill-posedness leads into

the finite dimensional space as well. And the most important application we're going

to look at in this chapter will be deconvolution or deblurring depending on the context in

which you're working. So if you think about mathematical imaging then deblurring will

be a term you're looking for but in a more general sense it's just something which is

called deconvolution. We consider the one dimensional case for simplicity so this is just in order

to keep the notation a bit less cumbersome, not the same ideas hold for 2D images for

example and also 3D data, 4D data, it doesn't really matter what kind of data you're looking

at. So we'll do everything one dimensional so that the notation is less clustered. Let

u be a function from R to R, one periodic, which means that u of x is u of x plus n for

all n in z. We also do this again just for simplicity. We can do everything for non-periodic

data as well but the mathematics is easier, the terms are less complicated. The issues

with non-periodic images will be artifacts at the boundary which we don't want to think

about for now. So you can generalize the whole setting to the more complicated case as well.

And equivalent, you could write u as being a map from let's say minus one half, one half

to R with u of minus one half is equal to u of one half. Because if that is the case,

if you have a function, I don't know what it looks like, maybe like that, then of course

you can continue it the same way and get a periodic function. I'm bad at drawing this

but you get the point, right? So if the function is defined on a set from minus one half to

one half, if the left endpoint and the right endpoint are the same, you can just glue them

together in order to get a periodic function on the whole real line. So that's how you

can do this. Sometimes we will think about u as being a function in this setting, sometimes

in this setting they're equivalent but sometimes it's easier to use one or the other one. And

that is something that we call the image. So I have an example for you. The example

will be, so ignore everything that you're seeing, just look at this picture here, so

this will be our image. You can think of this as being an intensity, so if it's zero then

it's maybe black and if it's one it's white. So this is an intensity of pixels along a

line, so it's a 1D image so to speak. And we will think about what blurriness means

and you can see that this will look something like that, so the interfaces, those jumps

will be less pronounced, everything will be smoothed down a bit, this will be the end

result. So the idea is that we take a true image and it's blurry in some sense and this

will be the output of the blurring operation. Now let's try to model the blurring operation.

For that we need a blurring or a convolution kernel and we call this psi zero from minus

one half to plus one half to r and we will take a specific example for such a blurring

kernel. You can exchange this for something could be different but we'll say well it's

zero for x outside some number a. a will be a number between minus one, sorry, will be

a number between zero and one half so it's strictly less than one half so think a might

be one fourth or something. This function is equal to zero outside of this threshold

a and inside this threshold it is ca times x plus a squared times x minus a squared.

How does it look like? So here we have minus one half and one half and a will be, well

maybe here, minus a plus a here. We know the function is zero here, zero here and this