Hi, we're going to take a closer look to the Tikhonov regularized solution and we will

see a slightly different viewpoint on how to interpret it.

The goal will be to view this solution, this reconstruction, as a solution of the so-called

normal equation and for that we consider a functional q lambda of u, which is the functional

which the Tikhonov regularization minimizes, so it's exactly this functional here.

So one half, or let's drop the one half for now, so a times u minus f squared plus lambda

times norm of u squared.

So the Tikhonov regularized solution is the unique minimizer of q lambda.

And we can compute this minimizer in a slightly different way by looking at the derivatives

of this functional.

So let's look at directional derivatives.

The directional derivative in direction w of q lambda at u is given by d by dt of q

lambda of u plus tw evaluated at t equal to zero.

So that's the definition of the directional derivative of q lambda in u in direction of

w.

And we can compute that.

This is, well it consists of two terms, d dt of au plus tw minus f squared plus dt of

u plus tw squared and everything evaluated at t equal to zero.

And these two terms we'll call time one and term two and looking at them, well let me

add a few pages first.

Okay, alright, so the first term looks like this.

We take the derivative, which is the derivative evaluated at t equal to zero of the inner

product of au plus tw minus f, au plus tw minus f.

And that can be multiplied out.

This is d by dt, t equal to zero of, well, the norm of au squared plus 2t au aw plus

t squared norm of aw squared plus, well no, minus 2t norm of au, oh that's one term missing,

2t, 2au and f minus 2t aw and f plus norm of f squared.

So take everything of that and take the derivative and take then t equal to zero and that is

the, let me check whether I've forgotten anything, but I think I haven't, should be alright.

Okay, so this becomes 2 times au aw plus 2t norm of aw squared minus 2aw f.

And we have to take t equal to zero, so this drops and so the middle term drops and we

get 2 times au minus f aw, that's the first term.

Second term is even a bit easier, so d by dt of u plus tw, u plus tw is equal to d by

dt of norm of u squared plus 2t uw plus t squared w squared, t is equal to zero and

well this is 2uw plus 2t squared w squared where t is equal to zero which is 2u times

w.

So this means the any minimizer of q lambda of u satisfies the following, so the minimizer

is, the minimizer is something we, let's just call it, let's call it ut lambda because it's

the Tiganov regularized solution and it satisfies that any direction derivative, in any direction

has to be zero.

So it satisfies that for all possible directions in our, I think, n, dw q lambda of u is equal

to zero which means that by what we just computed for all w, for all vectors w, we can drop

the two, what's the first term, au minus f aw plus uw is equal to zero, so au minus

f aw plus, I'll have, can't forget the lambda in front, plus lambda uw is equal to zero.

And we put this qualifier in here, t lambda and we can rewrite this by taking the transpose

of a and putting it on the other side, so this is equal to a transposed au t lambda

minus a transposed f and there's another term, plus lambda ut lambda comma w is equal to

zero.

And if that is zero for any directional vector w, then the left hand side has to be equal

to zero, this is kind of a variational argument here.