So today we will continue with studying the total variation model.

The original model is called TV regularization. So what we looked at was the functional

f squared plus alpha times r of u and r of u now can be chosen as

the semi norm BV or also TV of u and we have seen this is the supremum of all vector fields phi.

We can take them, it's actually okay if they are C1 and 0 at the boundary but we can take them in C

infinity and phi is equal to 1 almost everywhere of integral u divergence of phi and for example

if u is just a C1 function and the total variation is just the integral of the gradient.

Okay. Okay so we have understood this variation problem we have seen that the total variation

has some nice properties. In particular the total variation is finite if u is piecewise constant.

And this in 1D but we could have it in multiple dimension if u for example says of the form

constant Ci times the indicator function of some sub region omega i. Okay so think about

something like this you have omega 1 here omega 2 here omega 3 and whatever okay omega 4 this thing.

Okay and each of them has a distinct but uniform gray value so chi omega i is equal to 1 if x is in omega i and 0 else.

We can do basically the same computation as in 1D and compute integral u divergence of phi.

So this is then the sum Ci. Okay let's first write integral over omega indicator function of omega i times u.

Divergence of phi. So effectively this is 0 outside of omega i and 1 inside omega i so this is just the integral of divergence phi on the sub region omega i.

And now with Gauss theorem we have this is the sum 1 over m Ci integral on the boundary of omega i phi times n.

And now we know since phi is less or equal than 1 this implies phi times n is less or equal than 1 or Ci times phi dot n is less or equal.

We can do it without the absolute value. It's less or equal than the absolute value of Ci.

So we see that the total variation is definitely less or equal than the sum of the absolute values of Ci.

Okay and we have equality by choosing a vector field phi such that phi dot n is equal to the sine.

Of Ci on the boundary of omega i and then we take some suitable extension of the boundary.

Okay now the only problem is such phi in general will not be C infinity so this will not be smooth depending on what we have.

So for example if you take omega i like this inside the larger domain omega okay then we would have that the normal is like this.

So the normal field would be discontinuous in the corners right.

So if you take a phi so that this is true say okay Ci is for example 1 for C if Ci is 1 then we would take phi dot n should be equal to 1.

Which means effectively we have to choose phi is equal to n okay.

So we cannot do this exactly with having a differentiable we cannot even take a smooth phi.

However we can make a little trick we could say okay we cut out some region like I draw here is like with some size epsilon.

Okay so we say okay phi is equal to n say a distance epsilon to the corners.

Okay here it's just so it's just constant here in each of these and then we take some smooth interpolation here okay.

Okay so what we find then is indeed that integral of boundary omega i phi dot n is actually the integral of omega i.

Everywhere plus the integral here only on this epsilon neighborhood of the corners.

And here we take we have phi dot n minus 1 okay.

So away from the corners this is 1 anyway and on the corners okay we subtract 1 or close to the corners and we add the 1 here okay.

Okay now the interesting point is let's call this phi epsilon because it depends on the epsilon okay.

Now we could choose this so that this is less or equal than 1 in absolute value.

So this means this will converge to 0 as epsilon goes to 0.

Okay because we're integrating here over some very small segments of length epsilon or 2 epsilon.

Take all four of them it's 8 epsilon but it still it still goes to 0 so this integral we can bound by the absolute value we can bound by 8 epsilon.

If you have four corners so this goes definitely to 0.

So then we still see okay there is no maximum which we do not expect but if we take the supremum okay.

We cannot take epsilon equals to 0 but we can take epsilon arbitrarily small so in the limit in the supremum we still get this okay.

So we have this equality for quite general domains okay.

So it's a very natural thing and it's very nice to interpret from a geometric point of u of u.

So the total variation just basically counts the jumps.

I forgot something here of course of course here then I get the ci times the integral of 1.

So what we have here is the size of the jump how much the gray value is changing times the length of the boundary okay.

So if you have something like this this has a very long boundary somehow.

So it's penalized much more than for example a circle with the same volume which has a much smaller length of the boundary okay.