Hi, we'll continue by looking at the concrete examples in order to understand Hallimarr's

criteria better.

And the first example we are going to look at will be the following.

We assume that data is generated from parameters in the following form.

So we have in essence two identical data points.

We can also write this as y is 1, 1, 1, 1 times x, where this x is an R2 vector and

y of course again is an R2 vector.

So let's check whether this fulfills the three criteria for well-posed problems.

And we can see that the first requirement, which is surjectivity of the operator a, is

not fulfilled here.

So this operator here is not surjective, so we can see this for example like this.

If y is the vector 1, minus 1, then there is no solution x such that a x star is 1,

minus 1.

So that means, well the reason for that is of course that a is not onto or not surjective.

The range of a is all scalar multiples of the vector 1, 1.

So the first y1 and the second y2 have to be identical and then only then we are in

the range of a.

So one thing maybe is a bit confusing here, but why is this relevant?

We can see that y1 is x1 plus x2 and is the same as y2.

So from definition of the data we can see that y1 will always be y2, which means the

vector y will be in the range of a.

Because y1 will be equal to y2, so y will have form some constant times 1, 1.

So this will give us data that actually is in the range of a, so we don't have to worry

about this missing surjectivity of a.

So that's a good reasoning of course, so that makes sense, that is true.

So we should be worried.

At least that's the reasoning for that.

But it makes sense to even not allow this case to happen.

Usually data is acquired noisily.

That means that y will more likely look like something like ax plus epsilon1, epsilon2,

with epsilon1 not being the same error as epsilon2.

So those two lines correspond to two outputs of the measurement device and they will have

different measurement errors.

And if that is true, so if we have measurement noise in that specific form, and then y1 will

not be y2 if epsilon1 is not equal to epsilon2, which is highly probable.

So even in this setting, if the data is generated in the same way, so if the first component

of the data is generated in the same way as the second component, even a slight amount

of noise on those two components will drive the data outside of the range of a, and that

means that we cannot construct a solution x fitting the data anymore.

So for example this could be the result of measuring that with additive noise.

So the first noise component could be plus one, the second one could be minus one, and

this could happen for example.

So as we usually will have noise, this is a problematic setting.

So we see that the first Hadamard criterion is violated and thus the inverse problem is

ill-posed, at least in the presence of noise.

Example number two, this looks similar to the one before, but now we have just one data

point, so we have removed the copy of the second component here.

So the data is just the sum of x1 plus x2, and x1 and x2, this is the parameter, so we

could write this as 1, 1 times x1, x2, so this is the parameter, and this is the operator,

multiplication by this matrix here.