0
$\begingroup$

I am studying Fredholm theory and I'm not sure I grasp the importance of some aspects of it.

I have read the following 'stability of the index' theorem which states that if $D:X \to Y$ is a Fredholm operator between two Banach spaces, and $K:X \to Y$ is a compact operator, then $D+K$ is also a Fredholm operator and $$\text{index}(D+K) = \text{index}(D).$$

Ok, so I know that this means the operator $D+K$ has finite dimensional kernel and cokernel, and closed range, and it has the same index as the original operator $D$.

But what's so important about that?! Why would this be a useful thing to have when we want to solve some equation?

Also, this seems to be very specific, we need to first have a Fredholm operator, and then a compact operator, to use this theorem. Does this not massively limit the usefulness of this theorem in practise..or are Fredholm operators and compact operators very common in applications?

Does anyone know of a simple explicit example that illustrates the usefulness of this theorem?

$\endgroup$
0
$\begingroup$

An important example is the case $X=Y$ and $D=I=id_X$. Take a look in

https://en.wikipedia.org/wiki/Fredholm_integral_equation

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.