How Can I Calculate the Norm of the Operator \(I-L^{-1}K\)?

[sarrah1]

I have a linear integral operator

$K\psi=\int_{a}^{b} \,k(x,s) \psi(s) ds$

$L\psi=\int_{a}^{b} \,l(x,s) \psi(s) ds$

both are continuous

I know how to obtain the eigenvalues of each alone.

But how can I calculate the eigenvalues of the operator $I-{L}^{-1} K$ or at least the norm $||I-{L}^{-1} K||$
the reason I want to check if it is less than unity

thanks
Sarrah

 

[Ackbach]

Would it be helpful to mention that since $L(I-L^{-1}K)=L-K$, that therefore $\|L-K\| \le \|L\| \cdot \|I-L^{-1}K\|$?

 

[sarrah1]

thank you
but this gives a lower bound and I need an upper bound
thanks anyway for your help
sarrah

 

[Opalg]

I have a linear integral operator

$K\psi=\int_{a}^{b} \,k(x,s) \psi(s) ds$

$L\psi=\int_{a}^{b} \,l(x,s) \psi(s) ds$

both are continuous

I know how to obtain the eigenvalues of each alone.

But how can I calculate the eigenvalues of the operator $I-{L}^{-1} K$ or at least the norm $||I-{L}^{-1} K||$
the reason I want to check if it is less than unity

thanks
Sarrah

How about $\|I-{L}^{-1} K\| = \|{L}^{-1}(L-K)\| \leqslant \|{L}^{-1}\|\|L-K\|$ as a starting point? To show that this is small, you would need to have an estimate for $\|{L}^{-1}\|$ and also to know that $L$ is suitably close to $K$. Unless you have some extra condition saying that $\|L-K\|$ is small, I doubt whether it will necessarily be true that $\|I-{L}^{-1} K\| < 1.$

 

[sarrah1]

thank you very much Oplag

Again it's always you who run for help.

Your idea is smart in the sense you made it depending on $||L−K||$ and $||{L}^{-1}||$. How easy and simple.

HOWEVER although I can estimate a bound for $||L−K||$ i.e. to control this difference. But what about $||{L}^{-1}||$. This I can't tell anything about. I happened to have asked this question here, how can I calculate the inverse of a linear integral operator. If this is possible then half my journey is done. I am well versed in matrices and the inverse exists if the matrix is nonsingular. Does equally ${L}^{-1}$ has inverse eigenvalues of $L$ for instance like matrices. But how can I get this inverse? Even if the spectral radius is equal to 1/smallest eigenvalue in modulus of $L$ it will be less than $||{L}^{-1}||$ and not larger in order to obtain a bound for the latter.

Unfortunately $L$ and $K$ don't commute otherwise they can have I suppose the same eigenfunctions and I can work directly on the spectral radius of $||I-{L}^{-1}K||$

To elucidate more: My problem concerns the solution of a Fredholm 1st kind integral equation which is involved. Its kernel is $k(x,s)$ i.e. $f(x)=\int_{a}^{b} \,k(x,s)\psi(s) ds$ which I know. I am trying to substitute it by another kernel $l(x,s)$ which I can also choose. I have some algorithm which converges if $||I-{L}^{-1}K||$<1
thank you once more
Sarrah