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

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The discussion revolves around calculating the norm of the operator \(I - L^{-1}K\) and determining if it is less than unity. The user has established the definitions of the linear integral operators \(K\) and \(L\) and is seeking to find eigenvalues or bounds for the operator in question. Suggestions include using the relationship \(\|I - L^{-1}K\| = \|L^{-1}(L-K)\|\) to derive upper bounds, but the user struggles with estimating \(\|L^{-1}\|\). The problem is further complicated by the non-commutativity of \(L\) and \(K\), which hinders direct spectral analysis. The ultimate goal is to ensure that \(\|I - L^{-1}K\| < 1\) for the convergence of a specific algorithm related to a Fredholm integral equation.
sarrah1
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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
 
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Would it be helpful to mention that since $L(I-L^{-1}K)=L-K$, that therefore $\|L-K\| \le \|L\| \cdot \|I-L^{-1}K\|$?
 
thank you
but this gives a lower bound and I need an upper bound
thanks anyway for your help
sarrah
 
sarrah said:
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 $\|I-{L}^{-1} K\| < 1.$
 
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
 
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