On the spectral radius of bounded linear operators

[sarrah1]

Hi EVERYBODY:

General knowledge: The homogeneous linear Fredholm integral equation

$\mu\ \varPsi(x)=\int_{a}^{b} \,k(x,s) \varPsi(s) ds$ (1)

has a nontrivial solution if and only if $\mu$ is an eigenvalue of the integral operator $K$. By multiplying (1) by $k(x,s)$ and integrating while changing the order of integration one obtains

${\mu}^{2}\varPsi(x)=\int_{a}^{b} \,{[k(x,s)]}^{(2)}\varPsi(s) ds$

where ${[k(x,s)]}^{(2)}=\int_{a}^{b} \,{[k(x,t)]}^{(1)}{[k(t,s)]}^{(1)}dt$ , where ${[k(x,s)]}^{(1)}=k(x,s)$

after $n$ iterations ${\mu}^{n} \varPsi(x)=\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds$ (2)

where ${[k(x,s)]}^{(n)}=\int_{a}^{b} \,{[k(x,t)]}^{(1)}{[k(t,s)]}^{(n-1)}dt$

Now my Question:

Solving the non-homogeneous equation

${\varPsi}(x)=f(x)+\lambda\int_{a}^{b} \,k(x,s) {\varPsi}(s) ds$ , (where $\lambda$ is a parameter)

by Picard successive approximations, results in the following equation after $n$ iterations

${\varPsi}_{n+1}(x)=f(x)+\lambda\int_{a}^{b} \,k(x,s) {\varPsi}_{n}(s) ds$

to prove convergence (I can use norms to reach the classical condition $|\lambda|<1/||K||$) however I want to reach a condition based on the spectral radius of the operator $K$ i.e. $\rho(K)$ . So one can show by imitating the above that after $n$ iterations

${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}\int_{a}^{b} \,{[k(x,s)]}^{(n)} [{\varPsi}_{0}(s)-{\varPsi}(s)] ds$ (3)

if $|\mu|<1$ then Eq. (2) gives $ \lim_{{n}\to{\infty}}\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds=0$ , $\lor x$
can this help me equally in Eq. (3) by saying that convergence occurs if

$|\lambda| < 1/\rho(K)$

grateful
Sarrah

 

[Opalg]

${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}\int_{a}^{b} \,{[k(x,s)]}^{(n)} [{\varPsi}_{0}(s)-{\varPsi}(s)] ds$ (3)

if $|\mu|<1$ then Eq. (2) gives $ \lim_{{n}\to{\infty}}\int_{a}^{b} \,{[k(x,s)]}^{(n)} \varPsi(s) ds=0$ , $\lor x$
can this help me equally in Eq. (3) by saying that convergence occurs if

$|\lambda| < 1/\rho(K)$

grateful
Sarrah

If I understand this correctly, you can write equation (3) as ${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$. In that case, $$\|{\varPsi}_{n}(x)-{\varPsi}(x)\| \leqslant |{\lambda}|^{n}\|K^n\|\|{\varPsi}_{0}(s)-{\varPsi}(s)\|.\quad(4)$$ But $$\rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If $\rho(K) < 1/|\lambda|$, choose $\nu$ with $\rho(K) < \nu < 1/|\lambda|.$ Then $\|K^n\|^{1/n} < \nu$ for all sufficiently large $n$, so that $\|K^n\| < \nu^n.$ You can then conclude from $(4)$ that convergence occurs in $(3)$, because $\nu|\lambda| < 1$ and so $(\nu|\lambda|)^n \to0$ as $n\to\infty.$

 

[sarrah1]

If I understand this correctly, you can write equation (3) as ${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$. In that case, $$\|{\varPsi}_{n}(x)-{\varPsi}(x)\| \leqslant |{\lambda}|^{n}\|K^n\|\|{\varPsi}_{0}(s)-{\varPsi}(s)\|.\quad(4)$$ But $$\rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If $\rho(K) < 1/|\lambda|$, choose $\nu$ with $\rho(K) < \nu < 1/|\lambda|.$ Then $\|K^n\|^{1/n} < \nu$ for all sufficiently large $n$, so that $\|K^n\| < \nu^n.$ You can then conclude from $(4)$ that convergence occurs in $(3)$, because $\nu|\lambda| < 1$ and so $(\nu|\lambda|)^n \to0$ as $n\to\infty.$

I am most grateful Opalg.

Eq. (3) can -as you did - be written of course as
${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$
from which Eq. (4) also follows correctly

The problem is - as it seems - from me. What I understood from your proof is that the condition you wrote as $\nu|\lambda| < 1$ is another way of writing a sufficient condition. Because from (4) if $|\lambda ||K||<1$ it follows directly that ${\varPsi}_{n}\implies\varPsi$ in the limit.

my intention was to obtain a necessary condition based on the spectral radius though like we do in matrix theory. if $K$ is a matrix then
${T}^{-1}{K}^{n}T= diag({{\lambda}_{1}}^{n},...,{{\lambda}_{m}}^{n})$ (5)
so $\lambda\rho(K)<1$ becomes both a necessary and sufficient condition whereas $|\lambda|||K||<1$ becomes only a sufficient condition because $\rho(K)\le||K||$. I hope you got me.

So I thought there can be a relation like (5) for integral operators, something similar instead of $$\rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If or $\rho(K)\le||K||$

still very grateful
Sarrah

 

[sarrah1]

I am most grateful Opalg.

Eq. (3) can -as you did - be written of course as
${\varPsi}_{n}(x)-{\varPsi}(x)={\lambda}^{n}K^n({\varPsi}_{0}(s)-{\varPsi}(s))$
from which Eq. (4) also follows correctly

The problem is - as it seems - from me. What I understood from your proof is that the condition you wrote as $\nu|\lambda| < 1$ is another way of writing a sufficient condition. Because from (4) if $|\lambda ||K||<1$ it follows directly that ${\varPsi}_{n}\implies\varPsi$ in the limit.

my intention was to obtain a necessary condition based on the spectral radius though like we do in matrix theory. if $K$ is a matrix then
${T}^{-1}{K}^{n}T= diag({{\lambda}_{1}}^{n},...,{{\lambda}_{m}}^{n})$ (5)
so $\lambda\rho(K)<1$ becomes both a necessary and sufficient condition whereas $|\lambda|||K||<1$ becomes only a sufficient condition because $\rho(K)\le||K||$. I hope you got me.

So I thought there can be a relation like (5) for integral operators, something similar instead of $$\rho(K) = \lim_{n\to\infty}\|K^n\|^{1/n}.$$ If or $\rho(K)\le||K||$

still very grateful
Sarrah

Hi Oplag
Upon reading your post again, and your easy and sound proof, it seems to me there is no such necessary condition, only a sufficient one. Perhaps I didn't express myself well. I meant convergence is guaranteed if $|\lambda|\rho(K)<1$ ok, which is tight but could it be that $|\lambda| ||K||>1$ and convergence still exists ? based on the former condition spectral radius as $\rho(K)\le||K||$
many thanks
Sarrah