Can Linear Integral Operators Be Combined to Prove a Trivial Inequality?

[sarrah1]

I have a linear integral operator (related to integral equations)

$(Ky)(x)=\int_{a}^{b} \,k(x,s) y(s) ds$ and another one $(Ly)(x)=\int_{a}^{b} \,l(x,s) y(s) ds$ both are continuous

Before I proceed can I write:

$Ky=\int_{a}^{b} \,k(.,s) y(s) ds$ ? (I saw this notation in some books)

I also have $|b|.||L||<1 $ (b is a scalar)

Thus I need to prove that:

$||({(I-bL)}^{-1}.(K-L)||\le\frac{||K-L||}{1-|b|.||L||}$ #

I know it's correct since

$||{(I-bL)}^{-1}||\le \frac{1}{(1-|b|.||L||)} $ (from the geometric series theorem of bounded linear operators)

Thus

$||({(I-bL)}^{-1}.(K-L)||\le||{(I-bL)}^{-1}||.||K-L||$ (from the sub-multiplicative property)
$\le\frac{||K-L||}{1-|b|.||L||}$ #

I think it's correct of course and trivial.
is it?
many thanks
Sarrah

 

[Opalg]

I have a linear integral operator (related to integral equations)

$(Ky)(x)=\int_{a}^{b} \,k(x,s) y(s) ds$ and another one $(Ly)(x)=\int_{a}^{b} \,l(x,s) y(s) ds$ both are continuous

Before I proceed can I write:

$Ky=\int_{a}^{b} \,k(.,s) y(s) ds$ ? (I saw this notation in some books)

I also have $|b|.||L||<1 $ (b is a scalar)

Thus I need to prove that:

$||({(I-bL)}^{-1}.(K-L)||\le\frac{||K-L||}{1-|b|.||L||}$ #

I know it's correct since

$||{(I-bL)}^{-1}||\le \frac{1}{(1-|b|.||L||)} $ (from the geometric series theorem of bounded linear operators)

Thus

$||({(I-bL)}^{-1}.(K-L)||\le||{(I-bL)}^{-1}||.||K-L||$ (from the sub-multiplicative property)
$\le\frac{||K-L||}{1-|b|.||L||}$ #

I think it's correct of course and trivial.
is it?
many thanks
Sarrah

The answers to your questions are Yes and Yes. :)

 

[sarrah1]

thank you very much Oplag