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

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The discussion centers on the properties of linear integral operators defined by equations involving continuous functions. The user seeks clarification on whether a specific notation for the operator K is acceptable and confirms the condition |b|.||L||<1. They aim to prove an inequality involving the operators K and L, leveraging the geometric series theorem for bounded linear operators. The conclusion reached is that the proposed inequality is correct, and the notation used is validated by other sources. The responses affirm the user's understanding and correctness of the approach.
sarrah1
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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
 
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sarrah said:
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. :)
 
thank you very much Oplag
 

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