Solving Linear Integral Equation: Norm of Resolvent

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The discussion centers on the properties of linear integral operators, specifically the resolvent norm of the operator defined by $(Ky)(x)=\int_{a}^{b} k(x,s)y(s)ds$. It is established that if the condition $|b| \cdot ||K|| < 1$ holds, then the inequality $||(I-bK)^{-1}|| < \frac{1}{1 - |b| \cdot ||K||}$ is indeed correct. This conclusion is derived from the properties of bounded linear operators and their norms.

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sarrah1
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I have a linear integral operator (related to integral equations)
$(Ky)(x)=\int_{a}^{b} \,k(x,s)y(s)ds$
If $|b|. ||K||<1$ (b is a scalar)
can I say
$||(I-bK)-1||< 1 / (1-|b|.||K||)$
I think it's correct
is it?
thanks
 
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sarrah said:
I have a linear integral operator (related to integral equations)
$(Ky)(x)=\int_{a}^{b} \,k(x,s)y(s)ds$
If $|b|. ||K||<1$ (b is a scalar)
can I say
$||(I-bK)-1||< 1 / (1-|b|.||K||)$
I think it's correct
is it?
thanks
Yes, that is correct.

If $T$ is a bounded linear operator with $\|T\|<1$ and $x = (I-T)y$ then $\|x\| = \|y - T(y)\| \geqslant \|y\| - \|T(y)\| \geqslant \|y\| - \|T\|\|y\| = \|y\|(1-\|T\|).$ Therefore $\|(I-T)^{-1}(x)\| = \|y\| \leqslant \frac1{1-\|T\|}\|x\|.$ It follows that $\|(I-T)^{-1}\| \leqslant \frac1{1-\|T\|}$. Now all you have to do is to put $T = bK.$
 
extremely grateful
 

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