MHB Solving Linear Integral Equation: Norm of Resolvent

[sarrah1]

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

 

[Opalg]

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 $T = bK.$

 

[sarrah1]

extremely grateful