Given that $\cal{X}$ and $\cal{Y}$ are Banach spaces.
$A:\cal{X} \to \cal{Y}$ and $Y:\cal{Y}\to \cal{X}$ are linear bounded operators.
If $\rho(I_{\cal{X}} - YA)<1$, $\rho(I_{\cal{Y}} - AY)<1$ then $YA$ is invertible on the $R(YA)$, and $AY$ is invertible on the $R(AY)$where $\rho$ stands for the spectral radius and $R(A)$ denotes the range space of the operator $A$.
Therefore, $(YA)_{|R(YA)}^{-1}Y = Y (AY)_{|R(AY)}^{-1}$ ....$(1)$
I am not able to prove equality given by $(1) $ . I know that I have to show that their domain space is same and they attain the same values on the domain. I have checked that both the operators $(YA)_{|R(YA)}^{-1}Y and ~~~Y (AY)_{|R(AY)}^{-1}$ have same domain space $\cal{Y}$. However, I am not able to prove that they attain the same values on its domain.
Could anyone help me to clear my doubt? I would be very much thankful.
Thanks
In fact these two operators aren't defined on all of $\mathcal{Y}$. This is easiest to see with $Y(AY)|_{R(AY)}^{-1}$: certainly we can only make sense of this on $R(AY)$! So, consider some $AY(x)\in R(AY)$. $Y(AY)|_{R(AY)}^{-1}(AY(x))=Y(x)$ by the definition of $(AY)|_{R(AY)}^{-1}$.
We need to compare that to the left-hand side $(YA)|_{R(YA)}^{-1}Y(AY(x))$. First, observe that this is defined: $YAY(x)$ is definitely in $R(YA)$. Given that, can you finish showing $(YA)|_{R(YA)}^{-1}Y(AY(x))=Y(x)$, as desired?
EDIT: Here's some more detail about the domain. The two operators under discussion here are only defined on $R(AY)$ simply because $(AY)^{-1}$ is only defined on $R(AY)$. To apply $Y(AY)^{-1},$ I first have to apply $(AY)^{-1}$, and there's no way to do that to some $x$ that's not in the image of $AY$.