I am currently studying the original discrete/continous equivalence proof of the Lyapunov equation by Rice 1967.
- Continuous case: $A^\star L + L A = -C$ for $C \succcurlyeq 0$
- Discrete case: $A^\star L A - L = -C $ for $C \succcurlyeq 0$
The paper claims that the discrete case can be derived from the continuous case by plugging in the bilinear transformation $ A = (B + I) (B - I)^{-1}$ into $A^\star L + L A = -C$ so that we eventually get $B^\star L B - L = -\frac{1}{2} Y$ for $Y = (B^\star - I) C( B - I)$ after some elementary transformations.
I have tried to reach the same results:
$$ A^\star L + L A = -C \\ [(B + I) (B - I)^{-1}]^\star L + L (B + I) (B - I)^{-1} = -C \\ $$ Multiplying from the left by $(B^\star - I)$ and from the right by $(B-I)$ yields: $$ (B^\star - I)[(B + I) (B - I)^{-1}]^\star L (B-I) + (B^\star - I) L (B + I) (B - I)^{-1} (B-I) = -(B^\star - I) C (B-I) \\ \iff (B^\star - I) \left((B-I)^{-1} \right)^\star (B+I)^\star L (B-I) + (B^\star - I) L (B + I) (B - I)^{-1} (B-I) = -(B^\star - I) C (B-I) \\ \iff (B^\star - I) \left((B-I)^\star \right)^{-1} (B+I)^\star L (B-I) + (B^\star - I) L (B + I) (B - I)^{-1} (B-I) = -(B^\star - I) C (B-I) \\ \iff \underbrace{(B^\star - I) \left(B^\star-I \right)^{-1}}_{=I} (B+I)^\star L (B-I) + (B^\star - I) L (B + I) \underbrace{(B - I)^{-1} (B-I)}_{=I} = -(B^\star - I) C (B-I) \\ $$ Which could be put as: $$ (B+I)^\star L (B - I) + (B^\star - I) L (B +I) = -Y $$ Which steps am I missing to complete the proof?
P.S.: The paper mentions the following identities of which I can't verify the last step either: $$ A = (B + I) (B - I)^{-1} \\ = (B - I)^{-1} (B + I) \\ = I + 2 (B - I)^{-1} $$
We have $$ (B+I)^\star L (B - I) = B^\star LB-B^\star L+LB-B $$ and $$ ((B^\star - I) L (B +I) = B^\star LB+B^\star L-LB-B, $$ which gives $$ (B+I)^\star L (B - I) + (B^\star - I) L (B +I) = 2B^\star LB-2B=-Y. $$
For the last identity, note that $$ (B - I)^{-1} (B + I)= (B - I)^{-1} (B-I + 2I) = I + 2 (B - I)^{-1}. $$