I am reading a paper related to control theory and struggle to understand a matrix inequality derivation that is briefly introduced by the author:
Given that we have the following equations: $ x(k+1)=Ax(k)+Bu(k), $ and $A^\top PA-A=-Q,$ and Lyapunov difference function: $$ V(x(k+1))-V(x(k))=-x^\top(k)Qx(k)+2x^\top(k){A^\top}PBu(k)+u^\top(k)B^\top PBu(k), $$
how do we show that $$ V(x(k+1))-V(x(k))\leq-\frac{1}{2}\lambda_{min}(Q)x^2(k)+\left(\frac{2|A^\top PB|^2}{\lambda_{min}(Q)}+|B^\top PB|^2\right)u^2(k), $$ where $\lambda_{min}(Q)$ is the minimum eigenvalue of matrix $Q$?
A hint from the paper is that this is done by completing squares, but I have not figured out how exactly this can be done.
Edit: the paper is titled 'Input-to-state stability for discrete-time nonlinear systems' published in Automatica 2001. The question is related to Example 3.4 inside the paper.
I am not sure that inequality is correct, and probably there is a typo.
First, recall that $$x^\top y \le \frac{1}{2}x^\top x + \frac{1}{2}y^\top y.$$ I also assume that the following two inequalities are clear: $ -x^\top Q x \le -\lambda_{min}(Q)|x|^2$ and $u^\top B^\top P B u \le |B^\top P B| |u|^2$. Note that it should be $|u|^2$ and $|x|^2$ instead of $x^2$ and $u^2$ since the $(\cdot)^2$ is not defined for vectors.
Ok, now let us write $$ \begin{gathered}2x^\top A^\top P B u = \left(\sqrt{\lambda_{min}(Q)}x\right)^\top\left(\frac{2}{\sqrt{\lambda_{min}(Q)}}A^\top P B u\right) \\ \le \frac{1}{2}\lambda_{min}(Q)x^\top x + \frac{1}{2} u^\top B^\top P A \frac{4}{\lambda_{min}(Q)}A^\top P B u\end{gathered}\\ \le \frac{1}{2}\lambda_{min}(Q)|x|^2 + \frac{2}{\lambda_{min}(Q)}|A^\top P B|^2|u|^2.$$ Finally, we have $$\begin{gathered}-x^\top Q x + 2x^\top A^\top P B u + u^\top B^\top P B u \\ \le -\frac{1}{2}\lambda_{min}(Q)|x|^2 + \left(\frac{2}{\lambda_{min}(Q)}|A^\top P B|^2 + |B^\top P B|\right)|u|^2.\end{gathered}$$
The difference is that we have $|B^\top P B|$ instead of $|B^\top P B|^2$.
It is also worth noting that the Lypunov equation is $A^\top P A - P = -Q$, and not $A^\top P A - A = - Q$, as written by the authors.
Let us now show that the original claim is questionable. Let $x$ and $u$ be scalars (thus the matrices $Q$, $A$, $B$, and $P$ are scalars as well). Let $Q=1$, and take $x=0$ and $u=1$. Then the original claim implies $$B^2P \le 2(APB)^2 + \left(B^2P\right)^2$$ and $$1 \le 2A^2P+B^2P$$ that is not necessarily true for arbitrary (stable) $A$ and $B$.