Appendix A: Proofs
1.1 A.1 Proof of Lemma 1
Let \(u \in PC_{\infty }\), \(w_{n} \in PC_{\infty }\), \(w_{d} \in PC_{\infty }\), \( \alpha \in PS^{1}(\mathcal {A},T_0,\delta _{ \alpha })\), and \(x_0 \in \mathbb {R} ^n\) be arbitrary. The proof is organized into three steps:
-
1.
Motivated by Eq. (9), define matrices \( R_{e} \), \( A_e( \alpha ) \), and \( C_e( \alpha ) \), with \( R_{e} \) non-singular, so that \( R_{e} A_e( \alpha ) + L( \alpha ) C_e( \alpha ) = Q( \alpha ) R_{e} \).
-
2.
Use the equation from step 1 to bound a yet to be defined extended state \( x_e(t) \).
-
3.
Use the bound on \( x_e(t) \) to bound the plant’s state \( x(t) \).
Step 1
Let v be the number of columns of the constant full rank matrix \( R \) from Assumption 2. Let \( \bar{R} \in \mathbb {R} ^{v \times (v-n)}\) be any matrix so that \( R_{e} := \begin{bmatrix} R&\bar{R} \end{bmatrix} \) is non-singular, and then define
$$\begin{aligned} \begin{bmatrix} A_{12}( \alpha ) \\ A_{22}( \alpha ) \end{bmatrix} := R_{e} ^{-1} Q( \alpha ) \bar{R} , \end{aligned}$$
where \( A_{12}( \alpha ) \in \mathbb {R} ^{n \times (v-n)}\), \( A_{22}( \alpha ) \in \mathbb {R} ^{(v-n) \times (v-n)}\). Then, by the definition of \( R \), \( R_{e} \), \(A_{12}\), and \(A_{22}\),
$$\begin{aligned} \underbrace{ \begin{bmatrix} A( \alpha )&A_{12}( \alpha ) \\ 0&A_{22}( \alpha ) \end{bmatrix} }_{=:A_e(\alpha )} + R_{e} ^{-1} L( \alpha ) \underbrace{ \begin{bmatrix} C( \alpha )&0 \end{bmatrix} }_{=:C_e(\alpha )} = R_{e} ^{-1} Q( \alpha ) R_{e} . \end{aligned}$$
(45)
Step 2
Consider the control system
$$\begin{aligned} \begin{bmatrix} \dot{x}_{1}(t) \\ \dot{x}_{2}(t) \end{bmatrix}= & {} \begin{bmatrix} A( \alpha (t))&A_{12}( \alpha (t)) \\ 0&A_{22}( \alpha (t)) \end{bmatrix} \begin{bmatrix} x_{1}(t) \\ x_{2}(t) \end{bmatrix} + \begin{bmatrix} B( \alpha (t)) \\ 0 \end{bmatrix} u(t) + \begin{bmatrix} w_{d}(t) \\ 0 \end{bmatrix} \\ y_{e}(t)= & {} \begin{bmatrix} C( \alpha (t))&0 \end{bmatrix} \begin{bmatrix} x_{1}(t) \\ x_{2}(t) \end{bmatrix} + w_{n}(t) , \end{aligned}$$
where \( x_{1}(t) \in \mathbb {R} ^n\), \( x_{2}(t) \in \mathbb {R} ^{v-n}\). Define \(x_{e} :=(x_{1}, x_{2})\); then,
$$\begin{aligned} \dot{x}_{e}(t)= & {} \Bigg ( \begin{bmatrix} A( \alpha (t))&A_{12}( \alpha (t)) \\ 0&A_{22}( \alpha (t)) \end{bmatrix} + R_{e} ^{-1} L( \alpha (t)) \begin{bmatrix} C( \alpha (t))&0 \end{bmatrix} \Bigg ) x_{e}(t) \\&- R_{e} ^{-1} L( \alpha (t)) \Big ( y_{e}(t) - w_{n}(t) \Big ) + \begin{bmatrix} B( \alpha (t)) \\ 0 \end{bmatrix} u(t) + \begin{bmatrix} w_{d}(t) \\ 0 \end{bmatrix} . \end{aligned}$$
Using (45), we can write
$$\begin{aligned} \dot{x}_{e}(t)= & {} R_{e} ^{-1} Q( \alpha (t)) R_{e} x_e(t) - R_{e} ^{-1} L( \alpha (t)) y_{e}(t) + R_{e} ^{-1} L( \alpha (t)) w_{n}(t) \\&+ \begin{bmatrix} B( \alpha (t)) \\ 0 \end{bmatrix} u(t) + \begin{bmatrix} w_{d}(t) \\ 0 \end{bmatrix} . \end{aligned}$$
The solution to this differential equation, for \(t \ge 0\) and any \(x_{e}(0) \in \mathbb {R} ^{v}\), is
$$\begin{aligned} \begin{aligned} x_e(t)&= \Phi (t,0) x_{e}(0) + \int _{0}^{t} \Phi (t,\tau ) \Bigg ( \begin{bmatrix} B( \alpha (\tau )) \\ 0 \end{bmatrix} u(\tau ) - R_{e} ^{-1} L( \alpha (\tau )) y_{e}(\tau ) \Bigg ) \mathrm {d} \tau \\&\quad + \int _{0}^{t} \Phi (t,\tau ) \Bigg ( R_{e} ^{-1} L( \alpha (\tau )) w_{n}(\tau ) + \begin{bmatrix} w_{d}(\tau ) \\ 0 \end{bmatrix} \Bigg ) \mathrm {d} \tau , \end{aligned} \end{aligned}$$
where \( \Phi (t,0) \) is the state transition function of the unforced system
$$\begin{aligned} \dot{z}(t) = R_{e} ^{-1} Q( \alpha (t)) R_{e} z(t), \quad t \ge 0. \end{aligned}$$
Consider the change of coordinates \(p = R_{e} z\), where \( R_{e} \) is the non-singular matrix from Step 1. Then, \(\dot{p} = Q( \alpha (t)) p\) with \( Q( \alpha (t)) \in \mathcal {H}_{\infty }\) so that, by Proposition 2, there exist \(\gamma \ge 1\) and \(\lambda < 0\) such that
$$\begin{aligned} \left\| p(t) \right\| \le \gamma \mathrm {e} ^{\lambda t} \left\| p(0) \right\| , \quad t \ge 0. \end{aligned}$$
Again, following the proof of [40, Proposition 1], the constant \(\lambda \) can be taken to be the same as that in the filter (15). Then, there exists a constant \(c_{1}\) such that for all \(t \ge 0\), \(\left\| z(t) \right\| \le c_{1} \mathrm {e} ^{\lambda t} \left\| z(0) \right\| \), so we conclude that \(\left\| \Phi (t,0) \right\| \le c_{1} \mathrm {e} ^{\lambda t}\) for \(t \ge 0\). Using this bound on \( \Phi (t,0) \) in the expression for \(x_{e}(t)\), we get
$$\begin{aligned} \begin{aligned}&\left\| x_e(t) \right\| \\&\quad \le c \mathrm {e} ^{\lambda t} \left\| x_{e}(0) \right\| + c \int _{0}^{t} \mathrm {e} ^{\lambda (t - \tau )} \Big ( \left\| u(\tau ) \right\| + \left\| y_{e}(\tau ) \right\| \Big ) \mathrm {d} \tau + c \Big ( \left\| w_{n} \right\| _{\infty } + \left\| w_{d} \right\| _{\infty } \Big ), \end{aligned} \end{aligned}$$
(46)
with c defined as
$$\begin{aligned} c :=c_{1} \max \limits _{ \alpha \in \mathcal {A}} \Big \{1, \left\| R_{e} ^{-1} \right\| \left\| L( \alpha ) \right\| , \left\| B( \alpha ) \right\| , -\frac{\left\| R_{e} ^{-1} \right\| \left\| L( \alpha ) \right\| }{\lambda }, -\frac{1}{\lambda } \Big \}. \end{aligned}$$
Step 3
The subspace \(\{x_e: x_2 = 0\}\) is invariant for the extended system. Let x(t) be the solution of the plant’s ODE (14a) with initial condition x(0). Then, the initial condition \(x_{e}(0) = (x(0), 0)\) admits the solution \(x_{e}(t) = (x(t),0)\) and \(y_{e}(t) = y(t)\). Therefore, using (46),
$$\begin{aligned} \left\| x(t) \right\|= & {} \left\| x_e(t) \right\| \le c \mathrm {e} ^{\lambda t} \left\| x(0) \right\| + c \int _{0}^{t} \mathrm {e} ^{\lambda (t - \tau )} \Big ( \left\| u(\tau ) \right\| \\&+ \left\| y(\tau ) \right\| \Big ) \hbox {d}\tau + c \Big ( \left\| w_{n} \right\| _{\infty } + \left\| w_{d} \right\| _{\infty } \Big ). \end{aligned}$$
The solution v(t) of (15) equals the integral in the RHS of the above inequality, so we get the desired result. \(\square \)
1.2 A.2 Proof of Lemma 2
In the proofs of Lemmas 2 and 3, we utilize a crude bound on the maximum growth of the plant’s state over a single period.
Proposition 3
There exist constants \(\overline{T} > 0\) and \(c > 0\) so that for every \(T \in (0, \overline{T})\), \(w_{d} \in PC_{\infty }\), \( \alpha \in PS^{1}(\mathcal {A},T_0,\delta _{ \alpha })\), and \(x[k] \in \mathbb {R} ^n\), when the controller given by (15), (17)–(19), and (21), (22) is applied to the plant (14):
$$\begin{aligned}&\left\| x(t) - x[k] \right\| \\&\quad \le c T \Bigg ( v[k] + \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \left\| s[k] \right\| _{\infty } + \left\| w_{d} \right\| _{\infty } \Bigg ), \quad t \in [kT, (k+1)T]. \end{aligned}$$
Proof of Proposition 3
Let \(w_{d} \in PC_{\infty }\), \( \alpha \in PS^{1}(\mathcal {A},T_0,\delta _{ \alpha })\), and \(x[k] \in \mathbb {R} ^n\) be arbitrary. The solution x(t) to (14) with initial condition x[k] satisfies
$$\begin{aligned}&x(t) - x[k] \\&\quad = \int _{kT}^{t} \Big ( A( \alpha (\tau )) x[k] + B( \alpha (\tau )) u(\tau ) + w_{d}(\tau ) \Big ) \mathrm {d} \tau \\&\qquad + \int _{kT}^{t} A( \alpha (\tau )) \Big ( x(\tau ) - x[k] \Big ) \mathrm {d} \tau . \end{aligned}$$
Taking norms and substituting the expression for u(t) over \([kT, (k+1)T]\), we get the upper bound
$$\begin{aligned} \begin{aligned} \left\| x(t) - x[k] \right\|&\le T \Big ( \max \limits _{ \alpha \in \mathcal {A}}\left\| A( \alpha ) \right\| \left\| x[k] \right\| + \max \limits _{ \alpha \in \mathcal {A}}\left\| B( \alpha ) \right\| ( \max \limits _{ \alpha \in \mathcal {A}}\left\| H( \alpha ) \right\| + \rho ) \left\| z[k] \right\| \\&\quad + \max \limits _{ \alpha \in \mathcal {A}}\left\| B( \alpha ) \right\| ( \max \limits _{ \alpha \in \mathcal {A}}\left\| K( \alpha ) \right\| \left\| M \right\| + \rho ) \left\| r[k] \right\| \\&\quad + \max \limits _{ \alpha \in \mathcal {A}}\left\| B( \alpha ) \right\| \rho v[k] + \left\| w_{d} \right\| _{\infty } \Big ) \\&\quad + \max \limits _{ \alpha \in \mathcal {A}}\left\| A( \alpha ) \right\| \int _{kT}^{t} \left\| x(t) - x[k] \right\| \mathrm {d} \tau , \quad t \in [kT,(k+1)T]. \end{aligned} \end{aligned}$$
Using the definition of \(\bar{x}\), \(\bar{z}\), and s given in (24) and invoking the Bellman–Gronwall inequality, there exist constants \(c_{1} > 0\) and \(c_{2} > 0\) so that
$$\begin{aligned} \left\| x(t) - x[k] \right\| \le c_{1} T \Bigg ( v[k] + \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \left\| s[k] \right\| _{\infty } \Bigg ) \mathrm {e} ^{c_{2} T} + c_{1} T \left\| w_{d} \right\| _{\infty } \mathrm {e} ^{c_{2} T}. \end{aligned}$$
Then, for sufficiently small \(\overline{T}\), there exists a constant \(c > 0\) so that, for all \(T \in (0, \overline{T})\),
$$\begin{aligned} \left\| x(t) - x[k] \right\| \le c T \Bigg ( v[k] + \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \left\| s[k] \right\| _{\infty } + \left\| w_{d} \right\| _{\infty } \Bigg ). \end{aligned}$$
(47)
\(\square \)
Proof of Lemma 2
Fix \(\epsilon > 0\) and \( \overline{\delta } > 0\); let \(w_{n} \in PC_{\infty }\), \(w_{d} \in PC_{\infty }\), \( \alpha \in PS^{1}(\mathcal {A},T_0,\delta _{ \alpha })\), \(k \in \mathbb {Z} _+\), and \(x_0 \in \mathbb {R} ^n\) be arbitrary. We start with a claim.
Claim 1
There exist constants \(c > 0\) and \(\overline{T} > 0\), so that if \(T \in (0, \overline{T})\), \(v[k] + \left\| z[k] \right\| + \left\| r[k] \right\| > \epsilon \left\| x[k] \right\| \), and \( \alpha (t)\) is a.c. for \(t \in [kT, (k+1)T]\), then
$$\begin{aligned} \left\| C( \alpha ((k+1)T)) B( \alpha ((k+1)T)) - \widehat{CB}[k+1] \right\| \le c T + c \frac{ \left\| w_{d} \right\| _{\infty } }{\delta [k]} + c T^{-1} \frac{ \left\| w_{n} \right\| _{\infty } }{\delta [k]}. \end{aligned}$$
(48)
Proof of Claim 1
By hypothesis, \(v[k] + \left\| z[k] \right\| + \left\| r[k] \right\| > \epsilon \left\| x[k] \right\| \), which implies that \(\delta [k] \ne 0\). Then,
$$\begin{aligned} -2 h \widehat{CB}[k+1] \delta [k]= & {} y(kT+2h) - 2 y(kT+h) + y(kT) \nonumber \\= & {} \Big ( y_{nf}(kT+2h) - y_{nf}(kT+h) \Big ) \nonumber \\&- \Big ( y_{nf}(kT+h) - y_{nf}(kT) \Big ) \nonumber \\&+\, w_{n}(kT+2h) - 2 w_{n}(kT+h) + w_{n}(kT), \end{aligned}$$
(49)
where, with some abuse of notation, \( y_{nf}(t) := C(t) x(t) \) is the noise-free output signal. By this definition of \( y_{nf}(t) \), it follows that
$$\begin{aligned} \dot{y}_{nf}(t) = \Big (\dot{C}(t) + C(t)A(t) \Big ) x(t) + C(t) B(t) u(t) + C(t) w_{d}(t) \end{aligned}$$
for almost every \(t \in [kT, (k+1)T]\). Then, by the Fundamental Theorem of Calculus and the structure of the control signal (19), we have
$$\begin{aligned} y_{nf}(kT+h) - y_{nf}(kT)= & {} \int _{kT}^{kT+h} \Big ( \dot{C}(\tau ) + C(\tau )A(\tau ) \Big ) x(\tau ) \mathrm {d} \tau \\&+ \int _{kT}^{kT+h} C(\tau )B(\tau ) \mathrm {d} \tau \delta [k] \\&+ \int _{kT}^{kT+h} C(\tau )B(\tau ) \mathrm {d} \tau \bigg ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \bigg ) \\&+ \int _{kT}^{kT+h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau . \end{aligned}$$
In a similar fashion,
$$\begin{aligned} y_{nf}(kT+2h) - y_{nf}(kT+h)= & {} \int _{kT+h}^{kT+2h} \Big ( \dot{C}(\tau ) + C(\tau )A(\tau ) \Big ) x(\tau ) \mathrm {d} \tau \\&- \int _{kT+h}^{kT+2h} C(\tau )B(\tau ) \mathrm {d} \tau \delta [k] \\&+ \int _{kT+h}^{kT+2h} C(\tau )B(\tau ) \mathrm {d} \tau \\&\times \, \bigg ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \bigg ) \\&+ \int _{kT+h}^{kT+2h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau . \end{aligned}$$
Substituting the previous two expressions into (49) yields
$$\begin{aligned}&-2 h \widehat{CB}[k+1] \delta [k]\\&\quad = \int _{kT+h}^{kT+2h} \Big ( C(\tau ) A(\tau ) + \dot{C}(\tau ) \Big ) x(\tau ) \mathrm {d} \tau + \int _{kT+h}^{kT+2h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau \\&\qquad + \int _{kT+h}^{kT+2h} C(\tau ) B(\tau ) \mathrm {d} \tau \Big ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \Big ) \\&\qquad - \int _{kT}^{kT+h} C(\tau ) B(\tau ) \mathrm {d} \tau \Big ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \Big ) \\&\qquad -\int _{kT}^{kT+2h} C(\tau ) B(\tau ) \mathrm {d} \tau \delta [k] - \int _{kT}^{kT+h} \Big ( C(\tau ) A(\tau ) + \dot{C}(\tau ) \Big ) x(\tau ) \mathrm {d} \tau \\&\qquad - \int _{kT}^{kT+h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau + w_{n}(kT+2h) - 2 w_{n}(kT+h) + w_{n}(kT)\\&\quad = - 2 h C(kT+2h) B(kT+2h) \delta [k] \\&\qquad + \int _{kT}^{kT+2h} \Bigg ( C(kT+2h) B(kT+2h) - C(\tau ) B(\tau ) \Bigg ) \mathrm {d} \tau \delta [k] \\&\qquad + \int _{kT+h}^{kT+2h} \Bigg ( \Big ( C(\tau ) A(\tau ) + \dot{C}(\tau ) \Big ) x(\tau ) \\&\qquad - \Big ( C(\tau - h) A(\tau - h) + \dot{C}(\tau - h) \Big ) x(\tau - h) \Bigg ) \mathrm {d} \tau \\&\qquad + \int _{kT+h}^{kT+2h} \Bigg ( C(\tau ) B(\tau ) - C(\tau - h) B(\tau - h) \Bigg ) \mathrm {d} \tau \\&\qquad \times \Big ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \Big ) \\&\qquad + \int _{kT+h}^{kT+2h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau - \int _{kT}^{kT+h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau \\&\qquad + w_{n}(kT+2h) - 2 w_{n}(kT+h) + w_{n}(kT). \end{aligned}$$
Then, we have
$$\begin{aligned}&\left\| C(kT+2h) B(kT+2h) - \widehat{CB}[k+1] \right\| \\&\quad \le \frac{1}{2 h} \int _{kT}^{kT+2h} \left\| C(kT+2h) B(kT+2h) - C(\tau ) B(\tau ) \right\| \mathrm {d} \tau \\&\qquad + \frac{1}{2 h \delta [k]} \int _{kT+h}^{kT+2h} \left\| \Big ( C(\tau ) A(\tau ) + \dot{C}(\tau ) \Big ) x(\tau ) - \Big ( C(\tau - h) A(\tau - h) + \dot{C}(\tau - h) \Big ) x(\tau - h) \right\| \mathrm {d} \tau \\&\qquad + \frac{1}{2 h \delta [k]} \int _{kT+h}^{kT+2h} \left\| C(\tau ) B(\tau ) - C(\tau - h) B(\tau - h) \right\| \mathrm {d} \tau \Big ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \Big ) \\&\qquad + \frac{1}{2 h \delta [k]} \left\| \int _{kT+h}^{kT+2h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau - \int _{kT}^{kT+h} C(\tau ) w_{d}(\tau ) \mathrm {d} \tau \right\| \\&\qquad + \frac{1}{2 h \delta [k]} \left\| w_{n}(kT+2h) - 2 w_{n}(kT+h) + w_{n}(kT) \right\| . \end{aligned}$$
Utilizing order notation, Proposition 3, and applying Assumption 5 to boundFootnote 5 the Lipschitz continuous functions, we can write this concisely as
$$\begin{aligned}&\left\| C(kT+2h) B(kT+2h) - \widehat{CB}[k+1] \right\| \\&\quad = \mathcal {O}(T) + \mathcal {O}(T) \frac{v[k] + \left\| x[k] \right\| + \left\| z[k] \right\| + \left\| r[k] \right\| }{\delta [k]} \\&\qquad + \mathcal {O}(1) \frac{ \left\| w_{d} \right\| _{\infty } }{\delta [k]} + \mathcal {O}(T^{-1}) \frac{ \left\| w_{n} \right\| _{\infty } }{\delta [k]}. \end{aligned}$$
By hypothesis, we have
$$\begin{aligned} \frac{\left\| x[k] \right\| + v[k] + \left\| z[k] \right\| + \left\| r[k] \right\| }{\delta [k]} \le \Big ( \frac{1}{\epsilon } + 1 \Big ) \frac{\big ( v[k] + \left\| z[k] \right\| + \left\| r[k] \right\| \big )}{\rho ( v[k] + \left\| z[k] \right\| + \left\| r[k] \right\| )} = \frac{1}{\rho }\Big ( \frac{1}{\epsilon } + 1 \Big ), \end{aligned}$$
so for all \(T \in (0, \overline{T})\), with \(\overline{T}\) sufficiently small, we get the desired result:
$$\begin{aligned} \left\| C((k+1)T) B((k+1)T) - \widehat{CB}[k+1] \right\| = \mathcal {O}(T) + \mathcal {O}(1) \frac{ \left\| w_{d} \right\| _{\infty } }{\delta [k]} + \mathcal {O}(T^{-1}) \frac{ \left\| w_{n} \right\| _{\infty } }{\delta [k]} . \end{aligned}$$
\(\square \)
By hypotheses (i) and (iii) and Claim 1, the bound (48) holds. By hypothesis (ii) and the definition of \(\delta [k]\),
$$\begin{aligned} \frac{ \left\| w_{n} \right\| _{\infty } }{T \delta [k]} + \frac{ \left\| w_{d} \right\| _{\infty } }{\delta [k]} < \frac{1}{\rho c}. \end{aligned}$$
(50)
With \(c_{1}\) the constant from Claim 1, substituting (50) into (48) yields
$$\begin{aligned} \left\| C((k+1)T) B((k+1)T) - \widehat{CB}[k+1] \right\| < c_{1} \Bigg ( T + \frac{1}{\rho c} \Bigg ). \end{aligned}$$
(51)
We want to bound \(\Vert C( \alpha ((k+1)T) B( \alpha ((k+1)T) - \Pi _{\mathcal {F}}( \widehat{CB}[k+1] ) \Vert \), so we must account for the effect of the projection, \(\Pi _{\mathcal {F}}\). To ensure that \(\Pi _{\mathcal {F}}\) projects onto the interval of \(\mathcal {F}\) containing \(C((k+1)T)B((k+1)T)\), it is sufficient that the upper bound in (51) be less than half the minimum distance between intervals of \(\mathcal {F}\). By Assumptions 3 and 5, the image of \(\mathcal {A}\) under f has the form \(\mathcal {F} = [\underline{f}_{1}, \overline{f}_{1}] \cup \cdots \cup [\underline{f}_{q}, \overline{f}_{q}]\), where \(\overline{f}_{i} < \underline{f}_{i+1}\), \(i = 1, \ldots , q-1\), for some \(q \in \mathbb {N}\). Let \(d_{\min } :=\min \nolimits _{j} ( \underline{f}_{j+1} - \overline{f}_{j} ) \text {, } j = 1,\ldots , q-1\), and let \(T_1 :=\frac{1}{c_{1}} \min \{\frac{d_{\min }}{4}, \frac{ \overline{\delta } }{2 \ell } \}\), where \(\ell \) is the Lipschitz constant of \(f^{-1}\) on \(\mathcal {F}\). Then, from (51), for any \(c > \frac{c_{1}}{\rho } \max \{ \frac{4}{d_{\min }}, \frac{2 \ell }{ \overline{\delta } } \}\) and any \(T \in (0, T_1)\), we get
$$\begin{aligned} \left\| C((k+1)T) B((k+1)T) - \widehat{CB}[k+1] \right\| < \frac{c_{1} d_{\min }}{4 c_{1}} + \frac{c_{1} \rho d_{\min }}{4 c_{1} \rho } = \frac{d_{\min }}{2}, \end{aligned}$$
so it follows that
$$\begin{aligned}&\left\| C((k+1)T) B((k+1)T) - \Pi _{\mathcal {F}}( \widehat{CB}[k+1] ) \right\| \\&\quad \le \left\| C((k+1)T) B((k+1)T) - \widehat{CB}[k+1] \right\| . \end{aligned}$$
By Assumption 4, we have
$$\begin{aligned} \left\| \alpha ((k+1)T) - \hat{ \alpha } [k+1] \right\|\le & {} \ell \left\| C((k+1)T) B((k+1)T) - \Pi _{\mathcal {F}}( \widehat{CB}[k+1] ) \right\| \\< & {} c_{1} \ell \Bigg ( T + \frac{1}{\rho c} \Bigg ) < \frac{c_{1} \ell \overline{\delta } }{2 c_{1} \ell } + \frac{c_{1} \ell \rho \overline{\delta } }{2 c_{1} \ell \rho } = \overline{\delta } , \end{aligned}$$
where we have used the fact that \(\rho c > \frac{2 c_1 \ell }{\overline{\delta }}\). \(\square \)
1.3 A.3 Proof of Lemma 3
Let \(w_{n} \in PC_{\infty }\), \(w_{d} \in PC_{\infty }\), \( \alpha \in PS^{1}(\mathcal {A},T_0,\delta _{ \alpha })\), \(k \in \mathbb {Z}_{+}\), and \(x_0 \in \mathbb {R} ^n\) be arbitrary.
Proof of (iii)
In order to accomplish this, we proceed as follows:
-
Analyze all states at the sample points (Steps 1, 2, and 3),
-
Upper-bound a transformed closed-loop state (Step 4).
Step 1: Filter sample point analysis
At the sample points, the filter (15) satisfies
$$\begin{aligned} v[k+1] = \mathrm {e} ^{\lambda T} v[k] + \int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \left\| u(\tau ) \right\| \mathrm {d} \tau + \int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \left\| y(\tau ) \right\| \mathrm {d} \tau . \end{aligned}$$
By (19), the input term satisfies
$$\begin{aligned}&\int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \left\| u(\tau ) \right\| \mathrm {d} \tau \\&\quad = -\frac{1}{\lambda } \Big ( 1 - \mathrm {e} ^{\lambda h} \Big ) \Bigg [ \left\| H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] + \delta [k] \right\| \mathrm {e} ^{\lambda h} \\&\qquad + \left\| H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] - \delta [k] \right\| \Bigg ]\\&\quad \le -\frac{1}{\lambda } \Bigg (- \frac{\lambda T}{2} \Bigg ) \Bigg ( 2( \max \limits _{ \alpha \in \mathcal {A}}\left\| H( \alpha ) \right\| + \rho ) \left\| z[k] \right\| \\&\qquad + 2( \max \limits _{ \alpha \in \mathcal {A}}\left\| K( \alpha ) \right\| \left\| M \right\| + \rho ) \left\| r[k] \right\| + 2 \rho v[k] \Bigg ). \end{aligned}$$
Employing order notation, we can write this compactly as
$$\begin{aligned} \int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \left\| u(\tau ) \right\| \mathrm {d} \tau= & {} T \rho v[k] + \mathcal {O}(T) \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \mathcal {O}(T) \left\| s[k] \right\| _{\infty } , \end{aligned}$$
where \(\bar{x}\), \(\bar{z}\), and s are defined in (24). The output term satisfies
$$\begin{aligned}&\int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \left\| y(\tau ) \right\| \mathrm {d} \tau \\&\quad \le \int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \mathrm {d} \tau \overbrace{\left\| C( \alpha (kT)) x[k] \right\| }^{ = \mathcal {O}(1) \left\| x[k] \right\| } \\&\qquad + \int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \Big ( \underbrace{\left\| C( \alpha (\tau )) - C( \alpha (kT)) \right\| \left\| x[k] \right\| }_{ = \mathcal {O}(T) \left\| x[k] \right\| } \\&\qquad + \underbrace{\left\| C( \alpha (\tau )) \right\| \left\| x(\tau ) - x[k] \right\| }_{\text {bound using Proposition}~3} + \left\| w_{n} \right\| _{\infty } \Big ) \mathrm {d} \tau . \end{aligned}$$
Employing order notation, and utilizing Assumption 5 and Proposition 3, we can write this compactly as
$$\begin{aligned} \int _{kT}^{(k+1)T} \mathrm {e} ^{\lambda ((k+1)T - \tau )} \left\| y(\tau ) \right\| \mathrm {d} \tau= & {} \mathcal {O}(T) \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \mathcal {O}(T^2) v[k] + \mathcal {O}(T^2) \left\| s[k] \right\| _{\infty } \\&+ \mathcal {O}(T) \left\| w_{n} \right\| _{\infty } + \mathcal {O}(T^2) \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
Combining these upper bounds, for sufficiently small T there exist constants \(e_1 > 0\), \(\gamma _1 > 0\), \(\gamma _2 > 0\), and \(w_1 > 0\) such that, for all such T and all \(k \in \mathbb {Z} _+\),
$$\begin{aligned} v[k+1]\le & {} \big ( 1 + (\lambda + \rho )T + e_1 T^2 \big ) v[k] + \gamma _1 T \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \gamma _2 T \left\| s[k] \right\| _{\infty } \nonumber \\&+ w_1 T \left\| w_{n} \right\| _{\infty } + w_1 T^2 \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
(52)
Step 2: Controller sample point analysis
Starting with \( z[k+1] \), define
$$\begin{aligned} e_{z1} [k] :=T \Big ( F( \hat{ \alpha } [k]) - F( \alpha (kT)) \Big ) z[k] + T \Big ( G( \hat{ \alpha } [k]) - G( \alpha (kT)) \Big ) M r[k] , \end{aligned}$$
so that we can write
$$\begin{aligned} z[k+1] = \begin{bmatrix} 0&I + T F( \alpha (kT))&T G( \alpha (kT)) M \end{bmatrix} \begin{bmatrix} x[k] \\ z[k] \\ r[k] \end{bmatrix} + e_{z1} [k]. \end{aligned}$$
By Assumptions 5 and 6 (see Remark 1), there exists a constant \(\ell _1 > 0\) so that we have
$$\begin{aligned} \left\| e_{z1} [k] \right\| \le T \ell _1 \left\| \tilde{ \alpha } (kT) \right\| \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + T \ell _1 \left\| \tilde{ \alpha } (kT) \right\| \left\| s[k] \right\| _{\infty }, \end{aligned}$$
where \( \tilde{ \alpha } (kT) = \alpha (kT) - \hat{ \alpha } [k]\) denotes the parameter estimation error at time kT. By hypothesis \(\left\| \tilde{ \alpha } (kT) \right\| \le \overline{\delta } \), so
$$\begin{aligned} \left\| e_{z1} [k] \right\| \le T \ell _1 \overline{\delta } \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + T \ell _1 \overline{\delta } \left\| s[k] \right\| _{\infty }. \end{aligned}$$
(53)
We can treat \( r[k+1] \) similarly and define
$$\begin{aligned} \begin{aligned} e_{z2} [k] :=&- T \Big ( L( \hat{ \alpha } [k]) - L( \alpha (kT)) \Big ) C( \alpha (kT)) x[k] \\&+ T R \Big ( B( \hat{ \alpha } [k]) H( \hat{ \alpha } [k]) - B( \alpha (kT)) H( \alpha (kT)) \Big ) z[k] \\&+ T \Big ( ( Q( \hat{ \alpha } [k]) - Q( \alpha (kT)) ) + R ( B( \hat{ \alpha } [k]) K( \hat{ \alpha } [k]) - B( \alpha (kT)) K( \alpha (kT)) ) \Big ) r[k] \\&- T L( \hat{ \alpha } [k]) w_{n}[k], \end{aligned} \end{aligned}$$
so that
$$\begin{aligned} r[k+1]= & {} -T L( \alpha (kT)) C( \alpha (kT)) x[k] + T R B( \alpha (kT)) H( \alpha (kT)) z[k] \\&+ \Big (I + T \Big ( Q( \alpha (kT)) + R B( \alpha (kT)) K( \alpha (kT)) M \Big )\Big ) r[k] \\&+ e_{z2} [k]. \end{aligned}$$
Again, by Assumptions 5 and 6, there exists a constant \(\ell _2 > 0\) such that
$$\begin{aligned} \left\| e_{z2} [k] \right\| \le T \ell _2 \overline{\delta } \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + T \ell _2 \overline{\delta } \left\| s[k] \right\| _{\infty } + T \ell _2 \left\| w_{n} \right\| _{\infty } . \end{aligned}$$
(54)
Finally, we will use
$$\begin{aligned} e_{u} [k] :=\Big ( H( \hat{ \alpha } [k]) - H( \alpha (kT)) \Big ) z[k] + \Big ( K( \hat{ \alpha } [k]) - K( \alpha (kT)) \Big ) M r[k] , \end{aligned}$$
to write
$$\begin{aligned} \begin{aligned} u[k] - \delta (t)&= H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \\&= \begin{bmatrix} H( \alpha (kT))&K( \alpha (kT)) M \end{bmatrix} \begin{bmatrix} z[k] \\ r[k] \end{bmatrix} + e_{u} [k], \qquad t \in [kT, (k+1)T). \end{aligned} \end{aligned}$$
Again, by Assumptions 5 and 6, there exists a constant \(\ell _3 > 0\) such that
$$\begin{aligned} \left\| e_{u} [k] \right\| \le \ell _3 \overline{\delta } \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \ell _3 \overline{\delta } \left\| s[k] \right\| _{\infty }. \end{aligned}$$
(55)
Step 3: Plant sample point analysis
The value of the plant state at time \(t = (k+1)T\) is
$$\begin{aligned} x[k+1]= & {} x[k] + \int _{kT}^{(k+1)T} A( \alpha (\tau )) x(\tau ) \mathrm {d} \tau + \int _{kT}^{(k+1)T} B( \alpha (\tau )) u(\tau ) \mathrm {d} \tau \\&\quad + \int _{kT}^{(k+1)T} w_{d}(\tau ) \mathrm {d} \tau , \end{aligned}$$
so that, using the structure of the control signal (19),
$$\begin{aligned} x[k+1]= & {} (I + T A( \alpha (kT)) ) x[k] \\&+ T B( \alpha (kT)) \Bigg ( \begin{bmatrix} H( \alpha (kT))&K( \alpha (kT)) \end{bmatrix} \begin{bmatrix} z[k] \\ r[k] \end{bmatrix} + e_{u} [k] \Bigg ) \\&+ \int _{kT}^{(k+1)T} ( A( \alpha (\tau )) - A( \alpha (kT)) ) \mathrm {d} \tau x[k] \\&+ \int _{kT}^{(k+1)T} A( \alpha (\tau )) ( x(\tau ) - x[k]) \mathrm {d} \tau \\&+ \int _{kT}^{(k+1)T} ( B( \alpha (\tau )) - B( \alpha (kT)) ) \mathrm {d} \tau \Big ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \Big ) \\&+ \int _{kT}^{(k+1)T} B( \alpha (\tau )) \delta (\tau ) \mathrm {d} \tau + \int _{kT}^{(k+1)T} w_{d}(\tau ) \mathrm {d} \tau . \end{aligned}$$
Define
$$\begin{aligned} e_{p} [k]:= & {} \int _{kT}^{(k+1)T} ( A( \alpha (\tau )) - A( \alpha (kT)) ) \mathrm {d} \tau x[k] + \int _{kT}^{(k+1)T} A( \alpha (\tau )) ( x(\tau ) - x[k]) \mathrm {d} \tau \\&+ \int _{kT}^{(k+1)T} ( B( \alpha (\tau )) - B( \alpha (kT)) ) \mathrm {d} \tau \Big ( H( \hat{ \alpha } [k]) z[k] + K( \hat{ \alpha } [k]) M r[k] \Big ) \\&+ \int _{kT}^{(k+1)T} B( \alpha (\tau )) \delta (\tau ) \mathrm {d} \tau + \int _{kT}^{(k+1)T} w_{d}(\tau ) \mathrm {d} \tau , \end{aligned}$$
to be able to compactly write
$$\begin{aligned} \begin{aligned} x[k+1] =&\begin{bmatrix} I + T A( \alpha (kT))&T B( \alpha (kT)) H( \alpha (kT))&T B( \alpha (kT)) K( \alpha (kT)) M \end{bmatrix} \begin{bmatrix} x[k] \\ z[k] \\ r[k] \end{bmatrix} \\&+ T B( \alpha (kT)) e_{u} [k] + e_{p} [k]. \end{aligned} \end{aligned}$$
By Assumptions 5 and 6 (see Remark 1) and Proposition 3, there exists a constant \(\ell _4 > 0\) such that for small T
$$\begin{aligned} \left\| e_{p} [k] \right\| \le T^2 \ell _4 v[k] + T^2 \ell _4 \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + T^2 \ell _4 \left\| s[k] \right\| _{\infty } + T \ell _4 \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
(56)
Step 4: Closed-loop difference inequality
Combining the analysis of the plant and the controller, we obtain
$$\begin{aligned}&\begin{bmatrix} x[k+1] \\ z[k+1] \\ r[k+1] \end{bmatrix} \\&\quad = \left( I + T \begin{bmatrix} A( \alpha (kT))&B( \alpha (kT)) H( \alpha (kT))&B( \alpha (kT)) K( \alpha (kT)) M \\ 0&F( \alpha (kT))&G( \alpha (kT)) M \\ - L( \alpha (kT)) C( \alpha (kT))&R B( \alpha (kT)) H( \alpha (kT))&Q( \alpha (kT)) + R B( \alpha (kT)) K( \alpha (kT)) M \end{bmatrix} \right) \\&{\times } \begin{bmatrix} x[k] \\ z[k] \\ r[k] \end{bmatrix} + \begin{bmatrix} T B( \alpha (kT)) e_{u} [k] + e_{p} [k] \\ e_{z1} [k] \\ e_{z2} [k] \end{bmatrix} . \end{aligned}$$
In \((\bar{x}, \bar{z}, s)\)-coordinates, making use of (9) and (13), we have
$$\begin{aligned} \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \\ s[k+1] \end{bmatrix}= & {} \Bigg ( I + T \begin{bmatrix} P( \alpha (kT))&- \begin{bmatrix} X \\ Z \end{bmatrix} ^{-1} \begin{bmatrix} B( \alpha (kT)) K( \alpha (kT)) M \\ G( \alpha (kT)) M \end{bmatrix} \\ \begin{matrix} 0&0 \end{matrix}&Q( \alpha (kT)) \end{bmatrix} \Bigg ) \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \\ s[k] \end{bmatrix} \\&+ \begin{bmatrix} \begin{bmatrix} X \\ Z \end{bmatrix} ^{-1} \begin{bmatrix} T B( \alpha (kT)) e_{u} [k] + e_{p} [k] \\ e_{z1} [k] \end{bmatrix} \\ T R B( \alpha (kT)) e_{u} [k] + R e_{p} [k] - e_{z2} [k] \end{bmatrix} . \end{aligned}$$
To construct the decrescent norm, we define a difference inequality and apply two similarity transformations. We start by upper-bounding the transformed plant and controller states:
$$\begin{aligned} \left\| \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \end{bmatrix} \right\| _{1}\le & {} \left\| I + T P( \alpha (kT)) \right\| _{1} \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} \\&+ T \left\| \begin{bmatrix} X \\ Z \end{bmatrix} ^{-1} \begin{bmatrix} B( \alpha (kT)) K( \alpha (kT)) M \\ G( \alpha (kT)) M \end{bmatrix} \right\| _{1} \left\| s[k] \right\| _{1} \\&+ \left\| \begin{bmatrix} X \\ Z \end{bmatrix} ^{-1} \begin{bmatrix} T B( \alpha (kT)) e_{u} [k] + e_{p} [k] \\ e_{z1} [k] \end{bmatrix} \right\| _{1}. \end{aligned}$$
Next, we take advantage of the \(\mathcal {H}_{1}\) property that \( P( \alpha ) \) enjoys: By Proposition 1 there exists a constant \(\lambda _{1} < 0\) so that \(\left\| I + T P( \alpha ) \right\| _{1} \le 1 + \lambda _{1}T\) for all sufficiently small T; similarly, we can take advantage of the \(\mathcal {H}_\infty \) property that \(Q(\alpha )\) enjoys: There exists a constant \(\lambda _2 < 0\) so that \(\Vert I + TQ(\alpha )\Vert _\infty \le 1 + \lambda _2T\) for all sufficiently small T; clearly, we can choose \(\lambda _1\) and \(\lambda _2\) so that \(\lambda _2< \lambda _1 < 0\). Then, using (53), (55), and (56), there exist constants \(e_2 > 0\), \(\gamma _3 > 0\), and \(w_2 > 0\) such that for small T
$$\begin{aligned} \left\| \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \end{bmatrix} \right\| _{1}\le & {} \big ( 1 + \lambda _1 T + e_2 T ( \overline{\delta } + T) \big ) \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} \nonumber \\&+ \big ( \gamma _3 T + e_2 T ( \overline{\delta } + T) \big ) \left\| s[k] \right\| _{\infty } \nonumber \\&+ e_2 T^2 v[k] + w_2 T \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
(57)
In a similar fashion, we have an upper bound on \(s[k+1]\):
$$\begin{aligned} \begin{aligned} \left\| s[k+1] \right\| _{\infty }&\le \left\| I + T Q( \alpha (kT)) \right\| _{\infty } \left\| s[k] \right\| _{\infty } + T \left\| R B( \alpha (kT)) e_{u} [k] \right\| _{\infty } \\&\quad + \left\| R e_{p} [k] \right\| _{\infty } + \left\| e_{z2} [k] \right\| _{\infty }\\&\le (1 + \lambda _2 T) \left\| s[k] \right\| _{\infty } + T \left\| R B( \alpha (kT)) \right\| \left\| e_{u} [k] \right\| _{\infty } + \left\| R \right\| \left\| e_{p} [k] \right\| _{\infty } \\&\quad + \left\| e_{z2} [k] \right\| _{\infty }, \end{aligned} \end{aligned}$$
and using (54), (55), and (56), there exist constants \(e_3 > 0\) and \(w_3 > 0\) such that for small T
$$\begin{aligned} \left\| s[k+1] \right\| _{\infty }\le & {} \big ( 1 + \lambda _2 T + e_3 T ( \overline{\delta } + T) \big ) \left\| s[k] \right\| _{\infty } + e_3 T ( \overline{\delta } + T) \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} \nonumber \\&+ e_3 T^2 v[k] + w_3 T \left\| w_{n} \right\| _{\infty } + w_3 T \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
(58)
Now, we combine the bounds on the states. Recall that the controller parameter \(\rho \) was chosen to satisfy \(\rho \in (0, -\lambda )\), which means that \(\lambda + \rho < 0\); now fix
$$\begin{aligned} \bar{\lambda } \in \left( \max \{\lambda _1, \lambda _2, \lambda + \rho \}, 0\right) , \end{aligned}$$
which means that
$$\begin{aligned} \lambda _2< \lambda _1< \bar{\lambda } < 0. \end{aligned}$$
(59)
Combining the bounds given in (52), (57), and (58) yields, for small T:
$$\begin{aligned} \begin{bmatrix} v[k+1] \\ \left\| \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \end{bmatrix} \right\| _{1} \\ \left\| s[k+1] \right\| _{\infty } \end{bmatrix}\le & {} \begin{bmatrix} 1 + \bar{\lambda } T + e_1 T^2&\gamma _1 T&\gamma _2 T \\ e_2 T^2&1 + \lambda _{1}T + e_2 T ( \overline{\delta } + T)&\gamma _3 T + e_2 T ( \overline{\delta } + T) \\ e_3 T^2&e_3 T ( \overline{\delta } + T)&1 + \lambda _{2}T + e_3 T ( \overline{\delta } + T) \end{bmatrix} \begin{bmatrix} v[k] \\ \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} \\ \left\| s[k] \right\| _{\infty } \end{bmatrix} \\&+ \begin{bmatrix} w_1 T&w_1 T^2 \\ 0&w_2 T \\ w_3 T&w_3 T \end{bmatrix} \begin{bmatrix} \left\| w_{n} \right\| _{\infty } \\ \left\| w_{d} \right\| _{\infty } \end{bmatrix} . \end{aligned}$$
So, with \(E \in \mathbb {R} ^{3 \times 3}\) suitably chosen and T sufficiently small, we have
$$\begin{aligned} \begin{bmatrix} v[k+1] \\ \left\| \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \end{bmatrix} \right\| _{1} \\ \left\| s[k+1] \right\| _{\infty } \end{bmatrix}\le & {} \left( \underbrace{ \begin{bmatrix} 1 + \bar{\lambda } T&\gamma _1 T&\gamma _2 T \\ 0&1 + \lambda _{1} T&\gamma _3 T \\ 0&0&1 + \lambda _{2} T \end{bmatrix} }_{=: \Lambda } + T ( \overline{\delta } + T) E \right) \begin{bmatrix} v[k] \\ \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} \\ \left\| s[k] \right\| _{\infty } \end{bmatrix} \nonumber \\&+ \underbrace{ \begin{bmatrix} w_1 T&w_1 T^2 \\ 0&w_2 T \\ w_3 T&w_3 T \end{bmatrix} }_{ =: W(T)} \begin{bmatrix} \left\| w_{n} \right\| _{\infty } \\ \left\| w_{d} \right\| _{\infty } \end{bmatrix} . \end{aligned}$$
(60)
Next, we define three states as upper bounds of the above states at periods k and \(k+1\). This allows us to get equality, so we can solve and transform a difference equation rather than inequality. Define \(\psi :=(\psi _1, \psi _2, \psi _3) \in \mathbb {R} _+^3\) via
$$\begin{aligned}&\psi _1[k] :=v[k], \quad \psi _2[k] :=\left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1}, \quad \psi _3[k] :=\left\| s[k] \right\| _{\infty },\\&\psi [k+1] :=(\Lambda + T ( \overline{\delta } + T) E ) \psi [k] + W(T) \begin{bmatrix} \left\| w_{n} \right\| _{\infty } \\ \left\| w_{d} \right\| _{\infty } \end{bmatrix} . \end{aligned}$$
It is clear that \(v[k+1] \le \psi _1[k+1]\), \(\left\| \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \end{bmatrix} \right\| _{1} \le \psi _2[k+1]\), and \(\left\| s[k+1] \right\| _{\infty } \le \psi _3[k+1]\) because
$$\begin{aligned} \begin{bmatrix} v[k+1] \\ \left\| \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \end{bmatrix} \right\| _{1} \\ \left\| s[k+1] \right\| _{\infty } \end{bmatrix}\le & {} (\Lambda + T ( \overline{\delta } + T) E) \begin{bmatrix} v[k] \\ \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} \\ \left\| s[k] \right\| _{\infty } \end{bmatrix} + W(T) \begin{bmatrix} \left\| w_{n} \right\| _{\infty } \\ \left\| w_{d} \right\| _{\infty } \end{bmatrix} \\= & {} (\Lambda + T ( \overline{\delta } + T) E) \psi [k] + W(T) \begin{bmatrix} \left\| w_{n} \right\| _{\infty } \\ \left\| w_{d} \right\| _{\infty } \end{bmatrix} \\= & {} \psi [k+1]. \end{aligned}$$
Next, we perform two similarity transformations with the objective of diagonalizing the matrix \(\Lambda \) in (60). The first is a constant transformation of the form
$$\begin{aligned} V := \begin{bmatrix} 1&0&0 \\ 0&1&v_{23} \\ 0&0&1 \end{bmatrix} \end{aligned}$$
so that under similarity transformation we have
$$\begin{aligned}&V (\Lambda + T ( \overline{\delta } + T) E) V^{-1} \nonumber \\&= \begin{bmatrix} 1 + \bar{\lambda } T&\gamma _1 T&(\gamma _2 - v_{23} \gamma _1)T \\ 0&1 + \lambda _{1} T&(\gamma _3 + v_{23}(\lambda _{2} - \lambda _{1}))T \\ 0&0&1 + \lambda _{2} T \end{bmatrix} + T ( \overline{\delta } + T) V E V^{-1}. \end{aligned}$$
Choose \(v_{23} = \gamma _{3} / (\lambda _{1} - \lambda _{2})\) so that we are left with
$$\begin{aligned} V (\Lambda + T ( \overline{\delta } + T) E) V^{-1} = \begin{bmatrix} 1 + \bar{\lambda } T&\gamma _1 T&(\gamma _2 - v_{23} \gamma _1)T \\ 0&1 + \lambda _{1} T&0 \\ 0&0&1 + \lambda _{2} T \end{bmatrix} + T ( \overline{\delta } + T) V E V^{-1}. \end{aligned}$$
To complete the diagonalization of \(\Lambda \), consider a transformation of the form
$$\begin{aligned} Y := \begin{bmatrix} 1&\bar{Y} \\ 0&I \end{bmatrix} \in \mathbb {R} ^{3 \times 3}, \ \ \bar{Y} \in \mathbb {R} ^{1 \times 2}, \end{aligned}$$
so that, with \(\gamma _{4} :=\gamma _{2} - v_{23} \gamma _{1}\),
$$\begin{aligned}&Y V (\Lambda + T ( \overline{\delta } + T) E) V^{-1} Y^{-1}\\&\quad = \begin{bmatrix} 1 + \bar{\lambda } T&\begin{bmatrix} \gamma _1&\gamma _4 \end{bmatrix} T + \bar{Y} \begin{bmatrix} 1 + \lambda _{1} T&0 \\ 0&1 + \lambda _{2} T \end{bmatrix} - (1 + \bar{\lambda } T) \bar{Y} \\ \begin{bmatrix} 0 \\ 0 \end{bmatrix}&\begin{bmatrix} 1 + \lambda _{1} T&0 \\ 0&1 + \lambda _{2} T \end{bmatrix} \end{bmatrix} \\\\&\qquad + T ( \overline{\delta } + T) Y V E V^{-1} Y^{-1}. \end{aligned}$$
Choose \(\bar{Y} = \begin{bmatrix} \frac{\gamma _1}{ \bar{\lambda } - \lambda _1}&\frac{\gamma _4}{ \bar{\lambda } - \lambda _2} \end{bmatrix} \) to get
$$\begin{aligned} Y V (\Lambda + T ( \overline{\delta } + T) E) V^{-1} Y^{-1} = \begin{bmatrix} 1 + \bar{\lambda } T&0&0 \\ 0&1 + \lambda _{1} T&0 \\ 0&0&1 + \lambda _{2} T \end{bmatrix} + T ( \overline{\delta } + T) Y V E V^{-1} Y^{-1}. \end{aligned}$$
Defining \(N :=Y V\), we have
$$\begin{aligned} N \psi [k+1] = \left( \begin{bmatrix} 1 + \bar{\lambda } T&0&0 \\ 0&1 + \lambda _{1} T&0 \\ 0&0&1 + \lambda _{2} T \end{bmatrix} + T ( \overline{\delta } + T) N E N^{-1} \right) N \psi [k] + N W(T) \begin{bmatrix} \left\| w_{n} \right\| _{\infty } \\ \left\| w_{d} \right\| _{\infty } \end{bmatrix} . \end{aligned}$$
It follows from (59) that all elements of N are nonnegative, which can be used to prove that
$$\begin{aligned} \left\| N \psi [k+1] \right\| \ge \left\| N \begin{bmatrix} v[k+1] \\ \left\| \begin{bmatrix} \bar{x}[k+1] \\ \bar{z}[k+1] \end{bmatrix} \right\| _{1} \\ \left\| s[k+1] \right\| _{\infty } \end{bmatrix} \right\| = \left\| p [k+1] \right\| . \end{aligned}$$
Taking the \(\infty \)-norm of \( p [k+1]\), there exist constants \(\gamma > 0\) and \(c > 0\) such that for small T
$$\begin{aligned} \left\| p [k+1] \right\| _{\infty }\le & {} \left\| N \psi [k+1] \right\| _{\infty } \le \Big ( 1 + \bar{\lambda } T + \gamma T ( \overline{\delta } + T) \Big ) \left\| N \psi [k] \right\| _{\infty } \\&+ c T \left\| w_{n} \right\| _{\infty } + c T \left\| w_{d} \right\| _{\infty } \\= & {} \Big ( 1 + \bar{\lambda } T + \gamma T ( \overline{\delta } + T) \Big ) \left\| p [k] \right\| _{\infty } + c T \left\| w_{n} \right\| _{\infty } + c T \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
Now, choose \( \overline{\delta } > 0\) so that
$$\begin{aligned} \hat{\lambda } := \bar{\lambda } + 2 \gamma \overline{\delta } < 0; \end{aligned}$$
it follows that for small T, we have
$$\begin{aligned} \begin{aligned} \left\| p [k+1] \right\| _{\infty }&\le \Big ( 1 + \hat{\lambda } T \Big ) \left\| p [k] \right\| _{\infty } + c T \left\| w_{n} \right\| _{\infty } + c T \left\| w_{d} \right\| _{\infty } \\&\le \mathrm {e} ^{ \hat{\lambda } T} \left\| p [k] \right\| _{\infty } + c T \left\| w_{n} \right\| _{\infty } + c T \left\| w_{d} \right\| _{\infty } , \end{aligned} \end{aligned}$$
which means that (iii) holds. \(\square \)
Proof of (i)
By the definition of \( p \),
$$\begin{aligned} p (t) - p [k] = N \begin{bmatrix} | v(t) | - |v[k]| \\ \left\| \begin{bmatrix} \bar{x}(t) \\ \bar{z}(t) \end{bmatrix} \right\| _{1} - \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} \\ \left\| s(t) \right\| _{\infty } - \left\| s[k] \right\| _{\infty } \end{bmatrix} , \quad t \ge 0. \end{aligned}$$
Taking the 1-norm and using the reverse triangle inequality, for \(t \in [kT,(k+1)T)\) we get
$$\begin{aligned} \left\| p (t) - p [k] \right\| _{1}= & {} \mathcal {O}(1) | v(t) - v[k]| + \mathcal {O}(1) \left\| x(t) - x[k] \right\| \nonumber \\&+ \mathcal {O}(1) \left\| z(t) - z[k] \right\| + \mathcal {O}(1) \left\| r(t) - r[k] \right\| . \end{aligned}$$
(61)
The solution to (15) with initial condition v[k] is
$$\begin{aligned} v(t) = v[k] + \int _{kT}^{t} \lambda (v(\tau ) - v[k]) \mathrm {d} \tau + (t - kT) \lambda v[k] + \int _{kT}^{t} (\left\| u(\tau ) \right\| + \left\| y(\tau ) \right\| ) \mathrm {d} \tau . \end{aligned}$$
Rearranging and taking the absolute value, we have
$$\begin{aligned} |v(t) - v[k]| \le |\lambda | \int _{kT}^{t} |v(\tau ) - v[k]| \mathrm {d} \tau + T |\lambda | v[k] + \int _{kT}^{t} (\left\| u(\tau ) \right\| + \left\| y(\tau ) \right\| ) \mathrm {d} \tau , \end{aligned}$$
and by applying the Bellman–Gronwall inequality and using Proposition 3, it follows that
$$\begin{aligned} |v(t) - v[k]|= & {} \mathcal {O}(T) v[k] + \mathcal {O}(T) \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \mathcal {O}(T) \left\| s[k] \right\| _{\infty } \nonumber \\&+ \mathcal {O}(T) \left\| w_{n} \right\| _{\infty } + \mathcal {O}(T^2) \left\| w_{d} \right\| _{\infty } , \quad t \in [kT, (k+1)T). \end{aligned}$$
(62)
For \(t \in [kT, (k+1)T)\), we also have \(\left\| z(t) - z[k] \right\| = 0\) and \(\left\| r(t) - r[k] \right\| = 0\). Using these bounds on (61), along with Proposition 3 and (62), we get
$$\begin{aligned} \left\| p (t) - p [k] \right\| = \mathcal {O}(T) \left\| p [k] \right\| + \mathcal {O}(T) ( \left\| w_{n} \right\| _{\infty } + \left\| w_{d} \right\| _{\infty } ), \quad t \in [kT, (k+1)T). \end{aligned}$$
So, for \(t \in [kT, (k+1)T)\), there exists a constant \(c > 0\) so that for sufficiently small T.
$$\begin{aligned} \left\| p (t) - p [k] \right\| \le c T \left\| p [k] \right\| + c T \left\| w_{n} \right\| _{\infty } + c T \left\| w_{d} \right\| _{\infty } , \quad t \in [kT, (k+1)T). \end{aligned}$$
\(\square \)
Proof of (ii)
The bound (61) derived in the proof of part (i) remains valid. Additionally, using (17) we have
$$\begin{aligned} z[k+1] - z[k] = T F( \hat{ \alpha } [k]) z[k] + T G( \hat{ \alpha } [k]) r[k] , \end{aligned}$$
so taking the norm and employing order notation,
$$\begin{aligned} \left\| z[k+1] - z[k] \right\| = \mathcal {O}(T) \left\| \begin{bmatrix} \bar{x}[k] \\ \bar{z}[k] \end{bmatrix} \right\| _{1} + \mathcal {O}(T) \left\| s[k] \right\| _{\infty } = \mathcal {O}(T) \left\| p [k] \right\| . \end{aligned}$$
We can also upper-bound \(\left\| r[k+1] - r[k] \right\| \). We have, again from (17),
$$\begin{aligned} r[k+1] - r[k]= & {} T R B( \hat{ \alpha } [k]) H( \hat{ \alpha } [k]) z[k] \\&+ T \Big ( Q( \hat{ \alpha } [k]) + R B( \hat{ \alpha } [k]) K( \hat{ \alpha } [k]) M \Big ) r[k] \\&- T L( \hat{ \alpha } [k]) \Big ( C( \alpha [k]) x[k] + w_{n}[k] \Big ), \end{aligned}$$
so taking the norm and employing order notation,
$$\begin{aligned} \left\| r[k+1] - r[k] \right\| = \mathcal {O}(T) \left\| p [k] \right\| + \mathcal {O}(T) \left\| w_{n} \right\| _{\infty } . \end{aligned}$$
Applying these bounds to (61), along with Proposition 3 and (52), we have
$$\begin{aligned} \left\| p [k+1] - p [k] \right\| = \mathcal {O}(T) \left\| p [k] \right\| + \mathcal {O}(T) \left\| w_{n} \right\| _{\infty } + \mathcal {O}(T) \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
So there exists a constant \(c > 0\) so that, for all T sufficiently small,
$$\begin{aligned} \left\| p [k+1] \right\| \le (1 + c T) \left\| p [k] \right\| + c T \left\| w_{n} \right\| _{\infty } + c T \left\| w_{d} \right\| _{\infty } . \end{aligned}$$
\(\square \)
