Near optimal $\mathcal {H}_{\infty }$ performance in the decentralized setting

Abstract

In this paper, we consider the use of a linear periodic controller (LPC) for the control of linear time-invariant (LTI) plants in the decentralized setting with an $H_{\infty }$-performance criterion in mind. Here we show that if the plant is centrally stabilizable and detectable, the graph associated with the plant is strongly connected, and a technical condition on the relative degree holds, then we can design a decentralized LPC to provide a level of $H_{\infty }$ performance as close as desired to that provided by a given (stabilizing) LTI centralized controller; this will be the case even if the plant has an unstable decentralized fixed mode (DFM). This means, in particular, that under the listed conditions, we can design a decentralized LPC to provide near $H_{\infty }$-optimal centralized performance.

Appendix

Proof of Lemma 1

Fix $\bar{n}\in \mathbf{N}$ and $\tilde{h}\in (0,1)$. Let $t_0\in \mathbf{R}$, $x_0\in \mathbf{R}$, $h\in (0,\tilde{h})$, $\bar{u}\in \mathbf{R}$ and $\phi \in \mathbf{R}$ be arbitrary, and choose u(t) as indicated. If we solve the plant equation (1) and use the Cauchy–Schwarz inequality, then we obtain

$$\begin{aligned} \left\| x(t)-x_0 \right\|\le & {} \left\| ({\mathrm {e}}^{A(t-t_0)}-I)x_0 \right\| + \int _{t_0}^{t}\left\| e^{A(t-\tau )}B \right\| (\left\| \bar{u} \right\| +|\phi |) \mathrm{d} \tau \\&+\, \left( \int _{t_0}^{t}\left\| e^{A(t-\tau )}E \right\| ^2\mathrm{d}\tau \right) ^{\frac{1}{2}}\Vert r_{[ t_0 , t ]} \Vert _2 , \;\; t \in [ t_0, t_0 + 2 \bar{n}h ) . \end{aligned}$$

Clearly there exists a constant $\gamma _1>0$ such that

$$\begin{aligned} \left\| x(t)-x_0 \right\| \le \gamma _1h(\left\| x_0 \right\| +\left\| \bar{u} \right\| +|\phi |) + \gamma _1h^{\frac{1}{2}}\left\| r_{[t_0,t_0+2\bar{n}h]} \right\| _2,\;\; t\in [t_0,t_0+2\bar{n}h), \end{aligned}$$

(60)

which yields the second bound.

Using the fact that $\bar{y}_i(t) = v_i^{\mathrm {T}}y_i(t)$ for $t\in [t_0,t_0+\bar{n}h)$ we have

$$\begin{aligned} \bar{y}_i(t)= & {} v_i^{\mathrm {T}}C_2^i \left[ e^{A(t-t_0)}x_0+\int _0^{t-t_0}e^{A \tau }B ( \bar{u} + \bar{w}_j \phi ) \mathrm{d}\tau +\int _{t_0}^{t}e^{A(t-\tau )}Er(\tau ){\mathrm {d}\tau }\right] \\= & {} \sum _{k=0}^{\bar{n}}\frac{\bar{C}_2^iA^k(t-t_0)^k}{k!}x_0 +{\mathcal {O}}(h^{\bar{n}+1})x_0 + \sum _{k=0}^{\bar{n}-1}\frac{\bar{C}_2^iA^k(t-t_0)^{k+1}B}{(k+1)!}(\bar{u} +\bar{w}_j\phi )\\&+\,{\mathcal {O}}(h^{\bar{n}+1})(\bar{u}+\phi ) +\underbrace{\bar{C}_2^i\int _{t_0}^{t}e^{A(t-\tau )}Er(\tau )\mathrm{d}\tau }_{=: \mu _1(t)}\\= & {} \left[ \begin{array}{cccc} 1&t-t_0&\cdots&(t-t_0)^{\bar{n}}/{\bar{n}!} \end{array} \right] \times \left\{ \underbrace{ \left[ \begin{array}{c} \bar{C}_2^i \\ \bar{C}_2^iA \\ \vdots \\ \bar{C}_2^iA^{\bar{n}} \end{array} \right] }_ {=: {\mathcal O}_i} x_0 + \underbrace{ \left[ \begin{array}{c} 0 \\ \bar{C}_2^iB \\ \vdots \\ \bar{C}_2^iA^{\bar{n}-1}B \end{array} \right] }_{=: M_i} (\bar{u} +\bar{w}_j\phi ) \right\} \\&+{\mathcal {O}}(h^{\bar{n}+1}) x_0+{\mathcal {O}}(h^{\bar{n}+1})(\bar{u}+\phi )+ \mu _1(t). \end{aligned}$$

Before we proceed further, let us get a bound on the term $\mu _1(t)$. Using the Cauchy–Schwarz inequality, we have

$$\begin{aligned} \Vert {\mu _1(t)}\Vert \le \left( \int _{t_0}^{t_0 + \bar{n}h} \Vert \bar{C}_2^ie^{A(t-\tau )}E\Vert ^2\mathrm{d}\tau \right) ^{\frac{1}{2}}{\Vert r_{[t_0,t_0+\bar{n}h]}\Vert _2} , \;\; t \in [ t_0 , t_0 + \bar{n}h ] . \end{aligned}$$

Note that

$$\begin{aligned} rel.deg(\bar{C}_2^i(sI-A)^{-1}E) \ge rel.deg(C_2(sI-A)^{-1}E) = \eta _2, \end{aligned}$$

so we see that $\bar{C}_2 A^i E = 0$ for $i=0,1,\ldots , \eta _2-2$, so there exists a constant $\gamma _2$ so that

$$\begin{aligned} \Vert \bar{C}_2^i e^{A \theta } E \Vert \le \gamma _2 \theta ^{\eta _2-1} , \; \theta \in [ 0 , 2 \bar{n}h ] , \end{aligned}$$

which means that there exists a constant $\gamma _3$ so that

$$\begin{aligned} \left\| \mu _1(t) \right\| = \gamma _3 h^{\eta _2-{\frac{1}{2}}} {\Vert r_{[t_0,t_0+\bar{n}h]}\Vert _2}, \; t \in [ t_0 , t_0 + \bar{n}h ] . \end{aligned}$$

Now we sample $\bar{y}_i$ in the interval of $[t_0, t_0+\bar{n}h)$ and form $\bar{\mathcal Y}_i(t_0)$; we see that there exists a constant $\gamma _4$ and a function $\mu _2(h)$ so that

$$\begin{aligned} {{\bar{\mathcal Y}}}_i(t_0) = SH(h)[ {\mathcal O}_i x_0 + M_i (\bar{u} +\bar{w}_j\phi )]+ \mu _2(h) \end{aligned}$$

with

$$\begin{aligned} \left\| \mu _2(h) \right\|\le & {} \gamma _4 h^{\bar{n}+1}(\left\| x_0 \right\| +\left\| \bar{u} \right\| +|\phi |)+ \gamma _4 h^{\eta _2-{\frac{1}{2}}}\left\| r_{[t_0,t_0+\bar{n}h]} \right\| _2. \end{aligned}$$

(61)

If we analyse $\bar{y}_i(t)$ for $t\in [t_0+\bar{n}h, t_0+2\bar{n}h)$ in a similar fashion and form ${\bar{\mathcal Y}}_i(t_0+\bar{n}h)$, then we see that there exists a constant $\gamma _5$ and a function $\mu _3(h)$ so that

$$\begin{aligned} {\bar{\mathcal Y}}_i(t_0+\bar{n}h)= & {} SH(h)[ {\mathcal O}_i x(t_0+\bar{n}h) + M_i (\bar{u} -\bar{w}_j\phi ) ] + \mu _3(h) \end{aligned}$$

with

$$\begin{aligned} \left\| \mu _3(h) \right\|\le & {} \gamma _5 h^{\bar{n}+1}(\left\| x(t_0+\bar{n}h) \right\| +\left\| \bar{u} \right\| +|\phi |) + \gamma _5 h^{\eta _2-{\frac{1}{2}}}{\Vert r_{[t_0+\bar{n}h,t_0+2\bar{n}h]}\Vert _2}.\nonumber \\ \end{aligned}$$

(62)

Using the fact that $\bar{B}_j = B\bar{w}_j$ and subtracting ${\bar{\mathcal Y}}_i(t_0+\bar{n}h)$ from ${\bar{\mathcal Y}}_i(t_0)$ we obtain

$$\begin{aligned}&{\bar{\mathcal Y}}_i(t_0)- {\bar{\mathcal Y}}_i(t_0+\bar{n}h)-2SH(h) \underbrace{M_i \bar{w}_j}_{= M_{i,j}} \phi \\&\quad =SH(h){\mathcal O}_i [ x_0 - x(t_0+\bar{n}h)] + \mu _2(h)-\mu _3(h) , \end{aligned}$$

so

$$\begin{aligned}&H(h)^{-1} S^{-1} [ {\bar{\mathcal Y}}_i(t_0)- {\bar{\mathcal Y}}_i(t_0+\bar{n}h) ] - 2 M_{i,j} \phi \\&\quad = {\mathcal O}_i [ x_0 - x(t_0+\bar{n}h)] + H(h)^{-1}S^{-1} [ \mu _2(h)-\mu _3(h) ] =: \mu _4 (h). \end{aligned}$$

We can form a bound on $\left\| x_0 - x(t_0+\bar{n}h) \right\|$ using (60) and bounds on $\mu _2(h)$ and $\mu _3(h)$ using (61) and (62), respectively. Using the fact that $\Vert H(h)^{-1} \Vert = \frac{ \bar{n}!}{h^{\bar{n}}}$ and $S^{-1} = {\mathcal O} (1)$, we see that there exists a constant $\gamma _6$ so that

$$\begin{aligned}&\left\| \mu _4(h) \right\| \le \gamma _6 h (\left\| x_0 \right\| +\left\| \bar{u} \right\| +|\phi |)+ \gamma _6 h^{{\frac{1}{2}}}\left\| r_{[t_0,t_0+2\bar{n}h]} \right\| _2 \\&\quad +\, \gamma _6 h^{\eta _2 - \bar{n}-{\frac{1}{2}}}\left\| r_{[t_0,t_0+2\bar{n}h]} \right\| _2, \end{aligned}$$

which provides the first bound. $\square$

Proof of Lemma 2

We will model the proof on that of Lemma 2 of [16]. Here our objective is to show the existence of a LPC and we will not attempt to find one of the lowest order.

Since the controller is LPC, it is sufficient to look at what happens on the interval $[kT,(k+1)T)$. In channel i, we partition the state into three sub-states: $\psi _i^1$, $\psi _i^2$ and $\psi _i^3$. We use $\psi _i^1$ of dimension ${ql_i}$ to store $\{y_i(kT),y_i(kT+h),\ldots ,y_i(kT+(q-1)h)\}$. The second sub-state $\psi _i^2$ of dimension ${m_i}$ stores $\hat{\sqcap }_i[k]$. The last sub-state $\psi _i^3$ of dimension $\ell$ only comes into play in channel p: $\psi _p^3$ stores $\nu [k]$.

We limit our analysis to the interval of [0, T) as it can easily be extended due to the periodic nature of the controller. Let $e_i \in \mathbf{R}^q$ denote the ith normal vector. We set

$$\begin{aligned} (L_i^{11},M_i^1)[j] =\left\{ \begin{array}{ll} (0,e_1\otimes I_{r_i}) &{} j=0\\ (I,e_{j+1}\otimes I_{r_i})&{} j=1,\ldots ,q-1.\\ \end{array}\right. \end{aligned}$$

It is clear that for $j=0,1,\ldots ,q-1$, the vector $\left[ \begin{array}{c} \psi _i^1[j] \\ y_i(jh) \end{array} \right]$ contains all the elements $\{y_i(0),y_i(h),\ldots ,y_i(jh)\}$.

Now we turn to the second sub-state. Set

$$\begin{aligned} (L_i^{21},L_i^{22},M_i^2)[j] = \left\{ \begin{array}{ll} (0,I,0) &{} j=0,1,\ldots ,q-2\\ (L_i^{21}[q-1],0, \; M_i^2[q-1]) &{} j=q-1. \\ \end{array}\right. \end{aligned}$$

If we initialize $\psi _i^2[0]$ to be $\hat{\sqcap }_i [0]$, then we see that

$$\begin{aligned} \psi _i^2 [j] = \hat{\sqcap }_i [0] , \;\; j = 0,1,\ldots , q-1 . \end{aligned}$$

(63)

From the formula for $\hat{\sqcap }_i [1]$ given in the controller description, we see that it is a weighted sum of the elements of $\{ y_i (0) , \; y_i (h) ,\ldots , \; y_i ((q-1)h) \}$; using the fact that these elements are contained in $\left[ \begin{array}{c} \psi _i^1[q-1] \\ y_i ( ( q-1)h) \end{array} \right]$, it follows that we can define $L_i^{21}[q-1]$ and $M_i^2[q-1]$ so that

$$\begin{aligned} \psi _i^2 [q]= & {} L_i^{21}[q-1] \psi _i^1 [ q-1 ] + M_i^2[q-1] y_i ( q-1)h) = \hat{\sqcap }_i [1] . \end{aligned}$$

We conclude that (16) holds.

Now we turn to the third sub-state. With $q_p =2\bar{n}(l-l_p)$ we set $(L_i^{31},L_i^{33},M_i^3)[\cdot ] = 0$ for $i =1,\;2,\ldots ,\;p-1$ and

$$\begin{aligned} (L_p^{31},L_p^{33},M_p^3)[j] = \left\{ \begin{array}{ll} (0,I,0) &{} j=0,1,\ldots ,q_p-1\\ (L_p^{31}[q_p],\mathcal {F},M_p^3[q_p]) &{} j=q_p\\ (0,I,0) &{} j=q_p+1,\ldots ,q-1.\\ \end{array}\right. \end{aligned}$$

From the formula given for $\hat{y}_i (kT)$ in the controller description, we see that $\hat{y} (0)$ is a weighted sum of the elements of $\{ y_p (0) , \; y_p (h) , \cdots , \; y_p (q_p h) \}$; using the fact that these elements are contained in $\left[ \begin{array}{c} \psi _p^1 [q_p ] \\ y_p ( q_p h ) \end{array} \right]$, we can choose $L_p^{31}[q_p]$ and $M_p^3[q_p]$ such that

$$\begin{aligned} L_p^{31}[q_p]\psi _p^1[q_p] + M_p^3[q_p]y_p(q_ph) = \hat{y}(0). \end{aligned}$$

So if we initialize $\psi _p^3[0]=\nu [0]$, we see that

$$\begin{aligned} \psi _p^3[j]=\left\{ \begin{array}{ll} \nu [0],&{}j=0, 1, \cdots ,q_p\\ \nu [1],&{}j=q_p+1, \cdots , q-1 , \end{array}\right. \end{aligned}$$

as required.

At this point, we have defined $L_i$ and $M_i$ of $K_\mathrm{dec}$. It remains to define the time-varying matrices $Q_i$ and $R_i$ related to the output. To this end, we define $\phi _i[j]= \left[ \begin{array}{c} \psi _i[j] \\ y_i(jh) \end{array} \right]$. It is straightforward to verify that $\phi _i[j]$ contains $\{\hat{\sqcap }_i[0],y_i(0),\ldots ,y_i(jh)\}$ for $j=0,1,\ldots , q-1$; for the case of $i=p$, $\phi _i[j]$ also contains

$$\begin{aligned} \left\{ \begin{array}{ll} \nu [0],&{}\quad j=0, 1, \ldots ,q_p\\ \nu [1],&{}\quad j=q_p+1, \ldots , q-1 . \end{array}\right. \end{aligned}$$

• (i) For $j=0,\;1,\ldots ,\;q_p-1$, the control signal $u_i(jh)$ equals $\hat{\sqcap }_i[0]$ plus a linear combination of the elements of $y_i(0)$. Since $\hat{\sqcap }_i[0]$ and $y_i(0)$ are contained in $\phi _i [j]$, we see that $u_i(jh)$ is a linear combination of the elements of $\phi _i[j]$ for $j=0,1,\ldots ,q_p-1$.

• (ii) At $j=q_p$, $u_i(j) = \hat{\sqcap }_i[0]$ and $\hat{\sqcap }_i[0]$ is contained in $\psi _i[j]$, so it is a linear combination of the elements of $\phi _i(j)$.

• (iii) Let $j \in \{ q_p+1,\,\ldots ,\;q-1 \}$. The control signal $u_i(jh)$, $i=1,\ldots ,p-1$, equals $\hat{\sqcap }_i[0]$, which is contained in $\psi _i[j]$, so it is a linear combination of the elements of $\phi _i[j]$. The control signal $u_p(jh)$ equals $\hat{\sqcap }_p[0]$ plus a linear combination of the elements of $\sqcap [1] = H \nu [1] + J \hat{y} (0)$; but $\nu [1]$ lies in $\phi _p [j]$ and $\hat{y} (0)$ is a linear combination of the elements of $\phi _p [j]$, so $\sqcap [1]$ is a linear combination of the elements of $\phi _p[j]$.

Using the fact that $u(t) = u(jh)$ for $t\in [jh,(j+1)h)$, $j\in \mathbf{Z}^+$, we conclude that for $j=0,1,\ldots ,q-1$, the control signal $u_i(jh)$ is a linear combination of the elements of $\phi _i[j]$, which means that $Q_i$ and $R_i$ can be defined so that the controller (5) is identical to that of (9)–(14). $\square$

Proof of Lemma 3

Suppose that the controller (9)–(14) is applied to the plant (1), and let $x_0\in \mathbf{R}^n$, $\nu [0]\in \mathbf{R}^\ell$, $\hat{\sqcap }[0]\in \mathbf{R}^m$, $k\in \mathbf{Z}^+$ and $T>0$ be arbitrary. With the estimate $\hat{y}(kT)$ of y(kT) given by (10) we see that

$$\begin{aligned} \hat{y}(kT)-y(kT) = \left[ \begin{array}{c} \hat{y}_{1}(kT)-y_{1}(kT) \\ \hat{y}_{2}(kT)-y_{2}(kT) \\ \vdots \\ \hat{y}_{p-1}(kT)-y_{p-1}(kT) \\ 0 \end{array} \right] . \end{aligned}$$

We can use Lemma 1 to derive the estimation error of each individual quantity $[y_i (kT) ]_j$ for $i\in \{1, 2, \ldots , p-1\}$ and $j\in \{1, 2, \ldots , l_i\}$. More specifically, extending the error bound provided in (8) to the general case, we have

$$\begin{aligned} \left\| \hat{y}(kT)-y(kT) \right\|= & {} {\mathcal {O}}(T^{1-\delta })\max \limits _{\tau \in [kT,kT+2\bar{n}(l-l_p)h]}\left\| x(\tau ) \right\| +{\mathcal {O}}(T^{1-\delta })\left\| \hat{\sqcap }[k] \right\| \nonumber \\&+\, {\mathcal {O}}\left( T^{{\frac{1}{2}}-\delta }\right) \left\| r_{[kT,kT+2\bar{n}(l-l_p)h)} \right\| _2. \end{aligned}$$

(64)

We now obtain a bound on the first term of the RHS: for $t \in [kT, kT + 2 \bar{n}( l - l_p) h )$, we have

$$\begin{aligned} x(t)= & {} {\mathrm {e}}^{A(t-kT)}x(kT)+\int _{kT}^{t}e^{A(t-\tau )}Bu(\tau )\mathrm{d}\tau + \int _{kT}^{t}e^{A(t-\tau )}Er(\tau )\mathrm{d}\tau , \end{aligned}$$

(65)

so it is easy to see that

$$\begin{aligned} \left\| x(t) \right\|= & {} {\mathcal {O}}(1)\left\| x(kT) \right\| + \left( \int _{kT}^t{\mathcal {O}}(1)^2\mathrm{d}\tau \right) ^{\frac{1}{2}}\left\| r_k \right\| _2 \\&+\, \int _{kT}^{t}[ {\mathcal {O}}(1)\left\| \hat{\sqcap }[k] \right\| +{\mathcal {O}}({T^\delta })\left\| x(kT) \right\| ] \mathrm{d}\tau \\= & {} {\mathcal {O}}(1)\left\| x(kT) \right\| +{\mathcal {O}}(T)\left\| \hat{\sqcap }[k] \right\| +{\mathcal {O}}\left( T^{{\frac{1}{2}}}\right) \left\| r_k \right\| _2. \end{aligned}$$

Using this we can simplify (64):

$$\begin{aligned} \left\| \hat{y}(kT)-y(kT) \right\| = {\mathcal O}( T^{1 - \delta } ) \Vert x(kT) \Vert + {\mathcal O}( T^{1 - \delta } ) \Vert \hat{\sqcap } [k] \Vert + {\mathcal {O}}\left( T^{{\frac{1}{2}}-\delta }\right) \left\| r_k \right\| _2 , \end{aligned}$$

(66)

which yields (21).

Now let us find a conservative bound on the signal u(t). Observe that

$$\begin{aligned} \left\| u(t)-\hat{\sqcap }[k] \right\|= & {} {\mathcal {O}}(T^{\delta })\left\| y(kT) \right\| +{\mathcal {O}}(T^{\delta })\left\| \sqcap [k+1] \right\| \\= & {} {\mathcal {O}}(T^{\delta })\left\| x(kT) \right\| +{\mathcal {O}}(T^{\delta })\left\| \nu [k+1] \right\| \\&+\,{\mathcal {O}}(T^{\delta })\left\| \hat{y}(kT) \right\| ,\;t\in [kT, (k+1)T). \end{aligned}$$

Using (66) to bound $\hat{y}(kT)$ and the fact that $\nu [k+1] = \nu [k]+{\mathcal {O}}(T)\nu [k]+{\mathcal {O}}(T)\hat{y}(kT)$, this simplifies to

$$\begin{aligned} \left\| u(t)-\hat{\sqcap }[k] \right\|= & {} {\mathcal {O}}(T^{\delta })\left\| x(kT) \right\| +{\mathcal {O}}(T^{\delta })\left\| \nu [k] \right\| + {\mathcal {O}}(T)\left\| \hat{\sqcap }[k] \right\| \nonumber \\&+\, {\mathcal {O}}\left( T^{{\frac{1}{2}}}\right) \left\| r_k \right\| _2, \; t\in [kT,(k+1)T), \end{aligned}$$

(67)

which yields (19).

Now we return to the state x(t). Using (67) in (65), we can obtain a bound on $\Vert x(t) - x(kT) \Vert$:

$$\begin{aligned} \left\| x(t)-x(kT) \right\|= & {} {\mathcal {O}}(T)\left\| x(kT) \right\| + {\mathcal {O}}(T) \left\| \hat{\sqcap }[k] \right\| + {\mathcal {O}}(T^{1+\delta })\left\| \nu [k] \right\| \nonumber \\&+\, {\mathcal {O}}\left( T^{{\frac{1}{2}}}\right) \left\| r_k \right\| _2 \end{aligned}$$

(68)

for $t \in [kT, (k+1) T ) ,$ which yields Eq. (18).

We now perform a similar analysis as above to obtain a bound on the quantity $\hat{\sqcap }[k+1]-\sqcap [k+1]$. We first use Lemma 1 to obtain the estimation error of each element of $\hat{\sqcap }_i[k+1]_j$ for $i\in \{1,2,\ldots ,p-1\}$ and $j\in \{1,2,\ldots ,m_i\}$. More specifically, we use the bound in (8) with a minor change reflecting the fact that we are now probing with $T^\delta \bar{w}_p\sqcap _i[k+1]_j$ rather than $T^\delta \bar{w}_i[y_i(kT)]_j$. So (8) becomes (for $i=j=1$):

$$\begin{aligned}&{\mathcal {O}}(T^{1-\delta })(\left\| x(kT+(2\bar{n}(l-l_p)+1)h) \right\| +\left\| \hat{\sqcap }[k] \right\| )\\&\quad + \,{\mathcal {O}}(T)|\sqcap _1[k+1]_1| + {\mathcal {O}}\left( T^{{\frac{1}{2}}-\delta }\right) \left\| r_k \right\| _2. \end{aligned}$$

As a result, we end up with

$$\begin{aligned} \left\| \hat{\sqcap }[k+1]-\sqcap [k+1] \right\|= & {} {\mathcal {O}}(T^{1-\delta })\left\| \hat{\sqcap }[k] \right\| \nonumber \\&+\, {\mathcal {O}}(T^{1-\delta })\max \limits _{\tau \in [kT+(2\bar{n}(l-l_p)+1)h),(k+1)T]}\left\| x(\tau ) \right\| \nonumber \\&+\,{\mathcal {O}}(T)\left\| \sqcap [k+1] \right\| +{\mathcal {O}}\left( T^{{\frac{1}{2}}-\delta }\right) \left\| r_k \right\| _2. \end{aligned}$$

(69)

But it is easy to see that

$$\begin{aligned} \sqcap [k+1] = {\mathcal {O}}(1)v[k]+{\mathcal {O}}(1)\hat{y}(kT). \end{aligned}$$

After we use the bound on $\hat{y}(kT)$ given by (66) to simplify this, we substitute the resulting expression for $\sqcap [k+1]$ into (69) and use (68) to obtain a bound on $\max \limits _{\tau \in [kT,(k+1)T]}\left\| x(\tau ) \right\|$, yielding

$$\begin{aligned} \left\| \hat{\sqcap }[k+1]-\sqcap [k+1] \right\|= & {} {\mathcal {O}}(T^{1-\delta })\left\| x(kT) \right\| + {\mathcal {O}}(T^{1-\delta }) \left\| \hat{\sqcap }[k] \right\| \\&+\,{\mathcal {O}}(T)\left\| \nu [k] \right\| + {\mathcal {O}}\left( T^{{\frac{1}{2}}-\delta }\right) \left\| r_k \right\| _2, \end{aligned}$$

which yields (20). $\square$