Moving Horizon Estimation

Abstract

Nearly every model predictive control (MPC) algorithm is premised on knowledge of the system’s state. As a result, state estimation is vital to good MPC performance. Moving horizon estimation (MHE) is an optimization-based state estimation algorithm. Similar to MPC, it relies on the minimization of a sum of stage costs subject to a dynamic model. Unlike MPC, however, conditions under which MHE is robustly stable have been slow to emerge. Recently, several results have appeared about the robust stability of MHE. We generalize the result on the robust stability of MHE without a max-term presented in Müller (Automatica 79:306–314, 2017). using assumptions inspired by the result in Hu (Robust stability of optimization-based state estimation under bounded disturbances. ArXiv e-prints, 2017). Furthermore, we show that all systems that are covered by the assumptions used in those previous works satisfy a certain form of exponential incremental input/output-to-state stability.

Appendix

Proof (Proposition 1)

Without loss of generality, assume that x ₀ = 0 and that V (0) = 0. Because V (⋅ ) is Lipschitz continuous at the origin, there exists some δ > 0 and L > 0 such that if $\left \vert x\right \vert \leq \delta$ we have that $\left \vert V (x)\right \vert \leq L\left \vert x\right \vert$. Let $(\overline{s}(n))$ be a strictly increasing and unbounded sequence such that $\overline{s}(n)>\delta$ for all $n \in \mathbb{I}_{\geq 0}$. Define a sequence $(\tilde{M}(n))$ such that

$$\displaystyle{ \tilde{M}(n)\,:=\,\sup _{x\in \mathcal{C}}\left \vert V (x)\right \vert \quad \text{ subject to }\left \vert x\right \vert \leq \overline{s}(n) }$$

for all $n \in \mathbb{I}_{\geq 0}$. We have that $\tilde{M}(n)$ is finite for all $n \in \mathbb{I}_{\geq 0}$ because V (⋅ ) is locally bounded. Define another sequence $(M(n))$ such that

$$\displaystyle{ M(n)\,:=\,\max (\tilde{M}(n),L\delta ) }$$

Note that (M(n)) is a nondecreasing sequence. Now define a piecewise linear function

$$\displaystyle{ \tilde{\alpha }(s)\,:=\,\left \{\begin{array}{@{}l@{\quad }l@{}} Ls \quad &\text{if }s \in [0,\delta /2] \\ \frac{L\delta } {2} + (M(0) - L\delta /2)\frac{s-\delta /2} {\delta /2} \quad &\text{if }s \in (\delta /2,\delta ] \\ M(0) + (M(1) - M(0)) \frac{s-\delta } {\overline{s}(0)-\delta } \quad &\text{if }s \in (\delta,\overline{s}(0)] \\ M(n) + (M(n + 1) - M(n)) \frac{s-\overline{s}(n-1)} {\overline{s}(n)-\overline{s}(n-1)}\quad &\text{if }s \in (\overline{s}(n - 1),\overline{s}(n)] \end{array} \right. }$$

The function $\tilde{\alpha }(s)$ is continuous, nondecreasing, and $\tilde{\alpha }(0) = 0$. Furthermore, because it is piecewise-linear, it is locally Lipschitz. Because $\tilde{\alpha }(\overline{s}(n - 1)) = M(n)$, we also have that

$$\displaystyle{ \left \vert V (x)\right \vert \leq M(n) \leq \tilde{\alpha } (\left \vert x\right \vert )\quad \text{if }\left \vert x\right \vert \in (\overline{s}(n - 1),\overline{s}(n)] }$$

We have a similar bound for $\left \vert x\right \vert \in (\delta,\overline{s}(0)]$. Finally, from the Lipschitz bound, we have that $\left \vert V (x)\right \vert \leq \tilde{\alpha } (\left \vert x\right \vert )$ if x ≤ δ and thus for all $x \in \mathcal{ C}$. Finally, let $\alpha (s)\,:=\,s +\tilde{\alpha } (s)$. We have that α(⋅ ) is strictly increasing, continuous, zero at the origin, and asymptotically unbounded. Thus $\alpha (\cdot ) \in \mathcal{ K}_{\infty }$. Furthermore, α(⋅ ) is piecewise-linear and thus locally Lipschitz, and is therefore the required bound.

Proof (Proposition 2)

We prove this proposition by showing first that Statement 1 implies Statement 2, next that Statement 2 implies Statement 3, and then finally that Statement 3 implies Statement 4. Because Statement 4 is a restatement of Statement 1 with a particular form of the $\mathcal{L}$ function, the proof is then complete.

Proof (Statement 1 Implies Statement 2 (adapted from [1], Lemma 6))

Fix η ∈ (0, 1) and $\overline{s}> 0$. We seek to find T such that σ _x(s)ϕ(T) ≤ ηs for all $s \leq \overline{s}$. This condition is equivalent to ϕ(T) ≤ ηs∕σ _x(s) for all $s \in \left (0,\overline{s}\right ]$ and ϕ(T) ≤ ηlim_s↓0 s∕σ _x(s). Because σ _x(s) is Lipschitz continuous at zero, there exists some L > 0 and δ > 0 such that σ _x(s) ≤ Ls for 0 ≤ s ≤ δ. Thus, we have that s∕σ _x(s) ≥ s∕(sL) = 1∕L > 0 for s ≤ δ. Furthermore, s∕σ _x(s) > 0 and s∕σ _x(s) is continuous for all s > 0. Thus we have that $\inf _{s\leq \overline{s}}s/\sigma _{x}(s)\,:=\,\zeta> 0$. Finally, we require T such that ϕ(T) ≤ ηζ ≤ s∕σ _x(s) for $s \leq \overline{s}$. Because both η, ζ > 0 and $\phi (\cdot ) \in \mathcal{ L}$, there exists a T that fulfills this condition. Finally, because σ _x(s) is Lipschitz at the origin, we have that β(s, 0) is Lipschitz at s = 0.

Remark 3

Because the only place where s∕σ _x(s) might equal zero is at s = 0, this proof also implies that if there exists a single $\overline{s}> 0$, η ∈ (0, 1), and T > 0 such that σ _x(s)ϕ(T) ≤ ηs for all $s \in [0,\overline{s}]$, then we can find such a T for every $\overline{s}> 0$.

Proof (Statement 2 Implies Statement 3)

For brevity, we define $x_{1}(k) - x_{2}(k)\,:=\,\varDelta x(k)$, $\mathbf{w}_{1} -\mathbf{w}_{2}\,:=\,\varDelta \mathbf{w}$, and $\mathbf{y}_{1} -\mathbf{y}_{2}\,:=\,\varDelta \mathbf{y}$. Fix $\overline{s}> 0$ and choose T such that β(s, T) ≤ ηs for all $s \in [0,\overline{s}]$. First, we prove by induction in n that

$$\displaystyle{ \left \vert \varDelta x(k + nT)\right \vert \leq \eta ^{n}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k+nT-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+nT-1}) }$$

(4)

for all k ≥ 0 and n ≥ 1 if $\varDelta x(0) \leq \overline{s}$.

Base Case Suppose that $\varDelta x(0) \leq \overline{s}$. Then we have that

$$\displaystyle\begin{array}{rcl} \left \vert \varDelta x(k + T)\right \vert & \leq & \beta (\varDelta x(0),k + T) \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k+T-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+T-1}) {}\\ & \leq & \beta (\varDelta x(0),T) \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k+T-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+T-1}) {}\\ & \leq & \eta \left \vert \varDelta x(0)\right \vert \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k+T-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+T-1}) {}\\ \end{array}$$

by the fact that β(⋅ ) is nonincreasing in its second argument and by Statement 2.

Inductive Case Suppose that (4) holds for some $n \in \mathbb{I}_{\geq 0}$. We have that

$$\displaystyle{ \left \vert \varDelta x(k + nT)\right \vert \leq \eta ^{n}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k+nT-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+nT-1}) }$$

for all k ≥ 0. Suppose further that

$$\displaystyle{ \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k+nT-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+nT-1})> \overline{s} }$$

We then apply the original i-IOSS bound to obtain

$$\displaystyle\begin{array}{rcl} \left \vert \varDelta x(k + (n + 1)T)\right \vert \leq \beta (\left \vert \varDelta x(0)\right \vert,k + (n + 1)T)& \oplus & \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k+(n+1)T-1}) {}\\ & \oplus & \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+(n+1)T-1}) {}\\ \end{array}$$

Furthermore, we have that

$$\displaystyle{ \beta (\left \vert \varDelta x(0)\right \vert,k + nT) \leq \beta (\left \vert \varDelta x(0)\right \vert,T) \leq \eta \left \vert \varDelta x(0)\right \vert \leq \eta \overline{s} <\overline{s} }$$

Thus we have that the $\mathcal{KL}$ function bound is unnecessary, and therefore we have that

$$\displaystyle\begin{array}{rcl} \left \vert \varDelta x(k + (n + 1)T)\right \vert & \leq & \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k+(n+1)T-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+(n+1)T-1}) {}\\ & \leq & \eta ^{n+1}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k+(n+1)T-1}) {}\\ & & \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+(n+1)T-1}) {}\\ \end{array}$$

trivially. Now suppose that

$$\displaystyle{ \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k+nT-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+nT-1}) \leq \overline{s} }$$

Because we have that $\eta ^{n}\left \vert \varDelta x(0)\right \vert \leq \overline{s}$, we have that $\left \vert \varDelta x(k + nT)\right \vert \leq \overline{s}$ so we can apply the i-IOSS bound from $\varDelta x(k + nT)$ to obtain

$$\displaystyle\begin{array}{rcl} \left \vert \varDelta x(k + (n + 1)T)\right \vert & \leq & \beta (\left \vert \varDelta x(k + nT)\right \vert,T) \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{k+nT:k+(n+1)T-1}) {}\\ & & \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{k+nT:k+(n+1)T-1}) {}\\ & \leq & \eta \left \vert \varDelta x(k + nT)\right \vert \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{k+nT:k+(n+1)T-1}) {}\\ & & \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{k+nT:k+(n+1)T-1}) {}\\ & \leq & \eta ^{n+1}\left \vert \varDelta x(0)\right \vert \oplus \eta \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k+nT-1}) {}\\ & & \oplus \eta \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+nT-1}) \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{k+nT:k+(n+1)T-1}) {}\\ & & \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{k+nT:k+(n+1)T-1}) {}\\ & \leq & \eta ^{n+1}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k+(n+1)T-1}) {}\\ & & \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+(n+1)T-1}) {}\\ \end{array}$$

Thus we have that (4) for n implies (4) for n + 1, completing the proof by induction.

Now we bound $\left \vert \varDelta x(k)\right \vert$ for $k \in \mathbb{I}_{0:T-1}$. We have that

$$\displaystyle\begin{array}{rcl} \left \vert \varDelta x(k)\right \vert & \leq & \beta (\left \vert \varDelta x(0)\right \vert,k) \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) {}\\ & \leq & \beta (\left \vert \varDelta x(0)\right \vert,0) \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) {}\\ \end{array}$$

because β(⋅ ) is nonincreasing in its second argument. Because β(s, 0) is Lipschitz at s = 0, by Proposition 1 there exists some locally Lipschitz function $\bar{\alpha }(\cdot ) \in \mathcal{ K}_{\infty }$ such that $\beta (s,0) \leq \bar{\alpha } (s)$ for all $s \in \mathbb{R}_{\geq 0}$. Because $\bar{\alpha }(\cdot )$ is locally Lipschitz and β(s, 0) ≥ s, there exists some K ≥ 1 such that for all $s \in [0,\overline{s}]$, we have that $\bar{\alpha }(s) \leq Ks$. Thus we have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq K\left \vert \varDelta x(0)\right \vert \oplus \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

(5)

for all $\varDelta x(0)$ such that $\left \vert \varDelta x(0)\right \vert \leq \overline{s}$. Because we have that (4) holds for n ≥ 1 and k ≥ 0 and that (5) holds for n = 0 and k ≥ 0, we can combine these equations to write

$$\displaystyle{ \left \vert \varDelta x(k + nT)\right \vert \leq K\eta ^{n}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k+nT-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k+nT-1}) }$$

for all $\varDelta x$ such that $\left \vert \varDelta x\right \vert \leq \overline{s}$. We can then eliminate the index n using the floor function $\lfloor \cdot \rfloor$ to obtain the bound

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq K\eta ^{\lfloor k/T\rfloor }\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

Note that $\eta ^{\lfloor k/T\rfloor }\leq (1/\eta )\eta ^{k/T}$, so we have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq (K/\eta )\eta ^{k/T}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

Finally, let C : = K∕η and λ : = η ^1∕T. We have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq C\lambda ^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for all $\varDelta x(0)$ such that $\left \vert \varDelta x(0)\right \vert \leq \overline{s}$.

Proof (Statement 3 Implies 4)

We prove this statement in two steps. We first show that although both λ and C in Statement 3 depend on $\overline{s}$, the dependence of λ on $\overline{s}$ can be removed by increasing $C(\overline{s})$. We can remove this dependence because the term dependent on the initial conditions decays to be less than some smaller $\overline{s}$ within finite time. We then turn the function $C(\overline{s})s$ into a $\mathcal{K}_{\infty }$ function.

First, let $(\overline{s}(n))$ be a strictly increasing and unbounded sequence such that $\overline{s}(n)> 0$ for all $n \in \mathbb{I}_{1:\infty }$. By Statement 3, there exists sequences $(C(n))$ and $(\lambda (n))$ such that C(n) ≥ 1 and λ(n) ∈ (0, 1) and that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq C(n)\lambda (n)^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for all $\varDelta x(0)$ such that $\left \vert \varDelta x(0)\right \vert \leq \overline{s}(n)$ for all $n \in \mathbb{I}_{1:\infty }$. Without loss of generality, assume that (C(n)) and that (λ(n)) are nondecreasing. Let $\underline{\lambda }\,:=\,\lambda (1)$. We prove by induction that there exists a sequence $(\tilde{C}(n))$ such that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq \tilde{ C}(n)\underline{\lambda }^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for all $\varDelta x(0)$ such that $\left \vert \varDelta x(0)\right \vert \leq \overline{s}(n)$ and all $n \in \mathbb{I}_{1:\infty }$. The base case is given by Statement 3 for $\overline{s}(1)$.

Inductive Case Suppose for some n we have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq \tilde{ C}(n)\underline{\lambda }^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for all $\varDelta x(0)$ such that $\left \vert \varDelta x(0)\right \vert \leq \overline{s}(n)$. We also have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq C(n + 1)\lambda (n + 1)^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for all $\varDelta x(0)$ such that $\left \vert \varDelta x(0)\right \vert \leq \overline{s}(n + 1)$. Let

$$\displaystyle{ N\,:=\,\lceil \log _{\lambda (n+1)}\left ( \frac{\overline{s}(n)} {\overline{s}(n + 1)C(n + 1)}\right )\rceil }$$

in which $\lceil \cdot \rceil$ is the ceiling function. For all $\varDelta x(0)$ such that $\left \vert \varDelta x(0)\right \vert \leq \overline{s}(n + 1)$, we have that $C(n + 1)\lambda ^{N}\left \vert \varDelta x(0)\right \vert \leq \overline{s}(n)$. Suppose that $\gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) <C(n + 1)\lambda ^{N}\left \vert \varDelta x(0)\right \vert$. Then we have that $\left \vert \varDelta x(k)\right \vert \leq C(n + 1)\lambda ^{N}\left \vert \varDelta x(0)\right \vert$ alone, and thus we can apply the bound for all $\varDelta x(0) \leq \overline{s}(n)$ to obtain for all k ≥ N that

$$\displaystyle\begin{array}{rcl} \left \vert \varDelta x(k)\right \vert & \leq & \tilde{C}(n)\underline{\lambda }^{k-N}\left \vert \varDelta x(N)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{N:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{N:k-1}) {}\\ & \leq & \tilde{C}(n)\underline{\lambda }^{k-N}C(n + 1)\lambda (n + 1)^{N}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) {}\\ & & \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) {}\\ & =& \frac{\tilde{C}(n)C(n + 1)\lambda (n + 1)^{N}} {\underline{\lambda }^{N}} \underline{\lambda }^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) {}\\ & & \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) {}\\ \end{array}$$

Define

$$\displaystyle{ \tilde{C}(n + 1)\,:=\,\tilde{C}(n)C(n + 1)\left (\frac{\lambda (n + 1)} {\underline{\lambda }} \right )^{N} }$$

and we have the required bound. Now suppose that $\gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) \geq C(n + 1)\lambda ^{N}\left \vert \varDelta x(0)\right \vert$. Then, because the $\mathcal{KL}$ function is nonincreasing, we have that

$$\displaystyle\begin{array}{rcl} \left \vert \varDelta x(k)\right \vert & \leq & \gamma _{w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) {}\\ & \leq & \tilde{C}(n + 1)\underline{\lambda }^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) {}\\ \end{array}$$

which is the required bound. Both of these bounds apply for k ≥ N. For k < N, note that

$$\displaystyle\begin{array}{rcl} C(n + 1)\lambda (n + 1)^{k}& =& \frac{C(n + 1)\lambda (n + 1)^{k}} {\underline{\lambda }^{k}} \underline{\lambda }^{k} {}\\ & \leq & \frac{C(n + 1)\lambda (n + 1)^{N}} {\underline{\lambda }^{N}} \underline{\lambda }^{k} {}\\ & \leq & \tilde{C}(n + 1)\underline{\lambda }^{k} {}\\ \end{array}$$

because $\underline{\lambda }\leq \lambda (n + 1)$. Therefore we also have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq \tilde{ C}(n + 1)\underline{\lambda }^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for k < N and thus for all $k \in \mathbb{I}_{\geq 0}$. Thus the statement is proven for n + 1, and the first part of the proof is complete.

Next, define

$$\displaystyle{ \tilde{\alpha }(s)\,:=\,\left \{\begin{array}{@{}l@{\quad }l@{}} C(1)s\quad &\text{if }s \in [0,\overline{s}(1)]\\ C(n)s\quad &\text{if } s \in (\overline{s } (n - 1),\overline{s}(n)]\text{ for }n \geq 2 \end{array} \right. }$$

Note that this function is Lipschitz continuous at the origin and locally bounded. Furthermore, note that by construction we have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq \tilde{\alpha } (\left \vert \varDelta x(0)\right \vert )\underline{\lambda }^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for all $\varDelta x(0)$. By Proposition 1, there exists a locally Lipschitz function $\alpha (\cdot ) \in \mathcal{ K}_{\infty }$ such that $\tilde{\alpha }(s) \leq \alpha (s)$ for all $s \in \mathbb{R}_{\geq 0}$. Thus we have that

$$\displaystyle{ \left \vert \varDelta x(k)\right \vert \leq \alpha (\left \vert \varDelta x(0)\right \vert )\underline{\lambda }^{k}\left \vert \varDelta x(0)\right \vert \oplus \gamma _{ w}(\|\varDelta \mathbf{w}\|_{0:k-1}) \oplus \gamma _{y}(\|\varDelta \mathbf{y}\|_{0:k-1}) }$$

for all $\varDelta x(0)$, and so the result is established.

Proof (Proposition 3)

By Assumptions 2 and 3, the MHE problem has a solution $(\hat{x}(0),\widehat{\mathbf{d}})$. Denote the estimated state at time k within the MHE problem as $\hat{x}(k)$. By Assumption 3, we have that

$$\displaystyle{ \rho \underline{\gamma }_{p}(\left \vert \hat{e}_{p}\right \vert ) \oplus \underline{\gamma }_{s}(\|\widehat{\mathbf{d}}\|_{0:N-1}) \leq \rho \underline{\gamma }_{p}(\left \vert \hat{e}_{p}\right \vert ) +\underline{\gamma } _{s}(\|\widehat{\mathbf{d}}\|_{0:N-1}) \leq V _{N}(\hat{x}(0),\widehat{\mathbf{d}},\overline{x}) }$$

in which $\hat{e}_{p}\,:=\,\overline{x} -\hat{ x}(0)$. Furthermore, by optimality, we have that

$$\displaystyle\begin{array}{rcl} V _{N}(\hat{x}(0),\widehat{\mathbf{d}},\overline{x})& \leq & V _{N}(x(0),\mathbf{d},\overline{x}) {}\\ & \leq & \rho \overline{\gamma }_{p}(\left \vert e_{p}\right \vert ) + N\overline{\gamma }_{s}(\|\mathbf{d}\|_{0:N-1}) {}\\ & \leq & 2\rho \overline{\gamma }_{p}(\left \vert e_{p}\right \vert ) \oplus 2N\overline{\gamma }_{s}(\|\mathbf{d}\|_{0:N-1}) {}\\ \end{array}$$

Combining these bounds and rearranging, we obtain the following bounds

$$\displaystyle\begin{array}{rcl} \left \vert \hat{e}_{p}\right \vert & \leq & \underline{\gamma }_{p}^{-1}(2\overline{\gamma }_{ p}(\left \vert e_{p}\right \vert ) \oplus (2N/\rho )\overline{\gamma }_{s}(\|\mathbf{d}\|_{0:N-1})) \\ & =& \underline{\gamma }_{p}^{-1}(2\overline{\gamma }_{ p}(\left \vert e_{p}\right \vert )) \oplus \underline{\gamma }_{p}^{-1}((2N/\rho )\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1})){}\end{array}$$

(6)

$$\displaystyle\begin{array}{rcl} \|\widehat{\mathbf{d}}\|& \leq & \underline{\gamma }_{s}^{-1}(2\rho \overline{\gamma }_{ p}(\left \vert e_{p}\right \vert ) \oplus 2N\overline{\gamma }_{s}(\|\mathbf{d}\|_{0:N-1})) \\ & =& \underline{\gamma }_{s}^{-1}(2\rho \overline{\gamma }_{ p}(\left \vert e_{p}\right \vert )) \oplus \underline{\gamma }_{s}^{-1}(2N\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1})){}\end{array}$$

(7)

From the system’s i-IOSS bound, we have that

$$\displaystyle\begin{array}{rcl} \left \vert e(k)\right \vert & \leq & \beta (\left \vert e(0)\right \vert,k) \oplus \gamma _{d}(\|\mathbf{d} -\widehat{\mathbf{d}}\|_{0:N-1}) {}\\ & =& \beta (\left \vert \hat{x}(0) -\overline{x} + \overline{x} - x(0)\right \vert,k) \oplus \gamma _{d}(\|\mathbf{d} -\widehat{\mathbf{d}}\|_{0:N-1}) {}\\ & \leq & \beta (\left \vert e_{p}\right \vert + \left \vert \hat{e}_{p}\right \vert,k) \oplus \gamma _{d}(\|\mathbf{d}\|_{0:N-1} +\|\widehat{ \mathbf{d}}\|_{0:N-1}) {}\\ & \leq & \beta (2\left \vert e_{p}\right \vert \oplus 2\left \vert \hat{e}_{p}\right \vert,k) \oplus \gamma _{d}(2\|\mathbf{d}\|_{0:N-1} \oplus 2\|\widehat{\mathbf{d}}\|_{0:N-1}) {}\\ & & {}\\ \end{array}$$

We next substitute (6) and (7) into this expression.

$$\displaystyle\begin{array}{rcl} \left \vert e(k)\right \vert & \leq & \beta (2\left \vert e_{p}\right \vert \oplus 2\underline{\gamma }_{p}^{-1}(2\overline{\gamma }_{ p}(\left \vert e_{p}\right \vert )) \oplus 2\underline{\gamma }_{p}^{-1}((2N/\rho )\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1})),k) {}\\ & & \oplus \gamma _{d}(2\|\mathbf{d}\|_{0:N-1} \oplus 2\underline{\gamma }_{s}^{-1}(2\rho \overline{\gamma }_{ p}(\left \vert e_{p}\right \vert )) \oplus 2\underline{\gamma }_{s}^{-1}(2N\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1}))) {}\\ & =& \beta (2\left \vert e_{p}\right \vert,k) \oplus \beta (2\underline{\gamma }_{p}^{-1}(2\overline{\gamma }_{ p}(\left \vert e_{p}\right \vert )),k) \oplus \gamma _{d}(2\underline{\gamma }_{s}^{-1}(2\rho \overline{\gamma }_{ p}(\left \vert e_{p}\right \vert ))) {}\\ & & \oplus \beta (2\underline{\gamma }_{p}^{-1}((2N/\rho )\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1})),k) \oplus \gamma _{d}(2\|\mathbf{d}\|_{0:N-1}) {}\\ & & \oplus \gamma _{d}(\underline{\gamma }_{s}^{-1}(2N\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1}))) {}\\ \end{array}$$

Note that because $\underline{\gamma }_{p}(s) \leq \overline{\gamma }_{p}(s) \leq 2\overline{\gamma }_{p}(s)$, we have that $s \leq \underline{\gamma }_{p}^{-1}(2\overline{\gamma }_{p}(s))$. A similar argument follows for $\underline{\gamma }_{s}(\cdot )$ and $\overline{\gamma }_{s}(\cdot )$. Thus we have that the term $\beta (2\left \vert e_{p}\right \vert,k) \leq \beta (2\underline{\gamma }_{p}^{-1}(2\overline{\gamma }_{p}(\left \vert e_{p}\right \vert )),k)$ and the term $\gamma _{d}(2\|\mathbf{d}\|_{0:N-1}) \leq \gamma _{d}(\underline{\gamma }_{s}^{-1}(2N\overline{\gamma }_{s}(\|\mathbf{d}\|_{0:N-1})))$, so we can eliminate them from the maximization. Thus we have

$$\displaystyle\begin{array}{rcl} \left \vert e(k)\right \vert & \leq & \beta (2\underline{\gamma }_{p}^{-1}(2\overline{\gamma }_{ p}(\left \vert e_{p}\right \vert )),k) \oplus \gamma _{d}(2\underline{\gamma }_{s}^{-1}(2\rho \overline{\gamma }_{ p}(\left \vert e_{p}\right \vert ))) {}\\ & & \oplus \beta (2\underline{\gamma }_{p}^{-1}((2N/\rho )\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1})),k) \oplus \gamma _{d}(\underline{\gamma }_{s}^{-1}(2N\overline{\gamma }_{ s}(\|\mathbf{d}\|_{0:N-1}))) {}\\ \end{array}$$

for all k ≥ 0, which is the desired result.