Proof. Similar to the time-invariant case, select a non-negative function , its difference over a single time step for any *k ⩾ N* is

- (13)

Substitution of (12) and (33) into (14), it is obtained that

- (14)

To proceed further, note that from the delayed form of (12), it is possible to obtain

- (15)

multiplying both sides with *x*_{k − N + 1} and by using the fact that , (15) becomes

- (16)

Substitution of (11) and (16) into (14) and using the fact that , it is obtained that

- (17)

Thus, Δ*V*_{k} is non-increasing, implying that and are bounded and that there exists positive constants and such that and .

Applying (17) repeatedly for any *k* ∈ [*pN*,(*p* + 1)*N*] and noticing *k*_{0} = *k* − *pN*, it is obtained that

- (18)

Because *k*_{0} ∈ [0,*N*), and

when *k* ∞ , according to (14)

- (19)

Consider that *V*_{k} is non-negative, and is finite in the interval [0,*N*); thus, according to the convergence theorem of the sum of series, it is obtained that

- (20)

With the use of (20) and the sector condition on *ξ*_{k}, the Key Technical Lemma guarantees that *ξ*_{k} is bounded and consequently *α*_{k}*x*_{k − N + 1} 0 as *k* ∞ . Implying that there must exist a positive constant *ρ* such that max_{j ∈ [0,k]}{*α*_{j}*x*_{j − N + 1}}*⩽ ρ*. Then according to the definition of *α*_{k} in (8),

- (21)

Following the analysis in [15] and the bound on *ξ*_{k}, the maximum tracking error bound is found to be

- (22)

implying that . Because from (20), it is guaranteed that *α*_{k}*x*_{k − N + 1} 0 as *k* ∞ then the error bound as *k* ∞ is simply

- (23)

and *λ* can be selected small enough to ensure that the steady state error lim_{k ∞ } | *x*_{k + 1} | is as small as possible.□