Ulam stability of a successive approximation equation

Abstract

Let T be a bounded linear operator acting on a Banach space X. We obtain some results on Ulam stability for the linear difference equation $x_{n+1}=Tx_n+a_n$ associated with an iterative process for the linear equation $x-Tx=y.$ As applications, we get some stability results for the case when X is a finite-dimensional space and for the case when T is a Fredholm operator.

1 Introduction

One of the most important methods of determining solutions of an equation is the successive approximation method. It is connected with the existence of a fixed point of an operator and with a recurrent sequence converging to the fixed point. In this paper, we deal with Ulam stability of the difference equation which defines the sequence of successive approximations. In what follows, let ${\mathbb {K}}$ be one of the fields ${\mathbb {R}}$ of real numbers or ${\mathbb {C}}$ of complex numbers, X a Banach space over ${\mathbb {K}}$, and $T:X\rightarrow X$ a bounded linear operator. By ${\mathbb {N}}=\{0,1,2,\ldots \}$, we denote the set of all nonnegative integers.

Consider the equation:

$$\begin{aligned} x-Tx=y, \end{aligned}$$

(1)

where y is a given element in X. Under appropriate conditions on T, Eq. (1) admits a solution $x^*$ which is the limit of the sequence $(x_n)_{n\ge 0}$ defined by the difference equation:

$$\begin{aligned} x_{n+1}=Tx_n+y,\quad n\in {\mathbb {N}} \end{aligned}$$

(2)

for some $x_0\in X,$ called the sequence of successive approximations. In this connection, see also the early paper [24] by S. Reich.

In what follows, we consider the linear difference equation:

$$\begin{aligned} x_{n+1}=Tx_n+a_n,\quad n\in {\mathbb {N}}, \end{aligned}$$

(3)

where $(a_n)_{n\ge 0}$ is a given sequence in X. We study its Ulam stability which concerns the behavior of the solutions of the equation (3) under perturbations. For various results on difference equations, we refer the reader to [12, 13].

Definition 1.1

Equation (3) is called Ulam stable if there exists $L\ge 0$, such that for every $\varepsilon >0$ and any $(x_n)_{n\ge 0}$ in X satisfying the relation:

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert \le \varepsilon ,\quad n\in {\mathbb {N}}, \end{aligned}$$

(4)

there exists a sequence $(y_n)_{n\ge 0}$ in X, such that:

$$\begin{aligned}&y_{n+1}=Ty_n+a_n,\quad n\in {\mathbb {N}}, \end{aligned}$$

(5)

$$\begin{aligned}&\Vert x_n-y_n\Vert \le L\varepsilon ,\quad n\in {\mathbb {N}}. \end{aligned}$$

(6)

A sequence $(x_n)_{n\ge 0}$ which satisfies (4) for some $\varepsilon >0$ is called approximate solution of Eq. (3).

In other words, we say that Eq. (3) is Ulam stable if, for every approximate solution of it, there exists an exact solution close to it. The number L from (6) is called an Ulam constant of Eq. (3). In what follows, we will denote by $L_{R}$ the infimum of all Ulam constants of (3). If $L_{R}$ is an Ulam constant for (3), then we call it the best Ulam constant or the Ulam constant of Eq. (3). In general, the infimum of all Ulam constants of an equation is not an Ulam constant of that equation (see [7, 22]). If, in the above definition, the number $\varepsilon$ is replaced by a sequence of positive numbers $(\varepsilon _n)_{n\ge 0}$, we get the notion of generalized stability in Ulam sense. Let $S^{+}$ be the set of all sequences of nonnegative numbers, ${\mathcal {E}}\subseteq S^{+}$, and $U:{\mathcal {E}}\rightarrow S^{+}$ an operator.

Definition 1.2

We say that Eq. (3) is $({\mathcal {E}}$, U)-stable or generalized stable in Ulam sense, if for every sequence $\varepsilon =(\varepsilon _n)_{n\ge 0}$ in ${\mathcal {E}}$ and every sequence $(x_n)_{n\ge 0}$ in X satisfying:

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert \le \varepsilon _n, n\in {\mathbb {N}}, \end{aligned}$$

there exists a sequence $(y_n)_{n\ge 0}$ in X, such that:

$$\begin{aligned}&y_{n+1}=Ty_n+a_n, \quad n\in \mathbb {N,} \\&\Vert x_n-y_n\Vert \le (U \varepsilon )_n, \quad n\in {\mathbb {N}}. \end{aligned}$$

The problem of stability of functional equations was formulated by Ulam [25] in 1940 for the equation of the homomorphism of a metric group. The first answer to Ulam’s problem was given, a year later, by Hyers [14] for the Cauchy functional equation in Banach spaces. Since then, the topic was intensively studied by many authors, we can merely mention here a few papers on Ulam stability of functional equations as [2, 7, 15, 19]. Recall also the results obtained in [6, 16, 18] on Ulam stability of some second-order functional equations connected with Fibonacci and Lucas sequences.

Some results on Ulam stability for the linear difference equations in Banach spaces were obtained by Brzdek, Popa, and Xu in [8,9,10, 20]. Buse et al. [5, 11] proved that a discrete system $X_{n+1}=A X_n,$$n\in {\mathbb {N}},$ where A is a $m\times m$ complex matrix, is Ulam stable if and only if A possesses a discrete dichotomy. Recently, Baias and Popa obtained results on Ulam stability of linear difference equations of order one and two, and determined the best Ulam constant in [3, 4]. Popa and Rasa obtained an explicit representation of the best Ulam constant of some classical operators in approximation theory in [21, 23].

2 Main results

Recall first some classical results which will be used in the sequel. Let $T:X\rightarrow X$ be a linear and bounded operator and consider the geometric series:

$$\begin{aligned} \sum \limits _{n=0}^{\infty }T^n=I+T+T^2+\ldots \end{aligned}$$

(7)

Theorem 2.1

[17, Théorèm 1, Section 4.2]. For any linear and bounded operator $T:X\rightarrow X$, there exists:

$$\begin{aligned} lim\root n \of {\Vert T^n\Vert }=\rho . \end{aligned}$$

(8)

Moreover, the series (7) is absolutely convergent for $\rho <1$ and divergent for $\rho >1.$

Theorem 2.2

[17, Corollaire, Section 4.2]. The series (7) is absolutely convergent if and only if there exists $p\in {\mathbb {N}}$, such that:

$$\begin{aligned} \Vert T^p\Vert <1. \end{aligned}$$

(9)

In what follows, we present some results on Ulam stability and generalized Ulam stability for Eq. (3). The following lemma is useful in the sequel.

Lemma 2.3

If $(x_n)_{n\ge 0}$ satisfies Eq. (3), then:

$$\begin{aligned} x_n=T^nx_0+\sum \limits _{k=1}^{n} T^{n-k}a_{k-1},\quad n\ge 1. \end{aligned}$$

(10)

Proof

Induction on n. $\square$

The first result on generalized Ulam stability of (3) is contained in the next theorem.

Theorem 2.4

Suppose that T is an invertible operator and let $(\varepsilon _n)_{n\ge 0}$ be a sequence of positive numbers, such that the series:

$$\begin{aligned} \sum \limits _{n=1}^{\infty }\Vert T^{-n}\Vert \varepsilon _{n-1} \end{aligned}$$

(11)

is convergent. Then, for every sequence $(x_n)_{n\ge 0}$ in X satisfying:

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert \le \varepsilon _{n},\quad n\in {\mathbb {N}}, \end{aligned}$$

(12)

there exists a sequence $(y_n)_{n\ge 0}$ in X with the properties:

$$\begin{aligned}&y_{n+1}=Ty_n+a_n,\quad n\in {\mathbb {N}}, \end{aligned}$$

(13)

$$\begin{aligned}&\Vert x_n-y_n\Vert \le \sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k},\quad n\in {\mathbb {N}}. \end{aligned}$$

(14)

Moreover, if

$$\begin{aligned} \sup \limits _{n\ge 1}\frac{1}{\varepsilon _{n-1}}\sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k}<\infty , \end{aligned}$$

(15)

then the sequence $(y_n)_{n\ge 1}$ satisfying (13), (14) is unique.

Proof

Existence. Suppose that $(x_n)_{n\ge 0}$ satisfies relation (12) and let:

$$\begin{aligned} x_{n+1}-Tx_n-a_n=b_n,\quad n\ge 0. \end{aligned}$$

Then, $\Vert b_n\Vert \le \varepsilon _n,$$n\ge 0$, and taking into account (10), we get:

$$\begin{aligned} x_n=T^n(x_0+\sum \limits _{k=1}^{n}T^{-k}(a_{k-1}+b_{k-1})),\quad n\ge 1. \end{aligned}$$

Since:

$$\begin{aligned} \Vert T^{-n}b_{n-1}\Vert \le \Vert T^{-n}\Vert \cdot \Vert b_{n-1}\Vert \le \varepsilon _{n-1}\Vert T^{-n}\Vert ,\quad n\ge 1, \end{aligned}$$

it follows that the series:

$$\begin{aligned} \sum \limits _{n=1}^{\infty }T^{-n}b_{n-1} \end{aligned}$$

is convergent, according to the comparison test for series with positive terms. Let

$$\begin{aligned} \sum \limits _{n=1}^{\infty }T^{-n}b_{n-1}=s,\quad s\in X. \end{aligned}$$

Define the sequence $(y_n)_{n\ge 0}$ by the relation:

$$\begin{aligned} y_{n+1}=Ty_n+a_n,\quad n\ge 0,\quad y_0=x_0+s. \end{aligned}$$

Then, in view of Lemma 2.3, it follows:

$$\begin{aligned} y_n=T^n\left( y_0+\sum \limits _{k=1}^{n}T^{-k}a_{k-1}\right) ,\quad n\ge 1. \end{aligned}$$

Consequently:

$$\begin{aligned} x_n-y_n= & {} T^{n}\left( x_0-y_0+\sum \limits _{k=1}^{n}T^{-k}b_{k-1}\right) =T^{n}\left( -s+\sum \limits _{k=1}^{n}T^{-k}b_{k-1}\right) \\= & {} -T^{n}\left( \sum \limits _{k=0}^{\infty }T^{-n-k-1}b_{n+k}\right) =-\sum \limits _{k=0}^{\infty }T^{-k-1}b_{n+k},\quad n\ge 1. \end{aligned}$$

Hence:

$$\begin{aligned} \Vert x_n-y_n\Vert\le & {} \sum \limits _{k=0}^{\infty }\Vert T^{-k-1}b_{n+k}\Vert \le \sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \Vert b_{n+k}\Vert \\\le & {} \sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k},\quad n\in {\mathbb {N}}. \end{aligned}$$

Uniqueness. Suppose that for a sequence $(x_n)_{n\ge 0}$ satisfying (12), there exist two sequences $(y_n)_{n\ge 0},$$(z_n)_{n\ge 0}$ satisfying (13) and (14). Then:

$$\begin{aligned} \Vert y_n-z_n\Vert \le \Vert y_n-x_n\Vert +\Vert x_n-z_n\Vert \le 2\sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k},\quad n\ge 0. \end{aligned}$$

On the other hand, taking account of Lemma 2.3, it follows:

$$\begin{aligned} y_n-z_n=T^n(y_0-z_0) \text{ or } \text{ equivalently } y_0-z_0=T^{-n}(y_n-z_n). \end{aligned}$$

Hence:

$$\begin{aligned} \Vert y_0-z_0\Vert\le & {} \Vert T^{-n}(y_n-z_n)\Vert \le \Vert T^{-n}\Vert \Vert y_n-z_n\Vert \nonumber \\\le & {} 2\Vert T^{-n}\Vert \sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k}\nonumber \\= & {} 2\Vert T^{-n}\Vert \varepsilon _{n-1}\cdot \frac{1}{\varepsilon _{n-1}}\sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k},\quad n\ge 1. \end{aligned}$$

(16)

The convergence of the series (11) implies that:

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert T^{-n}\Vert \varepsilon _{n-1}=0. \end{aligned}$$

Therefore, according to (15) and (16), we get $y_0=z_0,$ and so $y_n=z_n,$ for all $n\in {\mathbb {N}}.$$\square$

Corollary 2.5

Suppose that T is an invertible operator and $(\varepsilon _n)_{n\ge 0}$ is a sequence of positive numbers, such that there exists $q\in (0,1)$ with the property:

$$\begin{aligned} \Vert T^{-1}\Vert \le q\frac{\varepsilon _n}{\varepsilon _{n+1}},\quad n\in {\mathbb {N}}. \end{aligned}$$

(17)

Then, for every sequence $(x_n)_{n\ge 0}$ in X satisfying the relation (12), there exists a sequence $(y_n)_{n\ge 0}$ in X with the properties (13) and:

$$\begin{aligned} \Vert x_n-y_n\Vert \le \frac{q}{1-q}\varepsilon _{n-1},\quad n\ge 1. \end{aligned}$$

Proof

The series $\sum \limits _{n=1}^{\infty }\Vert T^{-n}\Vert \varepsilon _{n-1}$ is convergent. Indeed:

$$\begin{aligned} \limsup \frac{\Vert T^{-n-1}\Vert \varepsilon _n}{\Vert T^{-n}\Vert \varepsilon _{n-1}}\le & {} \limsup \frac{\Vert T^{-n}\Vert \Vert T^{-1}\Vert \varepsilon _n}{\Vert T^{-n}\Vert \varepsilon _{n-1}}\\= & {} \limsup \frac{\varepsilon _n}{\varepsilon _{n-1}}\Vert T^{-1}\Vert \le q<1. \end{aligned}$$

Then, according to Theorem 2.4, for every sequence $(x_n)_{n\ge 0}$ satisfying (12), there exists a sequence $(y_n)_{n\ge 0}$:

$$\begin{aligned} y_{n+1}=Ty_n+a_n,\ n\in {\mathbb {N}}, y_0=x_0+s, \end{aligned}$$

such that

$$\begin{aligned} \Vert x_n-y_n\Vert \le \sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k},\quad n\in {\mathbb {N}}. \end{aligned}$$

Taking account of (17), we get:

$$\begin{aligned} \varepsilon _{n}\Vert T^{-1}\Vert&\le q \varepsilon _{n-1}\\ \varepsilon _{n+1}\Vert T^{-2}\Vert&\le \varepsilon _{n+1}\Vert T^{-1}\Vert \Vert T^{-1}\Vert \le q\varepsilon _n\Vert T^{-1}\Vert \le q^2\varepsilon _{n-1} \\ \ldots \\ \varepsilon _{n+k}\Vert T^{-k-1}\Vert&\le q^{k+1} \varepsilon _{n-1}. \end{aligned}$$

Hence:

$$\begin{aligned} \sum \limits _{k=0}^{\infty }\Vert T^{-k-1}\Vert \varepsilon _{n+k}\le & {} \left( \sum _{k=0}^{\infty }q^{k+1}\right) \varepsilon _{n-1}\\= & {} \frac{q}{1-q}\varepsilon _{n-1},\quad n\ge 1. \end{aligned}$$

The corollary is proved. $\square$

Corollary 2.6

Let T be an invertible operator with $\Vert T^{-p}\Vert <1$ for some $p\in {\mathbb {N}}.$ Then, Eq. (3) is Ulam stable with the Ulam constant:

$$\begin{aligned} L=\sum \limits _{n=1}^{\infty }\Vert T^{-n}\Vert . \end{aligned}$$

Proof

The condition $\Vert T^{-p}\Vert <1,$ leads to the convergence of the series $\sum \limits _{n=1}^{\infty }\Vert T^{-n}\Vert ,$ in view of Theorem 2.2. The conclusion of the corollary follows letting $\varepsilon _n=\varepsilon ,$$n\in {\mathbb {N}}$ and $\sum \nolimits _{n=1}^{\infty }\Vert T^{-n}\Vert =L$ in Theorem 2.4.

Uniqueness holds since for $\varepsilon _n=\varepsilon ,$$n\ge 0,$ the condition (15) is satisfied.

$\square$

Remark 2.7

If, in Corollary 2.6, we take $p=1,$ i.e., $\Vert T^{-1}\Vert <1,$ the conclusion holds with:

$$\begin{aligned} L=\frac{\Vert T^{-1}\Vert }{1-\Vert T^{-1}\Vert }. \end{aligned}$$

Indeed: $\sum \nolimits _{n=1}^{\infty }\Vert T^{-n}\Vert \le \sum \nolimits _{n=1}^{\infty }\Vert T^{-1}\Vert ^{n}=\frac{\Vert T^{-1}\Vert }{1-\Vert T^{-1}\Vert }.$

Similar results can be obtained replacing the condition on the operator $T^{-1}$ with conditions on T in the previous theorems and corollaries.

Theorem 2.8

Let $(\varepsilon _n)_{n\ge 0}$ be a sequence of positive numbers and suppose that there exists $q\in (0,1)$, such that:

$$\begin{aligned} \Vert T\Vert \le q\frac{\varepsilon _{n+1}}{\varepsilon _n},\quad n\in {\mathbb {N}}. \end{aligned}$$

(18)

Then, for every sequence $(x_n)_{n\ge 0}$ in X satisfying:

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert \le \varepsilon _{n},\quad n\in {\mathbb {N}}, \end{aligned}$$

(19)

there exists a sequence $(y_n)_{n\ge 0}$ in X with the properties:

$$\begin{aligned}&y_{n+1}=T y_n+a_n,\quad n\in {\mathbb {N}}, \nonumber \\&\Vert x_n-y_n\Vert \le \frac{1}{1-q}\varepsilon _{n-1},\quad n\ge 1. \end{aligned}$$

(20)

Proof

Let

$$\begin{aligned} x_{n+1}-Tx_n-a_n=b_n,\quad n\in {\mathbb {N}}, \end{aligned}$$

for some sequence $(x_n)_{n\ge 0}$ satisfying (19). Then, $\Vert b_n\Vert \le \varepsilon _n,$$n\in {\mathbb {N}},$ and according to Lemma 2.3, we get:

$$\begin{aligned} x_n=T^nx_0+\sum \limits _{k=1}^{n}T^{n-k}(a_{k-1}+b_{k-1}),\quad n\ge 1. \end{aligned}$$

Consider the sequence $(y_n)_{n\ge 0}$ given by (20) with $y_0=x_0,$ and then:

$$\begin{aligned} y_n=T^{n}x_0+\sum _{k=1}^{n} T^{n-k}a_{k-1}. \end{aligned}$$

(21)

Consequently:

$$\begin{aligned} \Vert x_n-y_n\Vert= & {} \Vert \sum \limits _{k=1}^{n}T^{n-k}b_{k-1}\Vert \le \sum \limits _{k=1}^{n}\Vert T^{n-k}b_{k-1}\Vert \\\le & {} \sum \limits _{k=1}^{n}\Vert T\Vert ^{n-k}\Vert b_{k-1}\Vert \le \sum \limits _{k=1}^{n}\Vert T\Vert ^{n-k}\varepsilon _{k-1},\quad n\ge 1. \end{aligned}$$

On the other hand, in view of (18), it follows:

$$\begin{aligned} \frac{\varepsilon _{n-1}}{\varepsilon _{k-1}}=\frac{\varepsilon _{n-1}}{\varepsilon _{n-2}}\cdot \frac{\varepsilon _{n-2}}{\varepsilon _{n-3}}\cdot \ldots \cdot \frac{\varepsilon _{k}}{\varepsilon _{k-1}}\ge \frac{1}{q^{n-k}}\Vert T\Vert ^{n-k},\quad n\ge k\ge 1, \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert\le & {} \sum \limits _{k=1}^{n}q^{n-k}\varepsilon _{n-1}\\= & {} (1+q+\ldots +q^{n-1})\varepsilon _{n-1}\\\le & {} \frac{1}{1-q}\varepsilon _{n-1},\quad n\ge 1. \end{aligned}$$

$\square$

Theorem 2.9

Suppose $\Vert T^{p}\Vert <1$ for some $p\in {\mathbb {N}}.$ Then, Eq. (3) is Ulam stable with the Ulam constant:

$$\begin{aligned} L=\sum \limits _{n=0}^{\infty }\Vert T^{n}\Vert . \end{aligned}$$

Proof

Let $\varepsilon >0$ and let $(x_n)_{n\ge 0}$ be a sequence in X satisfying:

$$\begin{aligned} x_{n+1}-Tx_n-a_n=b_n,\quad n\in {\mathbb {N}}, \end{aligned}$$

with $\Vert b_n\Vert <\varepsilon ,$$n\in {\mathbb {N}}.$ Then, in view of Lemma 2.3, we get:

$$\begin{aligned} x_n=T^nx_0+\sum \limits _{k=1}^{n}T^{n-k}(a_{k-1}+b_{k-1}),\quad n\ge 1. \end{aligned}$$

Define the sequence $(y_n)_{n\ge 0}$ by $y_{n+1}=Ty_n+a_n,$$n\in {\mathbb {N}},$$y_0=x_0.$ Then:

$$\begin{aligned} y_n=T^ny_0+\sum \limits _{k=1}^{n}T^{n-k}a_{k-1},\quad n\ge 1, \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert= & {} \Vert \sum \limits _{k=1}^{n}T^{n-k}b_{k-1}\Vert \le \sum \limits _{k=1}^{n}\Vert T^{n-k}\Vert \Vert b_{k-1}\Vert \\\le & {} \varepsilon \sum \limits _{k=1}^{n}\Vert T^{n-k}\Vert \le \varepsilon \sum \limits _{k=0}^{\infty }\Vert T^{k}\Vert =L\varepsilon ,\quad n\ge 1. \end{aligned}$$

$\square$

Remark 2.10

If, in Corollary 2.9, we take $p=1,$ i.e., $\Vert T\Vert <1,$ then the conclusion holds with:

$$\begin{aligned} L=\frac{1}{1-\Vert T\Vert }. \end{aligned}$$

Proof

Indeed, according to Corollary 2.9, we obtain:

$$\begin{aligned} \sum \limits _{n=0}^{\infty }\Vert T^{n}\Vert \le \sum \limits _{n=0}^{\infty }\Vert T\Vert ^{n}=\frac{1}{1-\Vert T\Vert }. \end{aligned}$$

$\square$

Finally, we present a nonstability result for Eq. (3). Taking into account that the stability results hold in general for $\Vert T\Vert <1$ or $\Vert T^{-1}\Vert <1,$ we will consider for nonstability results the case $\Vert T\Vert =1.$

Theorem 2.11

Suppose that $\Vert T\Vert =1$ and there exists $u_0\in X$, such that:

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert T^nu_0\Vert >0. \end{aligned}$$

(22)

Then, for every $\varepsilon >0$, there exists a sequence $(x_n)_{n\ge 0}$ in X satisfying:

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert \le \varepsilon ,\quad n\in {\mathbb {N}}, \end{aligned}$$

such that for every sequence $(y_n)_{n\ge 0}$ given by the recurrence:

$$\begin{aligned} y_{n+1}=Ty_n+a_n,\quad n\in {\mathbb {N}},\ y_0\in X, \end{aligned}$$

we have:

$$\begin{aligned} \sup \limits _{n\in {\mathbb {N}}}\Vert x_n-y_n\Vert =+\infty , \end{aligned}$$

i.e., Eq. (3) is not Ulam stable.

Proof

The sequence $(\Vert T^nu_0\Vert )_{n\ge 0}$ is decreasing (see Remark 2.14) and (22) shows that $\Vert T^nu_0\Vert >0,$$n\in {\mathbb {N}}.$ Let $\varepsilon >0$ and consider the sequence $(x_n)_{n\ge 0}$ defined by the relation:

$$\begin{aligned} x_{n+1}=Tx_n+a_n+\frac{T^{n+1}u_0}{\Vert T^{n+1}u_0\Vert }\varepsilon ,\quad n\in {\mathbb {N}}. \end{aligned}$$

Then, in view of Lemma 2.3, we get:

$$\begin{aligned} x_n=T^n x_0+\sum \limits _{k=1}^{n}T^{n-k}a_{k-1}+\varepsilon \left( \sum \limits _{k=1}^{n}\frac{1}{\Vert T^{k}u_0\Vert }\right) T^{n}u_0,\quad n\ge 1. \end{aligned}$$

On the other hand:

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert =\varepsilon ,\quad n\in {\mathbb {N}}, \end{aligned}$$

and hence, $(x_n)_{n\ge 0}$ is an approximate solution of Eq. (3). Let $(y_n)_{n\ge 0}$ be an arbitrary sequence in X,  $y_{n+1}=Ty_n+a_n,$$n\ge 0,$$y_0\in X.$ Then:

$$\begin{aligned} y_{n}=T^ny_0+\sum \limits _{k=1}^{n}T^{n-k}a_{k-1},\quad n\ge 1; \end{aligned}$$

therefore:

$$\begin{aligned} x_n-y_n=T^{n}(x_0-y_0)+\varepsilon \left( \sum \limits _{k=1}^{n}\frac{1}{\Vert T^{k}u_0\Vert }\right) T^{n}u_0,\quad n\in {\mathbb {N}}. \end{aligned}$$

The sequence $(T^{n}(x_0-y_0))_{n\ge 0}$ is bounded, since:

$$\begin{aligned} \Vert T^n(x_0-y_0)\Vert \le \Vert T^n\Vert \Vert x_0-y_0\Vert \le \Vert T\Vert ^n\Vert x_0-y_0\Vert =\Vert x_0-y_0\Vert , n\in {\mathbb {N}}. \end{aligned}$$

Taking account of

$$\begin{aligned} \lim \nolimits _{k\rightarrow \infty }\frac{1}{\Vert T^{k}u_0\Vert }>0, \end{aligned}$$

we get $\sum \limits _{k=1}^{\infty }\frac{1}{\Vert T^{k}u_0\Vert }=\infty .$

It follows:

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\Vert x_n-y_n\Vert= & {} \lim \limits _{n\rightarrow \infty }\Vert T^n(x_0-y_0)+\varepsilon \left( \sum \limits _{k=1}^{n}\frac{1}{\Vert T^ku_0\Vert }\right) T^nu_0\Vert \\\ge & {} \lim \limits _{n\rightarrow \infty }\left| \Vert T^n(x_0-y_0)\Vert -\varepsilon \left( \sum \limits _{k=1}^{n}\frac{1}{\Vert T^ku_0\Vert }\right) \Vert T^nu_0\Vert \right| \\= & {} +\infty . \end{aligned}$$

$\square$

Remark 2.12

Every linear and bounded operator T,  which has an eigenvalue $\lambda ,$$|\lambda |= 1,$ satisfies the condition (22).

Indeed, there exists $u_0\ne 0$, such that $Tu_0=\lambda u_0.$ Then, it is easy to check that $T^n u_0=\lambda ^n u_0,$ for all $n\in {\mathbb {N}},$ and the condition (22) is satisfied.

Remark 2.13

There exist operators T which do not satisfy condition (22). Indeed, if T is nilpotent, there exists $p\ge 1$, such that $T^p=0;$ therefore, $T^nu_0=0$ for all $n\ge p.$

Remark 2.14

Let $\Vert T\Vert =1,$$u_0\in X.$ Then:

$$\begin{aligned} \Vert T^{n+1}u_0\Vert =\Vert T(T^nu_0)\Vert \le \Vert T^n u_0\Vert , \end{aligned}$$

i.e., the sequence $(\Vert T^nu_0\Vert )_{n\ge 0}$ is decreasing and convergent. Suppose that there exists $p\in {\mathbb {N}}$, such that $\Vert T^p\Vert <1.$ Then:

$$\begin{aligned} \Vert T^{np}u_0\Vert =\Vert (T^p)^nu_0\Vert \le \Vert T^p\Vert ^n\Vert u_0\Vert , \end{aligned}$$

and so $\lim \limits _{n\rightarrow \infty }T^{np}u_0=0.$ This implies $\lim \limits _{n\rightarrow \infty }T^n u_0=0.$

Briefly, if $\lim \limits _{n\rightarrow \infty }T^n u_0\ne 0,$ as in Theorem 2.1, then $\Vert T^n\Vert =1,$ for all $n\in {\mathbb {N}}.$ An example is presented below.

Example 1

Let ${\mathcal {C}}[0,1]$ be the Banach space of continuous, real-valued functions defined on [0,1], endowed with the supremum norm. Consider the Bernstein operator:

$$\begin{aligned} B_m:{\mathcal {C}}[0,1]\rightarrow {\mathcal {C}}[0,1],\quad B_mf(x)=\sum \limits _{k=0}^{m}\left( {\begin{array}{c}m\\ k\end{array}}\right) x^k(1-x)^{m-k}f(\frac{k}{m}). \end{aligned}$$

Then, for a fixed m,  each operator $B_{m}^{n}$ is linear, positive, and reproduces the constant function 1. Therefore, $\Vert B_{m}^{n}\Vert =1,$$n\in {\mathbb {N}}.$ Moreover:

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }B_{m}^{n}f(x)=(1-x)f(0)+x f(1),\quad f\in {\mathcal {C}}[0,1] \end{aligned}$$

uniformly on [0, 1];  see, e.g, Example 3.2.7, with $b=0,$ in [1]. Hence $\lim \limits _{n\rightarrow \infty }B_{m}^{n}f=0$ if and only if $f(0)=f(1)=0.$

Consequently, for $T=B_m$, Eq. (3) is not Ulam stable.

On the other hand, the following result shows the stability of the equation (3) when T is a nilpotent operator.

Theorem 2.15

Let $T:X\rightarrow X$ be a nilpotent operator, i.e., there exists $p\ge 1$, such that $T^p=0.$ Then, for every $\varepsilon >0$ and every sequence $(x_n)_{n\ge 0}$ in X satisfying:

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert \le \varepsilon ,\quad n\in {\mathbb {N}}, \end{aligned}$$

(23)

there exists a sequence $(y_n)_{n\ge 0}$ in X with the properties:

$$\begin{aligned} y_{n+1}=Ty_n+a_n \\ \Vert x_n-y_n\Vert \le L\varepsilon ,\quad n\ge p, \end{aligned}$$

where $L=1+\Vert T\Vert +\cdots +\Vert T\Vert ^{p-1}.$

Proof

Let

$$\begin{aligned} x_{n+1}-Tx_n-a_n=b_n,\quad n\in {\mathbb {N}} \end{aligned}$$

for some sequence $(x_n)_{n\ge 0}$ satisfying (23). Then, in view of Lemma 2.3, we get:

$$\begin{aligned} x_n=T^nx_0+\sum \limits _{k=1}^{n}T^{n-k}(a_{k-1}+b_{k-1}),\quad n\ge 1. \end{aligned}$$

Since T is a nilpotent operator, we have:

$$\begin{aligned} x_n=\sum \limits _{k=1}^{p}T^{p-k}(a_{n-p+k-1}+b_{n-p+k-1}),\quad n\ge p. \end{aligned}$$

Define $(y_n)_{n\ge 0}$ by $y_{n+1}=Ty_n+a_n,$$n\in {\mathbb {N}},$$y_0=x_0.$ Then:

$$\begin{aligned} y_n=T^nx_0+\sum \limits _{k=1}^{n}T^{n-k}a_{k-1},\quad n\ge 1, \end{aligned}$$

or equivalently:

$$\begin{aligned} y_n=\sum \limits _{k=1}^{p}T^{p-k}a_{n-p+k-1},\quad n\ge p. \end{aligned}$$

Therefore:

$$\begin{aligned} x_n-y_n=\sum \limits _{k=1}^{p}T^{p-k}b_{n-p+k-1}=b_{n-1}+b_{n-2}T+\cdots +b_{n-p}T^{p-1},\quad n\ge p, \end{aligned}$$

and

$$\begin{aligned} \Vert x_n-y_n\Vert\le & {} \Vert b_n\Vert +\Vert b_{n-1}\Vert \Vert T\Vert +\ldots +\Vert b_{n-p}\Vert \Vert T^{p-1}\Vert \\\le & {} \varepsilon (1+\Vert T\Vert +\Vert T\Vert ^2+\cdots +\Vert T\Vert ^{p-1})\\= & {} \varepsilon L,\quad n\ge p. \end{aligned}$$

$\square$

3 Applications

Let $X={\mathbb {K}}^p$ be endowed with the Euclidean norm $(\Vert x\Vert =\sqrt{|x_1|^2+\cdots +|x_p|^2},$$x=(x_1,x_2,\ldots ,x_p)\in {\mathbb {K}}^p)$ and $T:{\mathbb {K}}^p\rightarrow {\mathbb {K}}^p,$$Tx=Ax,$ where A is a square matrix of order p with entries in ${\mathbb {K}}.$ Suppose that A is normal, denote by $\lambda _1,\ldots ,\lambda _p$ and $\Lambda _1,\ldots ,\Lambda _p$ the eigenvalues of A and $A^*A,$ respectively. Recall that $A^*$ denotes the conjugate transposed of A. Suppose that:

$$\begin{aligned} |\lambda _1|\le |\lambda _2|\le \cdots \le |\lambda _p| \text{ and } \Lambda _1\le \Lambda _2\le \cdots \le \Lambda _p. \end{aligned}$$

Then (see [17]) $\Vert T\Vert =\sqrt{\Lambda _p},$ and, if A is an invertible matrix, $\Vert T^{-1}\Vert =\frac{1}{\sqrt{\Lambda _1}}.$ Moreover, if A is a self-adjoint, invertible matrix, we obtain $\Vert T\Vert =|\lambda _p|,$$\Vert T^{-1}\Vert =\frac{1}{|\lambda _1|}.$ Consequently, we get the following result on Ulam stability.

Theorem 3.1

Let $\varepsilon >0$ and $(x_n)_{n\ge 0}$ be a sequence in ${\mathbb {K}}^p$ satisfying:

$$\begin{aligned} \Vert x_{n+1}-Ax_n-a_n\Vert \le \varepsilon ,\quad n\in {\mathbb {N}}. \end{aligned}$$

• i) If $\Lambda _p<1,$ then there exists a sequence $(y_n)_{n\ge 0}$ in ${\mathbb {K}}^p$, such that:

  $$\begin{aligned}&y_{n+1}=Ay_n+a_n,\\&\Vert x_n-y_n\Vert \le \frac{\varepsilon }{1-\sqrt{\Lambda _p}},\ n\in {\mathbb {N}}. \end{aligned}$$

• ii) If A is an invertible matrix and $\Lambda _1>1,$ then there exists a sequence $(y_n)_{n\ge 0}$ in ${\mathbb {K}}^p$, such that:

  $$\begin{aligned}&y_{n+1}=Ay_n+a_n, \\&\Vert x_n-y_n\Vert \le \frac{\varepsilon }{\sqrt{\Lambda _1}-1},\ n\in {\mathbb {N}}. \end{aligned}$$

Proof

The result follows from Remark 2.7 and Remark 2.10. $\square$

Remark 3.2

If A is a self-adjoint, invertible matrix Theorem 3.1 holds with $|\lambda _1|,|\lambda _p|$ instead of $\sqrt{\Lambda _1},$$\sqrt{\Lambda _p},$ respectively.

Consider the linear operator $T:L^2[a,b]\rightarrow L^2[a,b]$ defined by:

$$\begin{aligned} (Tx)(s)=\int _{a}^{b}K(s,t)x(t)dt, \end{aligned}$$

where $K:[a,b]\times [a,b]\rightarrow {\mathbb {R}}$ is symmetric, square-measurable and:

$$\begin{aligned} \int _a^b\int _a^b|K(s,t)|^2dsdt=L^2<\infty . \end{aligned}$$

Then, T is a continuous operator and $\Vert T\Vert =\frac{1}{|\lambda _1|},$ where $\lambda _1$ is the eigenvalue of K of least absolute value (see [17]). Then, for the linear difference equation:

$$\begin{aligned} x_{n+1}=Tx_n+a_n, x_0\in L^2[a,b],\quad n\in {\mathbb {N}}, \end{aligned}$$

where $(a_n)_{n\ge 0}$ is a sequence in $L^2[a,b]$; we get the following stability result.

Theorem 3.3

Suppose that $|\lambda _1|>1.$ Then, for every $\varepsilon >0$ and every $(x_n)_{n\ge 0}$ in $L^2[a,b]$ satisfying

$$\begin{aligned} \Vert x_{n+1}-Tx_n-a_n\Vert \le \varepsilon ,\quad n\in {\mathbb {N}}, \end{aligned}$$

there exists a sequence $(y_n)_{n\ge 0}$, such that:

$$\begin{aligned}&y_{n+1}=Ty_n+a_n, y_0\in L^2[a,b],\quad n\in {\mathbb {N}}, \\&\Vert x_n-y_n\Vert \le \frac{\varepsilon |\lambda _1|}{|\lambda _1|-1},\quad n\in {\mathbb {N}}. \end{aligned}$$

Proof

The result is a simple consequence of Remark 2.10$\square$