Rates of Convergence and Metastability for Chidume’s Algorithm for the Approximation of Zeros of Accretive Operators in Banach Spaces

Abstract

In this paper we give a quantitative analysis of an explicit iteration method due to C.E. Chidume for the approximation of a zero of an m-accretive operator $A:X\to 2^X$ in Banach spaces which does not involve the computation of the resolvent of A.

Keywords:

• Accretive operators
• explicit iteration method
• metastability
• proof mining
• rates of convergence
• uniformly convex Banach spaces
• uniformly smooth Banach spaces

2020 MATHEMATICS SUBJECT CLASSIFICATION:

• 47H04
• 47H06
• 47J25
• 03F10

1 Introduction

The approximation of zeros of monotone set-valued operators $A:X\to 2^X$ in Hilbert spaces X and—more generally—of accretive operators in Banach spaces is a central theme in both nonsmooth optimization as well as in the study of abstract Cauchy problems. The connection to optimization stems from the fact that the set of minimizers of proper l.s.c. convex functions $f:X\to (-\infty ,+\infty ]$ coincides with the zeros of $A:=\partial f,$ the so-called subdifferential of f.

The famous Proximal Point Algorithm (PPA) approximates zeros of A using the resolvent function (for $\gamma >0$) $$J_{\gamma A}:R(I+\gamma A)\to D(A),J_{\gamma A}(x):={(I+\gamma A)}^{-1}(x)$$which is a single-valued firmly nonexpansive function. If A is maximal monotone (such as $\partial f$) or m-accretive, then $R(I+\gamma A)=X$ and so $J_{\gamma A}$ is defined on the whole space X and it makes sense to consider for a sequence $({\gamma }_n)\subset (0,\infty )$ the iteration $$\mathrm{(}\text{PPA}\mathrm{)}:x_{n+1}:=J_{{\gamma }_{n}A}(x_n),x_1\in X.$$

Under suitable conditions on $({\gamma }_n),$ this algorithm $(x_n)$ converges weakly to a zero of A (provided that A has one, [1–3]) but—already in the case of Hilbert spaces—it in general fails to converge strongly (see [4, 5]).

Subsequently, so-called Halpern-type variants (HPPA) $$x_{n+1}={\alpha }_{n}u+(1-{\alpha }_n){(I+{\gamma }_{n}A)}^{-1}{x}_n({\gamma }_n>0,{\alpha }_n\in [0,1])$$of (PPA) have been considered (see e.g., [6–8]) which—under suitable conditions on $({\alpha }_n),({\gamma }_n)$ - do converge strongly even in Banach spaces which e.g. are uniformly smooth and at the same time uniformly convex (so e.g., for L^p with $1<p<\infty$).

One problem with both (PPA) and (HPPA) is that they involve the computation of $J_{\gamma A}$ and so the solution of an inverse problem. Hence the existence of strongly convergence algorithms based on explicit iterations of A itself instead of its resolvent is of interest. Such an algorithm $$(*)x_{n+1}=x_n-{\lambda }_n{u}_n-{\lambda }_n{\theta }_n(x_n-x_1),u\in A(x_n),{\lambda }_n,{\theta }_n\in (0,1)$$was given in [9] (in Hilbert space) and studied in certain Banach spaces in the case of single-valued A in [10] and in 2-uniformly smooth Banach spaces in [11] and for set-valued A and general uniformly smooth Banach spaces in [12]. The strong convergence of $(x_n)$ to a zero of A (under suitable conditions on $({\lambda }_n),({\theta }_n)$) is shown in [12] by reducing the situation to the following seminal result of Reich (see also [13] for another proof of an extension of this result):

Theorem 1.1.

[14] Let X be a real uniformly smooth Banach space and $A:X\to 2^X$ be m-accretive with $A^{-1}(0)\ne \varnothing$. Then $\underset{t\to \infty }{\mathrm{\text{lim}}}{J}_{\mathit{\text{tA}}}(x)$ exists and belongs to $A^{-1}(0).$

More specifically, Chidume shows that $$\left|\right|x_{n+1}-y_n\left|\right|\to 0,\text{where}{y}_n:=J_{{\theta }_n^{-1}A}(x_1).$$

In this paper we first extract from Chidume’s proof an explicit rate of convergence for $\left|\right|x_{n+1}-y_n\left|\right|\to 0$ (Theorem 3.1). It follows from general results in computability that even for $X:=\mathbb{R}$ there is in general no computable rate of convergence for neither $(x_n)$ nor $(y_n).$ However, what recently has been obtained is the next best thing, namely a rate of metastability in the sense of Tao [15, 16] for $(y_n)$ (see [17] which in turn builds on [18]): $$(+)\forall k\in \mathbb{N}\forall g:\mathbb{N}\to \mathbb{N}\exists n\le \Psi (k,g)\forall m,l\in [n,n+g(n)](\Vert y_m-y_l\Vert \le 2^{-k}).$$

Note that $(+),$ noneffectively, implies the Cauchy property and hence the convergence of $(y_n)$ but does not allow for an effective transformation of $\Psi$ into a rate of convergence for $(y_n).$

Our rate of convergence $\left|\right|x_{n+1}-y_n\left|\right|\to 0$ together with the rate of metastability $\Psi$ in $(+)$ can be combined into a rate $\Phi$ satisfying $$\begin{array}{c}\forall k\in \mathbb{N}\forall g:\mathbb{N}\to \mathbb{N}\exists n\le \Phi (k,g)\forall l,m\in [n,n+g(n)]\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}(\Vert x_l-x_m\Vert ,\Vert x_l-y_{l-1}\Vert \le 2^{-k})\end{array}$$

(Theorem 3.3).

This quantitative result can be seen as a finitization (in the sense of Tao [15]) of Chidume’s theorem as it mathematically trivially (though noneffectively) implies not only that $(x_n)$ is Cauchy and hence strongly convergent but also that $(x_n)$ converges to the same limit as $(y_n)$ converges to.

Definition 1.2.

[19, p. 99] Let X be a real Banach space. $J:X\to 2^{X^{*}}$ with $$J(x)=\{f\in X^{*}:〈x,f〉={\Vert x\Vert }^2,\Vert f\Vert =\Vert x\Vert \}$$is called the normalized duality mapping of X.

More information on (normalized) duality mappings can be found in [20, 21].

Definition 1.3.

([19, p. 128], and [12]) Let X be a Banach space and $A:X\to 2^X$ be a set-valued operator. The domain D(A), the range R(A) and the graph G(A) of A are defined as follows: $$\begin{array}{c}D(A)=\{x\in X:A(x)\ne \varnothing \},A(S)=\underset{x\in S}{\cup }A(x),\\ R(A)=A(X),G(A)=\{(x,u):x\in D(A),u\in A(x)\}.\end{array}$$

A is accretive if for all $\forall x,y\in X,u\in A(x),v\in A(y)\exists j(x-y)\in J(x-y)$ with $〈u-v,j(x-y)〉\ge 0.$

By Kato [22] this is equivalent to the statement that for all $s>0,x,y\in D(A),u\in A(x),v\in A(y)$ $$\Vert x-y\Vert \le \Vert x-y+s(u-v)\Vert .$$

A is m-accretive if for all t > 0 (or - equivalently - for some t > 0) $R(I+\mathit{\text{tA}})=X$.

$x\in X$ is a zero of A if $0\in A(x)$. $\mathit{\text{zer}} A$ denotes the set of all zeros of A.

Definition 1.4.

[23, p. 13] A Banach space X is called uniformly smooth if $$\forall \epsilon >0\exists \delta >0\forall x,y\in X \left(\Vert x\Vert =1\wedge \Vert y\Vert \le \delta \Rightarrow \Vert x+y\Vert +\Vert x-y\Vert \le 2+\epsilon \Vert y\Vert \right).$$

Remark 1.5.

It is well-known (see e.g., [23], p.13) that the definition above is equivalent to $$\underset{t\to 0}{\mathrm{\text{lim}}}\frac{{\rho }_X(t)}{t}=0,$$where $${\rho }_X(t)=\mathrm{\text{sup}}\left\{\frac{\Vert x+y\Vert +\Vert x-y\Vert }{2}-1:\Vert x\Vert =1,\Vert y\Vert \le t\right\}.$$

Definition 1.6.

X is called uniformly convex if $$\forall \epsilon \in (0,2]\exists \delta \in (0,1]\forall x,y\in X\left(\Vert x\Vert ,\Vert y\Vert \le 1\wedge \Vert x-y\Vert \ge \epsilon \Rightarrow \Vert \frac{1}{2}(x+y)\Vert \le 1-\delta \right).$$

As in [24], we call functions $\tau :(0,\infty )\to (0,\infty ),\eta :(0,2]\to (0,1]$ which provide a $\delta (\epsilon )$ in the definitions of uniform smoothness and uniform convexity moduli of uniform smoothness and uniform convexity respectively. Note that this differs from the terminology used in functional analysis where ρ_X is called the modulus of smoothness of X while it is not a modulus of smoothness in the sense of τ¹ and what is called the modulus of uniform convexity δ_X is a particular, namely the optimal, modulus of uniform convexity η in our sense.

Chidume assumes that A is bounded which is meant as ‘bounded on bounded sets’. As discussed in [25] this is equivalent to A possessing a uniform majorant $A^{*}:\mathbb{N}\to \mathbb{N}$ satisfying $$\forall x\in X\forall n\in \mathbb{N} \left(\Vert x\Vert \le n\Rightarrow \forall y\in A(x)(\Vert y\Vert \le A^{*}(n))\right).$$

By a majorant for a sequence ${(x_n)}_{n\in \mathbb{N}}$ in X we mean a sequence ${(p_n)}_{n\in \mathbb{N}}$ in $\mathbb{N}$ such that $p_n\ge \left|\right|x_n\left|\right|$ for all $n\in \mathbb{N}.$

Notation: Following [12], all our sequences start with the index $n\ge 1$ and we, therefore, use $\mathbb{N}:=\{1,2,3,\dots \}.$

2 Technical lemmas

In this section we collect some technical estimates for uniformly smooth Banach spaces which are essentially known but in some cases the values of certain constants had to be extracted from the literature.

Lemma 2.1.

(see [26, pp. 64–65], [27, p. 208]) Let X be a uniformly smooth Banach space and ρ_X be the function defined in Remark 1.5. Then for all $s,t\in \mathbb{R}$ with $s\ge t>0$ $$\frac{{\rho }_X(s)}{s^2}\le C\frac{{\rho }_X(t)}{t^2}\text{and}\frac{{\rho }_X(t)}{t}\le \frac{{\rho }_X(s)}{s},$$where $C=\frac{4{\tau }_0}{\sqrt{1+{\tau }_0^2}-1}{\prod }_{j=1}^{\infty }\left(1+\frac{15{\tau }_0}{4·2^j}\right),{\tau }_0=\frac{\sqrt{339}-18}{30}$.

Lemma 2.2.

(see [12, p. 36] and [27]) Let X be uniformly smooth. Then for all $x,y\in X$ and $j(x)\in J(x)$ $${\Vert x+y\Vert }^2\le {\Vert x\Vert }^2+2〈y,j(x)〉+D\mathrm{\text{max}}(\Vert x\Vert +\Vert y\Vert ,C){\rho }_X(\Vert y\Vert ),$$where $$\begin{array}{c}D=2\mathrm{\text{max}}(8,(40-16\sqrt{3})C),\\ C=\frac{4{\tau }_0}{\sqrt{1+{\tau }_0^2}-1}\prod _{j=1}^{\infty }\left(1+\frac{15{\tau }_0}{4·2^j}\right),\\ {\tau }_0=\frac{\sqrt{339}-18}{30}.\end{array}$$

Proof.

In [27, p. 208] it is shown that for all $p,q\in (1,\infty )$ with $\frac{1}{p}+\frac{1}{q}=1$ $${\Vert x+y\Vert }^p\le {\Vert x\Vert }^p+p〈y,j(x)〉+{\sigma }_p(x,y)$$with $${\sigma }_p(x,y)=p·l\underset{0}{\overset{1}{\int }}\frac{\mathrm{\text{max}}{(\Vert x+\mathit{\text{ty}}\Vert ,\Vert x\Vert )}^2}{t}{\rho }_X\left(\frac{t\Vert y\Vert }{\mathrm{\text{max}}(\Vert x+\mathit{\text{ty}}\Vert ,\Vert x\Vert )}\right)\mathit{\text{dt}},$$ where $$\begin{array}{c}l=\mathrm{\text{max}}\left(8,64C\frac{1}{K_q}\right),\\ K_q=4(2+\sqrt{3})\mathrm{\text{min}}\left\{\begin{array}{c}\mathrm{\text{min}}(1,\frac{1}{2}q(q-1)),(q-1)\mathrm{\text{min}}(1,\frac{1}{2}q),\hfill \\ (q-1)(1-{(\sqrt{3}-1)}^{\frac{q}{q-1}}),1-{(1+\frac{(2-\sqrt{3})q}{q-1})}^{1-q}\hfill \end{array}\right\}.\end{array}$$

For $p=q=2$ this yields $$\begin{array}{c}{\sigma }_2(x,y)=2\mathrm{\text{max}}(8,(40-16\sqrt{3})C)\underset{0}{\overset{1}{\int }}\frac{\mathrm{\text{max}}{(\Vert x+\mathit{\text{ty}}\Vert ,\Vert x\Vert )}^2}{t}{\rho }_X\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}\left(\frac{t\Vert y\Vert }{\mathrm{\text{max}}(\Vert x+\mathit{\text{ty}}\Vert ,\Vert x\Vert )}\right)\mathit{\text{dt}}.\end{array}$$

Distinguishing the cases $\frac{t}{\mathrm{\text{max}}(\Vert x+\mathit{\text{ty}}\Vert ,\Vert x\Vert )}\ge 1$ and $\frac{t}{\mathrm{\text{max}}(\Vert x+\mathit{\text{ty}}\Vert ,\Vert x\Vert )}\le 1,$ the respective inequalities from Lemma 2.1 imply that $$\begin{array}{c}{\Vert x+y\Vert }^2\le {\Vert x\Vert }^2+2〈y,j(x)〉\hfill \\ +2\mathrm{\text{max}}(8,(40-16\sqrt{3})C)\underset{0}{\overset{1}{\int }}\mathrm{\text{max}}\left(\Vert x+\mathit{\text{ty}}\Vert ,\Vert x\Vert ,\mathit{\text{Ct}}\right)\mathit{\text{dt}}{\rho }_X(\Vert y\Vert )\hfill \\ \le {\Vert x\Vert }^2+2〈y,j(x)〉+2\mathrm{\text{max}}(8,(40-16\sqrt{3})C)\mathrm{\text{max}}(\Vert x\Vert +\Vert y\Vert ,C){\rho }_X(\Vert y\Vert ).\hfill \end{array}$$

□

Lemma 2.3.

(see [28, p. 284]) Let X be a Banach space. Then for all $x,y\in X$ and for all $j(x+y)\in J(x+y)$ $${\Vert x+y\Vert }^2\le {\Vert x\Vert }^2+2〈y,j(x+y)〉.$$

Lemma 2.4.

[7, p. 243] Let ${({\rho }_n)}_{n\in \mathbb{N}},{({\gamma }_n)}_{n\in \mathbb{N}}$ be sequences of nonnegative real numbers, ${({\sigma }_n)}_{n\in \mathbb{N}}$ a sequence of real numbers and ${({\alpha }_n)}_{n\in \mathbb{N}}$ a sequence in $[0,1)$ such that for all $n\in \mathbb{N}$ $${\rho }_{n+1}\le (1-{\alpha }_n){\rho }_n+{\alpha }_n{\sigma }_n+{\gamma }_n.$$

If $$\sum _{n=1}^{\infty }{\alpha }_n=\infty (\text{or ‐ equivalently ‐}\prod _{n=1}^{\infty }(1-{\alpha }_n)=0),\underset{n\to \infty }{\mathrm{\text{lim}}\mathrm{\text{sup}}}{\sigma }_n\le 0\text{and}\sum _{n=1}^{\infty }{\gamma }_n<\infty ,$$then it follows that $$\underset{n\to \infty }{\lim }{\rho }_n=0.$$

We now give a quantitative version of Lemma 2.4. Similar versions have been used repeatedly in the context of proof mining e.g. in [29, 30]. For completeness, however, we give the proof for the particular formulation we need.

Lemma 2.5.

Let $({\rho }_n),({\gamma }_n),({\sigma }_n),({\alpha }_n)$ be as in the previous lemma and let $(p_n)\subset \mathbb{N}$ be a majorant for $({\rho }_n).$ Let ${\Phi }_1,{\Phi }_2,{\Phi }_3$ be rates witnessing quantitatively the conditions on $({\alpha }_n),({\sigma }_n),({\gamma }_n),$ i.e. $$\begin{array}{c}\forall k\in \mathbb{N}\forall N\in \mathbb{N}\left(\prod _{n=N}^{{\Phi }_1(k,N)}(1-{\alpha }_n)\le 2^{-k}\right),\\ \forall k\in \mathbb{N}\forall n\ge {\Phi }_2(k)\left({\sigma }_n\le 2^{-k}\right),\\ \forall k\in \mathbb{N} \left(\sum _{n={\Phi }_3(k)}^{\infty }{\gamma }_n\le 2^{-k}\right).\end{array}$$

Then $$\forall k\in \mathbb{N}\forall n\ge {\Phi }^{*}(k)({\rho }_n\le 2^{-k}),$$where ${\Phi }^{*}(k)=\mathrm{\text{max}}({\Phi }_1(k+\lceil {\hspace{0.17em}\mathrm{\text{log}}\hspace{0.17em}}_2{p}_N\rceil +1,N),N)+1$ and $N=\mathrm{\text{max}}({\Phi }_2(k+2),{\Phi }_3(k+2))$.

Proof.

Let $k\in \mathbb{N}$ be arbitrary and let $N:=\mathrm{\text{max}}({\Phi }_2(k+2),{\Phi }_3(k+2)).$

We prove by induction on n that for all $n\ge N$: $${\rho }_{n+1}\le \left(\prod _{i=N}^n(1-{\alpha }_i)\right){\rho }_N+\left(1-\prod _{i=N}^n(1-{\alpha }_i)\right)·2^{-k-2}+\sum _{i=N}^n{\gamma }_i.$$

The case n = N holds by assumption. Assume that the claim holds for $n\ge N.$ Then $$\begin{array}{c}{\rho }_{n+2}\le (1-{\alpha }_{n+1})\mathrm{}\left(\left(\prod _{i=N}^n(1-{\alpha }_i)\right)\mathrm{}{\rho }_N+\left(\mathrm{}1-\prod _{i=N}^n(1-{\alpha }_i)\mathrm{}\right)·2^{-k-2}+\mathrm{}\sum _{i=N}^n{\gamma }_i\mathrm{}\right)\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+{\alpha }_{n+1}{\sigma }_{n+1}+{\gamma }_{n+1}.\end{array}$$

Moving $(1-{\alpha }_{n+1})$ inside, using ${\alpha }_{n+1}\in [0,1]$ as well as ${\sigma }_{n+1}\le 2^{-k-2},$ the induction step follows.

Let now $n\ge \mathrm{\text{max}}({\Phi }_1(k+\lceil {\hspace{0.17em}\mathrm{\text{log}}\hspace{0.17em}}_2{p}_N\rceil +1,N),N)+1.$ Then - using ${\rho }_N\le p_N$ - ${\prod }_{i=N}^{n-1}(1-{\alpha }_i)\le \frac{2^{-k-1}}{p_N},(1-{\prod }_{i=N}^{n-1}(1-{\alpha }_{i-1}))\le 1$ and ${\sum }_{i=N}^{n-1}{\gamma }_i\le 2^{-k-2}$ imply ${\rho }_n\le 2^{-k}$. □

The following bound on the iterative sequence $(x_n)$ of Chidume’s algorithm is crucially used:

Lemma 2.6.

(see [12, p.37]) Let X be a uniformly smooth Banach space, $A:X\to 2^X$ be a bounded set-valued accretive operator with D(A) = X and $x^{*}\in \mathit{\text{zer}} A.$ Let ${({\lambda }_n)}_{n\in \mathbb{N}}$ and ${({\theta }_n)}_{n\in \mathbb{N}}$ be sequences in (0, 1) and $x_1\in X$ be arbitrary. Let C, D be as in Lemma 2.2 and let $L\in \mathbb{N}$ be such that $$\Vert x^{*}\Vert ,\Vert x^{*}-x_1\Vert \le L.$$

Let ${\gamma }_0,M_0,{(x_n)}_{n\in \mathbb{N}}$ be such that $$\begin{array}{c}{x}_{n+1}=x_n-{\lambda }_n{u}_n-{\lambda }_n{\theta }_n(x_n-x_1),u_n\in A(x_n),\\ M_0=\mathrm{\text{max}}\left\{1,\mathrm{\text{sup}}\{\Vert u+\theta (x-x_1)\Vert :\theta \in (0,1),u\in A(x),x\in X,\Vert x-x^{*}\Vert \le 2L\}\right\},\\ M^{*}=\mathrm{\text{sup}}\{D\mathrm{\text{max}}(\Vert x\Vert +\lambda M_0,C):\lambda \in (0,1),x\in X,\Vert x-x^{*}\Vert \le 2L\},\\ {\gamma }_0=\frac{1}{2}\mathrm{\text{min}}\left(1,\frac{L^2}{M^{*}{M}_0}\right).\end{array}$$

If for all $n\in \mathbb{N}$ $$\frac{{\rho }_X({\lambda }_n{M}_0)}{{\lambda }_n{M}_0}\le {\gamma }_0{\theta }_n,$$then ${(x_n)}_{n\in \mathbb{N}}$ is bounded by $3L.$

Proof.

[12, p.37] shows that under the conditions given, one has $\Vert x^{*}-x_n\Vert \le 2L,$ which implies the claim. □

In the following we give a more explicit and effective description of the bound on $(x_n)$ which avoids the use of $\mathrm{\text{sup}}$’s.

Corollary 2.7.

Let $A^{*}:\mathbb{N}\to \mathbb{N}$ be a uniform majorant of A witnessing that A is bounded on bounded sets, i.e. $$\forall n\in \mathbb{N}\forall (x,y)\in \mathrm{G}(A)(|\left|x\right||\le n\to |\left|y\right||\le A^{*}(n)).$$

Then the condition $$\frac{{\rho }_X({\lambda }_n{M}_0)}{{\lambda }_n{M}_0}\le {\gamma }_0{\theta }_n\text{with}{\gamma }_0=\frac{1}{2}\mathrm{\text{min}}\left(1,\frac{L^2}{M^{*}{M}_0}\right)$$in Lemma 2.6 can be replaced by $$\frac{{\rho }_X({\lambda }_n)}{{\lambda }_n}\le {\gamma }_0{\theta }_n,$$ with $${\gamma }_0=\frac{1}{2}\mathrm{\text{min}}\left(1,\frac{L^2}{D(A^{*}(3L)+5L)\mathrm{\text{max}}(A^{*}(3L)+8L,C)}\right)\frac{1}{C(A^{*}(3L)+5L)}.$$

Proof.

With Lemma 2.1 we get $$\frac{{\rho }_X({\lambda }_n{M}_0)}{{\lambda }_n{M}_0}\le C{M}_0\frac{{\rho }_X({\lambda }_n)}{{\lambda }_n}.$$

Together with the easy estimates $$M_0\le A^{*}(3L)+5L\text{and}{M}^{*}\le D\mathrm{\text{max}}(A^{*}(3L)+8L,C),$$the corollary follows. □

3 Main results

The next theorem gives an explicit and effective rate for the convergence of $\left|\right|x_{n+1}-y_n\left|\right|\to 0,$ where $(x_n)$ is the sequence generated by Chidume’s algorithm $(*)$ and $y_n:=J_{t_{n}A}(x_1)$ with $t_n:={\theta }_n^{-1}:$

Theorem 3.1.

Let X be a uniformly smooth Banach space, $A:X\to 2^X$ be a bounded set-valued m-accretive operator with D(A) = X with $\mathit{\text{zer}} A\cancel{=}\varnothing .$ Let $x^{*}\in \mathit{\text{zer}} A$ and let $A^{*}:\mathbb{N}\to \mathbb{N}$ be a uniform majorant of A. Let ${({\lambda }_n)}_{n\in \mathbb{N}}$ and ${({\theta }_n)}_{n\in \mathbb{N}}$ be sequences in (0, 1) and $x_1\in X$ arbitrary. Let ${(x_n)}_{n\in \mathbb{N}}$ and ${(y_n)}_{n\in \mathbb{N}}$ be sequences in A satisfying $$x_{n+1}=x_n-{\lambda }_n{u}_n-{\lambda }_n{\theta }_n(x_n-x_1),u_n\in A(x_n)\text{and}{t}_n={\theta }_n^{-1},y_n=J_{t_{n}A}(x_1).$$

Let $L\in \mathbb{N}$ be such that $\Vert x^{*}\Vert ,\Vert x^{*}-x_1\Vert \le L$, C and D as in Lemma 2.2, $\begin{array}{c}{M}_1=A^{*}(3L)+5L,M_2=D\mathrm{\text{max}}(M_1+5L,C),C^{*}=40L^2,{\sigma }_n=C^{*}\frac{\frac{{\theta }_{n-1}}{{\theta }_n}-1}{{\lambda }_n{\theta }_n}\\ {\gamma }_0=\frac{1}{2}\mathrm{\text{min}}(1,\frac{L^2}{D(A^{*}(3L)+5L)\mathrm{\text{max}}(A^{*}(3L)+8L,C)})\frac{1}{C(A^{*}(3L)+5L)}.\end{array}$ and

If then $$\prod _{n=1}^{\infty }(1-{\lambda }_n{\theta }_n)=0,\underset{n\to \infty }{\mathrm{\text{lim}}}{\sigma }_{n+1}=0,\sum _{n=1}^{\infty }(M_2{\rho }_X({\lambda }_n{M}_1))<\infty ,$$and $$\forall n\in \mathbb{N} \left({\theta }_n\ge {\theta }_{n+1},\frac{{\rho }_X({\lambda }_n)}{{\lambda }_n}\le {\gamma }_0{\theta }_n\right)$$ and the first three properties are quantitatively witnessed by ${\Phi }_1:{\mathbb{N}}^2\to \mathbb{N},{\Phi }_2,$ ${\Phi }_3:\mathbb{N}\to \mathbb{N},$ i.e. $$\begin{array}{c}\forall k\in \mathbb{N}\forall N\in \mathbb{N} \left(\prod _{n=N}^{{\Phi }_1(k,N)}(1-{\lambda }_{n+1}{\theta }_{n+1})\le 2^{-k}\right),\\ \forall k\in \mathbb{N}\forall n\ge {\Phi }_2(k)({\sigma }_{n+1}\le 2^{-k}),\\ \forall k\in \mathbb{N} \left(\sum _{n={\Phi }_3(k)}^{\infty }(M_2{\rho }_X({\lambda }_{n+1}{M}_1))\le 2^{-k}\right),\end{array}$$ then $$\forall k\in \mathbb{N}\forall n\ge {\Phi }^{*}(k,L,{\Phi }_1,{\Phi }_2,{\Phi }_3)\left(\Vert x_{n+1}-y_n\Vert \le 2^{-k}\right),$$ where $$\begin{array}{c}{\Phi }^{*}(k,L,{\Phi }_1,{\Phi }_2,{\Phi }_3)=\mathrm{\text{max}}({\Phi }_1(2k+\lceil {\hspace{0.17em}\mathrm{\text{log}}\hspace{0.17em}}_{2}25L^2\rceil +1,N),N)+1\text{and}\\ N=\mathrm{\text{max}}({\Phi }_2(2k+2),{\Phi }_3(2k+2)).\end{array}$$

Proof.

We follow closely the proof of Theorem 3.2 in [12]. First we show the boundedness of ${(y_n)}_{n\in \mathbb{N}}$ using the nonexpansivity of $J_{t_{n}A}$ and the fact that $x^{*}$ being a zero of A is a common fixed point of the resolvents $J_{t_{n}A}:$ $$\Vert y_n\Vert \le \Vert y_n-x^{*}\Vert +\Vert x^{*}\Vert =\Vert J_{t_{n}A}(x_1)-J_{t_{n}A}(x^{*})\Vert +\Vert x^{*}\Vert \le \Vert x_1-x^{*}\Vert +\Vert x^{*}\Vert \le 2L.$$

With Lemma 2.2, the boundedness of x_n (see Lemma 2.6 and Corollary 2.7) and y_n and the majorant $A^{*}$ of A together with Lemma 2.1 and the definition of $(x_n)$ it follows that $$\begin{array}{c}{\Vert x_{n+1}-y_n\Vert }^2={\Vert x_n-y_n-{\lambda }_n(u_n+{\theta }_n(x_n-x_1))\Vert }^2\\ \le {\Vert x_n-y_n\Vert }^2-2{\lambda }_n〈u_n+{\theta }_n(x_n-x_1),j(x_n-y_n)〉\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+D\mathrm{\text{max}}(\Vert x_n-y_n\Vert +{\lambda }_n\Vert u_n+{\theta }_n(x_n-x_1)\Vert ,C)\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}·{\rho }_X({\lambda }_n\Vert u_n+{\theta }_n(x_n-x_1)\Vert )\\ \le {\Vert x_n-y_n\Vert }^2-2{\lambda }_n〈u_n+{\theta }_n(x_n-x_1),j(x_n-y_n)〉\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+D\mathrm{\text{max}}(\Vert x_n\Vert +\Vert y_n\Vert +\Vert u_n\Vert +\Vert x_n\Vert +\Vert x_1\Vert ,C)\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}·{\rho }_X({\lambda }_n(\Vert u_n\Vert +\Vert x_n\Vert +\Vert x_1\Vert ))\\ \le {\Vert x_n-y_n\Vert }^2-2{\lambda }_n〈u_n+{\theta }_n(x_n-x_1),j(x_n-y_n)〉\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+D\mathrm{\text{max}}(3L+2L+A^{*}(3L)+3L+2L,C)\\ ·{\rho }_X({\lambda }_n(A^{*}(3L)+3L+2L))\\ ={\Vert x_n-y_n\Vert }^2-2{\lambda }_n〈u_n+{\theta }_n(x_n-x_1),j(x_n-y_n)〉\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+D\mathrm{\text{max}}(M_1+5L,C)·{\rho }_X({\lambda }_n{M}_1)\\ ={\Vert x_n-y_n\Vert }^2-2{\lambda }_n〈u_n+{\theta }_n(x_n-x_1),j(x_n-y_n)〉\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+M_2{\rho }_X({\lambda }_n{M}_1).\end{array}$$

As in [12, pp.38–39] one shows that $$〈u_n+{\theta }_n(x_n-x_1),j(x_n-y_n)〉\ge \frac{{\theta }_n}{2}{\Vert x_n-y_n\Vert }^2$$and (for $n\ge 2$) $$\Vert y_{n-1}-y_n\Vert \le \frac{{\theta }_{n-1}-{\theta }_n}{{\theta }_n}\Vert y_{n-1}-x_1\Vert ,$$ as well as $${\Vert x_n-y_n\Vert }^2\le {\Vert x_n-y_{n-1}\Vert }^2+2\Vert y_{n-1}-y_n\Vert \Vert x_n-y_n\Vert .$$

Combining these four inequalities we get (reasoning as in [12]) for all $n\ge 2$ $$\begin{array}{c}{\Vert x_{n+1}-y_n\Vert }^2\le {\Vert x_n-y_n\Vert }^2-2{\lambda }_n〈u_n+{\theta }_n(x_n-x_1),j(x_n-y_n)〉\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+M_2{\rho }_X({\lambda }_n{M}_1)\\ \le {\Vert x_n-y_n\Vert }^2-{\lambda }_n{\theta }_n{\Vert x_n-y_n\Vert }^2\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+M_2{\rho }_X({\lambda }_n{M}_1)\\ \le (1-{\lambda }_n{\theta }_n)({\Vert x_n-y_{n-1}\Vert }^2+2\Vert y_{n-1}-y_n\Vert \Vert x_n-y_n\Vert )\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+M_2{\rho }_X({\lambda }_n{M}_1)\\ \le (1-{\lambda }_n{\theta }_n){\Vert x_n-y_{n-1}\Vert }^2+2(1-{\lambda }_n{\theta }_n)\frac{{\theta }_{n-1}-{\theta }_n}{{\theta }_n}\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}·\Vert y_{n-1}-x_1\Vert \Vert x_n-y_n\Vert +M_2{\rho }_X({\lambda }_n{M}_1)\\ \le (1-{\lambda }_n{\theta }_n){\Vert x_n-y_{n-1}\Vert }^2+2(\frac{{\theta }_{n-1}}{{\theta }_n}-1)(\Vert y_{n-1}\Vert +\Vert x_1\Vert )\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}·(\Vert x_n\Vert +\Vert y_n\Vert )+M_2{\rho }_X({\lambda }_n{M}_1)\\ \le (1-{\lambda }_n{\theta }_n){\Vert x_n-y_{n-1}\Vert }^2+2\frac{{\lambda }_n{\theta }_n}{{\lambda }_n{\theta }_n}(\frac{{\theta }_{n-1}}{{\theta }_n}-1)(2L+2L)\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}·(3L+2L)+M_2{\rho }_X({\lambda }_n{M}_1)\\ =(1-{\lambda }_n{\theta }_n){\Vert x_n-y_{n-1}\Vert }^2+{\lambda }_n{\theta }_n\frac{\frac{{\theta }_{n-1}}{{\theta }_n}-1}{{\lambda }_n{\theta }_n}(40L^2)\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+M_2{\rho }_X({\lambda }_n{M}_1)\\ =(1-{\lambda }_n{\theta }_n){\Vert x_n-y_{n-1}\Vert }^2+{\lambda }_n{\theta }_n\frac{(\frac{{\theta }_{n-1}}{{\theta }_n}-1)}{{\lambda }_n{\theta }_n}{C}^{*}\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+M_2{\rho }_X({\lambda }_n{M}_1)\\ =(1-{\lambda }_n{\theta }_n){\Vert x_n-y_{n-1}\Vert }^2+{\lambda }_n{\theta }_n{\sigma }_n\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+M_2{\rho }_X({\lambda }_n{M}_1).\end{array}$$

With Lemma 2.5 applied to ${\rho }_n:=\left|\right|x_{n+1}-y_n|{|}^2$ and $p_n:=25L^2\ge {\rho }_n$ for all $n\ge 1$ and applying the square root we finally get $$\forall k\in \mathbb{N}\forall n\ge {\Phi }^{*}(k)\left(\Vert x_{n+1}-y_n\Vert \le 2^{-k}\right).$$

□

By Reich’s theorem, mentioned already in the introduction, one can show under the additional condition that $(t_n)$ diverges to $+\infty$, i.e. that $({\theta }_n)$ tends to 0, that $(y_n)$ strongly converges to a zero of A. An explicit rate of metastability witnessing this quantitatively has been computed (under the additional assumption that X is also uniformly convex) in the case of single-valued A in [18] and was recently generalized to the set-valued case in [17].

Theorem 3.2.

[17, Corollary 4.2] Let X be a Banach space which is both uniformly smooth and uniformly convex with respective moduli τ and $\eta .$ Let $A:X\to 2^X$ be m-accretive and $x^{*}$ such that $0\in A(x^{*}).$ Let $x\in X$ be arbitrary and $L\in \mathbb{N}$ such that $\Vert x-x^{*}\Vert \le L$ and ${(t_n)}_{n\in \mathbb{N}}\subseteq (0,\infty )$ with $$\underset{n\to \infty }{\mathrm{\text{lim}}}{t}_n=\infty$$and functions $\alpha ,\gamma :\mathbb{N}\to \mathbb{N}$ such that $$\forall n\in \mathbb{N}\forall m\ge \alpha (n)(t_m\ge n+1,t_n\le \gamma (n)).$$

Then there exists an explicit and fully effective rate ${\Psi }_{\tau ,\eta ,2L,\alpha ,\gamma }(k,g)$ of metastability for the sequence $y_n=J_{t_{n}A}(x)$ which only depends on $\tau ,\eta ,L,\alpha ,\gamma$ and $k,g:$² $$\forall k\in \mathbb{N}\forall g:\mathbb{N}\to \mathbb{N}\exists 1\le n\le {\Psi }_{\tau ,\eta ,2L,\alpha ,\gamma }(k,g)\forall l,m\in [n,n+g(n)](\Vert y_l-y_m\Vert \le 2^{-k})$$

As in [31, Theorem 2.8] we can now combine this with Theorem 3.1:

Theorem 3.3.

In addition to the assumptions made in Theorem 3.1 we assume that X is also uniformly convex with a modulus η and that $t_n:={\theta }_n^{-1}$ diverges to $+\infty$ and that we have functions $\alpha ,\gamma :\mathbb{N}\to \mathbb{N}$ with $$\forall n\in \mathbb{N}\forall m\ge \alpha (n)(t_m\ge n+1,t_n\le \gamma (n)).$$

Let ${\Psi }_{\tau ,\eta ,2L,\alpha ,\gamma }$ be the rate of metastability for $y_n:=J_{t_{n}A}(x_1)$ as in Theorem 3.2. Then $(x_n)$ converges to the limit of $(y_n)$ and we have the following explicit rate of metastability witnessing this fact: $$\begin{array}{c}\forall k\in \mathbb{N}\forall g:\mathbb{N}\to \mathbb{N}\exists 2\le n\le {\Psi }_{\tau ,\eta ,2L,\alpha ,\gamma }(k+1,g_k)+{\Phi }^{*}(k+2)+1\hfill \\ \forall l,m\in [n,n+g(n)](\Vert x_l-x_m\Vert ,\Vert x_l-y_{l-1}\Vert \le 2^{-k}),\hfill \end{array}$$where ${\Phi }^{*}$ is as in Theorem 3.1 and $g_k(n):=g(n+{\Phi }^{*}(k+2)+1)+{\Phi }^{*}(k+2).$

Proof.

For $(y_n)$ we know by Theorem 3.2 that $$\forall k\in \mathbb{N}\forall g:\mathbb{N}\to \mathbb{N}\exists 1\le n\le {\Psi }_{\tau ,\eta ,2L,\alpha ,\gamma }(k,g)\forall l,m\in [n,n+g(n)](\Vert y_l-y_m\Vert \le 2^{-k}).$$

Given k, g, we apply this to k + 1 and $g_k(n):=g(n+{\Phi }^{*}(k+2)+1)+{\Phi }^{*}(k+2),$ where ${\Phi }^{*}(k)$ is the rate from Theorem 3.1: $$\begin{array}{c}\exists 1\le n\le {\Psi }_{\tau ,\eta ,2L,\alpha ,\gamma }(k+1,g_k)\forall l,m\in [n,n+g(n+{\Phi }^{*}(k+2)+1)\\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+{\Phi }^{*}(k+2)](\Vert y_l-y_m\Vert \le 2^{-k-1}).\end{array}$$

Let n be as in the formula above and define $n\prime :=n+{\Phi }^{*}(k+2).$ Since $n\le n\prime ,$ we can restrict things to the smaller interval $[n\prime ,n\prime +g(n\prime +1)]=[n+{\Phi }^{*}(k+2),n+{\Phi }^{*}(k+2)+g(n+{\Phi }^{*}(k+2)+1)].$ By Theorem 3.1 we have for all l, m in this smaller interval that $$\begin{array}{c}\Vert x_{l+1}-x_{m+1}\Vert \le \Vert y_l-y_m\Vert +\Vert x_{l+1}-y_l\Vert +\Vert x_{m+1}-y_m\Vert \le \Vert y_l-y_m\Vert \\ \mathrm{\hspace{1em}}\mathrm{\hspace{1em}}+2^{-k-1}\le 2^{-k}.\end{array}$$

This means in total that for $n″:=n\prime +1$ $$\forall l,m\in [n″,n″+g(n″)](\Vert x_l-x_m\Vert ,||x_l-y_{l-1}||\le 2^{-k}).$$

By construction $$2\le n″\le n+{\Phi }^{*}(k+2)+1\le {\Psi }_{\tau ,\eta ,2L,\alpha ,\gamma }(k+1,g_k)+{\Phi }^{*}(k+2)+1$$which finishes the proof. □

The condition $$\sum _{n=1}^{\infty }{\rho }_X({\lambda }_n{M}_1)<\infty$$involves ρ_X which is defined as a supremum which may not be computable. The next proposition shows how we can replace this with a condition involving only the modulus τ instead of ${\rho }_X:$

Proposition 3.4.

Let X be a uniformly smooth Banach space with a modulus of uniform smoothness τ, ${({\lambda }_n)}_{n\in \mathbb{N}}$ be a sequence in (0, 1), $M_1>0$ some constant and C as in Lemma 2.1. Assume that τ is strictly increasing and $(0,1)\subseteq \tau (0,\infty )$ so that the inverse ${\tau }^{-1}$ of τ is defined on $(0,1).$ If for K > 0 $$\sum _{n=1}^{\infty }\left(\frac{1}{2}C{M}_1^2{\tau }^{-1}({\lambda }_n){\lambda }_n\right)\le K,$$then $$\sum _{n=1}^{\infty }{\rho }_X({\lambda }_n{M}_1)\le K.$$

Proof.

By Lemma 2.1 $${\rho }_X({\lambda }_n{M}_1)\le C{M}_1^2{\rho }_X({\lambda }_n)$$and by the definition of τ $$\forall \epsilon >0\forall x,y\in X\left(\Vert x\Vert =1\wedge \Vert y\Vert \le \tau (\epsilon )\Rightarrow \Vert x+y\Vert +\Vert x-y\Vert \le 2+\epsilon \Vert y\Vert \right).$$

Applying this to $\epsilon :={\tau }^{-1}({\lambda }_n)$ we see that $$\forall x,y\in X \left(\Vert x\Vert =1\wedge \Vert y\Vert \le {\lambda }_n\Rightarrow \Vert x+y\Vert +\Vert x-y\Vert \le 2+{\tau }^{-1}({\lambda }_n){\lambda }_n\right)$$

which in turn implies ${\rho }_X({\lambda }_n)\le \frac{1}{2}{\tau }^{-1}({\lambda }_n){\lambda }_n$ and so $${\rho }_X({\lambda }_n{M}_1)\le \frac{1}{2}C{M}_1^2{\tau }^{-1}({\lambda }_n){\lambda }_n$$from which the proposition follows. □

We now elaborate on an example from [12] for a special choice of the scalars ${\lambda }_n,{\theta }_n$ in the case of L^p with $1<p<\infty$ which satisfy the conditions in Chidume’s theorem and compute the corresponding rates ${\Phi }_1,{\Phi }_2,{\Phi }_3$ as well as $\alpha ,\gamma$ used in our bounds:

Example 3.5.

Let $X=L^p,p\in (1,\infty )$ and define $${\lambda }_n={(n+1)}^{-a}\text{and}{\theta }_n={(n+1)}^{-b}$$with $a+b<1$ and $$\begin{array}{c}a\in (\frac{1}{2},1),b\in (0,a)\text{for}p\ge 2\text{resp}.\\ a\in (\frac{1}{p},1),b\in (0,a(p-1))\text{for}p<2.\end{array}$$

Let $$N_0\ge \{\begin{array}{cc}\sqrt[a-b]{\frac{p-1}{2{\gamma }_0}}\hfill & p\ge 2\hfill \\ \sqrt[a(p-1)-b]{\frac{1}{p{\gamma }_0}}\hfill & p<2.\hfill \end{array}$$

Then ${({\lambda }_{N_0+n})}_{n\in \mathbb{N}},{({\theta }_{N_0+n})}_{n\in \mathbb{N}}$ satisfy the conditions in our Theorems 3.1 and (together with $\alpha ,\gamma$) Theorem 3.3 with $$\begin{array}{c}{\Phi }_1(k,N)=(N_0+N+1)2^k-N_0-2,\\ {\Phi }_2(k)=\mathrm{\text{max}}\left\{1,\lceil {(80L^{2}b{2}^k)}^{\frac{1}{1-a-b}}\rceil -N_0-2\right\},\\ {\Phi }_3(k)=\{\begin{array}{cc}\mathrm{\text{max}}\left\{1,\lceil {\left(\frac{2^{k-1}(p-1)M_1^2{M}_2}{2a-1}\right)}^{\frac{1}{2a-1}}\rceil -N_0-1\right\}\hfill & \text{if}p\ge 2,\hfill \\ \mathrm{\text{max}}\left\{1,\lceil {\left(\frac{2^k{M}_1^p{M}_2}{p(\mathit{\text{pa}}-1)}\right)}^{\frac{1}{\mathit{\text{pa}}-1}}\rceil -N_0-1\right\}\hfill & \text{if}p<2,\hfill \end{array}\\ \alpha (n)=\mathrm{\text{max}}\left\{1,{(n+1)}^{\frac{1}{b}}-N_0-1\right\},\\ \gamma (n)={(N_0+n+1)}^b.\end{array}$$

Proof.

First observe that we get (using the estimates for ρ_X with $X=L^p$ from [26, p.63] or [27, p.193]) $$\begin{array}{c}\frac{{\rho }_X({\lambda }_{N_0+n})}{{\lambda }_{N_0+n}}\le \frac{p-1}{2}{\lambda }_{N_0+n}\\ =\frac{p-1}{2}{(N_0+n+1)}^{-a}\\ ={\theta }_{N_0+n}\frac{p-1}{2}{(N_0+n+1)}^{b-a}\\ \le {\theta }_{N_0+n}\frac{p-1}{2}{N}_0^{b-a}\\ \le {\gamma }_0{\theta }_{N_0+n},\end{array}$$if $p\ge 2$ and $$\begin{array}{c}\frac{{\rho }_X({\lambda }_{N_0+n})}{{\lambda }_{N_0+n}}\le \frac{1}{p}{\lambda }_{N_0+n}^{p-1}\\ =\frac{1}{p}{(N_0+n+1)}^{-a(p-1)}\\ ={\theta }_{N_0+n}\frac{1}{p}{(N_0+n+1)}^{b-a(p-1)}\\ \le {\theta }_{N_0+n}\frac{1}{p}{N}_0^{b-a(p-1)}\\ \le {\gamma }_0{\theta }_{N_0+n},\end{array}$$ if $1<p<2.$ Next we have $$\begin{array}{c}\prod _{n=N}^{{\Phi }_1(k,N)}(1-{\lambda }_{N_0+n+1}{\theta }_{N_0+n+1})=\prod _{n=N}^{{\Phi }_1(k,N)}(1-\frac{1}{{(N_0+n+2)}^{a+b}})\\ =\prod _{n=N}^{{\Phi }_1(k,N)}(1-\frac{{(N_0+n+2)}^{1-a-b}}{N_0+n+2})\\ \le \prod _{n=N}^{{\Phi }_1(k,N)}\frac{N_0+n+1}{N_0+n+2}\\ =\frac{N_0+N+1}{N_0+{\Phi }_1(k,N)+2}\\ =2^{-k}.\end{array}$$

Let $n\ge {\Phi }_2(k).$ Then (using the Bernoulli inequality ${(1+x)}^b\le 1+\mathit{\text{bx}}$ with exponent $0\le b\le 1$ and $x\ge -1$) $$\begin{array}{c}{\sigma }_n=C^{*}\frac{\frac{{\theta }_{N_0+n}}{{\theta }_{N_0+n+1}}-1}{{\lambda }_{N_0+n+1}{\theta }_{N_0+n+1}}\\ =40L^2{(N_0+n+2)}^{a+b}({(1+\frac{1}{N_0+n+1})}^b-1)\\ \le 40L^{2}b\frac{{(N_0+n+2)}^{a+b}}{N_0+n+1}\\ =40L^{2}b\frac{N_0+n+2}{N_0+n+1}{(N_0+n+2)}^{a+b-1}\\ \le 80L^{2}b{(N_0+n+2)}^{a+b-1}\\ \le 2^{-k}.\end{array}$$

Ad ${\Phi }_3(k):$ Using again the aforementioned estimates for ρ_X we have (for $2\le p<\infty$) $$\begin{array}{c}(*)\sum _{n={\Phi }_3(k)}^{\infty }{M}_2{\rho }_X({\lambda }_{n+1}{M}_1)\le \sum _{n={\Phi }_3(k)}^{\infty }\frac{p-1}{2}{M}_1^2{M}_2{(N_0+n+2)}^{-2a}\\ \le \frac{p-1}{2}{M}_1^2{M}_2\sum _{n={\Phi }_3(k)+N_0+2}^{\infty }{n}^{-2a}.\end{array}$$

Let $s:=2a>1.$ Then for $M\in \mathbb{N},M\ge 2$ $$\sum _{n=M}^{\infty }{n}^{-s}\le {\int }_{M-1}^{\infty }\frac{1}{x^s}\mathit{\text{ds}}=\frac{1}{s-1}·\frac{1}{{(M-1)}^{s-1}}.$$

Hence for $$M\ge {\left(\frac{2^{k-1}(p-1)M_1^2{M}_2}{2a-1}\right)}^{\frac{1}{2a-1}}+1$$we get $$\sum _{n=M}^{\infty }{n}^{-2a}\le 2^{-k}·\frac{2}{(p-1)M_1^2{M}_2}.$$

Together with $(*)$ this yields $$\sum _{n={\Phi }_3(k)}^{\infty }{M}_2{\rho }_X({\lambda }_{n+1}{M}_1)\le 2^{-k}.$$

The case $1<p<2$ is treated analogously.

One easily verifies that $\alpha ,\gamma$ satisfy the requirements from Theorem 3.3 for our choice of scalars. □

Remark 3.6.

A rate similar to ${\Phi }_2$ above can be also obtained using R₄ from [32, Section 5], where a different argument is used.

Acknowledgments

This paper grew out of a Bachelor thesis [33] of the first author written under the supervision of the 2nd author.

Additional information

Funding

The second author was supported by the “Deutsche Forschungsgemeinschaft” Project DFG KO 1737/6-2.

Notes

1 The existence of a modulus τ is equivalent to the uniform smoothness of X while ρX is also defined for non-smooth Banach spaces and only the aforementioned limit statement expresses uniform smoothness.

2 In [17], $g:{\mathbb{N}}_0\to {\mathbb{N}}_0$ and so for $g:\mathbb{N}\to \mathbb{N}$ we have to apply the rate given in [17] to $g\prime (n):=g(\mathrm{\text{max}}(1,n))$ and to replace 0 by 1 in the original bound.