May 25, 2026
Firmly nonexpansive operators arise naturally as resolvents of monotone operators and as generalizations of projections and proximal mappings in convex optimization and fixed point theory. While their iterates are known to converge weakly to a fixed point, strong convergence is not guaranteed (Genel and Lindenstrauss, 1975).
In this paper, we provide a new explicit family of counterexamples in infinite-dimensional Hilbert spaces. In the harmonic case the Cesàro means of the iterates remain bounded away from the unique fixed point. A block-construction variant yields Cesàro means whose norms oscillate in the sense that their liminf is zero while their limsup is positive. These results show that von Neumann’s classical mean ergodic theorem for linear operators does not extend to Baillon’s nonlinear mean ergodic theorem even in the firmly nonexpansive setting, and they illustrate inherent limitations of averaging techniques in infinite-dimensional optimization.
2020 Mathematics Subject Classification: Primary 47H09, 47H10; Secondary 47A35, 47J25, 65K05, 90C25.
Baillon’s Nonlinear Mean Ergodic Theorem, Cesàro means, firmly nonexpansive mapping, Genel–Lindenstrauss example, Hilbert space, nonexpansive mapping, strong convergence, von Neumann’s Linear Mean Ergodic Theorem, weak convergence.
Throughout this paper,
with inner product \(\left\langle{\cdot},{\cdot}\right\rangle\) and induced norm \(\lVert\cdot\rVert\).
Suppose that \[\mathbf{T}\colon \mathbf{X}\to \mathbf{X} \;\;\text{is nonexpansive, with} \; \operatorname{Fix}\mathbf{T}\neq \varnothing.\] Finding a fixed point of \(\mathbf{T}\) is a central problem in optimization and variational analysis. Unfortunately, without any additional assumptions, iterating \(\mathbf{T}\) may not yield a solution (consider \(\mathbf{T}=-\operatorname{Id}\) with a starting point that is not the origin.) If, however, \(\mathbf{T}\) is \(\alpha\)-averaged nonexpansive, i.e., \(\mathbf{T}\) can be written as \((1-\alpha)\operatorname{Id}+\alpha \mathbf{N}\), with \(\alpha \in\left[0,1\right[\) and \(\mathbf{N}\colon \mathbf{X}\to \mathbf{X}\) nonexpansive, then the iterates of \(\mathbf{T}\) converge weakly to a point in \(\operatorname{Fix}\mathbf{T}\). An important special case is when \(\mathbf{T}\) is firmly nonexpansive, i.e., \(\tfrac{1}{2}\)-averaged nonexpansive. An obvious question is whether the convergence can fail to be strong. The answer is affirmative due to a now-classical example from 1975:
Example 1 (Genel-Lindenstrauss). (See [1].)
Suppose that \(\mathbf{X}=\ell^2\). Then there exist a bounded closed convex subset \(\mathbf{C}\) of \(\mathbf{X}\), a firmly nonexpansive mapping \(\mathbf{T}\colon \mathbf{C}\to \mathbf{C}\), and a starting point \(\mathbf{x}\in \mathbf{C}\) such that \(\mathbf{T}^nx\:{\rightharpoonup}\:\boldsymbol{0}\in \operatorname{Fix}\mathbf{T}\) but \(\inf_{n\geq 1}\|\mathbf{T}^n\mathbf{x}\|\geq \tfrac{1}{2}\).
Hence iterating \(\mathbf{T}\), even when \(\mathbf{T}\) is firmly nonexpansive, need not necessarily produce a sequence that converges strongly to a fixed point of \(\mathbf{T}\). It is thus tempting to consider Cesàro means of the iterates. Indeed, von Neumann’s linear mean ergodic theorem gives strong convergence in the linear case even when \(\mathbf{T}\) is merely assumed to be nonexpansive:
Fact 1 (von Neumann).
(See [2], and also [3], [4], [5], and [6].) Suppose that \(\mathbf{T}\colon \mathbf{X}\to\mathbf{X}\) is linear and nonexpansive. Then \[\frac{1}{n}\sum_{k=1}^n \mathbf{T}^{k-1}\mathbf{x}\to P_{\operatorname{Fix}\mathbf{T}}\mathbf{x}.\]
The nonlinear variant of 1 was provided in 1975 by Baillon:
Fact 2 (Baillon).
(See [7], and also [8].) Suppose that \(\mathbf{T}\colon\mathbf{X}\to\mathbf{X}\) is nonexpansive, with \(\operatorname{Fix}\mathbf{T}\neq\varnothing\). Then \[\frac{1}{n}\sum_{k=1}^n \mathbf{T}^{k-1}\mathbf{x}\:{\rightharpoonup}\: \text{ some point in \operatorname{Fix}\mathbf{T}.}\]
To the best of our knowledge, it is not known whether the convergence in 2 may fail to be strong.
The goal of this work is to settle this question. Our main results can be summarized as follows:
We will provide a new family of firmly nonexpansive mappings \(\mathbf{T}\) such that \((\mathbf{T}^n\mathbf{x})_{n\geq 1}\) converges weakly but not strongly (see 8). This is similar in spirit to [f:GL]; however, our construction is simpler.
For an incarnation of \(\mathbf{T}\) from R1, we will show that even the Cesàro means of the iterates fail to converge strongly and actually stay a positive distance away from the unique fixed point of \(\mathbf{T}\) (see 3).
We will also present another incarnation where one subsequence of the Cesàro means stays away from the unique fixed point of \(\mathbf{T}\) while another subsequence converges strongly to the fixed point (see [sec:ss:sumbiz]).
The remainder of this paper is organized as follows. In [sec:s:motiv], we provide the motivation for our abstract construction while realizations are discussed in [sec:s:constr]. In [sec:s:aux], we record a few inequalities that will be useful in later sections. We then introduce the underlying curve and mesh in [sec:s:curvmesh]. These will play a central role in [sec:s:seq], where we introduce the sequence of vectors that will eventually be the iterates. [sec:s:main] contains the main results R1 and R2 while R3 is presented in [s:bizarre].
The notation we employ is fairly standard and follows largely [9].
Suppose that \(\mathbf{T}\colon \mathbf{X}\to \mathbf{X}\) is a firmly nonexpansive mapping such that \(\operatorname{Fix}\mathbf{T}= \{\boldsymbol{0}\}\), with \(\mathbf{x}_{n} = \mathbf{T}^{n-1}\mathbf{x}_1\:{\rightharpoonup}\:\boldsymbol{0}\) but \(\mathbf{x}_n\not\to \boldsymbol{0}\). Because \(\mathbf{T}\) is firmly nonexpansive, we have \(\|\mathbf{T}\mathbf{x}_n -\mathbf{T}\boldsymbol{0}\|^2 \leq \left\langle{\mathbf{x}_n-\boldsymbol{0}},{\mathbf{T}\mathbf{x}_n - \mathbf{T}\boldsymbol{0}}\right\rangle\), i.e., \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:ant1} \endgroup \|\mathbf{x}_{n+1}\|^2 \leq \left\langle{\mathbf{x}_n},{\mathbf{x}_{n+1}}\right\rangle.\tag{1}\] Cauchy–Schwarz now implies that \(\|\mathbf{x}_{n+1}\|\leq\|\mathbf{x}_n\|\); hence, \((\|\mathbf{x}_n\|)_{n\geq 1}\) is decreasing and thus convergent. To avoid strong convergence, we must ensure that \(\lim_{n\to\infty}\|\mathbf{x}_n\| > 0\).
Now assume that \((\mathbf{u}(t))_{t\in\mathbb{R}_+}\) is a curve of unit vectors in \(\mathbf{X}\) that weakly converges to \(\boldsymbol{0}\) as \(t\to\infty\). We make the ansatz \[\mathbf{x}_n = \rho_n \mathbf{u}(t_n),\] where \((t_n)_{n\geq 1}\) increases to \(+\infty\) and \((\rho_n)_{n\geq 1}\) decreases to a \(\rho_\infty>0\). Then \(\mathbf{x}_n\:{\rightharpoonup}\:\boldsymbol{0}\), and \(\|\mathbf{x}_n\| = \rho_n \to \rho_\infty > 0\); consequently, \(\mathbf{x}_n\not\to \boldsymbol{0}\).
To make progress, we solve 1 with equality: \(\|\mathbf{x}_{n+1}\|^2 = \left\langle{\mathbf{x}_n},{\mathbf{x}_{n+1}}\right\rangle\) turns into \(\rho_{n+1}^2 = \rho_n \rho_{n+1} \left\langle{\mathbf{u}(t_n)},{\mathbf{u}(t_{n+1})}\right\rangle\) or \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:ant2} \endgroup \rho_{n+1} = \rho_n \left\langle{\mathbf{u}(t_n)},{\mathbf{u}(t_{n+1})}\right\rangle.\tag{2}\] Following the machine learning literature, we can think of \(\mathbf{u}(t)\) as a feature map, with kernel \(K\): \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:ant3} \endgroup \left\langle{\mathbf{u}(s)},{\mathbf{u}(t)}\right\rangle = K(s,t).\tag{3}\] Then 2 reads \[\rho_{n+1} = \rho_n K(t_n,t_{n+1}).\] We want the kernel \(K\) to satisfy \(K(t_n,t_{n+1})<1\) (to model the decrease of \(\rho_n\)) yet \[\prod_{n=1}^\infty K(t_n,t_{n+1}) > 0 \quad\text{(to model \rho_\infty>0)}.\] Suppose further that the kernel \(K\) is translation invariant, i.e., \(K(s,t) = \Phi(|s-t|)\) for some function \(\Phi\colon\mathbb{R}_+\to\mathbb{R}_+\). Writing \(d_n := t_{n+1}-t_n\), we thus want \(\sum_{n=1}^\infty d_n = +\infty\) (to guarantee \(t_n\to\infty\)) yet \[\prod_{n=1}^\infty \Phi(d_n) > 0;\] equivalently, \[\sum_{n=1}^\infty -\ln \Phi(d_n) < +\infty.\] The simplest way to achieve this is to assume that \(-\ln \Phi(d_n) = d_n^2\), i.e., \(\Phi(d_n) = \exp(-d_n^2)\), with \(\sum_{n=1}^\infty d_n^2 < +\infty\) but \(\sum_{n=1}^\infty d_n = +\infty\) which is possible when we consider the harmonic series. This leads us to \(\Phi(d) = \exp(-d^2)\), and so we are led to the Gaussian kernel (see also [10]) \[K(s,t) = \exp(-(s-t)^2).\] Then 3 turns into \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:ant4} \endgroup \left\langle{\mathbf{u}(s)},{\mathbf{u}(t)}\right\rangle = \exp(-(s-t)^2).\tag{4}\]
In [sec:s:constr], we will present two realizations for \((\mathbf{u}(t))_{t\in\mathbb{R}_+}\) that satisfy the inner product condition 4 .
In this section, we present two realizations for the curve \(\mathbf{u}\) that satisfies the inner product condition 4 , i.e., \[\left\langle{\mathbf{u}(s)},{\mathbf{u}(t)}\right\rangle = \exp(-(s-t)^2).\]
Suppose that \((\mathbf{e}_k)_{k\in\mathbb{N}}\) is an orthonormal sequence in \(\mathbf{X}\), and that \((\mathbf{e}_k)_{k\in\mathbb{N}}\) is part of an orthonormal basis of \(\mathbf{X}\). Now define \[(\forall t\in\mathbb{R}_+)\quad \mathbf{u}(t) := \sum_{k=0}^\infty u_k(t)\mathbf{e}_k, \quad \text{where}~ u_k(t) = \exp(-t^2)\sqrt{\frac{2^k}{k!}}t^k.\] Then \[\begin{align} \left\langle{\mathbf{u}(s)},{\mathbf{u}(t)}\right\rangle & = \sum_{k=0}^\infty u_k(s)u_k(t) = \sum_{k=0}^\infty \exp(-s^2)\sqrt{\frac{2^k}{k!}}s^k \exp(-t^2)\sqrt{\frac{2^k}{k!}}t^k\\ &= \exp(-s^2)\exp(-t^2)\sum_{k=0}^\infty \frac{(2st)^k}{k!} = \exp(-s^2)\exp(-t^2)\exp(2st)\\ &= \exp(-(s-t)^2), \end{align}\] as desired.
The most concrete3 version is \(\mathbf{X}=\ell_2\), with \(\mathbf{e}_k\) being the \(k\)th standard unit vector.
The following elegant realization and its derivation were suggested by ChatGPT 5.5. Suppose that \(\mathbf{X}=L^2(\mathbb{R})\), and define \(\mathbf{u}(t)\) by \[\mathbf{u}(t) \colon \mathbb{R}\to\mathbb{R}\colon r \mapsto \frac{1}{\sqrt[4]{\pi}}
\exp\Big(-\frac{(r-2t)^2}{2}\Big),\] which is a normalized translated Gaussian, with center at \(2t\).
Using the substitution \(u = r - s - t\), one obtains \[\begin{align} \langle \mathbf{u}(s),\mathbf{u}(t)\rangle &= \frac{1}{\sqrt{\pi}} \int_{\mathbb{R}} \exp\Big(-\frac{(r-2s)^2}{2}\Big) \exp\Big(-\frac{(r-2t)^2}{2}\Big)\,dr \\ &= \frac{1}{\sqrt{\pi}} \int_{\mathbb{R}} \exp\Big( -\frac{(r-2s)^2+(r-2t)^2}{2} \Big)\,dr\\ &= \frac{1}{\sqrt{\pi}} \int_{\mathbb{R}} \exp\big( -(r-s-t)^2-(s-t)^2 \big)\,dr\\ &= \frac{1}{\sqrt{\pi}} \exp(-(s-t)^2) \int_{\mathbb{R}} \exp\big( -(r-s-t)^2 \big)\,dr\\ &= \frac{1}{\sqrt{\pi}} \exp(-(s-t)^2) \int_{\mathbb{R}} \exp\big( -u^2 \big)\,du\\ &= \exp(-(s-t)^2). \end{align}\]
While \(\mathbf{X}=L^2(\mathbb{R})\) is separable, this construction is more elegant because it does not rely on working with a given orthonormal Schauder basis4 and the expansion of the exponential function.
We first prove a technical inequality for future use.
Proposition 3.
Let \(0 \leq x \leq 1/16\). Then \[\exp(2x+16x^2) \leq 1+32x.\]
. Note that \(80x + 512x^2 \leq 80/16 + 512(1/16)^2 = 7<30\). Hence \(80x^2+512x^3 \leq 30x\), and so \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{260515a} \endgroup (2x+16x^2)(1+32x) = 2x + 80x^2 + 512x^3 \leq 2x + 30x = 32x.\tag{5}\] Set \(w := 2x+16x^2\). Then \(0\leq w \leq 2/16+16(1/16)^2 = 3/16<1\) and we learn from 5 that \(w(1+32x) \leq 32x\), which we re-arrange to the \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:steak1} \endgroup \frac{1}{1-w} \leq 1 + 32x.\tag{6}\] On the other hand, since \(0\leq w < 1\), we have \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:steak2} \endgroup \exp(2x+16x^2) = \exp(w) = \sum_{n=0}^\infty \frac{w^n}{n!} \leq \sum_{n=0}^\infty w^n = \frac{1}{1-w}.\tag{7}\] Combining 6 with 7 , we obtain the conclusion. \(\hfill \quad \blacksquare\)
The next two results will be useful in later sections.
Lemma 1.
Suppose that two real numbers \(x,y\) satisfy \[0 \leq x \leq \frac{1}{16}\quad\text{and} \quad x+16x^2 \leq y.\] Then \[\exp(x)+\exp(-y) \leq 2.\]
. Because \(\exp(-\cdot)\) is decreasing, it is clear that \(\exp(x)+\exp(-y) \leq \exp(x)+\exp(-x-16x^2)\). It thus suffices to show that \[g(x) := \exp(x)+\exp(-x-16x^2) \leq 2.\] Note that \(g'(x) = \exp(x) - (1+32x)\exp(-x-16x^2)\). Hence \[g'(x) \leq 0 \Leftrightarrow \exp(2x+16x^2) \leq 1 + 32x,\] which holds true by 3. Therefore, \(g\) is decreasing on \([0,1/16]\), and so \(g(x) \leq g(0) = 2\), and we’re done. \(\hfill \quad \blacksquare\)
Lemma 2.
Let \(0\leq s_1 < s_2 <\cdots < s_n\) be real numbers that are separated at least by some \(\alpha\in\left]0,1\right]\): \[(\forall i\in\{1,\ldots,n-1\})\quad s_{i+1}-s_i \geq \alpha.\] Then \[\max_{i\in\{1,\ldots,n\}} \sum_{j=1}^n \exp\big(-(s_i-s_j)^2\big) \leq \frac{1+\sqrt{\pi}}{\alpha}.\]
. Let \(i\in\{1,\ldots,n\}\) be fixed. If \(j=i\pm k\), then \(|s_j - s_i| \geq k\alpha\), and so \[\sum_{j=1}^n \exp\big(-(s_i-s_j)^2\big) \leq 1 + 2\sum_{k=1}^\infty \exp\big(-(k\alpha)^2\big)\] On the other hand, a Riemann-sum argument and [13] show that \[\begin{align} \sum_{k=1}^\infty \exp\big(-(k\alpha)^2\big) &\leq \int_{0}^\infty \exp\big(-(t\alpha)^2\big) dt = \frac{1}{\alpha}\int_{0}^\infty \exp\big(-u^2\big) du = \frac{\sqrt{\pi}}{2\alpha}. \end{align}\] Altogether, we have \[\sum_{j=1}^n \exp\big(-(s_i-s_j)^2\big) \leq 1 + \frac{\sqrt{\pi}}{\alpha} \leq \frac{1+\sqrt{\pi}}{\alpha},\] which implies the conclusion. \(\hfill \quad \blacksquare\)
This section presents properties of the curve and introduces the mesh. The two objects are central in the construction in the next section.
For the rest of this paper, we assume that we are given a curve of unit vectors that satisfies
See [sec:s:constr] for realizations of such a curve.
Proposition 4. We have \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:david5} \endgroup \mathbf{u}(t)\:{\rightharpoonup}\:\boldsymbol{0}\quad\text{as}\quad t\to\infty\qquad{(1)}\] and \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:david6} \endgroup (\forall s,t \in\mathbb{R}_+)\quad \lVert\mathbf{u}(s)-\mathbf{u}(t)\rVert^2 = 2\big(1-\exp(-(s-t)^2)\big).\qquad{(2)}\]
. Set \(\mathbf{Y}:= \overline{\operatorname{span}}\,\{\mathbf{u}(t)\}_{t\in\mathbb{R}_+}\), and let \(\mathbf{y}\in \mathbf{Y}\) and \(\mathbf{z}\in \mathbf{Y}^\perp\). For fixed \(s\in\mathbb{R}_+\), we have \(\lim_{t\to\infty} \left\langle{\mathbf{u}(s)},{\mathbf{u}(t)}\right\rangle = \lim_{t\to\infty} \exp(-(s-t)^2) = 0\). This implies that \(\left\langle{\mathbf{y}},{\mathbf{u}(t)}\right\rangle \to 0\) as \(t\to\infty\), while \(\left\langle{\mathbf{z}},{\mathbf{u}(t)}\right\rangle\equiv 0\). Thus \(\left\langle{\mathbf{y}+\mathbf{z}},{\mathbf{u}(t)}\right\rangle \to 0\) as \(t\to\infty\), and ?? follows. Now let \(s,t\) be in \(\mathbb{R}_+\). Then \(\lVert\mathbf{u}(s)-\mathbf{u}(t)\rVert^2 = \lVert\mathbf{u}(s)\rVert^2 + \lVert\mathbf{u}(t)\rVert^2 - 2\left\langle{\mathbf{u}(s)},{\mathbf{u}(t)}\right\rangle = 2(1-\exp(-(s-t)^2))\). \(\hfill \quad \blacksquare\)
Remark 5. It is clear from ?? that the curve \(\mathbf{u}\) is continuous. One can show5 that \(\mathbf{u}\) is even continuously differentiable with \[\left\langle{\mathbf{u}'(t)},{\mathbf{u}(t)}\right\rangle = 0, \;\; \left\langle{\mathbf{u}'(s)},{\mathbf{u}'(t)}\right\rangle = \big(2-4(s-t)^2\big)\exp(-(s-t)^2), \;\; \|\mathbf{u}'(t)\| = \sqrt{2},\] and \[\|\mathbf{u}'(s)-\mathbf{u}'(t)\|^2 = 4-2\big(2-4(s-t)^2\big)\exp(-(s-t)^2).\] In the context of [sec:ss:ons], we have \[\mathbf{u}'(t) := \sum_{k=0}^\infty v_k(t)\mathbf{e}_k, \quad \text{where}~ v_k(t) = \exp(-t^2)\sqrt{\frac{2^k}{k!}}\big( kt^{k-1}-2t^{k+1} \big),\] where for \(k=0\) we set \(kt^{k-1} := 0\). In the context of [sec:ss:L2], we have \[\mathbf{u}'(t) \colon \mathbb{R}\to\mathbb{R}\colon r \mapsto \frac{1}{\sqrt[4]{\pi}} 2(r-2t) \exp\Big(-\frac{(r-2t)^2}{2}\Big).\]
From now on, we assume that \((d_n)_{n\geq 1}\) is a fixed sequence of step sizes or mesh increments \((d_n)_{n\geq 1}\) such that
Hence \((d_n)_{n\geq 1} \in \ell^2\smallsetminus\ell^1\), \(t_1 = 0\), and \((t_n)_{n\geq 1}\) is the mesh sequence we place over the nonnegative reals. We will use repeatedly that \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:tndn} \endgroup (\forall n\geq 1)\quad t_{n+1} - t_n = d_n.\tag{8}\] Note that we could also have started with the mesh sequence \((t_n)_{n\geq 1}\), and obtained the step sizes via \(d_n := t_{n+1}-t_n\). We will also assume from now on that \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \tag{9} \endgroup \begin{empheq}[box=\colorbox{myblue}{]}{equation} \begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \tag{10} \endgroup d_1 \leq \tfrac{1}{8} \end{empheq} and \begin{empheq}[box=\colorbox{myblue}{]}{equation} \begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \tag{11} \endgroup (\forall n\geq 1)\quad d_{n+1} \leq \frac{d_n}{1+64d_n^2}. \end{empheq}\] Consequently, \((d_n)_{n\geq 1}\) is strictly decreasing. Note that 11 is equivalent to \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:dmeshold} \endgroup (\forall n\geq 1)\quad \frac{1}{d_{n+1}}-\frac{1}{d_n} \geq 64d_n = 64(t_{n+1}-t_n),\tag{12}\] as well as to \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:dntninc} \endgroup (\forall n\geq 1)\quad \frac{1}{d_{n}} - 64t_n \leq \frac{1}{d_{n+1}} - 64t_{n+1}; \quad\text{consequently, \Big(\frac{1}{d_n} - 64t_n\Big)_{n\geq 1} is increasing}.\tag{13}\] Moreover, if one of the inequalities in 11 ,12 ,13 is an equality, then so are the other two.
Example 2 (harmonic mesh).
Let \(\delta \in \left]0,1/8\right]\) and suppose that \[(\forall n\geq 1)\quad d_n = \frac{\delta}{n}.\] Then [e:dmesh00] and 9 hold.
. Clearly, each \(d_n>0\), \(\sum_{k=1}^\infty d_k^2 = {\delta^2}\sum_{k=1}^\infty \frac{1}{k^2} < +\infty\) while \(t_n=\delta H_{n-1}\to +\infty\) where \(H_n\) denotes the \(n\)th harmonic number. Hence [e:dmesh00] holds. It is clear that 10 holds. Now let \(n\geq 1\). Then \[\begin{align} \frac{1}{d_{n+1}}-\frac{1}{d_n} &= \frac{n+1}{\delta} - \frac{n}{\delta} = \frac{1}{\delta} \geq 64 \delta \geq 64 \cdot \frac{\delta}{n} = 64d_n. \end{align}\] Hence 12 holds, and so does the equivalent 11 . \(\hfill \quad \blacksquare\)
Lemma 3. The following hold: \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:dmesh1} \endgroup (\forall 1\leq m<n)\quad \frac{1}{d_n}-\frac{1}{d_m} \geq 64(t_n-t_m),\qquad{(3)}\] \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:dmesh2} \endgroup (\forall n\geq 1)\quad d_nt_{n+1} < \frac{1}{32},\qquad{(4)}\] \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:meshcool} \endgroup (\forall 1\leq m<n)\quad \exp\big(2d_n(t_{n+1}-t_{m+1})\big)+ \exp\big(-2d_m(t_{n+1}-t_{m+1})\big) \leq 2.\qquad{(5)}\]
. ?? : Let \(1\leq m < n\). Then 12 , which is equivalent to 11 , implies that \[\begin{align} 64(t_n-t_m)&= \sum_{k=m}^{n-1} 64d_k \leq \sum_{k=m}^{n-1} \Big(\frac{1}{d_{k+1}}-\frac{1}{d_{k}}\Big) = \frac{1}{d_n}-\frac{1}{d_m}. \end{align}\]
?? : Let \(n\geq 1\). From 13 , which is equivalent to 11 , we know that \((1/d_n - 64t_n)_{n\geq 1}\) is increasing. Recalling 10 , we have in particular that \(8 \leq 1/d_1 -0 = 1/d_1 -64t_1 \leq 1/d_n -64t_n\). So \(64t_n<1/d_n\) and \(8\leq 1/d_n\). Therefore, \[d_nt_{n+1} = d_n(t_n + d_n) = d_nt_n + d_n^2 < \frac{1}{64} + \frac{1}{8^2} = \frac{1}{32}.\]
?? : Let \(1\leq m<n\) and abbreviate \(\delta := t_{n+1}-t_{m+1}<t_{n+1}\). Then 11 implies that \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:260516c} \endgroup d_n<d_m;\tag{14}\] thus, \[\delta = t_{n+1}-t_{m+1} = t_n-t_m+(t_{n+1}-t_n) - (t_{m+1}-t_m) = t_n-t_m+d_n - d_m < t_n-t_m.\] Combining this with ?? yields \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:260516b} \endgroup \frac{1}{d_n}-\frac{1}{d_m} > 64\delta.\tag{15}\] Now set \[\xi := 2d_n\delta \quad\text{and}\quad \eta := 2d_m\delta.\] Because \(\delta<t_{n+1}\) and ?? hold, we have \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:260516dd} \endgroup \begin{equation} 0 < \xi < 2d_nt_{n+1} \leq \frac{1}{16}. \end{equation} Using \ref{e:260516c} and \ref{e:260516b}, we estimate \begin{align} \eta-\xi &= 2\delta(d_m-d_n) = 2\delta d_nd_m\Big(\frac{1}{d_n}-\frac{1}{d_m}\Big) > 2\delta d_n^2 \cdot 64\delta = 32 \cdot 4d_n^2\delta^2 = 32\xi^2 > 16\xi^2. \end{align}\tag{16}\] Combining 16 with 1 yields \(\exp(\xi) + \exp(-\eta) \leq 2\), as announced. \(\hfill \quad \blacksquare\)
As explained in [sec:s:motiv], we now define the sequence of scalars \((\rho_n)_{n\geq 1}\) by
Proposition 6.
The sequence \((\rho_n)_{n\geq 1}\) satisfies \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:rho1} \endgroup (\forall n\geq 1)\quad \rho_n = \exp\Big(-\sum_{k=1}^{n-1} d_k^2\Big);\qquad{(6)}\] moreover,
. Because each \(d_n>0\), it is clear that \(\rho_{n+1}<\rho_n\). The identity ?? follows by induction. Since \(\sum_{k=1}^\infty d_k^2 < +\infty\), we have \(\rho_\infty > 0\) and \(\rho_n \to \rho_\infty\) as \(n\to\infty\). \(\hfill \quad \blacksquare\)
We are now ready to define the key sequence \((\mathbf{x}_n)_{n\geq 1}\) in this paper, namely
We also consider the smallest closed convex cone containing the corresponding set \(\{\mathbf{x}_n\}_{n\geq 1}\):
Proposition 7.
We have \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:kurve} \endgroup \mathbf{x}_n \:{\rightharpoonup}\:\boldsymbol{0}\;\;\text{and}\;\; \lVert\mathbf{x}_n\rVert =\rho_n \to \rho_\infty > 0.\qquad{(7)}\] Let \(1\leq m\leq n\). Then the following hold:
\(\left\langle{\mathbf{x}_m},{\mathbf{x}_n}\right\rangle = \rho_m\rho_n\exp(-(t_n-t_m)^2)\) and \(\|\mathbf{x}_n\|=\rho_n > \rho_{n+1} = \|\mathbf{x}_{n+1}\|\).
\(\left\langle{\mathbf{K}},{\mathbf{K}}\right\rangle \geq 0\), and so \(\mathbf{K}\) is an acute cone, i.e., \(\mathbf{K}\subseteq \mathbf{K}^{\oplus}\).
\(\left\langle{\mathbf{x}_{n+1}},{\mathbf{x}_n}\right\rangle = \|\mathbf{x}_{n+1}\|^2\).
\(\left\langle{\mathbf{x}_{n+1}-\mathbf{x}_n},{\mathbf{x}_{n+1}}\right\rangle = 0\).
. ?? follows from [xn], ?? , and [e:rho2]. [kurve1]: Clear from [xn] and [e:Kurve]. [kurve4]: This follows from [kurve1]. [kurve2]: Indeed, [kurve1] and [e:rho] yield \[\left\langle{\mathbf{x}_{n+1}},{\mathbf{x}_n}\right\rangle = \rho_{n+1}\rho_n\exp(-(t_{n+1}-t_n)^2) = \rho_{n+1}\rho_n\exp(-d_n^2) =\rho_{n+1}^2 = \|\mathbf{x}_{n+1}\|^2.\] [kurve3]: [kurve2] implies \(\left\langle{\mathbf{x}_{n+1}-\mathbf{x}_n},{\mathbf{x}_{n+1}}\right\rangle = \left\langle{\mathbf{x}_{n+1}},{\mathbf{x}_{n+1}}\right\rangle - \left\langle{\mathbf{x}_n},{\mathbf{x}_{n+1}}\right\rangle = \|\mathbf{x}_{n+1}\|^2 - \|\mathbf{x}_{n+1}\|^2 = 0\). \(\hfill \quad \blacksquare\)
For later convenience, and also motivated by ?? , we denote the weak limit of \((\mathbf{x}_n)_{n\geq 1}\) by
Lemma 4.
We have \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:schap} \endgroup (\forall 1\leq m < n \leq \infty)\quad 0\leq \left\langle{\mathbf{x}_{m+1}-\mathbf{x}_{n+1}},{(\mathbf{x}_{m}-\mathbf{x}_{m+1})-(\mathbf{x}_{n}-\mathbf{x}_{n+1})}\right\rangle.\qquad{(8)}\]
. Let \(1\leq m < n \leq \infty\). We will argue by cases. In both cases, we will work backward from the desired conclusion ?? by reversible algebraic manipulations.
Case 1: \(n=\infty\). Then \(\mathbf{x}_{n+1} = \mathbf{x}_\infty = \boldsymbol{0}\) (recall [e:bxinf]), and so ?? reduces to \(0\leq\left\langle{\mathbf{x}_{m+1}},{\mathbf{x}_{m}-\mathbf{x}_{m+1}}\right\rangle\), which holds with equality by 7[kurve3].
Case 2: \(n<\infty\). Using again 7[kurve3], we see that ?? is equivalent to \[0 \leq - \left\langle{\mathbf{x}_{m+1}},{\mathbf{x}_{n}-\mathbf{x}_{n+1}}\right\rangle - \left\langle{\mathbf{x}_{n+1}},{\mathbf{x}_{m}-\mathbf{x}_{m+1}}\right\rangle,\] and hence also to \[0 \leq 2\left\langle{\mathbf{x}_{m+1}},{\mathbf{x}_{n+1}}\right\rangle - \left\langle{\mathbf{x}_{m+1}},{\mathbf{x}_n}\right\rangle - \left\langle{\mathbf{x}_{n+1}},{\mathbf{x}_m}\right\rangle.\] In view of 7[kurve1], this is equivalent to \[0 \leq 2\rho_{m+1}\rho_{n+1}\exp(-(t_{n+1}-t_{m+1})^2) - \rho_{m+1}\rho_n\exp(-(t_n-t_{m+1})^2) - \rho_{n+1}\rho_m\exp(-(t_{n+1}-t_m)^2).\] In turn, using [e:rho], we can rewrite this condition as \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:david1} \endgroup \begin{align} 0 &\leq 2\rho_{m}\exp(-d_m^2)\cdot \rho_{n}\exp(-d_n^2)\cdot\exp(-(t_{n+1}-t_{m+1})^2) \\ &\quad - \rho_{m}\exp(-d_m^2)\cdot \rho_n\cdot \exp(-(t_n-t_{m+1})^2) \\ &\quad - \rho_{n}\exp(-d_n^2)\cdot \rho_m\cdot \exp(-(t_{n+1}-t_m)^2). \end{align}\tag{17}\] Abbreviate \(\delta := t_{n+1}-t_{m+1}\) and recall that \(t_{n+1} = t_n + d_n\) (see 8 ). Then \(t_n-t_{m+1} = \delta - d_n\) and \(t_{n+1}-t_m = \delta + d_m\). Hence 17 is equivalent to \[0 \leq \rho_m\rho_n \big(2\exp(-d_m^2-d_n^2-\delta^2) - \exp(-d_m^2-(\delta-d_n)^2) - \exp(-d_n^2-(\delta+d_m)^2)\big),\] and, after dividing by \(\rho_m\rho_n>0\) and expanding the squares, to \[\begin{align} 0 &\leq 2\exp(-d_m^2-d_n^2-\delta^2) - \exp(-d_m^2-\delta^2 + 2d_n\delta-d_n^2) -\exp(-d_n^2-\delta^2 - 2d_m\delta-d_m^2) \\ &= \exp(-d_m^2-d_n^2-\delta^2) \big(2 - \exp(2d_n\delta) - \exp(-2d_m\delta)\big). \end{align}\] Dividing by \(\exp(-d_m^2-d_n^2-\delta^2)>0\), we now face \[0 \leq 2 - \exp(2d_n\delta) - \exp(-2d_m\delta);\] however, this is just a re-arrangement of ?? . \(\hfill \quad \blacksquare\)
Lemma 5.
Let \(\mathbf{y}\in \mathbf{K}\), and suppose that \[\begin{align} \begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:david3} \endgroup (\forall 1\leq n< \infty)\quad 0 \leq \left\langle{\mathbf{x}_{n+1}-\mathbf{y}},{\mathbf{x}_{n}-\mathbf{x}_{n+1}}\right\rangle. \end{align}\qquad{(9)}\] Then \(\mathbf{y}=\boldsymbol{0}\).
. In view of 7[kurve3], we see that ?? is equivalent to \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:david4} \endgroup (\forall 1\leq n< \infty)\quad \left\langle{\mathbf{x}_n},{\mathbf{y}}\right\rangle \leq \left\langle{\mathbf{x}_{n+1}},{\mathbf{y}}\right\rangle,\tag{18}\] i.e., the sequence \((\left\langle{\mathbf{x}_n},{\mathbf{y}}\right\rangle)_{n\geq 1}\) is increasing. Moreover, \(\left\langle{\mathbf{x}_n},{\mathbf{y}}\right\rangle\to 0\) by ?? . Combining these two observations yields \((\forall n\geq 1)\) \(\left\langle{\mathbf{x}_n},{\mathbf{y}}\right\rangle \leq 0\); hence (recall [bK]), \[\mathbf{y}\in \mathbf{K}^\ominus.\] On the other hand, \(\mathbf{y}\in\mathbf{K}\) by assumption. Altogether, \(0\leq \|\mathbf{y}\|^2 = \left\langle{\mathbf{y}},{\mathbf{y}}\right\rangle \leq 0\) and we’re done. \(\hfill \quad \blacksquare\)
Lemma 6.
Let \(n\geq 1\), and let \(I\) be a nonempty subset of \(\{1,\ldots,n\}\). Set \(\delta := \max\big\{{|t_i-t_j|}~\big|~{i,j\in I}\big\}\). Then \[\bigg\|\frac{1}{n}\sum_{k=1}^n\mathbf{x}_k \bigg\| \geq \frac{\# I}{n}\rho_\infty \exp(-\delta^2/2)>0.\]
. By 7, we have \[(\forall i,j\in I)\quad \left\langle{\mathbf{x}_i},{\mathbf{x}_j}\right\rangle \geq \rho_\infty^2\exp(-\delta^2) > 0.\] Hence \[\begin{align} \bigg\|\frac{1}{n}\sum_{k=1}^n\mathbf{x}_k \bigg\|^2 &= \frac{1}{n^2}\sum_{i,j=1}^n \left\langle{\mathbf{x}_i},{\mathbf{x}_j}\right\rangle \geq \frac{1}{n^2}\sum_{i,j\in I} \left\langle{\mathbf{x}_i},{\mathbf{x}_j}\right\rangle \geq \frac{(\# I)^2}{n^2}\rho_\infty^2\exp(-\delta^2). \end{align}\] Now take the square root, and we’re done. \(\hfill \quad \blacksquare\)
We are now ready for our first main result, concerning regular iterates.
Theorem 8 (a firmly nonexpansive iteration that converges weakly but not strongly).
Set \[\mathbf{C}:= \overline{\operatorname{conv}}\,\{\mathbf{x}_n\}_{n\geq 2}.\] Then there exists a firmly nonexpansive operator \[\mathbf{T}\colon \mathbf{X}\to \mathbf{X}\] such that \[\overline{\operatorname{ran}}\,\mathbf{T}=\mathbf{C},\;\; \operatorname{Fix}\mathbf{T}= \{\boldsymbol{0}\}, \;\; (\forall n\geq 1)\; \mathbf{T}\mathbf{x}_n = \mathbf{x}_{n+1},\] and \[\mathbf{T}^n\mathbf{x}_1\:{\rightharpoonup}\:\boldsymbol{0}, \quad\text{ but } \quad \inf_{1\leq n<\infty} \|\mathbf{T}^n\mathbf{x}_1\|=\rho_\infty >0.\]
. Since \((\mathbf{x}_n)_{n\geq 2}\) lies in \(\mathbf{C}\), which is weakly closed, and since \(\mathbf{x}_n \:{\rightharpoonup}\:\boldsymbol{0}\) (by ?? ), we obtain \(\boldsymbol{0}\in \mathbf{C}\). Now set \[\mathbf{T}_0 \colon \{\mathbf{x}_n\}_{n\geq 1}\cup\{\mathbf{x}_\infty\} \to \{\mathbf{x}_n\}_{n\geq 2}\cup\{\mathbf{x}_\infty\}\colon \mathbf{x}_n\mapsto \mathbf{x}_{n+1}.\] Then \(\mathbf{T}_0\) is a well-defined bijection (recall 7[kurve1]), with \(\operatorname{Fix}\mathbf{T}_0 = \{\mathbf{x}_\infty\} = \{\boldsymbol{0}\}\). Combining 4 with [9], we see that \[\text{\mathbf{T}_0 is firmly nonexpansive.}\] By a refined version of the Kirszbraun-Valentine theorem (see [14]), there exists a firmly nonexpansive extension \(\mathbf{T}\colon \mathbf{X}\to \mathbf{X}\) of \(\mathbf{T}_0\), with the extra range localization property \(\operatorname{ran}\,\mathbf{T}\subseteq \overline{\operatorname{conv}}\,\operatorname{ran}\,\mathbf{T}_0 = \mathbf{C}\). Because \(\mathbf{T}\) extends \(\mathbf{T}_0\), we clearly have \(\operatorname{ran}\,\mathbf{T}_0 \subseteq \operatorname{ran}\,\mathbf{T}\) and so \(\overline{\operatorname{ran}}\,\mathbf{T}_0 \subseteq \overline{\operatorname{ran}}\,\mathbf{T}\). Because \(\mathbf{T}\) is firmly nonexpansive, hence maximally monotone, it follows that \(\overline{\operatorname{ran}}\,\mathbf{T}\) is convex (see, e.g., [9]). Altogether, \[\overline{\operatorname{ran}}\,\mathbf{T}= \overline{\operatorname{conv}}\,\operatorname{ran}\,\mathbf{T}_0 = \mathbf{C}.\] Clearly, \(\{\boldsymbol{0}\} = \operatorname{Fix}\mathbf{T}_0 \subseteq \operatorname{Fix}\mathbf{T}\). Conversely, let \(\mathbf{y}\in \operatorname{Fix}\mathbf{T}\). Then \(\mathbf{y}\in \overline{\operatorname{ran}}\,\mathbf{T}= \mathbf{C}\subseteq \mathbf{K}\). Because \(\mathbf{T}\) is firmly nonexpansive, we must have that ?? holds. Hence 5 implies that \(\mathbf{y}=\boldsymbol{0}\). Altogether, \(\operatorname{Fix}\mathbf{T}= \{\boldsymbol{0}\}\). Because \(\mathbf{T}\) extends \(\mathbf{T}_0\), we have \(\mathbf{T}\mathbf{x}_n = \mathbf{T}_0\mathbf{x}_n = \mathbf{x}_{n+1}\) for all \(n\geq 1\). Finally, ?? shows that \(\mathbf{T}^n\mathbf{x}_1 = \mathbf{x}_{n+1} \:{\rightharpoonup}\:\boldsymbol{0}\) but \((\|\mathbf{T}^n\mathbf{x}_1\|)_{n\geq 1}\) decreases to \(\rho_\infty>0\). \(\hfill \quad \blacksquare\)
Remark 9 (historical comments). Several comments on 8 are in order.
We already pointed out that Genel and Lindenstrauss [1] provided a similar construction (see [f:GL]). Their proof is much more geometric, and it also relies on the Kirszbraun-Valentine extension theorem.
Hundal [15] constructed a halfspace \(H\) and a nonempty closed cone \(K\) in \(\ell^2\) such that \(H\cap K\neq\varnothing\), and the sequence generated by iterating the composition of the projections \(P_KP_H\) fails to converge strongly to \(0\), the unique point in \(H\cap K\). While \(P_KP_H\) is an averaged nonexpansive mapping, it is unlikely to be firmly nonexpansive.
The first proximal point iteration that fails to converge strongly — even with flexibility in the parameters — is due to Güler [16]. In [17], Hundal’s example was re-interpreted as the iteration of a proximal (hence firmly nonexpansive) mapping.
The proofs of the examples constructed by Genel and Lindenstrauss, by Hundal, and by Güler appear to be significantly more complicated than the proof of 8. However, Güler’s construction gives a proximal mapping, rather than just a firmly nonexpansive one.
Remark 10 (\(\mathbf{T}\) viewed as a resolvent). Consider the operator \(\mathbf{T}\) from 8. Because \(\mathbf{T}\) is firmly nonexpansive, it must be the resolvent \(J_{\mathbf{A}} = (\mathbf{A}+\operatorname{Id})^{-1}\) of some maximally monotone operator \(\mathbf{A}\colon \mathbf{X}\rightrightarrows\mathbf{X}\). The Minty parametrization implies that \[\{(\mathbf{x}_{n+1},\mathbf{x}_n-\mathbf{x}_{n+1})\}_{n\geq 1} \cup \{(\boldsymbol{0},\boldsymbol{0})\} \subseteq \operatorname{gra}\mathbf{A}.\] Note that \(\boldsymbol{0}\in \mathbf{A}\boldsymbol{0}\) and \(\mathbf{x}_{n}-\mathbf{x}_{n+1}\in \mathbf{A}\mathbf{x}_{n+1}\). By 7[kurve3], \(\left\langle{\mathbf{x}_{n+1}-\boldsymbol{0}},{(\mathbf{x}_{n} -\mathbf{x}_{n+1}) - \boldsymbol{0}}\right\rangle = 0\). Hence \(\mathbf{A}\) is not strictly monotone. If \(\mathbf{A}\) were paramonotone, then it would follow that \(\boldsymbol{0}\in \mathbf{A}\mathbf{x}_{n+1}\) but this is false because \(\operatorname{zer}\mathbf{A}= \operatorname{Fix}\mathbf{T}= \{\boldsymbol{0}\}\). Hence \(\mathbf{A}\) is not paramonotone. Consequently, \(\mathbf{A}\) is not a subdifferential operator and \(\mathbf{T}=J_\mathbf{A}\) cannot be a proximal mapping.
We are now ready for our second main result: the incarnation of 8 through the harmonic mesh produces Cesàro means that do not converge strongly.
Example 3 (harmonic mesh and Cesàro means).
Suppose that the given mesh increments \((d_n)_{n\geq 1}\) come from the scaled harmonic mesh (see 2): let \(\delta \in \left]0,1/8\right]\), and assume that \[(\forall n\geq 1)\quad d_n = \frac{\delta}{n}.\] For the sequence \((\mathbf{x}_n)_{n\geq 1}\) defined in [xn], we saw in 8 that \((\mathbf{x}_n)_{n\geq 1} = (\mathbf{T}^{n-1}\mathbf{x}_1)_{n\geq 1}\) converges weakly to \(\boldsymbol{0}\) but \(\inf_{n\geq 1} \|\mathbf{T}^{n-1}\mathbf{x}_1\| = \rho_\infty > 0\). In fact, we have \[\|\mathbf{x}_1\| > \|\mathbf{x}_2\|> \cdots > \|\mathbf{x}_n\|\to \rho_\infty = \exp\Big(-\frac{\delta^2\pi^2}{6}\Big) > 0.\] Now consider the Cesàro means of \((\mathbf{x}_n)_{n\geq 1}\), i.e., \[\mathbf{y}_n := \frac{1}{n}\sum_{k=1}^n \mathbf{x}_k = \frac{1}{n}\sum_{k=1}^n \mathbf{T}^{k-1}\mathbf{x}_1.\] Then \((\mathbf{y}_n)_{n\geq 1}\) converges weakly to \(\boldsymbol{0}\) as well; moreover, \[\inf_{n\geq 1} \|\mathbf{y}_n\| \geq \frac{1}{2} \cdot \exp\Big(-\frac{\delta^2}{6}\big({\pi^2}+3\big)\Big) >0.\]
. [e:rho2] and, e.g., [18] yield \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:jackr8} \endgroup \rho_\infty = \exp\Big(-\sum_{k=1}^\infty d_k^2\Big) = \exp\Big(-\delta^2\sum_{k=1}^\infty \frac{1}{k^2}\Big) = \exp\Big(-\frac{\delta^2\pi^2}{6}\Big) > 0.\tag{19}\] We now turn to the Cesàro means. Note that \(\mathbf{y}_1 = \mathbf{x}_1\) and so [e:rho] and 7[kurve1] yield \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:ryobi} \endgroup \|\mathbf{y}_1\| = \|\mathbf{x}_1\| =\rho_1 = 1.\tag{20}\] Now let \(n\geq 2\), and consider the “upper-half” index set \[I_n := \big\{\lfloor n/2\rfloor+1,\ldots,n\big\}.\] Then6 \(\# I_n = \lceil n/2\rceil \geq n/2\). Let \(i,j\) be in \(I_n\) with \(i\leq j\). Then \((j-1)-i+1 = j-i \leq n - (\lfloor n/2\rfloor+1) = \lfloor (n-1)/2\rfloor\), and if \(i\leq k\leq j-1\), then \(\lfloor n/2\rfloor + 1 \leq k\) and so \(1/k \leq 1/(\lfloor n/2\rfloor + 1)\). It follows that \[\begin{align} t_j-t_i = \delta \sum_{k=i}^{j-1} \frac{1}{k} \leq \delta \cdot \lfloor (n-1)/2\rfloor \cdot \frac{1}{\lfloor n/2\rfloor + 1} <\delta. \end{align}\] We now deduce from 6 and 19 that \[\|\mathbf{y}_n\| \geq \frac{\# I_n}{n}\rho_\infty \exp(-\delta^2/2) \geq \frac{1}{2}\cdot \rho_\infty \exp(-\delta^2/2) = \frac{1}{2}\cdot \exp\Big(-\frac{\delta^2}{6}\big({\pi^2}+3\big)\Big) > 0,\] and we’re done. \(\hfill \quad \blacksquare\)
Remark 11. To the best of our knowledge, this is the first example of a firmly nonexpansive mapping \(\mathbf{T}\) such that its iterates \(\mathbf{T}^nx\:{\rightharpoonup}\:0\), but even the corresponding Cesàro means \(\frac{1}{n}\sum_{k=1}^n \mathbf{T}^{k-1}\mathbf{x}_1\) are bounded away from \(\boldsymbol{0}\) for some starting point \(\mathbf{x}_1\). In view of [6], the mapping \(\mathbf{T}\) is not odd. In the next section, we will present an even more bizarre example where one subsequence stays away from \(\boldsymbol{0}\) yet another converges to \(\boldsymbol{0}\) strongly.
In this section, we present a construction of a mesh, which was obtained with the help of ChatGPT 5.5, and whose corresponding firmly nonexpansive mapping \(\mathbf{T}\) has somewhat bizarre properties. It
all starts with our choice of block widths:
[box=]equation
(w_k)_k := (k+1)_k = (2,3,4,…).
We will iteratively construct a sequence of indices \((i(k))_{k\geq 1}\) with
[box=]equation 1 =: i(1) < i(2) < , I_k := {i(k), i(k)+1, …, i(k+1)-1}.
Let us now explain what happens in the first block \(I_1\). We let7 \[i(1) := 1,\;\; I_1 \gets \{1\},\;\; Q_1 \in \mathbb{N}\cap \left[8,+\infty\right[.\] Note that we only know \(i(1) = \min I_1\) and the eventually final block \(I_1\) will be constructed by enlarging \(I_1\). Right now, \(I_1\) is the current candidate for the first block. We now consider the first index \(n\) in the first candidate block: \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:firstn} \endgroup n \gets i(1) = 1\in I_1, \;\; t_1 := 0,\;\; d_1 := \frac{1}{Q_1} \leq \frac{1}{8}.\tag{21}\] We now set \(k \gets 1\), indicating the block counter. We are ready to move from \(n\) to \(n+1\), which will be explained in the next subsection.
Now suppose that \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \tag{22} \endgroup \begin{equation} \begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \tag{23} \endgroup \text{k and Q_k are given, i(k) is known, I_k is the current candidate block, } \end{equation} and that \begin{equation} \begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \tag{24} \endgroup \text{n\in I_k is given, and we know t_n and d_n.} \end{equation}\] (This is definitely true for \(k=1\) and \(n=i(1)=1\), see [sec:ss:k611n611].) We always set \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:tn431} \endgroup t_{n+1} := t_n + d_n.\tag{25}\] Next, we compare \(t_{n+1}\) to \(t_{i(k)} + w_k\).
Case 1 (staying in the block): \(t_{n+1} < t_{i(k)} + w_k\).
(For \(k=1\), this means \(t_{n+1} < 2\), which is definitely true for \(n=1\) because \(t_2 = t_1+d_1 = 0 + 1/Q_1 \leq
1/8\).) We then enlarge the candidate block and update the mesh increment according to \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else
\def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:stayn431} \endgroup
I_{k} \gets I_k \cup \{n+1\},\;\;
d_{n+1} := \frac{d_n}{1+64 d_n^2}.\tag{26}\] Keeping \(k\) unchanged, we now update \[n \gets n+1,\] and return to 22 .
Case 2 (leaving the block): \(t_{n+1} \geq t_{i(k)} + w_k\).
This condition signals that \(n\) is the last index of \(I_k\), and \(n+1\) is the first index of \(I_{k+1}\): \[I_k := \{i(k),\ldots,n\} \;\;\text{is complete},\quad i(k+1) := n+1, \quad I_{k+1} \gets \{i(k+1)\}.\] We update now the integer parameter \(Q_{k+1}\) by picking an integer such that
\[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else
\let\cref@currentlabel\cref@currentlabel \fi \label{e:Qk431} \endgroup
Q_{k+1} \geq \max\bigg\{i(k+1), 2^{k+2}w_{k+1},
\frac{1 + 64d^2_{i(k+1)-1} }{d_{i(k+1)-1}} \bigg\}.\tag{27}\] Note that \(n+1\in I_{k+1}\), that we know \(t_{n+1}\) (from 25 ), but this time we
update \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else
\let\cref@currentlabel\cref@currentlabel \fi \label{e:Qk43139} \endgroup
d_{n+1} := \frac{1}{Q_{k+1}}.\tag{28}\] We now update both counters \[k\gets k+1, \quad n\gets n+1\] before returning to 22 .
We have provided well-defined update rules. What is not yet resolved is whether we ever leave the first block, and whether the so-constructed mesh satisfies all required properties. We tackle these questions next.
Suppose that \(n\) and \(n+1\) are in \(I_k\). By the condition for being in the block (see Case 1 above), we have \[t_n < t_{i(k)} + w_k.\] By 26 and 21 , we have \[d_{n+1} = d_n/(1+64 d_n^2) < d_n < \cdots < d_{i(k)} = \frac{1}{Q_k},\] and (analogously to our discussion around 13 ) \[\frac{1}{d_{n}} -64 t_n = \frac{1}{d_{i(k)}} - 64 t_{i(k)} = Q_k - 64 t_{i(k)}.\] Hence altogether \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:jackr0} \endgroup Q_k \leq \frac{1}{d_n} = Q_k + 64(t_n-t_{i(k)}) < Q_k + 64 w_k.\tag{29}\] Thus \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:jackr1} \endgroup 0<\frac{1}{Q_k+64w_k} < d_n \leq \frac{1}{Q_k}.\tag{30}\] This lower bound, which is valid for all \(n\in I_k\) and independent of \(n\), guarantees us to escape block \(k\): indeed, if \(n_k\) is the smallest index in \(I_k\) such that \(t_{n_k+1} = t_{i(k)} + \sum_{m=i(k)}^{n_k} d_m \geq t_{i(k)}+w_k\), then \(I_k = \{i(k),\ldots,n_k\}\), \(i(k+1) = n_k+1\), and \[t_{n_k+1} \geq t_{i(k)} + w_k \geq w_k = k+1 \to\infty \quad\text{as k\to\infty.}\] Because \((t_n)_{n\geq 1}\) is clearly strictly increasing, we obtain \[\lim_{n\to\infty} t_n = \infty.\]
We have, in view of 30 and Case 1 (\(i(k+1)-1\in I_k\)), that \[w_k \leq \sum_{n\in I_k} d_n = t_{i(k+1)} - t_{i(k)} = d_{i(k+1)-1} + (t_{i(k+1)-1} - t_{i(k)}) <\frac{1}{Q_k} + w_k.\] Hence if \(d_n\in I_k\) and thus \(d_n\leq 1/Q_k\) by 30 , we get \[\sum_{n\in I_k} d_n^2 \leq \frac{1}{Q_k}\sum_{n\in I_k} d_n < \frac{1}{Q_k^2}+ \frac{w_k}{Q_k}.\] On the other hand, from 27 , we get \(Q_k \geq 2^{k+1}w_k \geq 2^{k+1}\) and so \[\sum_{n\in I_k} d_n^2 \leq \frac{1}{4^{k+1}} + \frac{1}{2^{k+1}}.\] Summing this over \(k\) gives the summability condition8 \(\sum_n d_n^2 <\infty\).
Combining the previous two subsections, we’ve verified [e:dmesh00]. Next, we saw 10 already in 21 . If \(n\) and \(n+1\) belong to \(I_k\), then 26 yields 11 . It remains to consider the case when \(n\in I_k\) and \(n+1\in I_{k+1}\). Then \(n=i(k+1)-1\) and \(n+1 = i(k+1)\); thus, 28 and 27 yield \[d_{n+1} = \frac{1}{Q_{k+1}} \leq \frac{d_{i(k+1)-1}}{1+64d^2_{i(k+1)-1}} = \frac{d_{n}}{1+64d^2_{n}},\] which is again 11 .
Altogether, we’ve shown that the mesh satisfies [e:dmesh00] and 9 . In particular, the conclusion of 8 holds true.
Now consider the Cesàro means of \((\mathbf{x}_n)_{n\geq 1}\): \[\mathbf{y}_n := \frac{1}{n}\sum_{k=1}^n \mathbf{x}_k = \frac{1}{n}\sum_{k=1}^n \mathbf{T}^{k-1}\mathbf{x}_1.\] We know that \((\mathbf{y}_n)_{n\geq 1}\) converges weakly to \(\boldsymbol{0}\), and in the next two subsections, we obtain the last main result R3 announced in [sec:s:intro].
Consider the block \(I_k = \{i(k),\ldots,i(k+1)-1\}\). We have \(t_{i(k)}<t_{i(k)}+w_k\) and \(t_{i(k+1)}\geq t_{i(k)}+w_k = t_{i(k)} + k+1 \geq t_{i(k)}+2\). Hence we well define \[j(k) := \min\big\{{n\in I_k}~\big|~{t_n \geq t_{i(k)}+1}\big\} \;\;\text{and}\;\; J_k := \{i(k),\ldots,j(k)\}\subseteq I_k \subseteq \{1,2,\ldots,j(k)\}.\] The corresponding mesh values start with \(t_{i(k)}\). By definition \(t_{j(k)-1}<t_{i(k)}+1\). Hence \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:jackr2} \endgroup t_{j(k)} = t_{j(k)-1} + d_{j(k)-1} < t_{i(k)}+1 + 1/Q_k \leq t_{i(k)}+1+1/8 = t_{i(k)}+9/8.\tag{31}\] Now let \(i,j\) be in \(J_k\). From 31 , \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:jackr11} \endgroup (\forall i,j\in J_k)\quad |t_i-t_j| \leq t_{j(k)} - t_{i(k)} < 9/8.\tag{32}\]
We now turn to the size of \(J_k\). First, we have the lower bound \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:Jklow} \endgroup \# J_k \geq Q_k\tag{33}\] because each mesh increment is at most \(1/Q_k\).
If \(t_n < t_{i(k)}+1\), then 29 yields \(1/d_n = Q_k+64(t_n-t_{i(k)})<Q_k+64\) and so \[d_n>\frac{1}{Q_k+64}.\] Hence after at most \(Q_k+64\) steps, we must have advanced from \(t_{i(k)}\) to a mesh point exceeding \(t_{i(k)}+1\). So we get the upper bounds \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:Jkup} \endgroup \# J_k \leq Q_k+65 \;\;\text{and}\;\; j(k) \leq i(k)+Q_k+64.\tag{34}\] We deduce from 33 and 34 that \[\begin{align} \begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:jackr10} \endgroup \frac{\# J_k}{j(k)} \geq \frac{Q_k}{i(k)+Q_k+64} \geq \frac{Q_k}{2Q_k+64}. \end{align}\tag{35}\]
Combining 32 , 35 , and 6, we obtain \[\|\mathbf{y}_{j(k)}\| \geq \frac{Q_k}{2Q_k+64} \cdot \rho_\infty \exp(-(9/8)^2/2)\] and therefore (because \(Q_k\to\infty\)) \[\varlimsup_{n\to\infty} \|\mathbf{y}_n\| \geq \varlimsup_{k\to\infty} \|\mathbf{y}_{j(k)}\| \geq \frac{\rho_\infty}{2}\exp(-81/128)>0.\]
In this subsection, we again consider the block \(I_k = \{i(k),\ldots,i(k+1)-1\}\), but this time we set for convenience \[j(k) := \max I_k = i(k+1)-1 \;\;\text{so that}\;\; I_k = \{i(k),\ldots,j(k)\}.\] Note that \[\# I_k = j(k)-i(k)+1 = i(k+1)-i(k).\] We now consider the Cesàro means over the block \(I_k\) alone: \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:zesaro} \endgroup \mathbf{z}_k := \frac{1}{\# I_k}\sum_{n\in I_k} \mathbf{x}_n.\tag{36}\] Because \(i(k+1)\) is the first index in block \(I_{k+1}\), it follows from the definition of leaving block \(I_k\) (see Case 2 above) that \[\sum_{n\in I_k} d_n =t_{j(k)+1} - t_{i(k)} = t_{i(k+1)} - t_{i(k)} \geq w_k.\] Recalling that \(d_n\leq 1/Q_k\) for \(n\in I_k\) (see 30 ), we conclude that \(w_k \leq \sum_{n\in I_k} d_n \leq (\# I_k)/Q_k\) and thus \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:jackr12} \endgroup \# I_k \geq w_k Q_k \geq w_k i(k).\tag{37}\] Next, we consider \(t_n,t_{n+1}\), where \(n,n+1\) are both in \(I_k\). By 30 , we have \[d_n = t_{n+1}-t_n > \frac{1}{Q_k+64w_k} =: \alpha_k.\] Now \(Q_k+64w_k \geq 8+64\cdot 2 = 136\), so \(\alpha_k \leq 1/136<1\). It thus follows from 2 that \[\max_{i\in I_k} \sum_{j\in I_k}\exp\big(-(t_i-t_j)^2\big) \leq \frac{{1}+\sqrt{\pi}}{\alpha_k} = \big({1}+\sqrt{\pi}\big)\big({Q_k+64w_k}\big).\] Using 7[kurve1], [e:rho2], 37 , and the facts that \(w_k=k+1\to\infty\) and \(Q_k \geq 2^{k+1}w_k\to \infty\), we obtain \[\begin{align} \|\mathbf{z}_k\|^2 &=\frac{1}{(\# I_k)^2}\sum_{i,j\in I_k}\left\langle{\mathbf{x}_i},{\mathbf{x}_j}\right\rangle =\frac{1}{(\# I_k)^2}\sum_{i,j\in I_k}\rho_i\rho_j\exp\big(-(t_i-t_j)^2\big)\\ &\leq\frac{1}{(\# I_k)^2}\sum_{i,j\in I_k}\exp\big(-(t_i-t_j)^2\big) \leq\frac{1}{\# I_k} ({1}+\sqrt{\pi})(Q_k+64w_k) \\ &\leq \frac{1}{w_kQ_k} ({1}+\sqrt{\pi})(Q_k+64w_k) = ({1}+\sqrt{\pi})\bigg(\frac{1}{w_k} + \frac{64}{Q_k}\bigg)\\ &\to 0. \end{align}\] Hence \[\begingroup \def\cref@currentlabel{} \ifx\current@theorem\relax\else \def\cref@currentlabel{\current@theorem} \fi \ifx\cref@currentlabel\undefined\else \let\cref@currentlabel\cref@currentlabel \fi \label{e:zzero} \endgroup \|\mathbf{z}_k\| \to 0.\tag{38}\] We now consider the (full) Cesàro mean \[\begin{align} \mathbf{y}_{j(k)} &= \frac{1}{j(k)}\sum_{n=1}^{j(k)} \mathbf{x}_n = \frac{1}{j(k)}\sum_{n=1}^{i(k)-1} \mathbf{x}_n + \frac{1}{j(k)}\sum_{n\in I_k} \mathbf{x}_n = \frac{1}{j(k)}\sum_{n=1}^{i(k)-1} \mathbf{x}_n + \frac{\# I_k}{j(k)} \frac{1}{\# I_k}\sum_{n\in I_k} \mathbf{x}_n\\ &= \frac{1}{j(k)}\sum_{n=1}^{i(k)-1} \mathbf{x}_n + \frac{\# I_k}{j(k)}\mathbf{z}_k. \end{align}\] Using \(\|\mathbf{x}_n\|\leq 1\) for all \(n\), 36 , the triangle inequality, 37 and 38 , we have \[\begin{align} \|\mathbf{y}_{j(k)}\| &\leq \frac{i(k)-1}{j(k)} + \frac{\# I_k}{j(k)}\|\mathbf{z}_k\| \leq \frac{i(k)-1}{\# I_k} + \frac{j(k)-i(k)+1}{j(k)}\|\mathbf{z}_k\|\\ &< \frac{i(k)}{\# I_k} + \|\mathbf{z}_k\| \leq \frac{1}{w_k} + \|\mathbf{z}_k\|\\ &\to 0. \end{align}\] Therefore, \[0\leq \varliminf_{n\to\infty} \|\mathbf{y}_n\| \leq \varliminf_{k\to\infty} \|\mathbf{y}_{j(k)}\| = 0.\]
To sum up, we have presented an example of a firmly nonexpansive mapping \(\mathbf{T}\colon \mathbf{X}\to \mathbf{X}\) with \(\operatorname{Fix}\mathbf{T}=\{\boldsymbol{0}\}\) such that for some \(\mathbf{x}_1\in \mathbf{X}\), the sequence of Cesàro means defined by \[\mathbf{y}_n = \frac{1}{n}\sum_{k=1}^n \mathbf{T}^{k-1}\mathbf{x}_1\] satisfies \[\mathbf{y}_n \:{\rightharpoonup}\:0,\;\; \varliminf_{n\to\infty} \|\mathbf{y}_n\| = 0,\;\; \text{ and } \;\;\varlimsup_{n\to\infty} \|\mathbf{y}_n\| > 0.\]
The research of HHB was partially supported by a Discovery Grant of the Natural Sciences and Engineering Research Council of Canada.
Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail: heinz.bauschke@ubc.ca.↩︎
Mathematics, University of British Columbia, Kelowna, B.C. V1V 1V7, Canada. E-mail: tung.tran@ubc.ca.↩︎
One can also consider \(\mathbf{X}=L_2[-1,1]\), with \(\mathbf{e}_k\) being a suitably normalized Legendre polynomial of degree \(k\) (see [11]), or \(\mathbf{X}=L_2[-\pi,\pi]\) with suitably normalized trigonometric functions (see [12]), or \(\mathbf{X}=L_2[0,1]\) with Walsh functions (see [12]).↩︎
An interesting Schauder basis for \(L^2(\mathbb{R})\) consists of the (suitably normalized) Hermite functions [11].↩︎
We omit the details because these formulas are not used in the rest of the paper.↩︎
The formulas involving floor and ceiling functions are proved by discussing parity (\(n\) is even or \(n\) is odd).↩︎
Following common usage in computer science, \(I_1 \gets \{1\}\) means that \(I_1\) is currently \(\{1\}\) but this set may be updated as the construction proceeds. In contrast, \(i(1) := 1\) signifies that \(i(1)\) is defined to be \(1\) and will stay that way.↩︎
The geometric series upper bound can be used to obtain a positive lower bound for \(\rho_\infty\) if needed.↩︎