Maximally monotone operators with ranges whose closures are not convex and an answer to a recent question by Stephen Simons

TL;DR

This paper investigates the convexity properties of ranges of certain maximally monotone operators, providing a negative answer to a question about the scalar value 4 and extending results to the Fitzpatrick-Phelps operator.

Contribution

It proves that the range closure of the sum of the Gossez operator and a multiple of the duality map is nonconvex for all scalars greater than or equal to 4, answering a recent open question.

Findings

Range closure is nonconvex for scalar ≥ 4.

Results apply to the Fitzpatrick-Phelps integral operator.

Provides an abstract framework for analyzing such operators.

Abstract

In his recent Proceedings of the AMS paper "Gossez's skew linear map and its pathological maximally monotone multifunctions", Stephen Simons proved that the closure of the range of the sum of the Gossez operator and a multiple of the duality map is nonconvex whenever the scalar is between 0 and 4. The problem of the convexity of that range when the scalar is equal to 4 was explicitly stated. In this paper, we answer this question in the negative for any scalar greater than or equal to 4. We derive this result from an abstract framework that allows us to also obtain a corresponding result for the Fitzpatrick-Phelps integral operator.

Equations

$$X$$

$$(\forall (x,x^{\ast })\in X\times X^{\ast })x^{\ast }\in Jx\iff ⟨x,x^{\ast }⟩=\Vert x{\Vert }^2=\Vert x^{\ast }{\Vert }^2.$$

$$\overline{\mathrm{span}}\mathrm{ran}(J-J)=X^{\ast }.$$

$$(\forall n\in \mathbb{N})(Jx{)}_n=\Vert x{\Vert }_1\mathrm{Sign}(x_n).$$

$$(\forall n\in \mathbb{N})|y_n|\le \Vert y{\Vert }_{\infty }=\Vert x{\Vert }_1$$

$$\Vert x{\Vert }_1^2$$

$$\le \sum _{n\in \mathbb{N}}|x_n||y_n|\le \sum _{n\in \mathbb{N}}|x_n|\Vert y{\Vert }_{\infty }=\Vert x{\Vert }_1\Vert y{\Vert }_{\infty }.$$

$$(\forall t\in [0,1])(Jx)(t)=\Vert x{\Vert }_1\mathrm{Sign}(x(t))\text{almost everywhere (a.e.). }$$

$$(\forall t\in [0,1])|y(t)|\le \Vert y{\Vert }_{\infty }=\Vert x{\Vert }_1\text{a.e.}$$

$$\Vert x{\Vert }_1^2$$

$$\le {\int }_0^1|x(t)||y(t)|dt\le {\int }_0^1|x(t)|\Vert y{\Vert }_{\infty }dt=\Vert x{\Vert }_1\Vert y{\Vert }_{\infty }.$$

$$r^{\ast }\in Ax+\lambda Jx.$$

$$\frac{\Vert r^{\ast }\Vert }{\Vert A\Vert +\lambda }\le \Vert x\Vert \le \frac{\Vert r^{\ast }\Vert }{\lambda }\text{and}\lambda \Vert x{\Vert }^2\le ⟨x,r^{\ast }⟩,$$

$$\text{A is skew}\Rightarrow \lambda \Vert x{\Vert }^2=⟨x,r^{\ast }⟩.$$

$$\big{(}Ax+\lambda{\,\operatorname{J}}x\big{)}\cap\operatorname{ball}(f^{*};\varepsilon)\neq\varnothing$$

$$\epsilon \le \frac{2\lambda \Vert f^{\ast }\Vert }{\Vert A\Vert +3\lambda }.$$

$$⟨x,f^{\ast }⟩\ge \lambda {\left(\frac{\Vert f^{\ast }\Vert -\epsilon }{\Vert A\Vert +\lambda }\right)}^2-\epsilon \left(\frac{\Vert f^{\ast }\Vert -\epsilon }{\Vert A\Vert +\lambda }\right).$$

$$\lambda \Vert x{\Vert }^2\le ⟨x,r^{\ast }⟩=⟨x,r^{\ast }-f^{\ast }⟩+⟨x,f^{\ast }⟩\le \Vert x\Vert \epsilon +⟨x,f^{\ast }⟩.$$

$$⟨x,f^{\ast }⟩\ge \lambda {\xi }^2-\epsilon \xi ,\text{where}\frac{\rho }{\alpha +\lambda }\le \xi \le \frac{\rho }{\lambda }.$$

$$\phi -\epsilon \le \rho \le \phi +\epsilon .$$

$$\epsilon \le 2\lambda l.$$

$$l\le \xi .$$

$$(Ax+\lambda Jx)\cap \mathrm{ball}(f^{\ast };\epsilon )\ne \varnothing \Rightarrow ⟨x,f^{\ast }⟩\le 3\epsilon$$

$$m(\lambda )=\frac{3{\lambda }^2+9\lambda +4-(\lambda +1)\sqrt{9{\lambda }^2+36\lambda +16}}{4\lambda +2}.$$

$$(2\lambda +1){\epsilon }^2-(3{\lambda }^2+9\lambda +4)\epsilon +\lambda \le 0;$$

$$\epsilon \ge \frac{3{\lambda }^2+9\lambda +4-(\lambda +1)\sqrt{9{\lambda }^2+36\lambda +16}}{4\lambda +2}=m(\lambda )>0.$$

$$\lambda \frac{(1-\epsilon {)}^2}{(1+\lambda {)}^2}-\epsilon \frac{1-\epsilon }{1+\lambda }\le 3\epsilon .$$

$$\lambda (1-\epsilon {)}^2-\epsilon (1-\epsilon )(1+\lambda )\le 3\epsilon (1+\lambda {)}^2$$

$$\iff (\lambda +{\epsilon }^2\lambda -2\lambda \epsilon )+(-\epsilon -\epsilon \lambda +{\epsilon }^2+{\epsilon }^2\lambda )\le 3\epsilon +3\epsilon {\lambda }^2+6\epsilon \lambda$$

$$\iff {\epsilon }^2(1+2\lambda )+\epsilon (-9\lambda -4-3{\lambda }^2)+\lambda \le 0,$$

$$(2\lambda +1){\xi }^2-(3{\lambda }^2+9\lambda +4)\xi +\lambda =0,$$

$$\frac{3{\lambda }^2+9\lambda +4\pm \sqrt{(3{\lambda }^2+9\lambda +4{)}^2-4(1+2\lambda )\lambda }}{2(1+2\lambda )}$$

Full text

2010 Mathematics Subject Classification: Primary 47H05, Secondary 46B20, 47A05

Keywords: Duality map, Fitzpatrick-Phelps operator, Gossez operator, maximally monotone operator, range, skew operator

1 Introduction

Throughout, we assume that

[TABLE]

where $X^{*}$ is the dual space of $X$. The duality mapping ${\,\operatorname{J}}$ of $X$ is the subdifferential operator of the function $\tfrac{1}{2}\|\cdot\|^{2}\colon X\to\mathbb{R}$; it satisfies

[TABLE]

Now let $A\colon X\to X^{*}$ be a bounded linear monotone operator, i.e., $(\forall x\in X)(\forall y\in X)$ $\left\langle{x-y},{Ax-Ay}\right\rangle\geq 0$. Then both $A$ and ${\,\operatorname{J}}$ are maximally monotone, and so is their sum $A+{\,\operatorname{J}}$ thanks to a result by Heisler (see, e.g., [8, Theorem 40.4]). If $X$ is a Hilbert space, then ${\,\operatorname{J}}=\operatorname{Id}$ and it is well known that the range ${\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})$ is equal to $X$. In striking contrast, it was shown very recently by Simons that for the so-called Gossez operator $G$ which acts on $\ell_{1}$, we have that $\overline{\operatorname{ran}}\,(G+\lambda{\,\operatorname{J}})$ is not convex for $0<\lambda<4$ (see Section 4 below.) It is also known that for the so-called Fitzpatrick-Phelps operator $F$, which acts on $L_{1}[0,1]$, the set $\overline{\operatorname{ran}}\,(F+1{\,\operatorname{J}})$ is not convex.

In this paper, we unify these results by providing an abstract result that allows us to deduce that no matter how $\lambda>0$ is chosen, neither $\overline{\operatorname{ran}}\,(F+\lambda{\,\operatorname{J}})$ nor $\overline{\operatorname{ran}}\,(G+\lambda{\,\operatorname{J}})$ is convex. This provides a negative answer to Simons’s [7, Problem 3.6].

Let us note here that neither $F$ nor $G$ is a subdifferential operator of a convex function; indeed, if $f\colon X\to\,\left]-\infty,+\infty\right]$ is convex, lower semicontinuous, and proper, then $f+\lambda\tfrac{1}{2}\|\cdot\|^{2}$ is supercoercive111Recall that $g\colon X\to\,\left]-\infty,+\infty\right]$ is supercoercive if $\lim_{\|x\|\to+\infty}g(x)/\|x\|=+\infty$. and hence $\overline{\operatorname{ran}}\,\partial(f+\lambda\tfrac{1}{2}\|\cdot\|^{2})=\overline{\operatorname{ran}}\,(\partial f+\lambda{\,\operatorname{J}})=X^{*}$ by Gossez’s [4, Corollaire 8.2].

The remainder of this note is organized as follows. In Section 2 we collect some auxiliary results for later use. Section 3 contains our main abstract results. Section 4 discusses the Gossez operator while Section 5 deals with the Fitzpatrick-Phelps operator.

2 Auxiliary results

In this section, we cover some technical results that will ease the proofs in subsequent sections.

Rugged Banach spaces

Definition 2.1**.**

(See also [1].) We say that $X$ is rugged if

[TABLE]

Example 2.2**.**

*(See [2, Remark 5.6] and also [1, Section 2.3.4].) The following Banach spaces are rugged:

(i) $\ell_{1}$and

(ii) $L_{1}[0,1]$.

Proof. First, we define $(\forall\xi\in\mathbb{R})$ $\operatorname{Sign}(\xi)=\{\operatorname{sign}(\xi)\}$, if $\xi\neq 0$; and $\operatorname{Sign}(\xi)=\left[-1,1\right]=\partial|\cdot|(0)$, otherwise.

(i): Let $x=(x_{n})_{n\in\mathbb{N}}\in\ell_{1}$, where $\mathbb{N}=\{1,2,\ldots\}$. We start by proving the following claim:

[TABLE]

Indeed, let $y=(y_{n})_{{n\in{\mathbb{N}}}}\in\ell_{\infty}$. If $(\forall{n\in{\mathbb{N}}})$ $y_{n}\in\lVert x\rVert_{1}\operatorname{Sign}(x_{n})$, then $\lVert x\rVert_{1}^{2}=\lVert y\rVert_{\infty}^{2}=\left\langle{x},{y}\right\rangle$ and hence $y\in{\,\operatorname{J}}x$. Conversely, suppose that $y\in{\,\operatorname{J}}x$. Then

[TABLE]

and

[TABLE]

Hence $\sum_{{n\in{\mathbb{N}}}}\lvert x_{n}\rvert(\lVert y\rVert_{\infty}-\lvert y_{n}\rvert)=0$; equivalently, $(\forall{n\in{\mathbb{N}}})$ $\lvert x_{n}\rvert(\lVert y\rVert_{\infty}-\lvert y_{n}\rvert)=0$. Now, if $x_{n}=0$ then (5) implies that $y_{n}\in\lVert x\rVert_{1}\operatorname{Sign}(x_{n})$. Alternatively, if $x_{n}\neq 0$ we must have $\lvert y_{n}\rvert=\lVert y\rVert_{\infty}$ which, in view of (6), implies that $y_{n}=\operatorname{sign}(x_{n})\lvert y_{n}\rvert=\operatorname{sign}(x_{n})\lVert y\rVert_{\infty}\in\operatorname{Sign}(x_{n})\lVert x\rVert_{1}$. Next, we show that $\ell_{1}$ satisfies (3). Clearly, $\{1,2\}\subseteq\mathbb{N}$, and we denote the corresponding canonical unit vectors in $\ell_{1}$ by $e_{1}$ and $e_{2}$, respectively. If $i\in\{1,2\}$ and $n\in\mathbb{N}\smallsetminus\{i\}$, then (4) yields $({\,\operatorname{J}}{e_{i}})_{i}=\{1\}$ and $({\,\operatorname{J}}{e_{i}})_{n}=[-1,1]$; consequently, $({\,\operatorname{J}}{e_{i}}-{\,\operatorname{J}}{e_{i}})_{i}=\{0\}$ and $({\,\operatorname{J}}{e_{i}}-{\,\operatorname{J}}{e_{i}})_{n}=[-2,2]$. Therefore, if $i\in\{1,2\}$ and $n\in\mathbb{N}\smallsetminus\{1,2\}$, then $({\,\operatorname{J}}{e_{1}}-{\,\operatorname{J}}{e_{1}}+{\,\operatorname{J}}e_{2}-{\,\operatorname{J}}e_{2})_{i}=[-2,2]$ while $({\,\operatorname{J}}{e_{1}}-{\,\operatorname{J}}{e_{1}}+{\,\operatorname{J}}e_{2}-{\,\operatorname{J}}e_{2})_{n}=[-4,4]$. It follows that $(\forall{n\in{\mathbb{N}}})$ $[-2,2]\subseteq({\,\operatorname{J}}{e_{1}}-{\,\operatorname{J}}{e_{1}}+{\,\operatorname{J}}e_{2}-{\,\operatorname{J}}e_{2})_{n}$, which in turn implies (3).

(ii): The proof parallels that of item (i); for completeness, we provide the details. Let $x\in L_{1}[0,1]$. We first claim that

[TABLE]

Indeed, let $y=L_{\infty}[0,1]$. If $(\forall t\in[0,1])$ $y(t)\in\lVert x\rVert_{1}\operatorname{Sign}(x(t))$ a.e., then $\lVert x\rVert_{1}^{2}=\lVert y\rVert_{\infty}^{2}=\left\langle{x},{y}\right\rangle$ and hence $y\in{\,\operatorname{J}}x$. Conversely, suppose that $y\in{\,\operatorname{J}}x$. Then

[TABLE]

and

[TABLE]

Hence $\int_{0}^{1}\lvert x(t)\rvert(\lVert y\rVert_{\infty}-\lvert y(t)\rvert)\,dt=0$; equivalently, $(\forall t\in[0,1])$ $\lvert x(t)\rvert(\lVert y\rVert_{\infty}-\lvert y(t)\rvert)=0$ a.e. Now if $x(t)=0$, then (8) implies that $y(t)\in\lVert x\rVert_{1}\operatorname{Sign}(x(t))$. Alternatively, if $x(t)\neq 0$, then we must have $\lvert y(t)\rvert=\lVert y\rVert_{\infty}$ which, in view of (9), implies that $y(t)=\operatorname{sign}(x(t))\lvert y(t)\rvert=\operatorname{sign}(x(t))\lVert y\rVert_{\infty}\in\operatorname{Sign}(x(t))\lVert x\rVert_{1}$ a.e. It remains to show that $L_{1}[0,1]$ satisfies (3). Set $A_{1}=[0,1/3]$ and $A_{2}=[2/3,1]$ and also $e_{1}=3\chi_{A_{1}}$ and $e_{2}=3\chi_{A_{2}}$, where $\chi_{B}$ denotes the characteristic function of a subset $B$ of $[0,1]$. If $i\in\{1,2\}$, $s\in A_{i}$ and $t\in[0,1]\smallsetminus A_{i}$, then $\lVert e_{i}\rVert_{1}=1$ and (7) yields $({\,\operatorname{J}}{e_{i}})(s)=\{1\}$ and $({\,\operatorname{J}}{e_{i}})(t)=[-1,1]$; consequently, $({\,\operatorname{J}}{e_{i}}-{\,\operatorname{J}}{e_{i}})(s)=\{0\}$ and $({\,\operatorname{J}}{e_{i}}-{\,\operatorname{J}}{e_{i}})(t)=[-2,2]$. Therefore, if $s\in A:=A_{1}\cup A_{2}$ and $t\in[0,1]\smallsetminus A$, then $({\,\operatorname{J}}{e_{1}}-{\,\operatorname{J}}{e_{1}}+{\,\operatorname{J}}e_{2}-{\,\operatorname{J}}e_{2})(s)=[-2,2]$ while $({\,\operatorname{J}}{e_{1}}-{\,\operatorname{J}}{e_{1}}+{\,\operatorname{J}}e_{2}-{\,\operatorname{J}}e_{2})(t)=[-4,4]$. It follows that $(\forall t\in[0,1])$ $[-2,2]\subseteq({\,\operatorname{J}}{e_{1}}-{\,\operatorname{J}}{e_{1}}+{\,\operatorname{J}}e_{2}-{\,\operatorname{J}}e_{2})(t)$ a.e., which in turn implies (3). $\hfill\quad\blacksquare$

Proposition 2.3**.**

Let $X$ be rugged, let $A\colon X\to X^{*}$ be a linear operator, and let $\lambda>0$. Then $\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})=X^{*}$.

Proof. We prove this by contradiction and thus assume that there exists $x^{*}\in X^{*}\smallsetminus(\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}}))$. The separation theorem yields $x^{**}\in X^{**}\smallsetminus\{0\}$ such that $\left\langle{x^{*}},{x^{**}}\right\rangle>\sup\left\langle{\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})},{x^{**}}\right\rangle\geq\sup\left\langle{{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})},{x^{**}}\right\rangle$. Because ${\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})$ is a balanced cone (i.e., closed under scalar multiplication), we deduce that $(\forall x\in X)$ $\left\langle{Ax+\lambda{\,\operatorname{J}}x},{x^{**}}\right\rangle=0$. Because $A$ is single-valued and $\lambda\neq 0$, it follows that $(\forall x\in X)$ $\left\langle{{\,\operatorname{J}}x-{\,\operatorname{J}}x},{x^{**}}\right\rangle=0$. Therefore, $\left\langle{X^{*}},{x^{**}}\right\rangle=\left\langle{\overline{\operatorname{span}}\,{\operatorname{ran}}\,({\,\operatorname{J}}-{\,\operatorname{J}})},{x^{**}}\right\rangle=0$ and thus $x^{**}=0$ which is absurd. $\hfill\quad\blacksquare$

Corollary 2.4**.**

Let $X$ be rugged, let $A\colon X\to X^{*}$ be a linear operator, and let $\lambda>0$. Then the following are equivalent:

1. (i)

$\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})=X^{*}$; 2. (ii)

$\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})$* is a subspace;* 3. (iii)

$\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})$* is a convex set.*

Proof. (See also [1, Proposition 15.3.8] for “(i)$\Leftrightarrow$(iii)”.) The implications “(i)$\Rightarrow$(ii)” and “(ii)$\Rightarrow$(iii)” are clear. “(iii)$\Rightarrow$(i)”: Assume that $\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})$ is convex. Then $\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})=\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})$. On the other hand, by Proposition 2.3, $\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})=X^{*}$. Altogether, we deduce that $\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})=X^{*}$. $\hfill\quad\blacksquare$

3 Main result

Lemma 3.1**.**

Let $A\colon X\to X^{*}$ be a bounded linear monotone operator. Let $\lambda>0$ and let $r^{*}\in X^{*}$. Suppose that $x\in X$ satisfies

[TABLE]

Then

[TABLE]

*and222Recall that $A\colon X\to X^{*}$ is skew if any of the following equivalent conditions holds: (i) $A^{*}|_{X}=-A$; (ii) $(\forall x\in X)(\forall y\in X)$ $\left\langle{x},{Ay}\right\rangle=-\left\langle{y},{Ax}\right\rangle$; (iii) $\pm A$ are monotone; (iv) $(\forall x\in X)$ $\left\langle{x},{Ax}\right\rangle=0$. *

[TABLE]

Proof. Let $x^{*}\in{\,\operatorname{J}}x$ satisfy $r^{*}=Ax+\lambda x^{*}$. Combining (2) and the triangle inequality yields $\lambda\|x\|=\lambda\|x^{*}\|=\|r^{*}-Ax\|\geq\|r^{*}\|-\|Ax\|\geq\|r^{*}\|-\|A\|\|x\|$. Hence $(\lambda+\|A\|)\|x\|\geq\|r^{*}\|$ which implies the first inequality in (11). On the other hand, from (2) and the monotonicity of $A$, we deduce that $\lambda\|x\|^{2}=\lambda\left\langle{x},{x^{*}}\right\rangle\leq\left\langle{x},{\lambda x^{*}+Ax}\right\rangle=\left\langle{x},{r^{*}}\right\rangle\leq\|x\|\|r^{*}\|$, which yields the remaining inequalities in (11) as well as (12). $\hfill\quad\blacksquare$

Theorem 3.2**.**

Let $A\colon X\to X^{*}$ be a bounded linear monotone operator. Let $\lambda>0$ and $\varepsilon\geq 0$. Suppose that $x\in X$ and $f^{*}\in X^{*}$ satisfy333We denote the closed ball centered at $f^{*}$ of radius $\varepsilon$ by $\operatorname{ball}(f^{*};\varepsilon)$.

[TABLE]

and

[TABLE]

Then

[TABLE]

Proof. Set $\varphi=\|f^{*}\|$ and let $r^{*}\in(Ax+\lambda{\,\operatorname{J}}x)\cap\operatorname{ball}(f^{*};\varepsilon)$. Combining this with (11), we obtain

[TABLE]

Next, set $\rho=\|r^{*}\|$, $\alpha=\|A\|$, and $\xi=\|x\|$. Then (16) and (11) yield

[TABLE]

Note that $\|f^{*}\|-\|f^{*}-r^{*}\|\leq\|r^{*}\|\leq\|f^{*}\|+\|r^{*}-f^{*}\|$ which implies

[TABLE]

Now, set $l=(\varphi-\varepsilon)/(\alpha+\lambda)\geq 0$. On the one hand, simple algebraic manipulations show that (14) is equivalent to $\varphi/(\alpha+3\lambda)\leq l$; consequently, $2\lambda\varphi/(\alpha+3\lambda)\leq 2\lambda l$. On the other hand, (14) states that $\varepsilon\leq 2\lambda\varphi/(\alpha+3\lambda)$. Altogether,

[TABLE]

Furthermore, (17) and (18) yield $\xi\geq\rho/(\alpha+\lambda)\geq(\varphi-\varepsilon)/(\alpha+\lambda)=l$; hence,

[TABLE]

Next, consider the function $g(\eta)=\lambda\eta^{2}-\varepsilon\eta$. Let $\eta\geq l$ and observe that (19) implies that $g^{\prime}(\eta)=2\lambda\eta-\varepsilon\geq 2\lambda l-\varepsilon\geq 0$. Hence, $g$ is increasing on $\left[l,+\infty\right[$. Combining (17) and (20), we learn that $\left\langle{x},{f^{*}}\right\rangle\geq g(\xi)\geq g(l)$. Hence $\left\langle{x},{f^{*}}\right\rangle\geq g(l)$, which is precisely (15). $\hfill\quad\blacksquare$

Definition 3.3** ((whs) condition).**

Let $A\colon X\to X^{*}$ be a bounded linear monotone operator, and let $f^{*}\in X^{*}$. We say that the (whs) condition444“whs” stands for wondrous half-space. holds if $\|A\|=1$, $\|f^{*}\|=1$, and for every $\lambda>0$ and every $\varepsilon>0$, the implication

[TABLE]

is true.

For every $\lambda>0$, we define for future convenience

[TABLE]

The function $m$ plays a role in the statement and proofs of this section. The next result will also establish the strict positivity555In passing, we note that $\lim_{\lambda\to 0^{+}}m(\lambda)=0$, $\lim_{\lambda\to+\infty}m(\lambda)=0$, $m$ is strictly increasing on $[0,\bar{\lambda}]$ and strictly decreasing on $\left[\bar{\lambda},+\infty\right[$, where $\bar{\lambda}=(3+\sqrt{15})/6\approx 1.14550$, and $m(\bar{\lambda})=(9-2\sqrt{15})/(12+2\sqrt{15})\approx 0.06349$. We omit the proofs as these properties are not needed in this paper. of $m$.

Corollary 3.4**.**

Let $A\colon X\to X^{*}$ be a bounded linear monotone operator with $\|A\|=1$, let $f^{*}\in X^{*}$ be such that $\|f^{*}\|=1$, let $\lambda>0$ and assume that there exist $x\in X$ and $\varepsilon>0$ such that $\varepsilon\leq 2\lambda/(1+3\lambda)$ and $(Ax+\lambda{\,\operatorname{J}}x)\cap\operatorname{ball}(f^{*};\varepsilon)\neq\varnothing$. Finally, assume that the (whs) condition holds. Then

[TABLE]

consequently,

[TABLE]

Proof. The (whs) condition implies that $\left\langle{x},{f^{*}}\right\rangle\leq 3\varepsilon$. On the other hand, (15) yields $\left\langle{x},{f^{*}}\right\rangle\geq\lambda(1-\varepsilon)^{2}/(1+\lambda)^{2}-\varepsilon(1-\varepsilon)/(1+\lambda)$. Altogether,

[TABLE]

In turn, (25) is equivalent to

[TABLE]

which is equivalent to (23). Now let’s view (23) as a quadratic inequality. Then $\varepsilon$ must lie in the closed interval given by the roots of the corresponding quadratic equation

[TABLE]

where the variable is $\xi$. The two roots of (27) are

[TABLE]

and both roots are positive. $\hfill\quad\blacksquare$

Theorem 3.5** (key result).**

Let $A\colon X\to X^{*}$ be a bounded linear monotone operator with $\|A\|=1$, let $f^{*}\in X^{*}$ be such that $\|f^{*}\|=1$, and let $\lambda>0$. Furthermore, assume the (whs) condition holds. Then either

[TABLE]

or

[TABLE]

In any case,

[TABLE]

Proof. Assume that the “either” alternative fails, i.e., $d(f^{*},{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}}))<2\lambda/(1+3\lambda)$. Then there exist $x_{n}\in X$ and $x_{n}^{*}\in{\,\operatorname{J}}x_{n}$ such that $\varepsilon_{n}:=\|f^{*}-(Ax_{n}+\lambda x_{n}^{*})\|\leq 2\lambda/(1+3\lambda)$ and $\varepsilon_{n}\downarrow d(f^{*},{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}}))$. Now (24) of Corollary 3.4 yields

[TABLE]

Hence the “or” case follows by letting $n\to+\infty$. Finally, we claim that

[TABLE]

It is straightforward but a bit tedious to verify that

[TABLE]

where

[TABLE]

It remains to show that $\tau>0$. Indeed,

[TABLE]

$\hfill\quad\blacksquare$

We are now ready for our abstract main result.

Corollary 3.6** (main result).**

Suppose that $X$ is rugged, let $A\colon X\to X^{*}$ be a bounded linear monotone operator with $\|A\|=1$, let $f^{*}\in X^{*}$ be such that $\|f^{*}\|=1$, and assume that the (whs) condition holds. Let $\lambda>0$, set $R=\overline{\operatorname{ran}}\,(A+\lambda{\,\operatorname{J}})$, and let $\alpha\in\mathbb{R}\smallsetminus\{0\}$. Then $R$ is not convex, $\overline{\operatorname{conv}}\,R=X^{*}$, and

[TABLE]

where

[TABLE]

Proof. Because $A$ and ${\,\operatorname{J}}$ are homogeneous, it follows that $\alpha R=R$. Hence $d(\alpha f^{*},R)=d(\alpha f^{*},\alpha R)=|\alpha|d(f^{*},R)\geq|\alpha|m(\lambda)$ by Theorem 3.5. The lack of convexity of $R$ and the fact that $\overline{\operatorname{conv}}\,R=X^{*}$ follow from Corollary 2.4 and Proposition 2.3, respectively. $\hfill\quad\blacksquare$

4 The Gossez operator revisited

In this section, we assume that

[TABLE]

and that

[TABLE]

is the Gossez operator [5] and [6]. It is easy to see that $G$ is a bounded linear operator with $\|G\|=1$ and that $G$ is skew, hence monotone.

We are now ready for the main result concerning the Gossez operator.

Theorem 4.1**.**

The Gossez operator $G$ satisfies the (whs) condition with $f^{*}=-(1,1,\ldots)\in\ell_{\infty}$; consequently, $(\forall\lambda>0)$ $\overline{\operatorname{ran}}\,(G+\lambda{\,\operatorname{J}})$ is not convex yet $\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(G+\lambda{\,\operatorname{J}})=\ell_{\infty}$.

Proof. Let $\lambda>0$, and let $\varepsilon>0$. Assume that $(x,r^{*})\in\ell_{1}\times\ell_{\infty}$ satisfies $r^{*}\in(Gx+\lambda{\,\operatorname{J}}x)\cap\operatorname{ball}(f^{*};\varepsilon)$. First, there exists $x^{*}\in{\,\operatorname{J}}x$ such that $r^{*}=Gx+\lambda x^{*}$. By definition of $G$, we have $(\forall{n\in{\mathbb{N}}})$ $r_{n}^{*}=-\sum_{k<n}x_{k}+\sum_{k>n}x_{k}+\lambda x_{n}^{*}$. Letting $n\to\infty$, we deduce that

[TABLE]

Second, because $r^{*}\in\operatorname{ball}(f^{*};\varepsilon)$, we have $\|r^{*}-f^{*}\|\leq\varepsilon$. Thus $(\forall{n\in{\mathbb{N}}})$ $|r_{n}^{*}+1|\leq\varepsilon$ and so $r_{n}^{*}\leq-1+\varepsilon$. Altogether, for all $n$ sufficiently large we have $\lambda x_{n}^{*}-r_{n}^{*}\leq\sigma+\varepsilon$

[TABLE]

Using (2) and (11), we estimate

[TABLE]

Adding (42) and (43), we obtain $0\leq\sigma+3\varepsilon$. Recalling the definition of $\sigma$ from (41), deduce that $\left\langle{x},{f^{*}}\right\rangle\leq 3\varepsilon$ and the (whs) condition holds. The conclusion now follows by applying Example 2.2(i) and Corollary 3.6. $\hfill\quad\blacksquare$

Remark 4.2** (a negative answer to a question posed by Stephen Simons).**

In [7], Stephen Simons proved that $\overline{\operatorname{ran}}\,(G+\lambda{\,\operatorname{J}})$ is not convex for $0<\lambda<4$. In [7, Problem 3.6] he asks whether $\overline{\operatorname{ran}}\,(G+4{\,\operatorname{J}})$ is convex. Theorem 4.1 not only provides a negative answer but also establishes, for every $\lambda>0$, the nonconvexity of $\overline{\operatorname{ran}}\,(G+\lambda{\,\operatorname{J}})$.

Remark 4.3** (the negative Gossez operator).**

The negative Gossez operator, $-G$, is much better behaved than $G$: indeed, combining [1, Example 14.2.2 and Theorem 15.3.7], we deduce that $(\forall\lambda>0)$ $\overline{\operatorname{ran}}\,(-G+\lambda{\,\operatorname{J}})=\ell_{\infty}$. (See also [2, Example 5.2].)

5 The Fitzpatrick-Phelps operator revisited

In this section, we assume that

[TABLE]

and that

[TABLE]

It is easy to see that $F$ is a bounded linear operator with $\|F\|=1$ and that $F$ is skew, hence monotone.

We are now ready for the main result concerning the Fitzpatrick-Phelps operator.

Theorem 5.1**.**

The Fitzpatrick-Phelps operator $F$ satisfies the (whs) condition with $f^{*}\equiv-1\in L_{\infty}[0,1]$; consequently, $(\forall\lambda>0)$ $\overline{\operatorname{ran}}\,(F+\lambda{\,\operatorname{J}})$ is not convex yet $\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(F+\lambda{\,\operatorname{J}})=L_{\infty}[0,1]$.

Proof. Let $\lambda>0$, and let $\varepsilon>0$. Assume that $(x,r^{*})\in L_{1}[0,1]\times L_{\infty}[0,1]$ satisfies

[TABLE]

First, there exists $x^{*}\in{\,\operatorname{J}}x$ such that $r^{*}=Fx+\lambda x^{*}$. By definition of $F$, $(\forall t\in[0,1])$ $r^{*}(t)=\int_{0}^{t}x(s)\,ds-\int_{t}^{1}x(s)\,ds+\lambda x^{*}(t)$. Thus, for every $t\in[0,1]$, we have $|r^{*}(t)+1|\leq\varepsilon$ and also

[TABLE]

In view of the (absolute) continuity of $t\mapsto\int_{0}^{t}x(s)\,ds$ (see, e.g., [9, Theorem 6.84]), we have for all $t>0$ sufficiently small,

[TABLE]

Now note that — using (2) and again (11) — we obtain

[TABLE]

Adding (48) and (49), we deduce that $0\leq 3\varepsilon-\left\langle{x},{f^{*}}\right\rangle$ and the (whs) condition thus holds. Finally, apply Example 2.2(ii) and Corollary 3.6. $\hfill\quad\blacksquare$

Remark 5.2**.**

Fitzpatrick and Phelps (see [3, Example 3.2]) showed directly that $\overline{\operatorname{ran}}\,(F+1{\,\operatorname{J}})$ is not convex. We extend their conclusion from $\lambda=1$ to any $\lambda>0$. We note that a referee pointed out to us that there is a gap in the proof of the nonconvexity of $\overline{\operatorname{ran}}\,(F+{\,\operatorname{J}})$ in the second paragraph of [3, page 64]; however, the work in this paper does not rely on their argument and thus provides an alternative (and simpler) proof.

Remark 5.3** (the negative Fitzpatrick-Phelps operator).**

The negative Fitzpatrick-Phelps operator, $-F$, satisfies the (whs) condition again with $f^{*}\equiv-1\in L_{\infty}[0,1]$. (The proof is similar with the only difference being that in the derivation of the counterpart of (48), we work with $t$ sufficiently close to $1$ rather than to [math].) Consequently, $(\forall\lambda>0)$ $\overline{\operatorname{ran}}\,(-F+\lambda{\,\operatorname{J}})$ is not convex yet $\overline{\operatorname{conv}}\,\overline{\operatorname{ran}}\,(-F+\lambda{\,\operatorname{J}})=L_{\infty}[0,1]$.

Acknowledgements

We thank a referee for her/his pertinent comments and suggestions which resulted in significant simplifications in some of the proofs. The research of HHB and XW was partially supported by Discovery Grants of the Natural Sciences and Engineering Research Council of Canada. The research of WMM was partially supported by the Natural Sciences and Engineering Research Council of Canada Postdoctoral Fellowship.