数理解析研究所講究録
第 1544 巻 2007 年 34-39

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バナッハ空間における極大単調作用素の連続性(Continuity of Maximal Monotone Operators in Banach Spaces)

玉川大学・工学部

豊田昌史 (Masashi Toyoda)

Faculty of Enginieering,Tamagawa UniversityIntroduction

1its dual space with pairing between in $E$
be a real Banach.space,. Let be an operator on $E$ to . The
denoted byandin
domain of is the set $D(T)=\{x\in E|Tx\neq\emptyset\}$ . The graph of is thesubset of $ExE^{*}$ given by $G(T)=\{(x,x^{*})\in E\cross E^{*}|x^{*}\in Tx\}$ . The range
is said to be monotone if for allof is the set $R(T)=\cup\{Tx|x\in D(T)\}$ .
$(x,x^{*}),$ $(y,y^{*})\in G(T)$ ,$\dot{L}etE$$E^{*}$$x^{*}$$E^{*}$$x$$2^{E^{*}}$$T$$\langle x,x^{*}\rangle$

$T$

$T$

$T$

$T$

$\langle x-y,x^{*}-y^{*}\rangle\geq 0$.

It is called maximal monotone if in addition, its graph $G(T)$ is not properly
contained in the graph of any other monotone operators on $E$ . The notion ofmaximal monotone operators allows one to treat convex minimization problems
as suitable problems associated with maximal monotone operators. In fact, let]. Thenbe a proper Iower semicontinuous convex function from to.
$f$

$f$

we$E$

$(-\infty, \infty$$obt\dot{u}n$$f(x_{0})=$ $o_{e\in B}^{\cdot}f(x)\Leftrightarrow 0\in\partial f(x_{0})$where$\partial f$

is the subdifferential of$f$defined by$\theta f(x)=\{x^{*}\in E^{*}|f(y)\geq\langle y-x,x^{*}\rangle+f\cdot(x), \forall y\in E\}$for all $x\in E;.see$ Problem 4.1.2 of [4]. According to the result of Rockafellar,
is maximal monotone; see Theorem 4.4.5 of [4].
the subdifferential
In this paper, we treat some continuity property (in particular, closedness)of maximal monotone operators.
$\partial f$

Closedness of m-accretive operators

2

A Banach space $E$ is said to be smooth if $\lim_{tarrow 0}(\Vert x+ty\Vert-\Vert x||)/t$ exists foris strictly
all $y\in U$ where $U=\{x\in E|||x\Vert=1\}$ . For example, if$x,$$E^{*}$

35
convex, then $E$ is smooth; see Problem 4.3.1 of [3]. The duality mappingfrom $E$ into
is defined by $J(x)=\{x^{*}\in E^{*}|\langle x, x^{*}\rangle=\Vert x\Vert^{2}=\Vert x^{*}\Vert^{2}\}$ for
all $x\in E$ . If $E$ is smooth, then is single valued; see Theorem 4.3.1 of [3]. An
operator on a smooth Banach space $E$ into
is said to be accretive if for all$J$$2^{E^{*}}$

$J$

$A$

$(x_{1},y_{1}),$$2^{E}$$(x_{2}, y_{2})\in G(A)$,$\langle y_{1}-y_{2},$$J(x_{1}-x_{2}))\geq 0$.

An accretive operator is called m-accretive if $R(I+\lambda A)=E$ for all.
It is known that in Hilbert spaees, an m-accretive operator is equivalent to a
maximal monotone operator; see Thmrem 3.5.4 of [4].
An operator
is called demiclosed if the following condition is satisfied:
if $x_{n}\in D(A)$ for $n=1,2,$$x_{n}arrow x$ and.if there are $y_{n}\in Ax_{n}$ such that, then. $y\in Ax$ . It is noted that we use the $symbolarrow \bm{t}darrow to$ denotethe convergence in the strong and the weak topology respectively. An operator
is called locally bounded at $x\in E$ if there is a neighborhood of such that$A(U)$ is bounded.
KenmochI[l] proved the following.
$A$$\lambda>\theta$$A$$\ldots,$$y_{n}arrow y$

$A$

$U$

$x$

Theorem (Kenmochi). Let $E$ be a reaf Banach space. Let
be uniformly
convex. Let be an accretive operator on $E$ towith open domain. Supposethat is demiclosed and $Ax$ is closed and convex for all $x\in D(A)$ and locallybounded at every point of $D(A)$ . Then is m-accretive.
$E^{*}$$A$$2^{E}$$A$$A$

3$\cdot$Closedness of maximal monotone operators

An operator on $E$ tois called demiclosed if the $f_{0}nowing$ condition issatisfied: if $x_{n}\in D(T)$ for $n=1,2,$$x_{n}arrow x$ and if there aresuch that, then $x^{*}\in Tx$ . For example, demicontinuous operators are
demiclosed in reflexive Banach spaces. In thiv section, we show some relation
between demiclosedness and maximality for monotone operators.
For our main result, we need the following result which was essentially provedby Showalter (see Remark of [2]) and here present the proof for the sake of$T$$2^{E}$$x_{n}^{*}\in Tx_{n}$$\ldots,$$x_{n}^{*}arrow x^{*}$

completeness.

Theorem (Showalter). Let $E$ be a Banach space and let be a monotone
operator on $E$ to. Let belong to the interior of $D(T)$ and $y\in\prime D(T)$ , andset $x_{t}=(1-t)x+ty$ for $t\in[0,1$ ) . $Then\cup\{Tx_{t}|t\in[0,1)\}$ is bounded in.
$T$$2^{B}$$x$$E^{*}$

Proof. Let $z\in E$ . We may show that
Let $t\in[0,1$ ) and. Choose$x_{t}^{*}\in Tx_{t}$is bounded.
monotone,
iswe have$\{\langle z,x_{t}^{*}\rangle|t\in i0,1), x_{t}^{*}\in Tx_{t}\}$$y^{*}\in Ty$. Since$T$$0\leq\langle y-x_{t},y^{*}-x_{t}^{*}\rangle=(1-t)(y-x,y^{*}-x^{*}\rangle$

So we obtain$\langle y-x, y^{*}\rangle\geq\langle y-x,x_{t}^{*}\rangle$.

.

36
Since

$x$

belongs to the interior of $D(T)$ , there exists some positive integer nosuch that$T(x+ \frac{1}{n_{0}})\neq\emptyset$

and$T(x- \frac{1}{n_{0}})\neq\emptyset$. Choose$z_{1}^{*} \in T(x+\frac{1}{n_{0}})$.

By monotonicity of , we have$T$

$0 \leq\langle x+\frac{1}{n_{0}}z-x_{t}, z_{1}^{*}-x_{t}^{*}\rangle=\langle\frac{1}{n_{0}}z-t(y-x), z_{1}^{*}-x_{t}^{*}\rangle$.

So we obtain$\langle\frac{1}{n_{0}}z-t(y-x), z_{1}^{*}\rangle+t(y-x,$$\geq\frac{1}{n_{0}}\langle z,x_{t}^{*}\rangle$$x_{t}^{*}\rangle$.

Therefore we have$\langle z,x_{t}^{*}\rangle\leq n_{0}(t(y-x,x_{t}^{*}\rangle+\frac{1}{n_{0}}\langle z\backslash ’ z_{1}^{*}\}-t\langle y-x, z_{1}^{*}\})$$\leq n_{0}(t(y-x,y^{*}\rangle+\frac{1}{n_{0}}\langle z, z_{1}^{*}\rangle-t\langle y-x, z_{1}^{*}\rangle)$$\leq n_{0}(|\langle y-x,y^{*}\rangle|+\frac{1}{n_{0}}(z,$Next choose$z_{2}^{*} \in T(x-\frac{1}{n_{0}})$. Since$T$$z_{i}^{*}\rangle$$+|(y-x,z_{1}^{*}\rangle$ $|)$.

is monotone, we also have.

$0 \leq\langle x-\frac{1}{n_{0}}z-x_{t}, z_{2}^{*}-\backslash x_{t}^{*}\rangle=\langle-\frac{1}{n_{0}}z-t(y-x), z_{2}^{*}-x_{t}^{*}\rangle$

It follows that.

$\langle-\frac{1}{n_{0}}z-i(y-x), z_{2}^{*}\rangle+t\langle y-x, x_{t}^{*}\rangle\geq-\frac{1}{n_{0}}\langle z, x_{t}^{l}\rangle$

So we have$- \langle z, x_{t}^{*}\rangle\leq.n_{0}(t\langle y-x, x_{t}^{*})-\frac{1}{n_{0}}(z, z_{2}^{*}\rangle-t\langle y-x,$ $z_{2}^{l}$)$)$$\leq n_{0}(t\langle y-x,y^{*}\rangle-\frac{1}{n_{0}}\langle z, z_{2}^{*}\rangle-t(y-x, z_{2}^{*}\rangle)$$\leq n_{0}(|\{y-x,y^{*}\rangle|.-\frac{1}{n_{0}}(z,$$z_{2}^{*}\rangle$$+|(y-x, z_{2}^{*}\rangle$ $|)$.

.
is bounded independently of $t\in[0,1$ ) andis bounded. The desired
result follows by the uniform boundedness theorem.

These show thatHence we obtain that$\langle z,x_{t}^{*}\rangle$$x_{t}^{r}\in Tx_{t}.$$\{\langle z, x_{t}^{*}\rangle|t\in[0,1), x_{t}^{*}\in Tx_{t}\}$$\square$Motivated by the result by Kenmochi, we obtain the following.

37Theorem A. Let $E$ be a reflexzve Banach space. Let be an opemtor on $E$ to. Suppose that $D(T)=E$ . Then is maximal monotone if and on. if isa monotone, demiclosed operator such that for all $x\in E,$ $Tx$ is a closed convex
set.
$T$

$2^{E^{*}}$$T$$ly$$T$

is maximal monotone. We show that is a mono
tone, demiclosed operator such that for all $x\in E,$ $Tx$ is a closed convex set. Asis maximal monotone, we have

Proof. Assume first that

$T$

$T$

$T$

$Tx= \bigcap_{(y,y)\in G(T)}\{x^{*}\in E^{*}|\langle x-y,x^{*}-y^{*}\rangle\geq 0\}$for all $x\in D(T)$ . Then $Tx$ is convex and weakstar closed because in the aboveintersection each set has these two properties. Since $E$ is reflexive, we have $Tx$
is a closed set.
$x_{n}arrow x$ and there are
Suppose that $x_{n}\in D(T)$ for $n=1,2,$$x^{*}\inTx$. We show that
. Let $(y,y^{*})\in G(T)$ . Since T.issuch thatmonotone, we have$x_{n}^{*}.\in Tx_{n}$$\ldots,$$x_{n}^{*}arrow x^{*}$$(y-x_{n},$ $y^{*}-x_{n}^{*}\rangle$ $\geq 0$.

So we obtain$(y-x,$$y^{*}-x^{*}\rangle$$=\langle y-x,y^{*}-x_{n}^{*}\rangle+\langle y-x,x_{n}^{*}-x^{*}\rangle$$=(y-x_{n},y^{*}-x_{n}^{*}\rangle+(x_{n}-x,y^{*}-x_{n}^{*}\rangle+(y-x,$ $x_{n}^{l}-x^{r}\rangle$$\geq\langle x_{n}-x,y^{*}-x_{n}^{*}\rangle+\langle y-x,x_{n}^{*}-x^{*}\rangle$$\geq-\Vert x_{n}-x\Vert\Vert y^{*}-x_{n}^{*}\Vert+\langle y-x,x_{n}^{*}-x^{*})$

This yields$\langle y-x,y^{*}-x^{*}\rangle\geq 0$.

.

Since is maximal monotone, $(x,x^{*})\in G(T)$ .
We now assume that is a monotone, demiclosed operator such that for all$x\in E,$ $Tx$ is a closed convex set. Then there exists a monotone operator
suchthat $G(T)\subset G(S)$ and $G(T)\neq G(S)$ . Let
be a point in $G(S)\backslash G(T)$ .
Since
andis weakly closed convex, there exists $z\in E$ satisfying
$T$

$T$

$S$

$(x_{0}, x_{0}^{*})$$x_{0}^{*}\not\in Tx_{0}$$Tx_{0}$$\langle z,x_{0}^{*}\rangle>\sup_{x^{*}\in \text{劣_{}0}}\langle z,x^{*}\rangle$.

(1) .

We put$x_{n}=(1- \frac{1}{n})x_{0}+\frac{1}{n}(x_{0}+.z)$

By Showalter’s Theorem,$\bigcup_{n=1}^{\infty}Tx_{n}$,$n=1,2,$is bounded. We choose$\ldots$.

$x_{n}^{*}\in Tx_{n}$Since $E$ is reflexive, there exists a subsequencesuch thatweakly to some
. Since is demiclosed, we have$(x_{n:}^{*})$$p^{*}\in E^{*}$$T$for each$(x_{\mathfrak{n}_{i}}^{*})$$p^{*}\in Tx_{0}$.

$n$.

converges

38
Since

$x_{0}^{*}\in Sx_{0},$$x_{n:}^{*}\in Sx_{n}$: and$S$

is monotone,$\langle x-x_{n},x_{0}^{*}-x_{n_{\dot{t}}}^{*}\rangle\geq 0$

SO that$\langle z,x_{0}^{*}-x_{n_{i}}^{*}\rangle\leq 0$

We see from this that$(z,x_{0}^{*}-p^{*}\rangle\leq 0$

So we obtain$\langle z,x_{0}^{*}\rangle\leq\langle z,p^{*}\rangle$,.

.
.

(2)

From (1) and (2), we see that.

$\langle z,x_{0}^{*}\rangle>\sup_{x\in Toe0}(z,x^{*}\rangle\geq\langle z,p^{*})\geq(z,x_{0}^{*}\rangle$

Hence we deduce a contradiction.

口Remark 1. Let $E$ be a $refl\infty cive$ Banach space. It is known that the duality
mapping is maximal monotone. We prove it using $Th\infty rem$ A. First, ismonotone and $Jx$ is a nonempty closed convex set for $x\in E$ ; see Theorem 4.2.1. We
andof [3], liUrthermore is demiclosed. Let. $x_{n}arrow x,$$semi\infty
ntinuou8$
in the wek
is lower
may show $x^{*}\in Jx$ . Since the norm oftopology, we have
$J$

$J$

$J$

$x_{n}^{*}\in Jx_{n}$$x_{n}^{*}arrow x^{*}$$E^{*}$$\Vert x^{*}\Vert\leq\lim_{r\iotaarrow}\inf_{\infty}\Vert x_{n}^{*}\Vert=\lim_{narrow}\inf_{\infty}\Vert x_{n}\Vert=\Vert x\Vert$.

On the other hand, we have.

$|(x,x^{*}\rangle-\Vert x_{n}\Vert^{2}|=|(x,x^{*}\rangle-\langle x_{n},x_{n}^{*}\rangle|\leq|\langle x,x^{*}-x_{n}^{*}\rangle|+|\langle x-x_{n},x_{n}^{*}\rangle|$

So we obtain that$\Vert x\Vert^{2}=\{x,x^{*}\rangle$$\Vert x\Vert^{2}=\Vert x^{*}\Vert^{2}$$|\langle x, x^{*}\rangle-\Vert x_{n}\Vert^{2}|arrow 0$$\leq\Vert x\Vert\Vert x^{l}.\Vert$. Then we have

, and we have$\Vert x\Vert\leq\cdot\Vert x^{*}\Vert$and hence $x^{*}\in Jx$ . By $Th\infty remA,$$\langle x,x^{*}\rangle=\backslash |.|x\Vert^{2}$. Therefore. .These imply that$J$$\langle x, x^{*}\rangle=$is maximal monotone.

Remark 2. Main result of this paper is related to our preceding work[5]. Let. We define psuedo-continuous at $x\in D(T)$be an operator on $E$ to, there exists a neighborhood
if for each weakstar open set with
Using this continuity, we obtain theof such that $z\in U$ implies
following.
$T$$2^{E}$

$T$

$V$

$x$

$U$

$V\subset\prime Tx$$Tz\cap V\neq\emptyset.$

be an operator on $E$.Theorem ([5]). Let $E$ be a real Banach space. Let. Suppose that $D(T)=E$ . Then is maximal monotone if and only iftois a monotone, psuedo-continuous operator such that for all $x\in E,$ $Tx$ is aweakstar closed convex set.
$T$

$2^{B^{*}}$

$T$

$T$

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References[1] N. Kenmochi, Accretive mappings in Banach spaces, Hiroshima Math. J. 2 (1972), 163-177.
[2] R. E. Showalter, Continuity of maximal monotone sets in Banach space, Proc, Amer.
Math. Soc. 42 (1973), 543-546.
[3] W. Takahashi, $Nonlinearfi\{mctional\backslash$ Analysis, Yokohama Publishers, Yokohama, 2000.
[4] W. Takahashi, Conv $ex$ Analysis and Approximation of Fixed Points, Yokohama Publishers, Yokohama, 2000. (in Japanese)[5] M. Toyoda, Continuity of set-valued monotone mappings, Proc. Internat. Conf. NonlinearAnal. Convex Anal. (Hirosaki, 2001), 501-505.