1

数理解析研究所講究録
第2065巻 2018年 1-8

Resolvents of convex functions and the shrinking
projection method on geodesic spaces
凸関数のリゾルベントと測地距離空間における収縮射影法
東邦大学理学部木村泰紀
Yasunori Kimura

Department of Information Science
Faculty of Science
Toho University

Abstract

We apply the shrinking projection method to the convex minimization prob‐
lems on a complete geodesic space by using the notion of resolvents for convex
functions. We consider the calculation errors for metric projections onto theclosed convex sets, and obtain the upper bound of the error for the limit ofapproximation sequence. We also discuss the choice of the coefficient of convex

combination used in the iterative process.

1

Introduction

In the fixed point theory, the approximation methods for fixed points is one of themost important branches in the study of nonexpansive and other nonlinear mappings
as well as the existence of fixed points. We will focus on the shrinking projection
method, which was originally proved in 2008 by Takahashi, Takeuchi, and Kubota [3].

Theorem 1 (Takahashi, Takeuchi, and Kubota [3]). Let H be a real Hilbert space andC a nonempty closed convex subset of H.

Let T : C\rightarrow C be a nonexpansive mapping
such that the set Fix T of fixed points of T is nonempty. Let \{$\alpha$_{n}\} be a sequence in[0 , 1[ with $\alpha$_{0}=\displaystyle \sup_{n\in \mathrm{N} $\alpha$_{n} < 1 . For a given point u\in H , generate a sequence \{x_{n}\}by the following iterative scheme: x_{1}\in C, C_{1}=C , andy_{n}=$\alpha$_{n}x_{n}+(1-$\alpha$_{n})Tx_{n},C_{n+1}=\{z\in C : \Vert y_{n}-z\Vert \leq\Vert x_{n}-zx_{n+1}=P_{C_{n+1}}u\cap C_{n},2for n\in \mathbb{N} . Then, \{x_{n}\} converges strongly to P_{\mathrm{F}\mathrm{i}\mathrm{x}T}u\in C , where P_{K} is the metricprojection of H onto a nonempty closed convex subset K of H.
We note that the original result deals with the family of nonexpansive mappings.
In this paper, we will apply this method to the convex minimization problems oncomplete CAT(I) spaces by using the notion of resolvents for convex functions. The
concept of resolvent on complete CAT(I) spaces was recently proposed by Kimura andKohsaka [2]. We consider the calculation errors for metric projections onto the closed
convex sets, and obtain the upper bound of the error for the limit of approximation
sequence. We also discuss the choice of the coefficient of convex combination used in
the iterative process.

2

Preliminaries

Let X be a metric space. For x, y\in X and l\geq 0 , we call a mapping c : [0, l]\rightarrow X\mathrm{a}geodesic with endpoints x, y if it satisfies that c(0)=x, c(l)=y , and d(c(t), c(s))=|t-s| for every t, s \in [0, l] . If a geodesic exists for every x, y \in X, X is called
a geodesic space. In what follows, we assume that a geodesic is always unique forevery choice endpoints. The image of a geodesic c with endpoints x, y\in X is called
a geodesic segment joining x and y , and we denote it by [x, y] . For x, y \in X and$\alpha$\in[0 , 1 ] , we define z= $\alpha$ x\oplus(1- $\alpha$)y as a unique point z\in [x, y] such thatd(z, y)= $\alpha$ d(x, y) , d(x, z)=(1- $\alpha$)d(x, y).

Let X be a geodesic space such that d(u, v) < $\pi$/2 for all u, v \intriangle forx, y, z \inX

is defined as \triangle(x, y, z)=X.

A geodesic[y, z]\cup[z, x]\cup [x, y] . A CAT(I)

space is usually defined by using the notion of comparison triangle defined in themodel space having the curvature 1. For the details, see [1]. We remark that we may
characterize a geodesic space X to be a CAT(I) space by that the inequality\cos d( $\alpha$ x\oplus(1- $\alpha$)y, z)\sin d(x, y)\geq\cos d(x, z)\sin( $\alpha$ d(x, y))+\cos d(y, z)\sin((1- $\alpha$)d(x, y))holds for every x, y, z\in X and $\alpha$\in[0 , 1 ] . In particular, since \sin td(x, y)\leq t\sin d(x, y)
for t\in[0 , 1 ] , it holds that\cos d( $\alpha$ x\oplus(1- $\alpha$)y, z)\geq $\alpha$\cos d(x, z)+(1- $\alpha$)\cos d(y, z)
for any $\alpha$\in[0 , 1 ] and x, y, z in a CAT(I) space.
Let C be a nonempty closed convex subset of a complete CAT(I) space X such thatd(u, v)< $\pi$/2 for all u, v\in X . For any x\in X , there exists a unique y_{x}\in C such thatd(x, y_{x})=d(x, C)=\displaystyle \inf_{y\in C}d(x, y) . We define the metric projection P_{C} : X\rightarrow C by
P_{C}x=y_{x} for x\in X.

Let X be a complete CAT(I) space and let f : X\rightarrow ] -\infty, +\infty ] be a proper lower
semicontinuous convex function. Then, for x\in X , the function g_{x} : X\rightarrow ] -\infty, +\infty ]3defined by g_{x}(y)=f(y)+\tan d(y, x) sin d(y_{x}, x) has the unique minimizer. We definea resolvent J_{f} : X\rightarrow X by

J_{f}x=\displaystyle \arg\min_{y\in X}g_{x}(y)=\arg\min_{y\in X}(f(y)+\tan d(y, x)\sin d(y, x))
for x\in X . It is known that the set Fix J_{f}=\{z\in X : J_{f}z=z\} of fixed points of J_{f}\displaystyle \arg\min_{y\in X}f(y) of minimizers of f on X . We also know thatthe resolvent operator is firmly spherically nonspreading in the sense thatcoincides with the set(\cos d(J_{f}x, x)+\cos d(J_{f}y, y))\cos^{2}d(J_{f}x, J_{f}y)\geq 2\cos d(J_{f}x, y)\cos d(x, J_{f}y)for all x, y\in X . In particular, it is quasinonexpansive;d(J_{f}x, z)\leq d(x, z)
for all x \in X and z\in Fix J_{f} . For a nonempty closed convex subset C\subset X , if f is
the indicator function i_{C} of C defined by

i_{C}(x)=\left\{ begin{aray}{l \infty&(x\inC),\ 0&(x\not\inC), \end{aray}\right.
then, J_{f} is the metric projection onto C . For more details of the resolvent operators,see [2].

3

Approximation of the minimizer of the function

In this section, we prove an approximation theorem to a solution to a convex mini‐

mization problem for a convex function defined on a complete CAT(I) space.
Theorem 2. Let X be a complete CAT(I) space such that d(u, v) < $\pi$/2 for anyu, v\in X and that a subset \{z\in X:d(v, z)\leq d(u, z)\} is convex for any u, v\in X . Let
f : X\rightarrow]-\infty, +\infty] be a proper lower semicontinuous convex function such that the set
S=\displaystyle \arg\min_{x\in X}f(x) of its minimizers is nonempty. Let \{$\alpha$_{n}\} be a sequence in [0 , 1[with $\alpha$_{0} =\displaystyle \sup_{n\in \mathrm{N} $\alpha$_{n} < 1, \{$\epsilon$_{n}\} be a sequence in [ 0, $\pi$/4[, and $\epsilon$_{0} =\displaystyle \lim\sup_{n\rightarrow\infty}$\epsilon$_{n}.
Let J_{f} be the resolvent for f . For given u\in X , generate a sequence \{x_{n}\} \subset X and\{C_{n}\} as follows: x_{1}=u, C_{1}=X , andy_{n}=$\alpha$_{n}x_{n}\oplus(1-$\alpha$_{n})J_{f}x_{n},C_{n+1}=\{z\in X : d(y_{n}, z)\leq d(x_{n}, z)\}\cap C_{n},x_{n+1}\in C_{n+1} such that \cos d(u, x_{n+1})\geq\cos d(u, C_{n+1})\cos$\epsilon$_{n+1},
for each n\in \mathrm{N} . Then

\displaystyle \lim_{n\rightar ow}\sup_{\infty}d(x_{n}, J_{f}x_{n})\leq \frac{2$\epsilon$_{0} {1-$\alpha$_{0}

4

and

f(p)\displaystyle \leq\lim_{n\rightar ow}\inf_{\infty}f(J_{f}x_{n})\leq\lim_{n\rightar ow}\sup_{\infty}f(J_{f}x_{n})\displaystyle \leq f(p)+ $\pi$(\sec^{2}\frac{2$\epsilon$_{0} {1-$\alpha$_{0} +\mathrm{I})\sin\frac{$\epsilon$_{0} {1-$\alpha$_{0} .
Moreover, if $\epsilon$_{0}=0 , then \{x_{n}\} converges to P_{S}u , where P_{S} is the metric projectionof X onto S.
Proof. We first show that each C_{n} is a closed convex subset containing S by induction.
It is obvious that C_{1} =X satisfies the conditions. Suppose that C_{k} is a nonempty
closed convex subset of X and S\subset C_{k} for fixed k\in \mathrm{N} . To show S\subset C_{k+1} , let z\in S.
Then, since J_{f} is quasinonexpansive with Fix J_{f}=S , we have that\cos d(y_{k}, z)=\cos d($\alpha$_{k}x_{k}\oplus(1-$\alpha$_{k})J_{f}x_{k}, z)\geq$\alpha$_{k}\cos d(x_{k}, z)+(1-$\alpha$_{k})\cos d(J_{f}x_{k}, z)=\cos d(x_{k}, z),which implies d(y_{k}, z)\leq d(x_{k}, z) , and hence z\in S\subset C_{k+1} . We also get that C_{k+1} isclosed from the continuity of the metric d . The convexity of C_{k+1} is obtained fromthe assumption of the space. We also know that, there exists at least one point y\in C_{k}such that \cos d(u, y)\geq\cos d(u, C_{k})\cos$\epsilon$_{k} . In fact, the point P_{C_{k}}u\in X satisfies that\cos d(u, P_{C_{k}}u)=\cos d(u, C_{k})\geq\cos d(u, C_{k})\cos$\epsilon$_{k}.
Consequently, we have that C_{n} is a closed convex subset containing S for all n\in \mathbb{N}and thus \{x_{n}\} is well defined. Since x_{n}\in C_{n} , we have that\cos d(u, x_{n})\geq\cos d(u, C_{n})\cos$\epsilon$_{n}for every n\in \mathrm{N}.
Let p_{n}=P_{C_{n}}u , where P_{C_{n}} is the metric projection of X onto C_{n} for n\in \mathrm{N} , and let

C_{0}=\displaystyle \bigcap_{n=1}^{\infty}C_{n} . Then, since each C_{n} is convex, we have that $\alpha$ p_{n}\oplus(1- $\alpha$)x_{n} \in C_{n}for $\alpha$\in]0, 1[. Therefore,\sin d(p_{n}, x_{n})\cos d(p_{n}, u)\geq\sin d(p_{n}, x_{n})\cos d( $\alpha$ p_{n}\oplus(1- $\alpha$)x_{n}, u)\geq\sin( $\alpha$ d(p_{n},x_{n}))\cos d(p_{n}, u)+\sin((1- $\alpha$)d(p_{n}, x_{n}))\cos d(x_{n}, u)and thus we have that\sin d(p_{n}, x_{n})-\sin( $\alpha$ d(p_{n}, x_{n}))2 \displaystyle \sin( 1- $\alpha$)d(p_{n},x_{n}) \frac{\cos d(x_{n},u)}{\cos d(p_{n},u)}5=2\displaystyle \cos(\frac{1- $\alpha$}{2}d(p_{n}, x_{n}) \sin(\frac{1- $\alpha$}{2}d(p_{n}, x_{n}) \frac{\cos d(x_{n},u)}{\cos d(p_{n},u)}.
We also have that\displaystyle \sin d(p_{n}, x_{n})-\sin( $\alpha$ d(p_{n}, x_{n}) =2\cos(\frac{1+ $\alpha$}{2}d(p_{n},x_{n}) \sin(\frac{1- $\alpha$}{2}d(p_{n}, x_{n}).

Suppose that p_{n}\neq x_{n} . Then these inequalities imply that\displaystyle \cos(\frac{1+ $\alpha$}{2}d(p_{n}, x_{n}) \geq\cos(\frac{1- $\alpha$}{2}d( p_{n}, x_{n}) \frac{\cos d(x_{n},u)}{\cos d(p_{n},u)}.
Tending $\alpha$\rightarrow 1 , we have that\displaystyle \cos d(p_{n},x_{n})\geq\frac{\cos d(x_{n},u)}{\cos d(p_{n},u)}=\frac{\cos d(x_{n},u)}{\cos d(u,C_{n}) \geq\cos$\epsilon$_{n},and it follows that d(p_{n}, x_{n}) \leq $\epsilon$_{n} for every n \in \mathb {N} . If p_{n}is trivially true.

Thus it holds for all
Since p_{n+1}(1-$\alpha$_{n})d(J_{f}x_{n}, x_{n}) , and $\alpha$_{0}=\displaystyle \sup_{n\in \mathrm{N} $\alpha$_{n}<1 , we have thatn\in\mathbb{N} .

=\inx_{n} ,

this inequality
C_{n+1}, d(y_{n}, x_{n})=d(J_{f}x_{n}, x_{n})=\displaystyle \frac{\mathrm{I} {1-$\alpha$_{n} d(y_{n}, x_{n})\leq\underline{1}(d(y_{n},p_{n+1})+d(p_{n+1}, x_{n}))1-$\alpha$_{0}\leq\underl
ine{2}d(x_{n},p_{n+1})
1-$\alpha$_{0}\leq\underl
ine{2}(d(x_{n},p_{n})+d$\omega$_{n},p_{n+1})
1-$\alpha$_{0}\displaystyle \leq \frac{2}{1-$\alpha$_{0} ($\epsilon$_{n}+d(p_{n},p_{n+1})
for every n\in \mathrm{N} . Tending n\rightarrow\infty , we obtain that\displaystyle \lim_{n\rightar ow}\sup_{\infty}d(J_{f}x_{n}, x_{n})\leq\frac{2$\epsilon$_{0} {1-$\alpha$_{0} .
Let p=P_{S}u\in S and z_{ $\alpha$}= $\alpha$ J_{f}x_{n}\oplus(1- $\alpha$)p for $\alpha$\in ] 0 , 1[. Then, we have thatf(J_{f}x_{n})+\tan d(J_{f}x_{n}, x_{n})\sin d(J_{f}x_{n}, x_{n})\leq f(z_{ $\alpha$})+\tan d(z_{ $\alpha$}, x_{n})\sin d(z_{ $\alpha$}, x_{n})\leq $\alpha$ f(J_{f}x_{n})+(1- $\alpha$)f(p)+\tan d(z_{ $\alpha$}, x_{n})\sin d(z_{ $\alpha$}, x_{n})Since \tan t\sin t=\sec t-\cos t=1/\cos t-\cos t , we have that

(1- $\alpha$)(f(J_{f}x_{n})-f(p)).

6\leq\tan d(z_{ $\alpha$}, x_{n})\sin d(z_{ $\alpha$}, x_{n})-\tan d(J_{f}x_{n}, x_{n})\sin d(J_{f}x_{n}, x_{n})= (\displaystyle \frac{1}{\cos d(z_{ $\alpha$},x_{n})}-\frac{1}{\cos d(J_{f}x_{n},x_{n})}) -(\cos d(z_{ $\alpha$}, x_{n})-\cos d(J_{f}x_{n}, x_{n})= (\displaystyle \frac{1}{\cos d(z_{ $\alpha$},x_{n})\cos d(J_{f}x_{n},x_{n})}+1)(\cos d(J_{f}x_{n}, x_{n})-\cos d(z_{ $\alpha$}, x_{n})=E_{ $\alpha$,n}(\cos d(J_{f}x_{n}, x_{n})-\cos d(z_{ $\alpha$}, x_{n}where

E_{ $\alpha$,n}=\displaystyle \frac{1}{\cos d(z_{ $\alpha$},x_{n})\cos d(J_{f}x_{n},x_{n})}+1.
Let D_{n}=d(J_{f}x_{n},p) for n\in \mathrm{N} . If D_{n_{0}}=0 for some n_{0}\in \mathrm{N} , then

J_{f}x_{n_{0}}=p\in S= Fix J_{f}.
This implies that x_{n}=x_{n_{0}}=p for all n\geq n_{0} and hence \{x_{n}\} converges to p=P_{\mathcal{S}}u,
the conclusions of the theorem obviously hold. Therefore, we suppose that D_{n} >0
for all n\in \mathrm{N} . We have that(\cos d(J_{f}x_{n}, x_{n})-\cos d(z_{ $\alpha$}, x_{n}))\sin D_{n}=\cos d(J_{f}x_{n}, x_{n})\sin D_{n}-\cos d( $\alpha$ J_{f}x_{n}\oplus(1- $\alpha$)p, x_{n})\sin D_{n}\leq\cos d(J_{f}x_{n}, x_{n})\sin D_{n}\cos d(J_{f}x_{n}, x_{n})\sin( $\alpha$ D_{n})-\cos d(p, x_{n})\sin((1- $\alpha$)D_{n})=\cos d(J_{f}x_{n}, x_{n})(\sin D_{n}-\sin( $\alpha$ D_{n}))-\cos d(p, x_{n})\sin((1- $\alpha$)D_{n})\leq(\sin D_{n}-\sin( $\alpha$ D_{n}))-\cos d(p, x_{n})\sin((1- $\alpha$)D_{n})‐\displaystyle \leq 2\cos\frac{(1+ $\alpha$)D_{n} {2}\sin\frac{(1- $\alpha$)D_{n} {2}-\cos d(p, x_{n})\sin( 1- $\alpha$)D_{n}),and therefore(f(J_{f}x_{n})-f(p) \displaystyle \frac{\sin D_{n} {D_{n}=\displaystyle \frac{E_{ $\alpha$,n} {(1- $\alpha$)D_{n} (\cos d(J_{f}x_{n}, x_{n})-\cos d(z_{ $\alpha$}, x_{n}) \sin D_{n}\displaystyle \leq\frac{E_{ $\alpha$,n} {(1- $\alpha$)D_{n} 2\cos\frac{(1+ $\alpha$)D_{n} {2}\sin\frac{(1- $\alpha$)D_{n} {2}-\cos d(p, x_{n})\sin( 1- $\alpha$)D_{n})=E_{ $\alpha$,n}(\displaystyle \frac{\sin( 1- $\alpha$)D_{n}/2)}{(1- $\alpha$)D_{n}/2}\cos\frac{(1+ $\alpha$)D_{n} {2}-\frac{\sin( 1- $\alpha$)D_{n})}{(1- $\alpha$)D_{n} \cos d(p, x_{n}).

Since E_{ $\alpha$,n}\rightarrow 1/\cos^{2}d(J_{f}x_{n}, x_{n})+1=\sec^{2}d(J_{f}x_{n}, x_{n})+1 as $\alpha$\uparrow 1 , we have that(f(J_{f}x_{n})-f(p) \displaystyle \frac{\sin D_{n} {D_{n}7\leq (sec2 d(J_{f}x_{n}, x_{n})+1 ) (\cos D_{n}-\cos d(p, x_{n}))=(\sec^{2}d(J_{f}x_{n}, x_{n})+1)(\cos d(J_{f}x_{n},p)-\cos d(x_{n},p)).

Further we have that\cos d(J_{f}x_{n},p)-\cos d(x_{n},p)=2\displaystyle \sin\frac{d(x_{n},p)+d(J_{f}x_{n},p)}{2}\sin\frac{d(x_{n},p)-d(J_{f}x_{n},p)}{2}\displaystyle \leq 2\sin\frac{d(x_{n},p)-d(J_{f}x_{n},p)}{2}\displaystyle \leq 2\sin\frac{d(J_{f}x_{n},x_{n})}{2}.
Since \sin D_{n}/D_{n}\geq 2/ $\pi$ for all n\in \mathbb{N} , we get that\displaystyle \frac{2}{ $\pi$}(f(J_{f}x_{n})-f(p) \leq(f(J_{f}x_{n})-f(p) \frac{\sin D_{n} {D_{n}\displaystyle \leq 2(\sec^{2}d(J_{f}x_{n}, x_{n})+1)\sin\frac{d(J_{f}x_{n},x_{n})}{2}.
Tending n\rightarrow\infty , we have that\displaystyle \frac{2}{ $\pi$} (\lim_{n\rightar ow}\sup_{\infty}f(J_{f}x_{n})-f(p)\displaystyle \leq 2\lim_{n\rightar ow}\sup_{\infty}(\sec^{2}d(J_{f}x_{n}, x_{n})+1)\sin\frac{d(J_{f}x_{n},x_{n})}{2}\displaystyle \leq 2(\sec^{2}\frac{2$\epsilon$_{0} {1-$\alpha$_{0} +1)\sin\frac{$\epsilon$_{0} {1-$\alpha$_{0} .
Hence we obtain thatf(x_{0})-f(p)\displaystyle \leq\lim_{n\rightar ow}\sup_{\infty}f(J_{f}x_{n})-f(p)\leq $\pi$(\sec^{2}\frac{2$\epsilon$_{0} {1-$\alpha$_{0} +\mathrm{I})\sin\frac{$\epsilon$_{0} {1-$\alpha$_{0} ,and therefore

f(p)\displaystyle \leq\lim_{n\rightar ow}\inf_{\infty}f(J_{f}x_{n})\leq\lim_{n\rightar ow}\sup_{\infty}f(J_{f}x_{n})\displaystyle \leq f(p)+ $\pi$(\sec^{2}\frac{2$\epsilon$_{0} {1-$\alpha$_{0} +1)\sin\frac{$\epsilon$_{0} {1-$\alpha$_{0} ,which is the desired result.

For the latter part of the theorem, suppose $\epsilon$_{0}0 . Then, since d(x_{n},p_{n}) \leq $\epsilon$
and \displaystyle \lim_{n\rightar ow\infty}$\epsilon$_{n} =$\epsilon$_{0} =0 , we get that both \{x_{n}\} and \{p_{n}\} converges to p_{0}=P_{C_{0}}u=

Moreover, since\displaystyle \lim_{n\rightar ow}\sup_{\infty}d(J_{f}x_{n}, x_{n})\leq\frac{2$\epsilon$_{0} {1-$\alpha$_{0} =0,8\{J_{f}x_{n}\} also converges to p_{0} . Using the lower semicontinuity of f , we have thatf(p)\leq f(p_{0})\displaystyle \leq\lim_{n\rightar ow}\inf_{\infty}f(J_{f}x_{n})\leq f(p)+ $\pi$ (sec20 + 1) \sin 0
=f(p)

.

S . Since S \subset C_{0} and p_{0}Therefore p_{0} \in \displaystyle \arg\min_{x\in X}f(x)p_{0}=P_{S}u , which completes the proof.
==

P_{C_{0}}u , we have that\squareLet us consider the values\displaystle\frac{2$\epsilon$_{0} 1-$\alpha$_{0}and$\pi$(\displaystyle\sec^{2}\frac{2$\epsilon$_{0} {1-$\alpha$_{0} +1)\sin\frac{$\epsilon$_{0} {1-$\alpha$_{0}of the upper bounds for \displaystyle \lim\sup_{n\rightarrow\infty}d(x_{n}, J_{f}x_{n}) and \displaystyle \lim\sup_{n\rightarrow\infty}|f(p)-f(J_{f}x_{n})|=\displaystyle \lim\sup_{n\rightarrow\infty}|\min_{x\in X}f(x)-f(J_{f}x_{n})| . It is easy to see that these values are increasing

with respect to $\alpha$_{0}\in[0, 1[. Therefore, if we wish to make them as small as possible,

the best choice is $\alpha$_{0} to be 0 . Letting simply $\alpha$_{n} =0 for all n\in \mathrm{N} , we obtain thefollowing iterative scheme: x_{1}=u, C_{1}=X , andC_{n+1}=\{z\in X : d(J_{f}x_{n}, z)\leq d(x_{n}, z)\}\cap C_{n},x_{n+1}\in C_{n+1} such that \cos d(u, x_{n+1})\geq\cos d(u, C_{n+1})\cos$\epsilon$_{n+1},
for each n\in \mathrm{N} . Then, by the main result, we obtain that \displaystyle \lim\sup_{n\rightarrow\infty}d(x_{n}, J_{f}x_{n}) \leq
2$\epsilon$_{0} andf(p)\displaystyle \leq\lim_{n\rightar ow}\inf_{\infty}f(J_{f}x_{n})\leq\lim_{n\rightar ow}\sup_{\infty}f(J_{f}x_{n})\leq f(p)+ $\pi$ (sec2 (2$\epsilon$_{0})+1 ) \sin$\epsilon$_{0},which is much simpler consequence and the best choice of the coefficients \{$\alpha$_{n}\} fromthe view of error estimate.

Acknowledgment. This work was supported by JSPS KAKENHI Grant Number
15\mathrm{K}05007.

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