# Theory Analysis for the Convergence of Kernel-Regularized Online Binary Classification Learning Associated with RKBSs

Authors: Lin Liu; Xiaoling Pan; Baohuai Sheng
Journal: Journal of Mathematics
 (2023)

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## Abstract

It is known that more and more mathematicians have paid their attention to the field of learning with a Banach space since Banach spaces may provide abundant inner-product structures. We give investigations on the convergence of a kernel-regularized online binary classification learning algorithm in the setting of reproducing kernel Banach spaces (RKBSs), design an online iteration algorithm with the subdifferential of the norm and the logistic loss, and provide an upper bound for the learning rate, which shows that the online learning algorithm converges if RKBSs satisfy 2-uniform convexity.


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## Body

## 1. Introduction

Theory analysis for the convergence of binary classification learning algorithms has always been one of the most central problems in learning theory, and many experts have paid their attention to this topic (see e.g., [1–13]).A binary classification algorithm produces a binary classifierC:X⟶Y which divides the input space X into two classes represented by Y=−1,1. The classifier C gives a prediction y∈Y for each point x∈X (a vector x∈Rn with n components corresponding to n practical measurements). A real-valued function f:X⟶R can be used to produce a classifier Cx=sgnfx where(1)sgnfx=1,iffx≥0,−1,iffx<0.In many cases, we borrow the samples or observations with the kernel-regularized off-line classification algorithms to produce a classifier from a hypothesis function space called a reproducing kernel Hilbert space (RKHS). Letρ be a probability distribution on Z=X×Y and z=zt=xt,ytt=1T∈ZT be a set of random samples drawn independently (i.i.d) according to ρ. Let X be a compact subset in the Euclidean space. The RKHS associated with a Mercer kernel Kxy=Kx,y is a Hilbert space HK,⋅K consisting of all the continuous real functions defined on X, satisfying (see e.g., [14])(2)fx=f,KxK,f∈HK,x∈X.The kernel-regularized batch off-line binary classification learning algorithm associated with a classification lossϕ and RKHS HK is defined as (see e.g., [3, 10–12, 15, 16])(3)fz,λ=argminf∈HK1T∑t=1Tϕytfxt+λ2fK2,where λ is the penalty parameter, ϕ is a binary classification loss which is often a differential convex function defined on −∞,+∞ satisfying ϕ′0<0,ϕ′t≤0t≥0, and limt⟶+∞ϕt=0 or mint∈−∞,+∞ϕt=ϕ1 (see e.g., [3, 10, 11, 17, 18]). Then, the classifier Cx can be obtained by taking Cx=sgnfz,λx.For the regression functionfρ,(4)fρx=∫Xydρy|x=Py=1|x−Py=−1|x,x∈X.We define the Bayes rule (see e.g., [19]) as(5)fcx≔sgnfρx=1,ifPy=1|x≥Py=−1|x,−1,ifPy=1|x<Py=−1|x.We define the misclassification probability of a classifierf:X⟶Y as(6)Rf≔Probfx≠y=∫XPy≠fx|xdρXx.Then, the aim of learning theory is to bound the misclassification error in probability (see e.g., [3, 16])(7)Rsgnfz,λ−Rfc.Many mathematicians have conducted extensive research in this field (see e.g., [3, 10–13, 16]).In the case of batch learning, we need to test all samples in each training. When the amount of data is large or new sample points are added, the learning ability of batch learning decreases significantly. Online learning is one of the most effective approaches raised for analyzing and processing big data in various applications, such as communication, electronics, and other fields (see, e.g., [20–23]). The performance of kernel-based regularized online learning algorithms has been investigated, and their effectiveness has been verified (see e.g., [24–26]). Unlike off-line learning algorithms, online learning algorithms process the observations one by one, and the output is adjusted in time according to the results of last learning. For example, a way of obtaining learning sequence through the observation z=xt,ytt=1T is the following online iterations (see e.g., [13, 27])(8)f1=0,ft+1=ft−ηtϕ′ytftxtytKxt+λft,where ηt is called the step size, λ>0 is the regularization parameter, and the sequence ft satisfies ft∈HK,t∈NT+1=1,2,…,T+1. Algorithm (8) corresponds to the match off-line binary learning model (3).The geometry properties of the Hilbert space are well understood, and the bilinearity of the inner product makes a thoroughgoing analysis possible. However, the simplicity structure of the Hilbert space has limitations since much data do not come with distance induced from an inner product. Hilbert spaces also have some limitations, all Euclidean spaces of the same basis cardinality are isometrically isomorphic, and there is only one inner-product space. On the other hand, the Banach space may have richer geometry structures and various distances, can provide a more natural notion of distance between data points, and is more suitable for describing complicated data (see e.g., [28]). The reproducing property theory for the Banach space has been investigated (see e.g., [29–33]), and corresponding kernel-regularized regression learning has been defined and investigated by many mathematicians (see e.g., [4, 34–37]).A question then arises, can we design an online classification learning algorithm corresponding to the kernel-regularized binary classification in the setting of RKBSs? From the literature, we know that such kinds of discussions are rare. On the other hand, we find that the Banach geometry properties and skills have been used to design descent iteration algorithms for solving the minimization of Tikhonov functionals in Banach spaces (see, e.g., [38]) and have also been used to design sharp approximation methods in approximation theory (see [39–41]). So it is possible for us to design online learning algorithms to generate binary classifiers. This is the main motivation for writing this manuscript.We denote byB the Banach space with a dual space B∗ and norm ⋅B. For f∈B and f∗∈B∗, we write f,f∗B=f∗f.A reproducing kernel Banach space (RKBS)B on X is a reflexive Banach space of real functions on X, and there exists a unique function K:X×X⟶R called the reproducing kernel of B such that(9)Kx,⋅∈B,K⋅,x∈B∗,x∈X,and(10)fx=f,K⋅,xB,f∗x=Kx,⋅,f∗B,f∗∈B∗,x∈X.WhenB is an RKHS, K is indeed the reproducing kernel in the usual sense (see e.g., [14]).SinceB is a reflective Banach space, we have(11)f,f∗≤fBf∗B∗,f∈B,f∗∈B∗.We express our idea through the logistic loss as follows (see e.g., [42–44]):(12)ϕt=log1+e−t,t∈R.The batch kernel-regularized off-line learning algorithm corresponding toϕ is(13)fz,λ=argminf∈B1T∑t=1Tϕytfxt+λ2fB2,where B is an RKBS associated with a reproducing kernel Kx,y. The integral-type model corresponding to scheme (13) is(14)fλϕ=arginff∈BEρϕf+λ2fB2,where Eρϕ is the generalization error defined as(15)Eρϕ=∫Zϕyfxdρ.We shall design a kind of kernel-regularized online learning algorithm associating the lossϕ in the setting that B is an RKBS with respect to a kernel Kx,y on X×X and use the Banach geometry properties of B to measure the learning rate. For this purpose, we need some notions of Banach geometry theory and skills.LetB be a Banach space and Ff:B⟶R∪∓∞ be a convex function. Then,(16)∂Ff=f∗∈B∗:Ff′−Ff≥f′−f,f∗B,∀f′∈B≠∅,and we call ∂Ff the subdifferential of Ff at f∈B.LetB,⋅B be a normed space and Ff:B⟶R∪∓∞ be a real-valued functional. We say F is Gateaux differentiable at f0∈B if there is ξ∈B∗ such that for any g∈B, there holds(17)limt⟶0Ff0+tg−Ff0t=g,ξB,and we write ∇GFf0=ξ.For a Banach spaceB,⋅B, we define the modulus of convexity and smoothness as(18)δBε=inf1−x+y2:x=y=1andx−y=ε,0≤ε≤2,and(19)ρBt=sup12x+y+x−y−1:x=1,y=t,t>0.We sayB is uniformly convex if δBε>0 for all 0<ε≤2 and uniformly smooth if limt↓0ρBt/t=0 (see e.g., [45, 46]). Let p,q>1 be real numbers. We say B is p-uniformly convex (resp.q-uniformly smooth) if there is a constant c>0 such that δBε≥cεp (resp. ρBt≤ctq).It is shown by [47] that the Banach space B is p-uniformly convex if and only if B∗ is q-uniformly smooth; B is q-uniformly smooth if and only if B∗ is p-uniformly convex where p≥2,1<q≤2,1/p+1/q=1.We defineJθfB=∂1/θfBθ and Jθf∗B∗=∂1/θf∗B∗θ. Then, the uniform convexity and uniform smoothness of B and B∗ can be described by JθfB and Jθf∗B∗, respectively (see e.g., [46, 48–51]). If B=Hk is an RKHS, then J2fB=J2f∗B∗=f. This fact encourages us to design an online learning algorithm with the help of J2fB and J2f∗B∗. We modify the kernel-regularized online learning algorithm (8) according to the subgradient method (see Algorithm 1 in Section 6.2) and define following iteration algorithm:(20)f1=0,ft+1∗=ft∗−ηt−yt1+expytftxtK⋅,xt+λft∗ft+1∈J2ft+1∗B∗,,ft∗∈J2ftB,where t∈NT+1=1,2,…,T+1,ft∗∈J2ftB, and ft+1∈J2ft+1∗B∗ are chosen arbitrarily. The excess misclassification error (7) is(21)RsgnfT+1−Rfc.Algorithm 1: (The subgradient method) (see section 14.3 of [59] or see section 2 of [60]).
Step 1: Given an initial pointx∈,k≔1.Step 2: Computefxk,gk∈∂fxk.Step 3: Choose step sizeαk>0 and set xk+1=xk−αkgk/gk,k≔k+1,go to Step2.Sinceϕ′t=−1/1+et and ϕ″t=et/1+et2, we have ϕ′0<0 and ϕ″0>0. By Theorem 34 in the Appendix A of [3], we have a constant cϕ>0 such that for any measurable function f, there holds(22)Rsgnf−Rfc≤cϕEρϕf−Eρϕfρϕ.Therefore, to estimate error (21), we need to bound the excess generalization error(23)EρϕfT+1−Eρϕfρϕ,where fρϕ is the minimizer of the generalization error(24)fρϕ=arginfEρϕf:fis measurable onX.We can decompose the error (23) as(25)EρϕfT+1−Eρϕfρϕ≤EρϕfT+1−Eρϕfλϕ+Eρϕfλϕ−Eρϕfρϕ≤H+Dfρϕ,λ,where(26)Dfρϕ,λ=inff∈BEρϕf−Eρϕfρϕ+λ2fB2,it is a K-function in learning theory and its decay is determined by the approximation ability of B, and(27)H=EρϕfT+1−Eρϕfλϕ,which is called the sample, its convergence rate will be bounded later.We shall first give the convergence rate estimate forH and then show the convergence rate for (25) under the assumption that Dfρϕ,λ has a decay of power.The main contribution lies in extending the online iteration learning algorithm from the RKHS setting into an RKBS setting first time and giving investigation on the performance.The organization of this paper is as follows: In Section2, we provide some assumptions on Dfρϕ,λ and the kernel Kx,y, respectively, based on which we give the main results of the present manuscript, among which there is an upper bound for error H and an explicit convergence rate for (21). Two kinds of RKBSs satisfying the assumptions in the present paper are given in Section 3. In Section 4, we provide some auxiliary lemmas. Theorems and corollaries are shown in Section 5. Section 6 shows an appendix which provides some inequalities for Banach spaces.In what follows, we shall writeA=OB if there is a constant C>0 such that A/B≤C. We say A∼B if A=OB and B=OA.


## 2. The Results

To give the upper bound estimate for error (23), we provide two assumptions.Assumption 1.
TheK-functional Dfρϕ,λ has the decay of power, i.e.,(28)Dfρϕ,λ=Oλβ,λ⟶0+,where 0<β≤1.Assumption 2.
K⋅,xB∗ and Kx,⋅B are uniformly continuous about x on X, and there is a constant M>0 such that(29)supx∈XK⋅,xB∗≤M,supx∈XKx,⋅B≤M.
In Assumption1, Dfρϕ,λ denotes the approximation ability of RKBS B whose convergence rate has been investigated by many papers (see e.g., [52–54]); it is a usual assumption in learning theory (see e.g., [11, 12, 36]).
We point here that the RKBSs which satisfy Assumption2 are existent; in Section 3, we provide two kinds of RKBSs whose reproducing kernels satisfy Assumption 2. Also, by the results of Lemma 9 in [35], we know if Assumption 2 holds, then both B,⋅B and B∗,⋅B∗ are compact spaces, so fz,λ and fλϕ are existent and unique.


### 2.1. The Convergence Rate

We now give some bounds for the convergence of the online iteration algorithm.Theorem 1.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumptions 1 and 2. Assume that the sequences ηt satisfy ∑t=1∞ηt=+∞ and limt⟶+∞ηt=0. Then,(30)EfT+1−fλϕB⟶0,T⟶+∞.We now give an error bound forfT+1−fλϕB.Theorem 2.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumption 2. Let fλϕ be defined as in (14) and ft:t=1,…,T+1 be the sequence defined by algorithm (20). If ηt=1/Ct−θ with 0<λ≤1,0<θ≤1 and C≥λ+4kM2+1, then(31)EfT+1−fλϕB2≤Dλ+9NT1−θC21−θ21−θexp−cλ1−2θ−1C1−θT+11−θ+18NλcCT−θ,if0<θ<1,Dλ+8NCC−cλT+1−cλC,ifθ=1,where Dλ=2Dfρϕ,λ/λ,N=4kM2+1,k is the constant defined in the convex inequality (140).Under some assumptions, error (31) can be bounded explicitly.Corollary 1.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumption 2. Let ft:t=1,…,T+1 be the sequence produced by the algorithm (20). For any θ∈0,1,0<a≤min1−θ,θ, we take λ=T−α. If the step size is chosen as ηt=1/Ct−θ, then(32)EfT+1−fλϕB2≤Cc,k,C,M,θ×1Tθ−α,where Cc,k,C,M,θ is a constant depending only on c,k,θ,M and C.Finally, we can give an error upper bound forRsgnfT+1−Rfc in mathematical expectation.Corollary 2.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumptions 1 and 2. Let ft:t=1,…,T+1 be the sequence produced by the algorithm (20). For θ∈0,1,0<a≤min1−θ,θ, and λ=T−α, we choose ηt=1/Ct−θ. Then,(33)ERsgnfT+1−Rfc=O1Tθ−α/2+1Tβα.


### 2.2. Comments

It looks that the form of the convergence for the online learning in the present paper is the same as that of some published papers (see e.g., [13, 24, 25, 27]). There are essential differences. First, our investigations are in the setting of RKBSs. Note that a 2-uniformly convex Banach space has the closest properties with a Hilbert space. So it is very natural that the form of the iteration sequences and the performance may be the same as that in a RKHS setting.Second, it is hopeful that the geometry property of the RKBS hypothesis influences the learning rate (this fact has been proved in the case of match off-line learning, see [34]). So it will be an interesting topic for us to design online iteration algorithms according to the geometry parameters of the RKBS hypothesis, for example, the modulus of convexity δBε and the parameters p,θ,q in the convex inequality (see them in Section 6), and investigate the performance. From this view, we can say that the present paper is only a beginning.Third, the choicesft∗∈J2ftB and ft+1∈J2ft+1∗B∗ are arbitrary. It will be interesting if the choices can be finished according to some rules.


## 2.1. The Convergence Rate

We now give some bounds for the convergence of the online iteration algorithm.Theorem 1.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumptions 1 and 2. Assume that the sequences ηt satisfy ∑t=1∞ηt=+∞ and limt⟶+∞ηt=0. Then,(30)EfT+1−fλϕB⟶0,T⟶+∞.We now give an error bound forfT+1−fλϕB.Theorem 2.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumption 2. Let fλϕ be defined as in (14) and ft:t=1,…,T+1 be the sequence defined by algorithm (20). If ηt=1/Ct−θ with 0<λ≤1,0<θ≤1 and C≥λ+4kM2+1, then(31)EfT+1−fλϕB2≤Dλ+9NT1−θC21−θ21−θexp−cλ1−2θ−1C1−θT+11−θ+18NλcCT−θ,if0<θ<1,Dλ+8NCC−cλT+1−cλC,ifθ=1,where Dλ=2Dfρϕ,λ/λ,N=4kM2+1,k is the constant defined in the convex inequality (140).Under some assumptions, error (31) can be bounded explicitly.Corollary 1.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumption 2. Let ft:t=1,…,T+1 be the sequence produced by the algorithm (20). For any θ∈0,1,0<a≤min1−θ,θ, we take λ=T−α. If the step size is chosen as ηt=1/Ct−θ, then(32)EfT+1−fλϕB2≤Cc,k,C,M,θ×1Tθ−α,where Cc,k,C,M,θ is a constant depending only on c,k,θ,M and C.Finally, we can give an error upper bound forRsgnfT+1−Rfc in mathematical expectation.Corollary 2.
LetB be a 2-uniformly convex RKBS whose reproducing kernel Kx,y satisfies Assumptions 1 and 2. Let ft:t=1,…,T+1 be the sequence produced by the algorithm (20). For θ∈0,1,0<a≤min1−θ,θ, and λ=T−α, we choose ηt=1/Ct−θ. Then,(33)ERsgnfT+1−Rfc=O1Tθ−α/2+1Tβα.


## 2.2. Comments

It looks that the form of the convergence for the online learning in the present paper is the same as that of some published papers (see e.g., [13, 24, 25, 27]). There are essential differences. First, our investigations are in the setting of RKBSs. Note that a 2-uniformly convex Banach space has the closest properties with a Hilbert space. So it is very natural that the form of the iteration sequences and the performance may be the same as that in a RKHS setting.Second, it is hopeful that the geometry property of the RKBS hypothesis influences the learning rate (this fact has been proved in the case of match off-line learning, see [34]). So it will be an interesting topic for us to design online iteration algorithms according to the geometry parameters of the RKBS hypothesis, for example, the modulus of convexity δBε and the parameters p,θ,q in the convex inequality (see them in Section 6), and investigate the performance. From this view, we can say that the present paper is only a beginning.Third, the choicesft∗∈J2ftB and ft+1∈J2ft+1∗B∗ are arbitrary. It will be interesting if the choices can be finished according to some rules.


## 3. Examples

We now construct two kinds of RKBS kernel functions which are 2-uniformly convex, and their reproducing kernels satisfy Assumption2.Letα,β>−1 be two given real numbers. wα,βx=cα,β1−xα1+xβ is the generalized Jacobi weight function satisfying ∫−11wα,βxdx=1. The weighted space Lwα,βp consists of the real functions on −1,1 for which the norms(34)fp,wα,β=∫−11fxPwα,βxdx1/p,1≤p<+∞,esssupx∈−1,1fx,p=+∞,are finite. Denote by pkα,βx the Jacobi orthonormal polynomial of order k which is normalized by(35)∫−11pkα,βxpmα,βxwα,βxdx=δk,m,pmα,β1=1.We have(36)akα,βf=∫−11fxpkα,βxwα,βxdx,k=0,1,2,….For givenϕ∈Lwα,βp, we define generalized symmetric translations Kϕ;x,y as(37)Kϕ;x,y≔Kα,βϕ;x,y=∑k=0+∞akα,βϕpkα,βxpkα,βy,x,y∈−1,1.Then, the bivariate functionKϕ;x,y is a Mercer kernel on −1,1×−1,1.Proposition 1.
Let1<p≤2≤q<+∞ and q=p/p−1. Assume that ϕ∈Lwα,β1 satisfies akα,βϕ>0, both akα,βϕq−1k=0+∞∈l1 and akα,βϕp−1k=0+∞∈l1 hold, and there exists a constant M>0 such that pkα,βx≤M. We define(38)Bα,β=f∈Lwα,βp:∑k=0+∞akα,βfqakα,βϕ<+∞,fx=∑k=0+∞akα,βfpkα,βx.With the norm,(39)fBα,β=∑k=0+∞akα,βfqakα,βϕ1/q.We define(40)Bα,β,∗=g∈Lwα,βq:∑k=0+∞akα,βgpakα,βϕ<+∞,gu=∑k=0+∞akα,βgpkα,βu.With the norm,(41)gBα,β,∗=∑k=0+∞akα,βgpakα,βϕ1/p.Also, we define a continuous bilinear form onBα,β×Bα,β,∗ as(42)f,gBα,β=∑k=0+∞akα,βfakα,βgakα,βϕ,f∈Bα,β,g∈Bα,β,∗.Then,Kϕ;x,y is a reproducing kernel for Bα,β and Bα,β,∗, i.e.,(43)fx=f,Kϕ;⋅,xB,gy=K⋅ϕ;y,⋅,gB,f∈Bα,β,g∈Bα,β,∗.Moreover, there are the following results:(i)
Bα,β is 2-uniformly smooth, and Bα,β,∗ is 2-uniformly convex.(ii)
Kϕ;⋅,xBα,β,∗ is uniformly continuous about x; K.ϕ;y,⋅Bα,β is uniformly continuous about y.(iii)
There holds(44)Kϕ;⋅,xBα,β,∗≤M∑k=0+∞akα,βϕp−11/p<+∞.And(45)K.ϕ;y,⋅Bα,β≤M∑k=0+∞akα,βϕq−11/q<+∞.Proof.
The proofs can be found from the Section Appendix of [36].
Letα>1/2 and 1≤p≤+∞ be real numbers, and L1R+,dμα denotes the space of all measurable real functions on R+=0,+∞ such that(46)fp,α=∫R+fxpdμα1/p<+∞,1≤p<+∞,esssupx∈R+fx<+∞,p=+∞,where dμαx=x2α+1/2αΓα+1dx,α>1/2. The function jαz is defined on R+ by(47)jαz=Γα+1∑n=0+∞−1nz/22nn!Γn+α+1,z∈R+,which is called the normalized Bessel function of the first kind and order α, where Γα+1 is the Gamma function. For f∈L1R+,dμα, we define the Fourier–Bessel transform Fαf by(48)Fαfλ=∫R+fxjαλxdμα,λ∈R+.
Forϕ∈L1R+,dμα, we define(49)Kϕ;x,u=∫R+Fαϕλjαλxjαλudμα,u,x∈R+.Proposition 2.
Let1<p≤2<q<+∞ and q=p/p−1 . Assume that ϕ∈L1R+,dμα satisfies Fαϕ>0, both Fαϕq−1∈L1R+,dμα and Fαϕp−1∈L1R+,dμα hold.We define(50)Bϕ=f∈LpR+,dμα∩C∗R:FαfFαϕ1/q∈LqR+,dμα,fx=∫R+Fαfλjαλxdμα.With the norm, fBϕ=∫R+Fαfλq/Fαϕλdμα1/q,(51)B∗ϕ=g∈LqR+,dμα∩C∗R:FαgFαϕ1/p∈LpR+,dμα,gu=∫R+Fαgλjαλudμα,and with the norm,(52)gB∗ϕ=∫R+FαgλpFαϕλdμα1/p.We define a bilinear form onBϕ×B∗ϕ as(53)f,gBϕ=∫R+FαfλFαgλFαϕλdμα,f∈Bϕ,g∈B∗ϕ.Then,Kxϕ;x,u is a reproducing kernel for B and B∗, i.e.,(54)fx=f,Kϕ;⋅,xBϕ,gy=Kϕ;y,⋅,gBϕ,f∈Bϕ,g∈B∗ϕ.Moreover, there are the following results:(i)
Bϕ is 2-uniformly smooth, and B∗ϕ is 2-uniformly convex.(ii)
Kϕ;⋅,xB∗ϕ is uniformly continuous about x; Kϕ;y,⋅Bϕ is uniformly continuous about y.(iii)
There hold inequalities(55)Kϕ;⋅,xB∗ϕ≤∫R+Fαϕλp−11/p<+∞.And(56)Kϕ;y,⋅Bϕ≤∫R+Fαϕλq−11/q<+∞.Proof.
The proofs can be found from the Section Appendix of [36].


## 4. Lemmas

To prove the results in Section2, we provide here some lemmas which are proved in Section 5.Lemma 1.
Letϕ be a convex loss function, and for any f∈B, there holds(57)Eρϕf−Eρϕfλϕ≤Mf−fλϕB.Lemma 2 (see remark 4.6.1 in [55] and line 15 in pages 1128 in [46]).
The setsJ2fB and J2f∗B∗ defined in Section 2 have the following expressions:(58)J2fB=f∗∈B∗:f,f∗=fB2=f∗B∗2,and(59)J2f∗B∗=f∈B:f,f∗=fB2=f∗B∗2.Lemma 3.
Letft:t=1,…,T+1 be the sequence produced by the algorithm (20) and us take ftλ∗=−yt/1+expytftxtK⋅,xt+λft∗. Then,(60)ηtftλ∗=ft∗−ft+1∗,ft∗∈J2ftB,ftλ∗∈J2ftλB.ft+1∗−fλϕ∗∈B∗. Since J2ftλB=∂1/2ftλB2, we have(61)ftλ∗B∗2=ftλB2.Also, sinceJ2ft+1−fλϕB=∂1/2ft+1−fλϕB2, then(62)ft+1−fλϕB2=ft+1∗−fλϕ∗B∗2.Proof.
(61) and (62) can be deduced from (58) and (59)Lemma 4.
Letft:t=1,…,T+1 be the sequence produced by the algorithm (20). If ηt≤1/λ for any t=1,…,T+1, then there holds(63)ftB=ft∗B∗≤Mλ.Lemma 5.
Letfλϕ be defined as in (14). Then, there exists j2fλϕ∈J2fλϕB such that(64)−∫Zy1+expyfλϕxK⋅,xdρ+λj2fλϕ⋅=0.Lemma 6.
Letfλϕ be defined as in (14) and c be a constant in inequality (138) in Appendices. Then, for any f∈B, there holds(65)cλf−fλϕB2≤Eρϕf+λ2fB2−Eρϕfλϕ+λ2fλϕB2,∀f∈B.Lemma 7.
LetB be a 2-uniformly convex RKBS, and therefore, B∗ is a 2- uniformly smooth RKBS. Let ft:ft∈B,t∈NT+1 be the sequence produced by the algorithm (20) and k be the constant in inequality (140) in Appendices. If λ>0,ηt≤1/λ, then(66)Ez1,z2,…,zTfT+1−fλϕB2≤4kM2∑t=1Tηt2∏j=t+1T1−2cηjλ+2Dfρϕ,λλ∏t=1T1−2cηtλ.Furthermore, there holds(67)Ez1,z2,…,zTfT+1−fλϕB2≤4kM2∑t=1Tηt2exp−2cλ∑j=t+1Tηj+2Dfρϕ,λλexp−2cλ∑t=1Tηt.Lemma 8 (see (56) in Lemma 4 of [13]).
For any0≤t<T and 0<θ≤1, there holds(68)∑j=t+1Tj−θ≥11−θT+11−θ−t+11−θ,ifθ<1,logT+1−logt+1,ifθ=1.Lemma 9 (see Lemma 5 of [13]).
Let0<ν≤1 and 0<θ≤1. Then, ∑t=1T−11/t2θexp−ν∑j=t+1Tj−θ is bounded by(69)18νTθ+9T1−θ1−θ21−θexp−ν1−2θ−11−θT+11−θ,0<θ<1,81−θT+1−ν,θ=1.


## 5. Proofs

We now give proofs for the lemmas, theorems, and corollaries in the present paper.Proof of Lemma 1.
We denoteCX as the space of continuous functions on X and(70)f∞=supx∈Xfx.
Combining (11) and (29) with the reproducing property, we know that(71)fx=f,K⋅,x≤fBK⋅,xB∗≤MfB.
Therefore,(72)f∞≤MfB,for allf∈B.
According to the median value theorem, there existsξ between yfT+1x and yfλϕx such that(73)ϕyfT+1x−ϕyfλϕx=ϕ′ξfT+1x−fλϕx=−11+expξfT+1x−fλϕx≤fT+1x−fλϕx.
Due to the reproducing property and (29), we have(74)fT+1x−fλϕx=fT+1−fλϕ,K⋅,xB≤fT+1−fλϕBK⋅,xB∗≤MfT+1−fλϕB.
Since∫Zdρ=1, we have(75)EρϕfT+1−Eρϕfλϕ≤∫ZϕyfT+1x−ϕyfλϕxdρ≤∫ZfT+1x−fλϕxdρ≤MfT+1−fλϕB,(57) thus holds.Proof of Lemma 4.
We prove it by induction. According to the definition of the algorithm (20), we know ft∗∈J2ftB. By (58), we have ft∗B∗2=ftB2, where ft∗∈B∗,t∈NT+1. Since f1=0, then f1∗B∗2=f1B2=0, and the definition of the Banach space of functions, we have f1∗=0. The case t=1 is trivial since f1∗=0. We suppose that (63) holds true for ft∗. We now consider ft+1∗. It can be expressed as(76)ft+1∗=ft∗+ηtyt1+expytftxtK⋅,xt−ληtft∗=1−ληtft∗+ηtyt1+expytftxtK⋅,xt.
Under Assumption2, there is(77)ft+1∗B∗≤1−ληtft∗B∗+Mηtyt1+expytftxt.
Since,(78)yt1+expytftxt≤1.The assumption that ft∗B∗≤M/λ, we have(79)ft+1∗B∗≤1−ληtft∗B∗+Mηt≤1−ληtMλ+Mηt=Mλ.
Also, there holdsft+1B=ft+1∗B∗≤M/λ. (63) thus holds.Proof of Lemma 5.
Sincefλϕ is the solution of (14), there holds(80)0∈∂fEρϕf+λ2fB2|f=fλϕ.
Note thatEρϕf and fB2 are convex functions, we have(81)∂fEρϕf+λ2fB2|f=fλϕ=∂fEρϕf|f=fλϕ+∂fλ2fB2|f=fλϕ=0.
Then, there isj2fλϕ∈J2fλϕB such that(82)−∫Zy1+expyfλϕxK⋅,xdρ+λj2fλϕ⋅=0,(64) thus holds.Proof of Lemma 6.
Since B is 2-uniformly convex RKBS, for anyf∈B, by inequality (138), we have(83)λ2fB2−λ2fλϕB2≥f−fλϕ,λj2fλϕ⋅B+λcf−fλϕB2.
Therefore,(84)Eρϕf+λ2fB2−Eρϕfλϕ+λ2fλϕB2≥f−fλϕ,−∫Zy1+expyfλϕxK⋅,xdρB+f−fλϕ,λj2fλϕ⋅B+λcf−fλϕB2=f−fλϕ,−∫Zy1+expyfλϕxK⋅,xdρ+λj2fλϕ⋅B+λcf−fλϕB2.
Combining (64) with the above inequality, we have(85)Eρϕf+λ2fB2−Eρϕfλϕ+λ2fλϕB2≥λcf−fλϕB2.(65) then holds.Proof of Lemma 7.
By Lemma3, we have(86)ftλ∗=−yt1+expytftxtK⋅,xt+λft∗.Also, we know that ηtftλ∗=ft∗−ft+1∗. Since B∗ is a 2-uniformly smooth RKBS, by inequality (140) and equality (62), we have(87)ft+1−fλϕB2=ft+1∗−fλϕ∗B∗2=ft∗−fλϕ∗−ηtftλ∗B∗2≤ft∗−fλϕ∗B∗2+kηt2ftλ∗B∗2+2ηtfλϕ−ft,ftλ∗B.
We take(88)A=fλϕ−ft,−yt1+expytftxtK⋅,xt+λft∗B.
Since B is a 2-uniformly convex RKBS, by inequality (138), we have that(89)λfλϕ−ft,ft∗B≤λ2fλϕB2−ftB2.
It follows that(90)A=λfλϕ−ft,ft∗−fλϕ−ft,yt1+expytftxtK⋅,xtB≤λ2fλϕB2−ftB2−fλϕ−ft,yt1+expytftxtK⋅,xtB=λ2fλϕB2−ftB2−yt1+expytftxtfλϕxt−ftxt≤λ2fλϕB2−ftB2+ϕytfλϕxt−ϕytftxt=ϕytfλϕxt+λ2fλϕB2−ϕytftxt+λ2ftB2,where we have used the fact that ϕ is a convex function and(91)ϕytfλϕxt−ϕytfλxt≥ϕ′ytfλxtytfλϕxt−ytfλxt=−ytfλϕxt−fλxt1+expytfλxt.
Sinceft depends on z1,z2,…,zt−1 but not on zt, it follows that(92)Ez1,z2,…,ztA≤Ez1,z2,…,zt−1Eztϕytfλϕxt+λ2fλϕB2−Eztϕytftxt+λ2ftB2=Eρϕfλϕ+λ2fλϕB2−Ez1,z2,…,zt−1Eρϕft+λ2ftB2.
Combining (87) and (92), we have(93)Ez1,z2,…,ztft+1−fλϕB2=Ez1,z2,…,ztft+1∗−fλϕ∗B∗2≤Ez1,z2,…,zt−1ft∗−fλϕ∗B∗2+kηt2Ez1,z2,…,ztλft∗−yt1+expytftxtK⋅,xtB∗2+2ηtEz1,z2,…,zt−1Eρϕfλϕ+λ2fλϕB2−Eρϕft+λ2ftB2.
According to (65), we have(94)Eρϕfλϕ+λ2fλϕB2−Eρϕft+λ2ftB2≤−cλft−fλϕB2.
Combining (61), (62), (93), and (94), we arrive at(95)Ez1,z2,…,ztft+1−fλϕB2=Ez1,z2,…,ztft+1∗−fλϕ∗B∗2≤Ez1,z2,…,zt−1ft∗−fλϕ∗B∗2+kηt2Ez1,z2,…,ztλft∗−yt1+expytftxtK⋅,xtB∗2−2cληtEz1,z2,…,zt−1ft−fλϕB2=Ez1,z2,…,zt−1ft−fλϕB2+kηt2Ez1,z2,…,ztλft−yt1+expytftxtKxt,⋅B2−2cληtEz1,z2,…,zt−1ft−fλϕB2=1−2cληtEz1,z2,…,zt−1ft−fλϕB2+kηt2Ez1,z2,…,ztλft−yt1+expytftxtKxt,⋅B2.
By (63), we know(96)λft−yt1+expytftxtKxt,⋅B≤λftB+M−yt1+expytftxt≤λ×Mλ+M=2M.
Substituting (96) into (95), we obtain(97)Ez1,z2,…,ztft+1−fλϕB2≤1−2cληtEz1,z2,…,zt−1ft−fλϕB2+4kM2ηt2.
Applying the above relation iteratively fort=T,T−1,…,1, we have(98)Ez1,z2,…,zTfT+1−fλϕB2≤1−2cληTEz1,z2,…,zT−1fT−fλϕB2+4kM2ηT2≤1−2cληT1−2cληT−1Ez1,z2,…,zT−2fT−1−fλϕB2+4kM2ηT−12+4kM2ηT2≤……≤4kM2∑t=1Tηt2∏j=t+1T1−2cηjλ+∏t=1T1−2cηtλEf1−fλϕB2.
Sincef1=0, and according to the definition of Dfρϕ,λ, we have fλϕB2≤2Dfρϕ,λ/λ. It follows that(99)Ez1,z2,…,zTfT+1−fλϕB2≤4kM2∑t=1Tηt2∏j=t+1T1−2cηjλ+2Dfρϕ,λλ∏t=1T1−2cηtλ,(66) thus holds. According to the inequality 1−u≤e−u, for any u>0, we obtain (67) following from (66).Proof of Theorem 1.
It is easy to see that(100)∏t=1T1−2cηtλ≤exp−2cλ∑t=1Tηt.
By Assumption1, there is(101)2Dfρϕ,λλ∏t=1T1−2cηtλ≤2Dfρϕ,λλexp−2cλ∑t=1Tηt≤2λ1−βexp−2cλ∑t=1Tηt⟶0T⟶+∞.
Therefore, for anyε>0, there exists T1>0 such that for T≥T1 holds(102)2Dfρϕ,λλ∏t=1T1−2cηtλ≤ε.
According to the assumption,limt⟶∞ηt=0, for every t≥tε , we have(103)∑t=1Tηt2∏j=t+1T1−2cηjλ=∑t=1tεηt2∏j=t+1T1−2cηjλ+∑t=tε+1Tηt2∏j=t+1T1−2cηjλ.
Sincetε is fixed, there exists some T2∈N, and for every T≥T2, it holds that(104)∑j=tε+1Tηj≥∑j=tε+1T2ηj≥12cλlogtε4λ2ε.
For any1≤t≤tε, we have(105)∏j=t+1T1−2cηjλ≤exp−∑j=t+1T2cηjλ≤exp−∑j=tε+1T2cηjλ≤4λ2εtε.
According to the assumptions forηt, there exists an integer t0 such that ηtλ≤1/2 for all t≥t0; hence,(106)∑t=1tεηt2∏j=t+1T1−2cηjλ≤4λ2εtε∑t=1tεηt2≤ε.
We know that there exists sometε∈N such that ηt≤2cλε and(107)∑t=tε+1Tηt2∏j=t+1T1−2cηjλ≤2cλε∑t=tε+1Tηt∏j=t+1T1−2cηjλ.
Since2cηtλ=1−1−2cηtλ, we denote ∏j=T+1T1−2cηjλ=1.(108)2cλ∑t=tε+1Tηt∏j=t+1T1−2cηjλ=∑t=tε+1T∏j=t+1T1−2cηjλ−∏j=tT1−2cηjλ=1−∏j=tε+1T1−2cηjλ≤1.
From (106) and (108), we know that for any ε>0, there exists T2>0 such that for T≥T2, there holds(109)∑t=1Tηt2∏j=t+1T1−2cηjλ≤ε+ε=2ε.
LetT3=maxT1,T2, then by (66), (102), and (109), we have(110)Ez1,z2,…,ztfT+1−fλB2≤8kM2+1ε,whenever T>T3.Thus, we have (30).Proof of Theorem 2.
According to (93), we need to know more concrete expression for(111)λft∗−yt1+expytftxtK⋅,xtB∗2.
By (61) and (143) and Assumption 2, we have(112)λft∗−yt1+expytftxtK⋅,xtB∗2=λft−yt1+expytftxtKxt,⋅B2≤2λ2ftB2+2yt1+expytftxt2Kxt,⋅B2≤2λ2ftB2+2M2yt1+expytftxt2.
Whenx≥0, we have(113)0<e−x1+e−2x≤log1+e−x≤e−x≤1.
Since1/1+e−x≤e−x/1+e−2x≤log1+e−x and log1+e−x2≤2log1+e−x, we have(114)λft−yt1+expytftxtKxt,⋅B2≤2λ2ftB2+4M2ϕytftxt≤2λftB2+4M2ϕytftxt=4×λ2ftB2+4M2ϕytftxt≤4M2+1ϕytftxt+λ2ftB2.
By the definition offλϕ, we can see(115)Eρϕfλϕ+λ2fλϕB2≤Eρϕ0=∫Zlog2dρ≤1.
Substituting (114) into (93), we get(116)Ez1,z2,…,ztft+1−fλϕB2=Ez1,z2,…,ztft+1∗−fλϕ∗B∗2≤Ez1,z2,…,zt−1ft∗−fλϕ∗B∗2+4M2+1kηt2Ez1,z2,…,ztϕytftxt+λ2ftB2+2ηtEz1,z2,…,zt−1Eρϕfλϕ+λ2fλϕB2−Eρϕft+λ2ftB2=Ez1,z2,…,zt−1ft−fλϕB2+4M2+1kηt2Ez1,z2,…,zt−1Eρϕft+λ2ftB2+2ηtEz1,z2,…,zt−1Eρϕfλϕ+λ2fλϕB2−Eρϕft+λ2ftB2.
According to (115), we have(117)Ez1,z2,…,zt−1Eρϕft+λ2ftB2≤Ez1,z2,…,zt−1Eρϕft+λ2ftB2−Eρϕfλϕ+λ2fλϕB2+1.
Combining (116) with (117), then(118)Ez1,z2,…,ztft+1−fλϕB2≤Ez1,z2,…,zt−1ft−fλϕB2+2ηt1−2M2+1kηtEz1,z2,…,zt−1Eρϕfλϕ+λ2fλϕB2−Eρϕft+λ2ftB2+4M2+1kηt2.
Based on the assumptions aboutηt, we know that(119)1−2M2+1kηt≥12.
Also, by (65), we have(120)Eρϕfλϕ+λ2fλϕB2−Eρϕft+λ2ftB2≤−cλft−fλϕB2.
We take(121)B=2ηt1−2M2+1kηtEz1,z2,…,zt−1Eρϕfλϕ+λ2fλϕB2−Eρϕft+λ2ftB2.
Then, by (120), we have(122)B≤−cληtEz1,z2,…,zt−1ft−fλϕB2.
Combining (116) with (122), then(123)Ez1,z2,…,ztft+1−fλϕB2≤1−cληtEz1,z2,…,zt−1ft−fλϕB2+4M2+1kηt2.
We denoteN=4kM2+1. For t=T,T−1,…,1, we apply the iterative relation (123) repeatedly, and we have(124)Ez1,z2,…,zTfT+1−fλϕB2≤1−cληTEz1,z2,…,zT−1fT−fλϕB2+NηT2≤1−cληT1−cληT−1Ez1,z2,…,zT−2fT−1−fλϕB2+NηT−12+NηT2≤…≤1−cληT1−cληT−1…1−cλη1Ef1−fλϕB2+Nη12=N∑t=1Tηt2∏j=t+1T1−cηjλ+∏t=1T1−cηtλEfλϕB2≤N∑t=1Tηt2∏j=t+1T1−cηjλ+2Dfρϕ,λλ∏t=1T1−cηtλ.
For anyu≥0, we know that inequality 1−u≤e−u holds, and (124) implies that(125)Ez1,z2,…,zTfT+1−fλϕB2≤N∑t=1Tηt2exp−cλ∑j=t+1Tηj+2Dfρϕ,λλexp−cλ∑t=1Tηt.
We take(126)I1=2Dfρϕ,λλexp−cλ∑t=1Tηt=2Dfρϕ,λλexp−cλC∑t=1Tt−θ.We have(127)I2=N∑t=1T1Ct−θ2exp−cλC∑j=t+1Tj−θ=NC2∑t=1Tt−2θexp−cλC∑j=t+1Tj−θ.
Then, by (125) and the assumption about the step size ηt, we have(128)Ez1,z2,…,zTfT+1−fλϕB2≤I1+I2.
Now, we estimateI1 and I2. By (68), we obtain the following estimate for I1(129)I1≤Dλexp−cλC1−2θ−11−θT+11−θ,if0<θ<1,DλT+1−cλ/C,ifθ=1.
On the other hand, by (69) with v=cλ/C,s=θ, we have(130)∑t=1T−1t−2θexp−cλC∑j=t+1Tj−θ≤18cλ/CTθ+9T1−θ1−θ21−θexp−cλ1−2θ−1C1−θT+11−θ,if0<θ<1,81−cλ/CT+1−cλ/C,ifθ=1.
Therefore,(131)∑t=1Tt−2θexp−cλC∑j=t+1Tj−θ≤∑t=1T−1t−2θexp−cλC∑j=t+1Tj−θ+T−2θ≤18CλcTθ+9T1−θ1−θ21−θexp−cλ1−2θ−1C1−θT+11−θ+T−2θ,if0<θ<1,8CC−cλT+1−cλ/C+T−2,ifθ=1,(131) yields the following estimate for I2:(132)I2≤NC218CλcTθ+9T1−θ1−θ21−θexp−cλ1−2θ−1C1−θT+11−θ+T−2θif0<θ<1,NC28CC−cλT+1−cλ/C+T−2,ifθ=1.
SinceT−2θ≤T−θ and λ≤C, the conclusion can be estimated by combining (128), (129), and (132).Proof of Corollary 1.
For anyθ∈0,1,0<α<min1−θ,θ, by (31) with λ=T−α, and the fact(133)Dfρϕ,λ≤Eρϕ0≤1,we have(134)EfT+1−fλϕB2≤2Tα+36kM2+1C21−θ21−θT1−θexp−c1−2θ−1C1−θT1−θ−α+72kM2+1cCTα−θ.
Since for anyν>0,c>0,η>0, there exists L>0 such that exp−cTν≤LT−η; hence, the first term on the right-hand side of (134) decays in the form of O1/Tη for any large η>0.
However, the second term on the right-hand side of (134) is bounded by O1/Tθ−α. Consequently, there exists a constant Cc,k,C,M,θ depending only on c,k,θ,M and C such that(135)Ez1,z2,…,zTfT+1−fλϕB2≤Cc,k,C,M,θ×1Tθ−α,(32) thus holds.Proof of Corollary 2.
Collecting decomposition (25), (28) together with (32), we have (33).


## 6. Appendix: Further Results on Convex Analysis

We give some known inequalities in Banach spaces and the optimization algorithm about the subgradient.


### 6.1. Some Inequalities

By (3.2) in Corollary1 of [46], we know B is p-uniformly convex if and only if there is a positive cp>0 such that for all f,g∈B and all jpf∈JpfB, there holds(136)1pfBp−gBp≥f−g,jpgB+cpf−gBp.By (ii)′ in Corollary 1’ of [46], we know B is θ-uniformly smooth if and only if there is a constant cθ>0 such that for all f,g∈B and all jθg∈JθgB, there holds(137)1θfBθ−gBθ≤f−g,jθgB+cθf−gBθ.In particular,B is a 2-uniformly convex space if and only if there exists a constant c>0 such that for all jg∈J2gB, there holds(138)12fB2−gB2≥f−g,j2gB+cf−gB2,∀f,g∈B.Byii′ in Corollary 1’ of [46], we know B is q-uniformly smooth, and there is a constant k>0 such that for all f,g∈B and all jqf∈JqfB, there holds(139)f+gBq≤fBq+qg,jqfB+kgBq.In particular,B is a 2-uniformly smooth space if and only if there exists a constant k>0 such that for all jf∈J2fB, there holds(140)f+gB2≤fB2+2g,jfB+kgB2.By (29) in Lemma 2.1 of [56], we know B is a uniformly convex Banach space with the moduli of convexity of power type q≥2. There is a constant d>0 such that(141)tf+1−tgBq+dWqtf−gBq≤tfBq+1−tgBq,for all f,g∈B and t∈0,1, where(142)Wqt=t1−tq+1−tq.Combining (141) with (142) and taking t=1/2,q=2, we have(143)f+gB2≤2fB2+2gB2,f∈B,g∈B.It is known that all the Hilbert spaces, and for1<p<+∞, the Banach spaces lp,Lp and (Sobolev spaces) Wmp all are both uniformly convex and uniformly smooth (see e.g., [50, 57, 58]).


### 6.2. The Subgradient Method

Letfx be a convex function defined on the Euclidean space. Then, the minimization problem minx∈Rdfx has a solution x∗, i.e., fx∗=minx∈Rdfx if and only if 0∈∂fx∗. To obtain approximation x∗, one uses the classical subgradient descent.We observe thatgk∈∂fxk is chosen arbitrarily.


## 6.1. Some Inequalities

By (3.2) in Corollary1 of [46], we know B is p-uniformly convex if and only if there is a positive cp>0 such that for all f,g∈B and all jpf∈JpfB, there holds(136)1pfBp−gBp≥f−g,jpgB+cpf−gBp.By (ii)′ in Corollary 1’ of [46], we know B is θ-uniformly smooth if and only if there is a constant cθ>0 such that for all f,g∈B and all jθg∈JθgB, there holds(137)1θfBθ−gBθ≤f−g,jθgB+cθf−gBθ.In particular,B is a 2-uniformly convex space if and only if there exists a constant c>0 such that for all jg∈J2gB, there holds(138)12fB2−gB2≥f−g,j2gB+cf−gB2,∀f,g∈B.Byii′ in Corollary 1’ of [46], we know B is q-uniformly smooth, and there is a constant k>0 such that for all f,g∈B and all jqf∈JqfB, there holds(139)f+gBq≤fBq+qg,jqfB+kgBq.In particular,B is a 2-uniformly smooth space if and only if there exists a constant k>0 such that for all jf∈J2fB, there holds(140)f+gB2≤fB2+2g,jfB+kgB2.By (29) in Lemma 2.1 of [56], we know B is a uniformly convex Banach space with the moduli of convexity of power type q≥2. There is a constant d>0 such that(141)tf+1−tgBq+dWqtf−gBq≤tfBq+1−tgBq,for all f,g∈B and t∈0,1, where(142)Wqt=t1−tq+1−tq.Combining (141) with (142) and taking t=1/2,q=2, we have(143)f+gB2≤2fB2+2gB2,f∈B,g∈B.It is known that all the Hilbert spaces, and for1<p<+∞, the Banach spaces lp,Lp and (Sobolev spaces) Wmp all are both uniformly convex and uniformly smooth (see e.g., [50, 57, 58]).


## 6.2. The Subgradient Method

Letfx be a convex function defined on the Euclidean space. Then, the minimization problem minx∈Rdfx has a solution x∗, i.e., fx∗=minx∈Rdfx if and only if 0∈∂fx∗. To obtain approximation x∗, one uses the classical subgradient descent.We observe thatgk∈∂fxk is chosen arbitrarily.



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