June 11, 2026
This paper is devoted to discussing the linear tensor product problems in the worst case setting. We consider algorithms that use finitely many evaluations of arbitrary continuous linear functionals. We investigate algebraic \((s,t)\)-weak tractability (ALG-\((s,t)\)-WT) under the absolute error criterion in the case \(\lambda_1>1\), where \(\lambda_1\) is the square of the univariate maximal singular value. We solve the problem by giving the necessary and sufficient conditions for ALG-\((s,t)\)-WT on univariate singular values and fill the gap left open.
In the field of information-based complexity (IBC), the computational hardness of solving a multivariate problem \(S = \{S_d\}_{d \in \mathbb{N}}\) is measured by its information complexity \(n(\varepsilon, S_d)\), which is defined as the minimal number of information evaluations required to approximate the solution to within an accuracy \(\varepsilon > 0\). Tractability of multivariate problems studies how the information complexity \(n(\varepsilon, S_d)\) depends on the dimension \(d\) and the reciprocal of the accuracy \(\varepsilon\). There are two kinds of tractability based on polynomial convergence and exponential convergence. The algebraic tractability (ALG tractability) describes how \(n(\varepsilon, d)\) behaves as a function of \(d\) and \(\varepsilon^{-1}\), while the exponential tractability (EXP tractability) does as a function of \(d\) and \(1+\ln\varepsilon^{-1}\). Recently the study of ALG and EXP tractabilities has attracted much interest, and a great number of interesting results have been obtained (see [1]–[16] and the references therein).
This note is devoted to discussing ALG tractability of linear tensor product problems in the worst case setting. For such problems with respect to the absolute and normalized error criteria, necessary and sufficient conditions for various notions of ALG tractability on univariate singular values were obtained in [4], [8], [13], [14], and those for EXP tractability were obtained in [11], [12], [17]. However, we notice that there is only one case left open. More precisely, for the absolute error criterion and in the case \(\lambda_1>1\), necessary and sufficient condition for ALG-\((s,t)\)-WT were not obtained, where \(\lambda_1\) is the square of the univariate maximal singular value.
In this note we investigate ALG-\((s, t)\)-WT for the absolute error criterion in the case \(\lambda_1>1\). We solve the problem by giving a necessary and sufficient condition for ALG-\((s,t)\)-WT and fill the gap left open.
Linear tensor product problems are important special cases of linear multivariate problems over Hilbert spaces. A linear tensor product problem is defined as the \(d\)-folded tensor product of a single univariate linear problem, see [8].
We consider a compact linear operator \(S_1 : H_1\to G_1\), where \(H_1\) and \(G_1\) are two Hilbert spaces, and without loss of generality we assume \(\dim H_1=\infty\). Then \(W_1 := S_1^* S_1: H_1 \to H_1\) is a compact self-adjoint non-negative operator, where \(S^*\) is the adjoint operator of the linear operator \(S\). The eigenvalues of \(W_1\) are \(\lambda_j,\;j\in{\mathbb{N}}\), which satisfy \[\lambda_1=\lambda_2=\cdots=\lambda_m>\lambda_{m+1}\ge\lambda_{m+2}\ge\dots\ge0,\, m\in {\mathbb{N}}.\] Since \(S_1\) is compact, it follows that \(\lim\limits_{j\to\infty}\lambda_j=0\). Without loss of generality, we assume \(\lambda_2>0\).
For \(d\in \Bbb N\), the tensor product operator \(S_{d}\) is defined to be the \(d\)-fold tensor product of the compact linear operator \(S_1\), i.e., \[S_{d}=\bigotimes_{i=1}^d S_{1}: H_{d}\to G_d,\] where \(H_{d}\) and \(G_d\) are the Hilbert spaces given by \[H_{d}=\bigotimes_{i=1}^dH_{1},\;\; G_d=\bigotimes_{i=1}^d G_1.\] We say that \(S = \{S_d\}_{d\in{\mathbb{N}}}\) is a linear tensor product problem. Clearly, \(W_{d}:=S_{d}^*S_{d}: H_{d}\to H_{d}\) is a compact linear operator and the eigenvalues of \(W_{d}\) are \(\lambda_{d,{\boldsymbol{j}}},\;{{\boldsymbol{j}}\in{\mathbb{N}}^d}\), satisfying \[\lambda_{d,{\boldsymbol{j}}}=\prod_{k=1}^d \lambda_{j_k}.\] Assume that \(\{\lambda_{d,j}\}_{j\in\Bbb N}\) is the non-increasing rearrangement of \(\{\lambda_{d,{\boldsymbol{j}}}\}_{{\boldsymbol{j}}\in \Bbb N^d}\). Then \[\lambda_{d,1}\ge \lambda_{d,2}\ge \dots \ge 0, \;\;\; \lambda_{d,1}=\lambda_1^d,\]and \(\{\lambda_{d,j}\}_{j\in\Bbb N}\) is the squares of the ordered singular values of \(S_d\).
Consider the problem of approximating \(S_d\) by algorithms \(A_{n,d}\) using at most \(n\) information evaluations from the class \(\Lambda^{\rm all}\) consisting of all continue functionals on \(H_d\). The form of \(A_{d,n}\) is \[A_{n,d}(f)=\phi_{n,d}(L_1(f),L_2(f),\dots,L_d(f)),\] where \(L_j\in \Lambda^{\rm all }=H_d^*\) and \(\phi_{n,d}:\mathbb{R}^n \to G_d\) is any mapping. The \(n\)-th minimal worst case error of \(S_d\) is \[e(n,d)=\inf_{A_{n,d}}\sup_{\substack{f \in H_{d}\\ \|f\|_{H_{d} } \le 1}}\|S_{d}(f)-A_{n,d}(f)\|_{G_d}.\] For \(n=0\), the initial error \(e(0,d)\) is denoted by \[e(0,d)=\sup_{\substack{ f \in H_{d} \\ \|f\|_{H_{d}} \le 1}}\|S_{d}(f)\|_{G_d}.\] It is well known that for the linear tensor product problem \(S=\{S_d\}\) in the worst case setting, the \(n\)-th minimal worst case error is related to the \((n+1)\)-th singular value of the operator \(S_d\), i.e., \[e(n,d)=\sqrt{\lambda_{d,n+1}} \;\;{\rm and }\;\; e(0,d)=\sqrt{\lambda_{d,1}}=\lambda_1^{d/2}.\]
The information complexity for the normalized error criterion (NOR) or the absolute error criterion (ABS) is defined by \[n^X(\varepsilon,S_{d})= \min\{\,n\in {\mathbb{N}}: e(n,d) \le \varepsilon {\rm CRI}_d\,\},\] where \(X\in\{{\rm ABS,\;NOR}\}\), and \[{\rm CRI}_d=\left\{ \begin{array}{ll} 1 &ifX={\rm ABS},\\ e(0,d) &ifX={\rm NOR}. \end{array}\right.\] It follows \[n^{\rm ABS}(\varepsilon,S_d)=n^{\rm NOR}({\varepsilon}/{\lambda_1^{d/2}},S_d),\;\;\;n^{\rm ABS}(\lambda_1^{1/2}\varepsilon,S_1)=n^{\rm NOR}(\varepsilon,S_1).\] The information complexity can be rewritten as \[\begin{align} \label{1461} n^X(\varepsilon,S_{d})&=\min\{\,n\in {\mathbb{N}}: \lambda_{d,n+1}\le \varepsilon^2 {\rm CRI}^2_d\,\}\notag\\ &= \big|\big\{(j_1,\dots,j_d) \in {\mathbb{N}}^d : \prod_{k=1}^d\lambda_{j_k}> \varepsilon^2 {\rm CRI}^2_d \big\}\big|, \end{align}\tag{1}\] where \(| A |\) represents the cardinality of the set \(A\).
Various notions of ALG or EXP tractability are defined in terms of how the information complexity \(n^X(\varepsilon,S_{d})\) depends on \(d\) and \(\varepsilon^{-1}\), or alternatively on \(d\) and \(1+\ln \varepsilon^{-1}\). In this note, we consider ALG-\((s,t)\)-weak tractability (ALG-\((s,t)\)-WT). For fixed \(s, t> 0\), the problem \(S\) is said to be ALG-\((s,t)\)-weakly tractable if \[\lim\limits_{\varepsilon^{-1} + d \to \infty} \frac{\ln n^{X}(\varepsilon, S_d)}{\varepsilon^{-s}+d^t} = 0.\] Roughly speaking, ALG-\((s, t)\)-WT means that the information complexity is neither exponential in \(d^t\), nor in \(\varepsilon^{-s}\).
Similarly, for fixed \(s, t> 0\), the problem \(S\) is said to be EXP-\((s,t)\)-weakly tractable (EXP-\((s,t)\)-WT) if \[\lim\limits_{\varepsilon^{-1} + d \to \infty} \frac{\ln n^{X}(\varepsilon, S_d)}{(1+\ln\varepsilon^{-1})^s +d^t} = 0.\]
For the linear tensor product problem \(S\) in the worst case setting, the information complexity \(n^{X}(\varepsilon,S_d )\) is fully determined by the univariate singular values of \(S_1\). For NOR or ABS, Siedlecki and Weimar in [14] gave the necessary and sufficient conditions for ALG-\((s,t)\)-WT. However, for ABS and in the case \(\lambda_1>1\), they provided separate necessary and sufficient conditions for ALG-\((s,t)\)-WT. See the following ALG-\((s,t)\)-WT results of the linear tensor product problem \(S\) with \(\lambda_2>0\) for ABS and for the class \(\Lambda^{\rm all}\).
Let \(\lambda_1<1\). Then \(S\) is ALG-\((s,t)\)-WT if and only if \[\label{1462} \lim_{n\to\infty}\frac{\lambda_n}{\ln^{-2/s}n}=0.\tag{2}\]
Let \(\lambda_1=1\) and
assume that \(m=1\). Then \(S\) is ALG-\((s,t)\)-WT if and only if \[\lim_{n\to\infty}\frac{\lambda_n}{\ln^{-2/s}n}=0.\]
assume that \(m>1\). Then \(S\) is ALG-\((s,t)\)-WT if and only if \[t>1 \quad \text{and} \quad \lim_{n\to\infty}\frac{\lambda_n}{\ln^{-2/s}n}=0.\]
Let \(\lambda_1>1\) and define \(S_1':=\frac{1}{\sqrt{\lambda_1}} S_1\). Then ALG-\((s,t)\)-WT of \(S\) implies \[t>1 \quad \text{and} \quad \lim_{\varepsilon^{-1}+d\to\infty} \frac{\max_{\ell=1,\ldots,d}\limits \left[ \ell \cdot \ln n^{\mathrm{abs}}\!\left( ( \varepsilon/ \lambda_1^{d/2} )^{1/\ell},\, S'_1 \right) \right]}{\varepsilon^{-s}+d^{t}} = 0.\] Moreover, the conditions \[\quad t>1 \quad \text{and} \quad \lim_{\varepsilon^{-1}+d\to\infty} \frac{\ln d \cdot \max_{\ell=1,\dots,d}\limits \left[ \ell \cdot \ln n^{\mathrm{abs}}\!\left( ( \varepsilon/\lambda_1^{d/2} )^{1/\ell},\, S'_1 \right) \right]}{\varepsilon^{-s}+d^{t}} = 0\] are sufficient for \(S\) to be ALG-\((s,t)\)-WT.
Although Siedlecki and Weimar did not establish necessary and sufficient conditions on \(\{\lambda_j\}\) for ALG-\((s,t)\)-WT under the ABS setting with \(\lambda_1>1\), they gave in [14] that a necessary condition for ALG-\((s,t)\)-WT is \[\lim_{j\rightarrow\infty}\frac{(\ln\frac{1}{\lambda_j})^t}{\ln j}=\infty.\] Clearly, this necessary condition is much stronger than the decay condition 2 which characterizes ALG-\((s,t)\)-WT in the case \(\lambda_1\le1\) or for NOR.
In this note, we investigate ALG-\((s,t)\)-WT for ABS in the case \(\lambda_1>1\) and obtain a necessary and sufficient condition for ALG-\((s,t)\)-WT. We show that the above necessary condition together with the condition \(t>1\) are also sufficient for ALG-\((s,t)\)-WT. Our main result can be formulated as follows.
Theorem 1. Consider a linear tensor product problem \(S=(S_d)_{d\in{\mathbb{N}}}\) with \(\lambda_2>0\) in the worst case setting for the absolute error criterion and for the class \(\Lambda^{\rm all}\). Assume that \(\lambda_1>1\). Then \(S\) is ALG-\((s,t)\)-WT if and only if \(t>1\) and \[\label{1463} \lim_{j\rightarrow\infty}\frac{(\ln\frac{1}{\lambda_j})^t}{\ln j}=\infty.\qquad{(1)}\]
Remark 1. It is interesting to see that ALG-\((s,t)\)-WT only involves the parameter \(t\) in the case \(\lambda_1>1\) and for ABS. The authors in [12] investigated EXP-tractability of the linear tensor product problem \(S=\{S_d\}_{d\in{\mathbb{N}}}\) with \(\lambda_2>0\) in the worst case setting for the class \(\Lambda^{\rm all}\). They obtained that in the case \(\lambda_1> 1\) and for ABS, \(S\) is EXP-\((s,t)\)-WT if and only if \(t>1\) and \[\lim_{j\rightarrow\infty}\frac{(\ln\frac{1}{\lambda_j})^{\min(s,t)}}{\ln j}=\infty.\]By Theorem 1, we obtain that for ABS and in the case \(\lambda_1> 1\), \(S\) is ALG-\((s,t)\)-WT if and only if \(S\) is ALG-\((t,t)\)-WT, and if and only if \(S\) is EXP-\((s,t)\)-WT with \(s>t\).
The paper is organized as follows. In Section 2 we give some preliminaries in the worst case setting. In Section 3, we give the proof of Theorem 1.
For \(\varepsilon \in (0,1)\) and \(\lim\limits_{j\to\infty}\lambda_j=0\), define \[j(\varepsilon) :=n^{\rm NOR}(\varepsilon,S_1)= \max\{j \in {\mathbb{N}}: \lambda_j > \varepsilon^2\lambda_1\}.\] We put \(j(\varepsilon)=0\) for \(\varepsilon\ge1\). Then \(j(\varepsilon)\) is well defined and always finite. Furthermore, \(j(\varepsilon)\) goes to infinity if and only if all \(\lambda_j\)’s are positive.
Consider the normalized problem \(S'=\{S'_d\}\). Let \(S'_1=\frac{1}{\sqrt{\lambda_1}}S_1\). The ordered eigenvalues \(\lambda'_j,\;{j\in {\mathbb{N}}}\) of \(W'_1:= (S'_1)^ * \cdot S'_1\) satisfy \(\lambda'_j=\frac{\lambda_j}{\lambda_1}\) and \[1=\lambda'_1=\cdots=\lambda'_m>\lambda'_{m+1}\ge\lambda'_{m+2}\ge\dots\ge0.\] Let \(S'_d=\bigotimes\limits_{i=1}^{d}S'_1\) be the \(d\)-fold tensor product operator of \(S_1'\). Then the eigenvalues of \(W'_{d}:=(S'_{d})^*\cdot S'_{d}\) are \[\lambda'_{d,{\boldsymbol{j}}}=\prod_{k=1}^d \lambda'_{j_k},\;{\boldsymbol{j}}\in \Bbb N^d.\] Assume that \(\{\lambda'_{d,j}\}_{j\in\Bbb N}\) is the non-increasing rearrangement of \(\{\lambda'_{d,{\boldsymbol{j}}}\}_{{\boldsymbol{j}}\in\Bbb N^d}\). That is \[1=\lambda'_{d,1}\ge \lambda'_{d,2}\ge \dots \ge 0.\] The information complexity of \(S'\) is given by \[n^{X}(\varepsilon,S'_d):= \Big|\big\{{\boldsymbol{j}}\in {\mathbb{N}}^d : \lambda'_{j_1} \lambda'_{j_2}\cdots \lambda'_{j_d}> \varepsilon^2 {\rm CRI}^2_d \big\}\Big|.\] It is obvious that \[n^{\rm ABS}(\varepsilon,S'_d)=n^{\rm NOR}(\varepsilon,S'_d)=n^{\rm NOR}(\varepsilon,S_d), \;n^{\rm ABS}(\varepsilon,S'_1)=n^{\rm NOR}(\varepsilon,S_1)=j(\varepsilon).\]
Lemma 2. Let \(\{\lambda_j\}_{j\in{\mathbb{N}}}\) be a non-increasing sequence and \(\lim\limits_{j\rightarrow\infty}\lambda_j=0\). Then ?? holds if and only if \[\label{2462} \lim_{\varepsilon\to 0}\frac{\ln j(\varepsilon)}{(\ln\varepsilon^{-1})^t}=0.\qquad{(2)}\]
Proof. If \(\lambda_{j}=0\) for some \(j_0\in{\mathbb{N}}\), then \(\lambda_j=0\) for \(j\ge j_0\) and hence \(j(\varepsilon)\le j_0\). It follows that ?? and ?? hold in this case. Without loss of generality, we assume that \(\{\lambda_j\}\) is a positive sequence.
We first show that ?? implies ?? . Assume that ?? holds. For every \(\eta\in(0,1)\), there exists an \(\varepsilon_0\in(0,1)\) such that \[\frac{\ln j(\varepsilon)}{(\ln \varepsilon^{-1})^t}<\eta \;\;\;{\rm for}\; \varepsilon\in(0,\varepsilon_0).\] This implies \[j(\varepsilon)<e^{\eta (\ln\varepsilon^{-1})^t} \le \lceil e^{\eta (\ln\varepsilon^{-1})^t}\rceil :=v(\varepsilon),\] where \(\lceil x\rceil\) is the ceiling of a real number \(x\). From the definition of \(j(\varepsilon)\), we have \(j(\varepsilon)+1\le v(\varepsilon)\) and \[\lambda_{v(\varepsilon)}\le \lambda_{j(\varepsilon)+1}\le \varepsilon^2\lambda_1.\] Let \(v(\varepsilon)=k\). If we vary \(\varepsilon\in(0,\varepsilon_0)\), then \(k\) can take the values \[k= \lceil e^{\eta (\ln\varepsilon_0^{-1})^t} \rceil,\lceil e^{\eta (\ln\varepsilon_0^{-1})^t}\rceil+1,\cdots.\] Since \(k\le e^{\eta (\ln\varepsilon^{-1})^t}+1\), we get \[\varepsilon^{2}\le e^{-2\big(\frac{\ln (k-1)}{\eta}\big)^{1/t}}.\] Hence, \[\lambda_{k}\le \varepsilon^2\lambda_1\le e^{-2\big(\frac{\ln (k-1)}{\eta}\big)^{1/t}}\cdot\lambda_1\;\; \;{\rm for}\; k\ge \lceil e^{\eta (\ln\varepsilon_0^{-1})^t}\rceil.\] This yields \[\begin{align} \frac{(\ln\frac{1}{\lambda_{k}})^t}{\ln k}\; &\ge\frac{\big((\frac{ 2\ln(k-1)}{\eta})^{1/t}-\ln\lambda_1\big)^t}{\ln k}\\&\ge \frac{\frac{2\ln(k-1)}{\eta}\big(1-\frac{\ln \lambda_1\cdot t}{(2 \ln(k-1)/\eta)^{1/t}}\big)}{\ln k}\to \frac{2}{\eta} \;\;\;{\rm as \;}k\to\infty, \end{align}\]where in the last inequality we used the fact that \((a-b)^t\ge a^t(1-\frac{b}{a}t)\) for \(a\ge b> 0\) and \(t>1\). Since \(\eta\) can be arbitrary small, we obtain ?? .
Next, if \(\eqref{1463}\) holds, then for every \(M>0\) there exists a \(J>0\) such that \[\big(\ln\frac{1}{\lambda_j}\big)^t\geq M\ln j \;\;\;{\rm for}\;j\geq J.\] This implies \(\lambda_j\le e^{-(M\ln j)^{\frac{1}{t}}}\). We solve the inequality \(\varepsilon^2\lambda_1\ge e^{-(M\ln j)^{\frac{1}{t}}},\) and obtain the solution \[j\ge e^{\frac{(2\ln\varepsilon^{-1}-\ln\lambda_1)^t}{M}}.\] This concludes that if \[j\ge \max\{J,\; e^{\frac{(2\ln\varepsilon^{-1}-\ln\lambda_1)^t}{M}}\},\] then \[\lambda_j\le e^{-(M\ln j)^{\frac{1}{t}}}\le \varepsilon^2\lambda_1.\] From the definition of \(j(\varepsilon)\), we obtain \[j(\varepsilon)\le \max\{J,\; e^{\frac{(2\ln\varepsilon^{-1}-\ln\lambda_1)^t}{M}}\}.\] Taking the logarithm yields that for every \(\delta\in(0,s)\), \[\frac{\ln j(\varepsilon)}{(\ln \varepsilon^{-1})^t}\le \frac{\max\{\ln J,\;\frac{(2\ln\varepsilon^{-1}-\ln\lambda_1)^t}{M}\}}{(\ln \varepsilon^{-1})^t}\to \frac{2^t}{M} \;\;\;{\rm as \;}\varepsilon\to 0.\] Since \(M\) can be arbitrary large, we obtain ?? .
Lemma 2 is proved. ◻
The number of \(k\)-element subsets of an \(n\)-element set is called the number of combinations of \(n\) objects taken \(k\) at a time, and is denoted by \[\binom{n}{k}=\frac{n!}{k!(n-k)!},\] and the number of permutations of \(n\) objects taken \(k\) at a time, denoted \(P(n,k)\), is given by \[P(n,k)=\binom{n}{k}\cdot k!=\frac{n!}{(n-k)!}.\]
Lemma 3. We have \[\label{2463} n^{\rm NOR}(\varepsilon,S_d')\le P\big(d,{a_{\varepsilon}(d)}\big)m^{d-a_{\varepsilon}(d)}\prod_{k=1}^{a_{\varepsilon}(d)}j(\varepsilon^{\frac{1}{k}}),\qquad{(3)}\] where \(a_{\varepsilon}(d):=\min\big\{d, \big\lceil\frac{\ln\varepsilon^{-2}}{\ln(\lambda'_{m+1})^{-1}}\big\rceil-1\big\}\).
Proof. For \(\varepsilon\in(0,1)\), set \[A(\varepsilon)=\big\{{\boldsymbol{j}}\in {\mathbb{N}}^d : \lambda'_{j_1} \lambda'_{j_2}\cdots \lambda'_{j_d}> \varepsilon^2 \big\}=\big\{{\boldsymbol{j}}\in {\mathbb{N}}^d : \lambda_{j_1} \lambda_{j_2}\cdots \lambda_{j_d}> \varepsilon^2 \lambda_1^{d}\big\}.\] By 1 we get \[n^{\rm NOR}(\varepsilon,S'_d)=n^{\rm NOR}(\varepsilon,S_d)= \big|A(\varepsilon)\big|.\] We have \(\lambda'_j=\lambda_j/\lambda_1\) and \[1=\lambda'_1=\cdots=\lambda'_m>\lambda'_{m+1}\ge\lambda'_{m+2}\ge\cdots\ge0.\]
If \(\lambda'_{m+1}=0\), then \[a_{\varepsilon}(d)=0, \;\;n^{\rm NOR}(\varepsilon,S'_d)=m^d,\] and ?? holds. Otherwise, suppose that \(\lambda'_{m+1}>0\). Set \[u=|\{k\in [d] : j_k>m\}|\] for \({\boldsymbol{j}}\in A(\varepsilon)\). Then \[(\lambda'_{m+1})^{u} \ge\prod_{k=1}^d\lambda'_{j_k}>\varepsilon^2.\] It implies that \[u\le \big\lceil\frac{\ln\varepsilon^{-2}}{\ln(\lambda'_{m+1})^{-1}}\big\rceil-1.\] Then we have \[u\le \min\big\{d, \big\lceil\frac{\ln\varepsilon^{-2}}{\ln(\lambda'_{m+1})^{-1}}\big\rceil-1\big\}:=a_{\varepsilon}(d),\]i.e., for \({\boldsymbol{j}}\in A(\varepsilon)\) there are at most \(a_{\varepsilon}(d)\) indices \(j_k\) which satisfy \(j_k>m\). Hence there are at least \(d-a_{\varepsilon}(d)\) indices \(j_k\) which satisfy \(j_k\le m\) and hence \(\lambda_{j_k}'=1\). There are \(\binom{d}{a_{\varepsilon}(d)}m^{d-a_{\varepsilon}(d)}\) ways to select \(d-a_{\varepsilon}(d)\) indices \(j_k\) that satisfy \(j_k\le m\).
For the remaining \(a_{\varepsilon}(d)\) indices, let \(j_{1,\max}\) be the largest index of the eigenvalues in the product \(\prod\limits_{k=1}^d\lambda'_{j_k}\). Since \[\lambda'_{j_{1,\max}}\ge \prod\limits_{k=1}^d\lambda'_{j_k}>\varepsilon^2,\] we obtain that \[j_{1,\max}\le \max\big\{j:\lambda'_j>{\varepsilon^2}\big\} =j(\varepsilon).\] Let \(j_{k,\max}\) be the \(k\)-th largest index of the eigenvalues in the product \(\prod\limits_{k=1}^d\lambda'_{j_k},\) \(k=2,\cdots,a_\varepsilon(d)\). Since \[(\lambda'_{j_{k,\max}})^k\ge \prod_{i=1}^k \lambda'_{j_{i,\max}}\ge \prod_{k=1}^d\lambda'_{j_k} >\varepsilon^2,\] we obtain \[j_{k,\max}\le \max\big\{j:\lambda'_j>\varepsilon^{\frac{2}{k}}\big\} =j(\varepsilon^{\frac{1}{k}}) .\] Then for the remaining \(a_{\varepsilon}(d)\) indices, if \(k=1,\dots, a_\varepsilon(d)\), only the \(k\)-th largest indices \[j_{k,\max}\in\big\{1, 2, \dots, j(\varepsilon^{\frac{1}{k}})\big\},\;\] may belong to \(A(\varepsilon)\) and there are at most \(j(\varepsilon^{\frac{1}{k}})\) ways to select \(j_{k,\max}\). Also there are \(a_{\varepsilon}(d)!\) arrangements for these \(a_{\varepsilon}(d)\) indices. Therefore, we obtain \[n^{\rm NOR}(\varepsilon,S_d')\le \binom{d}{a_{\varepsilon}(d)}m^{d-a_{\varepsilon}(d)} \Big(\prod_{k=1}^{a_{\varepsilon}(d)}j(\varepsilon^{\frac{1}{k}})\Big) a_{\varepsilon}(d)!.\] Hence ?? holds. Lemma 3 is proved. ◻
Necessity. The necessity for ALG-\((s,t)\)-WT was obtained in [14]. We give the proof for the sake of readers’ convenience. Assume that ALG-\((s,t)\)-WT holds for ABS. Since \(\lambda_1>1\), from [8], we know that \(S\) suffers from the curse of dimensionality, and there exist constants \(c>0\) and \(a>0\), such that for all \(\varepsilon\in(0,1)\), \[n^{\rm ABS}(\varepsilon, S_d)\ge c(1+a)^d.\] Hence we have \(t>1\). Otherwise \[\frac{\ln n^{\rm ABS}(1/2,S_d)}{2^s+d^t}\ge \frac{\ln c+d\ln(1+a)}{2^s+d^t}\] does not tend to zero as \(d\to\infty\) which leads to a contradiction. Consider the normalized problem \(S'=\{S'_d\}\), then we have \(S'_d=\frac{1}{\sqrt{\lambda^d_1}}S_1\). The information complexity of the problem \(S\) can be rewritten as \[n^{\rm ABS}(\varepsilon,S_d):= \Big|\big\{{\boldsymbol{j}}\in {\mathbb{N}}^d : \lambda'_{j_1} \lambda'_{j_2}\cdots \lambda'_{j_d}> \frac{\varepsilon^2}{\lambda_1^d} \big\}\Big|=n^{\rm ABS }(\varepsilon/\lambda_1^\frac{d}{2},S'_d).\] Then we have \[\begin{align} n^{\rm ABS}(\varepsilon,S_d)&\ge \big|\big\{j_1 \in {\mathbb{N}}: \lambda'_{j_1}> \frac{\varepsilon^2}{\lambda_1^d} \big\}\big|= n^{\rm ABS}\Big(\varepsilon/\lambda_1^{\frac{d}{2}},S'_1\Big)=j(\varepsilon/\lambda_1^{\frac{d}{2}}). \end{align}\] Taking \(\varepsilon=\frac{1}{2}\), we have \[0=\lim_{\varepsilon^{-1}+d\to \infty}\frac{\ln n^{\rm ABS}(\varepsilon,d)}{\varepsilon^{-s}+d^t}=\lim_{d\to \infty}\frac{\ln n^{\rm ABS}(\frac{1}{2},d)}{2^s+d^t}\ge \lim_{d\to \infty}\frac{\ln j(1/(2\lambda_1^{\frac{d}{2}}))}{d^t}.\] Thus \[\lim_{d\to \infty}\frac{\ln j(1/(2\lambda_1^{\frac{d}{2}}))}{d^t}=0.\] Let \(\varepsilon=1/(2\lambda_1^{\frac{d}{2}})\), we have \(d=\frac{\ln1/2+\ln\varepsilon^{-1}}{\ln \lambda_1}\). This concludes that \[\lim_{\varepsilon\to 0}\frac{\ln j(\varepsilon)}{(\ln \varepsilon^{-1})^t}=0,\]and by Lemma 3 we get ?? .
Sufficiency. Assume that ?? holds and \(t>1\), we show that \(S\) is ALG-\((s,t)\)-WT for ABS. Consider the normalized problem \(S'=(S'_d)\), \[n^{\rm ABS}(\varepsilon,S_d)=n^{\rm NOR}(\varepsilon/\lambda_1^\frac{d}{2},S'_d).\] By Lemma 3, we have \[\begin{align} n^{\rm NOR}(\varepsilon,S'_d)\le P\big(d,{a_{\varepsilon}(d)}\big)m^{d-a_{\varepsilon}(d)}\prod_{k=1}^{a_{\varepsilon}(d)}j(\varepsilon^{\frac{1}{k}})\le d!\, m^d \prod_{k=1}^{d}j(\varepsilon^{\frac{1}{k}}),\label{3461} \end{align}\tag{3}\] where \[\; a_{\varepsilon}(d):=\min\big\{d,\big\lceil\frac{\ln\varepsilon^{-2}}{\ln(\lambda'_{m+1})^{-1}}\big\rceil-1\big\}.\] For \(x\in(0,1)\), let \(\ln j(x)=b(x) (\ln x^{-1})^t\). Since ?? holds, by Lemma 2 we have ?? . Then there exists a constant number \(M\) such that \[\lim\limits_{x\to 0^+}b(x)= 0, \;\;{\rm and}\;\; 0\le b(x)\le M \;{\rm for\;all }\;x\in(0, 1).\] Taking the logarithm in 3 we get \[\begin{align} \label{3462} {\ln n^{\rm ABS}(\varepsilon,S_d)}&\le {\ln d!+d\ln m+\sum_{k=1}^{d}\ln j\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)}\notag \\ &\le d\ln d +d\ln m+ \sum_{k=1}^{d}b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\Big(\frac{1}{k}(\ln\varepsilon^{-1}+\frac{d}{2}\ln \lambda_1)\Big)^t \notag \\ &\le d\ln (md) +\sum_{k=1}^{d}b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\frac{1}{k^t}\Big((2\ln \varepsilon^{-1})^t+d^t(\ln \lambda_1)^t\Big)\notag\\ &\le d\ln (md) +2^tM\sum_{k=1}^{d}\frac{(\ln \varepsilon^{-1})^t}{k^t} +(\ln \lambda_1)^td^t\sum_{k=1}^{d}b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\frac{1}{k^t}\notag\\ &=: I_1+2^tM I_2+(\ln \lambda_1)^t I_3, \end{align}\tag{4}\] where in the third inequality we used the inequality \((a+b)^t\le 2^t(a^t+b^t)\) for \(a,b\ge 0,\;t>1\). Since \(t>1\), we have \[\begin{align} \lim_{\varepsilon^{-1}+d \rightarrow \infty} \frac{I_1}{ \varepsilon^{-s}+d^t }=\lim_{\varepsilon^{-1}+d \rightarrow \infty} \frac{d\ln (md)}{\varepsilon^{-s}+d^t } = 0, \end{align}\] and \[\begin{align} \lim_{\varepsilon^{-1}+d \rightarrow \infty} \frac{I_2}{ \varepsilon^{-s}+d^t }=\lim_{\varepsilon^{-1}+d \rightarrow \infty} \frac{\sum\limits_{k=1}^{d}\frac{1}{k^t}\cdot (\ln \varepsilon^{-1})^t}{ \varepsilon^{-s}+d^t }\le \zeta(t) \lim_{\varepsilon^{-1}+d \rightarrow \infty} \frac{(\ln \varepsilon^{-1})^t}{ \varepsilon^{-s}+d^t } = 0, \end{align}\] where \(\zeta(t)=\sum_{k=1}^\infty\frac{1}{k^t}\) is the Riemann-Zeta function and is finite for \(t>1\).
Since the series \(\sum_{k=1}^{\infty}\frac{1}{k^t}\) is convergent for \(t>1\), we get for any \(\delta>0\), there exists a constant \(N_\delta\in{\mathbb{N}}\) which depends on \(\delta\), such that \[\sum_{k=N_\delta+1}^{\infty}\frac{1}{k^t}<\delta.\] Then \[\begin{align} I_3&= \sum\limits_{k=1}^{N_\delta}b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big) \frac{1}{k^t}\cdot d^t +\sum\limits_{k=N_\delta+1}^{\infty} b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\frac{1}{k^t}\cdot d^t\\ &\le \sum\limits_{k=1}^{N_\delta} b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\frac{1}{k^t}\cdot d^t+M\delta\cdot d^t. \end{align}\] For the above \(\delta>0\), since \(\lim\limits_{x\to0}b(x)=0\), there exists a \(\delta_1\in(0,1)\) such that for \(x\in (0,\delta_1)\), it holds \[b(x)<\delta.\] Since \(\lambda_1>1\), we get that \[\lim_{d+\varepsilon^{-1}\to\infty}\varepsilon/\lambda_1^{\frac{d}{2}}=0.\] Hence there exists a number \(X>0\) such that \(\varepsilon/\lambda_1^{\frac{d}{2}}<\delta_1^{N_\delta}\) whenever \(d+\varepsilon^{-1}>X\). It follows that \[b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\le \delta, \;\;{\rm for }\;\;k=1, 2, \dots , N_\delta.\] Hence\[\sum\limits_{k=1}^{N_\delta} b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\frac{1}{k^t} \cdot d^t \le \delta\zeta(t)\, d^t.\] We obtain that for \(d+\varepsilon^{-1}>X\), \[\begin{align} \frac{I_3}{\varepsilon^{-s}+d^t }&=\frac{d^t\sum\limits_{k=1}^{d}b\big((\varepsilon/\lambda_1^{\frac{d}{2}})^{\frac{1}{k}}\big)\frac{1}{k^t}}{\varepsilon^{-s}+d^t } \\ & \le \delta\zeta(t) \cdot \frac{d^t}{\varepsilon^{-s}+d^t }+M\delta\cdot\frac{d^t}{\varepsilon^{-s}+d^t }\\ &\le \delta(\zeta(t)+M). \end{align}\] Since \(\delta\) can be arbitrary small, we conclude \[\lim_{\varepsilon^{-1}+d \rightarrow \infty} \frac{I_3}{d^t + \varepsilon^{-s}}=0.\] Going back to 4 we have \[\begin{align} \lim_{\varepsilon^{-1}+d \rightarrow \infty} \frac{\ln n^{\rm ABS}(\varepsilon, S_{d})}{\varepsilon^{-s}+d^t } \le \lim_{\varepsilon^{-1}+d \rightarrow \infty}\frac{I_1+2^tMI_2+ (\ln \lambda_1)^tI_3}{\varepsilon^{-s}+d^t }=0. \end{align}\]
Theorem 1 is proved. \(\hfill\Box\)