Tractability of Multivariate Integration in Hybrid Function Spaces

Abstract

We consider tractability of integration in reproducing kernel Hilbert spaces which are a tensor product of a Walsh space and a Korobov space. The main result provides necessary and sufficient conditions for weak, polynomial, and strong polynomial tractability.

Appendix: The Proof of Theorem 3

Proof

We show the result by an inductive argument. We start our considerations by dealing with the case where $d_1=d_2=1$. According to Algorithm 1, we have chosen $g_1=1\in G_{b,m}$ and $z_1\in Z_N$ such that $e^2_{(1,1),{\varvec{\alpha }},{\varvec{\gamma }}}(g_1,z_1)$ is minimized as a function of $z_1$. In the following, we denote the points generated by $(g,z)\in G_{b,m}\times Z_N$ by $(x_n (g), y_n (z))$.

According to Eq. (10), we have

$$ e^2_{(1,1),{\varvec{\alpha }},{\varvec{\gamma }}}(g_1,z_1)= e^2_{1,\alpha _1,\gamma ^{(1)}}(1)+\theta _{(1,1)}(z_1), $$

where $e^2_{1,\alpha _1,\gamma ^{(1)}}(1)$ denotes the squared worst-case error of the polynomial lattice rule generated by 1 in the Walsh space ${\mathscr {H}}(K^{\mathrm{Wal}}_{1,\alpha _1,\gamma ^{(1)}})$, and where

$$\begin{aligned}&\theta _{(1,1)}(z_1):=\frac{\gamma _{1}^{(2)}}{N^2} \sum _{n,n'=0}^{N-1} \left( 1+\gamma _1^{(1)}\sum _{k_1\in {\mathbb {N}}}\frac{\mathrm{wal}_{k_1}(x_{n,1}(1) \ominus x_{n',1}(1))}{b^{\alpha _1 \lfloor \log _b k_1\rfloor }}\right) \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times \sum _{l_1\in {\mathbb {Z}}\setminus \{0\}} \frac{\mathrm {e}_{l_1}(y_{n,1}(z_1)-y_{n',1}(z_1))}{\left| l_1\right| ^{\alpha _2}}. \end{aligned}$$

By results in [2], we know that

$$\begin{aligned} e^2_{1,\alpha _1,\gamma ^{(1)}}(1) \le \frac{2}{N}\left( 1+\gamma _1^{(1)}\mu (\alpha _1)\right) . \end{aligned}$$

(11)

Then, as $z_1$ was chosen to minimize the error,

$$\begin{aligned}&{\theta _{(1,1)}(z_1)\le \frac{1}{\phi (N)} \sum _{z\in Z_N} \theta _{(1,1)}(z)}\\&=\frac{\gamma _{1}^{(2)}}{N^2} \sum _{n,n'=0}^{N-1} \left( 1+\gamma _1^{(1)}\sum _{k_1\in {\mathbb {N}}}\frac{\mathrm{wal}_{k_1}(x_{n,1}(1) \ominus x_{n',1}(1))}{b^{\alpha _1 \lfloor \log _b k_1\rfloor }}\right) \\&\,\,\times \frac{1}{\phi (N)} \sum _{z\in Z_N} \sum _{l_1\in {\mathbb {Z}}\setminus \{0\}} \frac{\mathrm {e}_{l_1}(y_{n,1}(z)-y_{n',1}(z))}{\left| l_1\right| ^{\alpha _2}}\\&\le \gamma _{1}^{(2)} \left( 1+\gamma _1^{(1)}\mu (\alpha _1)\right) \varSigma _B, \end{aligned}$$

where

$$\begin{aligned} \varSigma _B:= & {} \frac{1}{N^2}\sum _{n=0}^{N-1}\sum _{n'=0}^{N-1} \left| \frac{1}{\phi (N)} \sum _{z\in Z_N} \sum _{l_1\in {\mathbb {Z}}\setminus \{0\}} \frac{\mathrm {e}^{2\pi \mathtt {i}(n-n')zl_1/N}}{\left| l_1\right| ^{\alpha _2}}\right| \\= & {} \frac{1}{N}\sum _{n=1}^{N} \left| \frac{1}{\phi (N)} \sum _{z\in Z_N} \sum _{l_1\in {\mathbb {Z}}\setminus \{0\}} \frac{\mathrm {e}^{2\pi \mathtt {i}n z l_1/N}}{\left| l_1\right| ^{\alpha _2}}\right| , \end{aligned}$$

since the inner sum in the second line always has the same value. We now use [16, Lemmas 2.1 and 2.3] and obtain $\varSigma _B\le 4\zeta (\alpha _2)N^{-1}$, where we used that N has only one prime factor. Hence we obtain

$$\begin{aligned} \theta _{(1,1)}(z_1)\le \frac{\gamma _{1}^{(2)}}{N}\left( 1+\gamma _1^{(1)}\mu (\alpha _1)\right) 4\zeta (\alpha _2). \end{aligned}$$

(12)

Combining Eqs. (11) and (12) yields the desired bound for $(g_1,z_1)$.

Let us now assume $d_1\in [s_1]$ and $d_2\in [s_2]$ and that we have already found generating vectors ${\varvec{g}}_{d_1}^*$ and ${\varvec{z}}_{d_2}^*$ such that the bound in Theorem 3 is satisfied.

In what follows, we are going to distinguish two cases: In the first case, we assume that $d_1<s_1$ and add a component $g_{d_1+1}$ to ${\varvec{g}}_{d_1}^*$, and in the second case, we assume that $d_2<s_2$ and add a component $z_{d_2+1}$ to ${\varvec{z}}_{d_2}^*$. In both cases, we will show that the corresponding bounds on the squared worst-case errors hold.

Let us first consider the case where we start from $({\varvec{g}}_{d_1}^*,{\varvec{z}}_{d_2}^*)$ and add, by Algorithm 1, a component $g_{d_1+1}$ to ${\varvec{g}}_{d_1}^*$. According to Eq. (10), we have

$$\begin{aligned} e^2_{(d_1+1,d_2),{\varvec{\alpha }},{\varvec{\gamma }}}(({\varvec{g}}_{d_1}^*,g_{d_1+1}),{\varvec{z}}_{d_2}^*)=e^2_{(d_1,d_2),{\varvec{\alpha }},{\varvec{\gamma }}}({\varvec{g}}_{d_1}^*,{\varvec{z}}_{d_2}^*)+ \theta _{(d_1+1,d_2)}(g_{d_1+1}), \end{aligned}$$

where

$$\begin{aligned}&{\theta _{(d_1+1,d_2)}(g_{d_1+1})}\\&:=\frac{\gamma _{d_1 +1}^{(1)}}{N^2} \sum _{n,n'=0}^{N-1} \left[ \prod _{j=1}^{d_1}\left( 1+\gamma _j^{(1)}\sum _{k\in {\mathbb {N}}}\frac{\mathrm{wal}_{k}(x_{n,j}(g_j) \ominus x_{n',j}(g_j))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\right) \right] \\&\quad \times \left[ \prod _{j=1}^{d_2}\left( 1+\gamma _j^{(2)}\sum _{l\in {\mathbb {Z}}\setminus \{0\}}\frac{\mathrm {e}_{l}(y_{n,j}(z_j) - y_{n',j}(z_j))}{\left| l\right| ^{\alpha _2}}\right) \right] \\&\quad \times \sum _{k\in {\mathbb {N}}}\frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g_{d_1+1}) \ominus x_{n',d_1+1}(g_{d_1+1}))}{b^{\alpha _1\lfloor \log _b k \rfloor }}. \end{aligned}$$

However, by the assumption, we know that

$$\begin{aligned} e^2_{(d_1,d_2),{\varvec{\alpha }},{\varvec{\gamma }}}({\varvec{g}}_{d_1}^*,{\varvec{z}}_{d_2}^*) \le \frac{2}{N} \prod _{j=1}^{d_1}\left( 1+\gamma _j^{(1)}2\mu (\alpha _1)\right) \prod _{j=1}^{d_2}\left( 1+\gamma _j^{(2)}4\zeta (\alpha _2)\right) . \end{aligned}$$

(13)

Furthermore, as $g_{d_1+1}$ was chosen to minimize the error,

$$\begin{aligned}&{\theta _{(d_1+1,d_2)}(g_{d_1+1})\le \frac{1}{N} \sum _{g\in G_{b,m}} \theta _{(d_1+1,d_2)}(g)}\\&\le \gamma _{d_1 +1}^{(1)}\left[ \prod _{j=1}^{d_1}\left( 1+\gamma _j^{(1)}\mu (\alpha _1)\right) \right] \left[ \prod _{j=1}^{d_2}\left( 1+\gamma _j^{(2)}2\zeta (\alpha _2)\right) \right] \varSigma _C, \end{aligned}$$

where

$$\begin{aligned} \varSigma _C:= & {} \frac{1}{N^2}\sum _{n,n'=0}^{N-1}\left| \frac{1}{N} \sum _{g\in G_{b,m}}\sum _{k\in {\mathbb {N}}}\frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g) \ominus x_{n',d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\right| \\= & {} \frac{1}{N^2}\sum _{n=0}^{N-1}\sum _{n'=0}^{N-1}\left| \frac{1}{N} \sum _{g\in G_{b,m}}\sum _{k\in {\mathbb {N}}}\frac{\mathrm{wal}_{k}(x_{n\ominus n',d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\right| \\= & {} \frac{1}{N}\sum _{n=0}^{N-1}\left| \frac{1}{N} \sum _{g\in G_{b,m}}\sum _{k\in {\mathbb {N}}} \frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\right| , \end{aligned}$$

where we used the group structure of the polynomial lattice points (see [4, Sect. 4.4.4]) in order to get from the first to the second line and where we again used that the inner sum in the second line always has the same value. We now write

$$\begin{aligned} \varSigma _C=&\frac{1}{N}\sum _{k\in {\mathbb {N}}}\frac{1}{b^{\alpha _1\lfloor \log _b k \rfloor }} +\frac{1}{N}\sum _{n=1}^{N-1}\left| \frac{1}{N} \sum _{g\in G_{b,m}}\sum _{k\in {\mathbb {N}}} \frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\right| \\ =&\frac{\mu (\alpha _1)}{N} +\frac{1}{N}\sum _{n=1}^{N-1}\left| \frac{1}{N} \sum _{g\in G_{b,m}}\sum _{k\in {\mathbb {N}}} \frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\right| . \end{aligned}$$

Let now $n\in \{1,\ldots ,N-1\}$ be fixed, and consider the term

$$\begin{aligned} \varSigma _{C,n} :=&\frac{1}{N} \sum _{g\in G_{b,m}}\sum _{k\in {\mathbb {N}}} \frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\\ =&\sum _{\begin{array}{c} k\in {\mathbb {N}}\\ k\equiv 0 (N) \end{array}}\frac{1}{N} \sum _{g\in G_{b,m}} \frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}+\sum _{\begin{array}{c} k\in {\mathbb {N}}\\ k\not \equiv 0 (N) \end{array}}\frac{1}{N} \sum _{g\in G_{b,m}} \frac{\mathrm{wal}_{k}(x_{n,d_1+1}(g))}{b^{\alpha _1\lfloor \log _b k \rfloor }}\\ =:&\varSigma _{C,n,1}+\varSigma _{C,n,2}. \end{aligned}$$

By results in [2],

$$\varSigma _{C,n,1}=\sum _{\begin{array}{c} k\in {\mathbb {N}}\\ k\equiv 0 (N) \end{array}}\frac{1}{b^{\alpha _1\lfloor \log _b k \rfloor }}= \frac{\mu (\alpha _1)}{b^{m\alpha }}\le \frac{\mu (\alpha _1)}{N}.$$

Furthermore,

$$\begin{aligned} \varSigma _{C,n,2}= & {} \sum _{\begin{array}{c} k\in {\mathbb {N}}\\ k\not \equiv 0 (N) \end{array}} \frac{1}{b^{\alpha _1\lfloor \log _b k \rfloor }}\frac{1}{N} \sum _{g\in G_{b,m}} \mathrm{wal}_{k}(x_{n,d_1+1}(g))\\= & {} \sum _{\begin{array}{c} k\in {\mathbb {N}}\\ k\not \equiv 0 (N) \end{array}}\frac{1}{b^{\alpha _1\lfloor \log _b k \rfloor }}\frac{1}{N} \sum _{g=0}^{b^m-1}\mathrm{wal}_{k}\left( \frac{g}{b^m}\right) , \end{aligned}$$

where we used that

$$\begin{aligned} \sum _{g\in G_{b,m}}\mathrm{wal}_{k}(x_{n,d_1+1}(g))=&\sum _{g\in G_{b,m}}\mathrm{wal}_{k} \left( \nu _m \left( \frac{n(x)g(x)}{f(x)}\right) \right) \\ =&\sum _{g\in G_{b,m}}\mathrm{wal}_{k} \left( \nu _m \left( \frac{g(x)}{f(x)}\right) \right) =\sum _{g=0}^{b^m-1}\mathrm{wal}_{k}\left( \frac{g}{b^m}\right) , \end{aligned}$$

since $n\ne 0$ and since g takes on all values in $G_{b,m}$, and f is irreducible. However, $\sum _{g=0}^{b^m-1}\mathrm{wal}_{k}\left( \frac{g}{b^m}\right) =0$ and so $\varSigma _{C,n,2}=0$. This yields $\left| \varSigma _{C,n}\right| \le \mu (\alpha _1) N^{-1}$ and $\varSigma _C \le 2\mu (\alpha _1)N^{-1}$, which in turn implies

$$\begin{aligned} \theta _{(d_1+1,d_2)}(g_{d_1+1})\le&\frac{2\gamma _{d_1 +1}^{(1)}\mu (\alpha _1)}{N} \prod _{j=1}^{d_1}\left( 1+\gamma _j^{(1)}\mu (\alpha _1)\right) \prod _{j=1}^{d_2}\left( 1+\gamma _j^{(2)}2\zeta (\alpha _2)\right) . \end{aligned}$$

Combining the latter result with Eq. (13), we obtain

$$\begin{aligned} e^2_{(d_1+1,d_2),{\varvec{\alpha }},{\varvec{\gamma }}}(({\varvec{g}}_{d_1}^*,g_{d_1+1}),{\varvec{z}}_{d_2}^*))\le&\frac{2}{N} \prod _{j=1}^{d_1+1}\left( 1+2\gamma _j^{(1)}\mu (\alpha _1)\right) \prod _{j=1}^{d_2}\left( 1+\gamma _j^{(2)}4\zeta (\alpha _2)\right) . \end{aligned}$$

The case where we start from $({\varvec{g}}_{d_1}^*,{\varvec{z}}_{d_2}^*)$ and add, by Algorithm 1, a component $z_{d_2+1}$ to ${\varvec{z}}_{d_2}^*$ can be shown by a similar reasoning. We just sketch the basic points: According to Eq. (10), we have

$$\begin{aligned} e^2_{(d_1,d_2+1),{\varvec{\alpha }},{\varvec{\gamma }}}({\varvec{g}}_{d_1}^*,({\varvec{z}}_{d_2}^*,z_{d_2+1}))= e^2_{(d_1,d_2),{\varvec{\alpha }},{\varvec{\gamma }}}({\varvec{g}}_{d_1}^*,{\varvec{z}}_{d_2}^*)+\theta _{(d_1,d_2+1)}(z_{d_2+1}), \end{aligned}$$

where $e^2_{(d_1,d_2),{\varvec{\alpha }},{\varvec{\gamma }}}({\varvec{g}}_{d_1}^*,{\varvec{z}}_{d_2}^*)$ satisfies (13) and where

$$\begin{aligned} \theta _{(d_1,d_2+1)}(z_{d_2+1})\le \gamma _{d_2 +1}^{(1)}\left[ \prod _{j=1}^{d_1}\left( 1+\gamma _j^{(1)}\mu (\alpha _1)\right) \right] \left[ \prod _{j=1}^{d_2}\left( 1+\gamma _j^{(2)}2\zeta (\alpha _2)\right) \right] \varSigma _D, \end{aligned}$$

with

$$\begin{aligned} \varSigma _D=\frac{1}{N}\sum _{n=0}^{N-1}\left| \frac{1}{\phi (N)} \sum _{z\in Z_N} \sum _{l\in {\mathbb {Z}}\setminus \{0\}}\frac{\mathrm {e}^{2\pi \mathtt {i}nzl/N}}{\left| l\right| ^{\alpha _2}}\right| \le \frac{4\zeta (\alpha _2)}{N}, \end{aligned}$$

according to [16, Lemmas 2.1 and 2.3]. This implies

$$\theta _{(d_1,d_2+1)}(z_{d_2+1})\le \frac{\gamma _{d_2 +1}^{(1)}4\zeta (\alpha _2)}{N} \prod _{j=1}^{d_1}\left( 1+\gamma _j^{(1)}\mu (\alpha _1)\right) \prod _{j=1}^{d_2}\left( 1+\gamma _j^{(2)}2\zeta (\alpha _2)\right) .$$

Combining these results we obtain the desired bound. $\square$