Lattice rules in non-periodic subspaces of Sobolev spaces

Abstract

We investigate quasi-Monte Carlo (QMC) integration over the s-dimensional unit cube based on rank-1 lattice point sets in weighted non-periodic Sobolev spaces $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}})$ and their subspaces of high order smoothness $\alpha >1$, where $\varvec{\gamma }$ denotes a set of the weights. A recent paper by Dick, Nuyens and Pillichshammer has studied QMC integration in half-period cosine spaces with smoothness parameter $\alpha >1/2$ consisting of non-periodic smooth functions, denoted by $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {cos}})$, and also in the sum of half-period cosine spaces and Korobov spaces with common parameter $\alpha$, denoted by $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {kor}+\mathrm {cos}})$. Motivated by the results shown there, we first study embeddings and norm equivalences on those function spaces. In particular, for an integer $\alpha$, we provide their corresponding norm-equivalent subspaces of $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}})$. This implies that $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {kor}+\mathrm {cos}})$ is strictly smaller than $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}})$ as sets for $\alpha \ge 2$, which solves an open problem by Dick, Nuyens and Pillichshammer. Then we study the worst-case error of tent-transformed lattice rules in $\mathcal {H}\left( K_{2,\varvec{\gamma },s}^{\mathrm {sob}}\right)$ and also the worst-case error of symmetrized lattice rules in an intermediate space between $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {kor}+\mathrm {cos}})$ and $\mathcal {H}(K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}})$. We show that the almost optimal rate of convergence can be achieved for both cases, while a weak dependence of the worst-case error bound on the dimension can be obtained for the former case.

Appendices

Proof of Lemma 3

In order to prove Lemma 3, we need the following result.

Lemma 5

For $k\in \mathbb {N}$, we have

$$\begin{aligned} \sin (2\pi k\phi (x)) = \frac{8}{\pi }\sum _{\begin{array}{c} \ell =1\\ \ell :\mathrm {odd} \end{array}}^{\infty }\frac{k}{4k^2-\ell ^2}\cos (2\pi \ell x). \end{aligned}$$

Proof

Let $\ell \in \mathbb {N}_0$. Since $\phi$ is given by

$$\begin{aligned} \phi (x)={\left\{ \begin{array}{ll} 2x &{} \text {if }x\in [0,1/2], \\ 2-2x &{} \text {otherwise}, \\ \end{array}\right. } \end{aligned}$$

it is an easy exercise to check that

$$\begin{aligned} \int _{0}^{1}\sin (2\pi k\phi (x))\sin (2\pi \ell x) \mathrm {d}x = 0, \end{aligned}$$

and

$$\begin{aligned} \int _{0}^{1}\sin (2\pi k\phi (x))\cos (2\pi \ell x) \mathrm {d}x&= \int _{0}^{1}\sin (2\pi kx)\cos (\pi \ell x)\mathrm {d}x \\&= \frac{1}{2}\int _{0}^{1}\left( \sin ((2k+\ell )\pi x)+\sin ((2k-\ell )\pi x)\right) \mathrm {d}x \\&= {\left\{ \begin{array}{ll} 0 &{} \text {for even }\ell , \\ \displaystyle \frac{4}{\pi }\frac{k}{4k^2-\ell ^2} &{} \text {for odd }\ell . \end{array}\right. } \end{aligned}$$

Thus the Fourier series of $\sin (2\pi k\phi (\cdot ))$ is given by

$$\begin{aligned} \sin (2\pi k\phi (x))&= 2\sum _{\ell =1}^{\infty }\cos (2\pi \ell x)\int _{0}^{1}\sin (2\pi k\phi (y))\cos (2\pi \ell y) {\mathrm {d}{y}} \\&= \frac{8}{\pi }\sum _{\begin{array}{c} \ell =1\\ \ell :\mathrm {odd} \end{array}}^{\infty }\frac{k}{4k^2-\ell ^2}\cos (2\pi \ell x). \end{aligned}$$

Hence we complete the proof. $\square$

Now we are ready to prove Lemma 3.

Proof of Lemma 3

By definition, we have

$$\begin{aligned} -1+K_{2,1,1}^{\mathrm {sob}}(\phi (x),\phi (y))&= b_1(\phi (x))b_1(\phi (y))+b_2(\phi (x))b_2(\phi (y))-\tilde{b}_4(\phi (x) \nonumber \\ {}&\quad -\phi (y)). \end{aligned}$$

(6)

Using the Fourier series of a triangle wave provided in [12, Chapter 1, 1.444], we have

$$\begin{aligned} b_1(\phi (x))= \phi (x)-\frac{1}{2} = -\frac{4}{\pi ^2}\sum _{\begin{array}{c} k=1\\ k:\mathrm {odd} \end{array}}^{\infty }\frac{\cos (2\pi kx)}{k^2}. \end{aligned}$$

Thus the Fourier series of the first term of (6) is given by

$$\begin{aligned} b_1(\phi (x))b_1(\phi (y)) = \frac{1}{\pi ^4}\sum _{\begin{array}{c} k,\ell =1\\ k,\ell :\mathrm {odd} \end{array}}^{\infty }\frac{16}{k^2\ell ^2}\cos (2\pi kx)\cos (2\pi \ell y). \end{aligned}$$

Using the Fourier series of $b_2$ as shown in (2) and the equality $\cos (2\pi k\phi (x))=\cos (4\pi kx)$ which holds for any $k\in \mathbb {N}$ and $x\in [0,1]$, we have

$$\begin{aligned} b_2(\phi (x))= \frac{1}{2\pi ^2}\sum _{k=1}^{\infty }\frac{\cos (2\pi k\phi (x))}{k^2}=\frac{2}{\pi ^2}\sum _{\begin{array}{c} k=1\\ k:\mathrm {even} \end{array}}^{\infty }\frac{\cos (2\pi kx)}{k^2}. \end{aligned}$$

Thus the Fourier series of the second term is given by

$$\begin{aligned} b_2(\phi (x))b_2(\phi (y)) = \frac{1}{\pi ^4}\sum _{\begin{array}{c} k,\ell =1\\ k,\ell :\mathrm {even} \end{array}}^{\infty }\frac{4}{k^2\ell ^2}\cos (2\pi kx)\cos (2\pi \ell y). \end{aligned}$$

Finally let us consider the third term of (6). Using the Fourier series of $\tilde{b}_4$ and the equality $\cos (2\pi k\phi (x))=\cos (4\pi kx)$, we have

$$\begin{aligned} \tilde{b}_4(\phi (x)-\phi (y))&= \frac{-2}{(2\pi )^4}\sum _{k=1}^{\infty }\frac{\cos (2\pi k(\phi (x)-\phi (y)))}{k^4} \\&= -\frac{2}{\pi ^4}\sum _{\begin{array}{c} k=1\\ k:\mathrm {even} \end{array}}^{\infty }\frac{\cos (2\pi kx)\cos (2\pi ky)}{k^4}\\&\quad -\,\frac{1}{8\pi ^4}\sum _{k=1}^{\infty }\frac{\sin (2\pi k\phi (x))\sin (2\pi k\phi (y))}{k^4}. \end{aligned}$$

Using Lemma 5, the second term of the last expression is given by

$$\begin{aligned}&-\frac{1}{8\pi ^4}\sum _{k=1}^{\infty }\frac{\sin (2\pi k\phi (x))\sin (2\pi k\phi (y))}{k^4} \nonumber \\&\quad = -\frac{8}{\pi ^6}\sum _{k=1}^{\infty }\frac{1}{k^2}\sum _{\begin{array}{c} \ell ,m=1\\ \ell ,m:\mathrm {odd} \end{array}}^{\infty }\frac{1}{(4k^2-\ell ^2)(4k^2-m^2)}\cos (2\pi \ell x)\cos (2\pi m y) \nonumber \\&\quad = -\frac{8}{\pi ^6}\sum _{\begin{array}{c} \ell ,m=1\\ \ell ,m:\mathrm {odd} \end{array}}^{\infty }\cos (2\pi \ell x)\cos (2\pi m y)\sum _{k=1}^{\infty }\frac{1}{k^2(4k^2-\ell ^2)(4k^2-m^2)}. \end{aligned}$$

(7)

Noting that the equalities

$$\begin{aligned} \sum _{k=1}^{\infty }\frac{1}{4k^2-\ell ^2} = \frac{1}{2\ell ^2}\quad \text {and}\quad \sum _{k=1}^{\infty }\frac{1}{(4k^2-\ell ^2)^2} = \frac{1}{4\ell ^2}\left( \frac{\pi ^2}{4}-\frac{2}{\ell ^2} \right) . \end{aligned}$$

hold for any positive odd integer $\ell$, the inner sum of (7) can be evaluated as follows. In case of $\ell = m$, we have

$$\begin{aligned} \sum _{k=1}^{\infty }\frac{1}{k^2(4k^2-\ell ^2)^2}&= \frac{1}{\ell ^2}\sum _{k=1}^{\infty }\left( \frac{1}{\ell ^2}\left( \frac{1}{k^2}-\frac{4}{4k^2-\ell ^2} \right) + \frac{4}{(4k^2-\ell ^2)^2}\right) \\&= \frac{1}{\ell ^2}\left( \frac{1}{\ell ^2}\left( \frac{\pi ^2}{6}-\frac{2}{\ell ^2} \right) + \frac{1}{\ell ^2}\left( \frac{\pi ^2}{4}-\frac{2}{\ell ^2} \right) \right) = \frac{1}{\ell ^4}\left( \frac{5}{12}\pi ^2-\frac{4}{\ell ^2} \right) . \end{aligned}$$

Otherwise if $\ell \ne m$, we have

$$\begin{aligned}&\sum _{k=1}^{\infty }\frac{1}{k^2(4k^2-\ell ^2)(4k^2-m^2)} \\&\quad = \frac{1}{\ell ^2 m^2}\sum _{k=1}^{\infty }\left( \frac{1}{k^2}+\frac{4}{\ell ^2-m^2}\left( \frac{m^2}{4k^2-\ell ^2}-\frac{\ell ^2}{4k^2-m^2}\right) \right) \\&\quad = \frac{1}{\ell ^2 m^2}\left( \frac{\pi ^2}{6}+\frac{4}{\ell ^2-m^2}\left( \frac{m^2}{2\ell ^2}-\frac{\ell ^2}{2m^2}\right) \right) = \frac{1}{\ell ^2m^2}\left( \frac{\pi ^2}{6}-\frac{2}{\ell ^2}-\frac{2}{m^2}\right) . \end{aligned}$$

By substituting these results on the Fourier series into (6), the result of the lemma follows. $\square$

Proof of Theorem 3

In what follows, for a function f defined over $[0,1]^2$, we define a function $\mathrm {sym}[f]$ by

$$\begin{aligned} \mathrm {sym}[f(x,y)] := \frac{f(x,y)+f(1-x,y)+f(x,1-y)+f(1-x,1-y)}{4}. \end{aligned}$$

Since we have

$$\begin{aligned} \int _{[0,1]^s}\int _{[0,1]^s}K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}(\mathrm {odd}+\alpha )}(\varvec{x},\varvec{y})\mathrm {d}\varvec{x}{\mathrm {d}\varvec{y}}=1 \quad \text {and}\quad \int _{[0,1]^s}K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}(\mathrm {odd}+\alpha )}(\varvec{x},\varvec{y}){\mathrm {d}\varvec{y}} = 1, \end{aligned}$$

for any $\varvec{x}\in [0,1]^s$, it follows from (3) that

$$\begin{aligned}&\left( e^{\mathrm {wor}}\left( P_{N,\varvec{z}}^{\mathrm {sym}};\mathcal {H}\left( K_{\alpha , \varvec{\gamma },s}^{\mathrm {sob}(\mathrm {odd}+\alpha )}\right) \right) \right) ^2 \nonumber \\&\quad = -\,1+ \frac{1}{\left( 2^sN\right) ^2}\sum _{\varvec{x},\varvec{y}\in P_{N,\varvec{z}}^{\mathrm {sym}}}K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}(\mathrm {odd}+\alpha )}\left( \varvec{x},\varvec{y}\right) \nonumber \\&\quad = -\,1+ \frac{1}{N^2}\sum _{\varvec{x},\varvec{y}\in P_{N,\varvec{z}}}\frac{1}{2^{2s}}\sum _{v,w\subseteq 1:s}K_{\alpha ,\varvec{\gamma },s}^{\mathrm {sob}(\mathrm {odd}+\alpha )}\left( \mathrm {sym}_v\left( \varvec{x}\right) , \mathrm {sym}_w\left( \varvec{y}\right) \right) \nonumber \\&\quad =\frac{1}{N^2}\sum _{\varvec{x},\varvec{y}\in P_{N,\varvec{z}}}\sum _{\emptyset \ne u\subseteq 1:s}\gamma _u \prod _{j\in u}\left[ -1 + \mathrm {sym}\left[ K_{\alpha ,1,1}^{\mathrm {sob}(\mathrm {odd}+\alpha )}\left( x_j ,y_j\right) \right] \right] . \end{aligned}$$

(8)

As we assume that $\alpha$ is even, we have

$$\begin{aligned}&-1+\mathrm {sym}[K_{\alpha ,1,1}^{\mathrm {sob}(\mathrm {odd}+\alpha )}(x ,y)] \nonumber \\&\quad = \mathrm {sym}\left[ \sum _{\begin{array}{c} \tau =1 \\ \tau :\mathrm {odd} \end{array}}^{\alpha } b_{\tau }(x)b_{\tau }(y) + b_{\alpha }(x)b_{\alpha }(y) - \tilde{b}_{2\alpha }(x-y) \right] \nonumber \\&\quad = \sum _{\begin{array}{c} \tau =1 \\ \tau :\mathrm {odd} \end{array}}^{\alpha } \mathrm {sym}\left[ b_{\tau }(x)b_{\tau }(y)\right] + \mathrm {sym}\left[ b_{\alpha }(x)b_{\alpha }(y)\right] - \mathrm {sym}\left[ \tilde{b}_{2\alpha }(x-y) \right] . \end{aligned}$$

(9)

Since $b_\tau (x) = -b_\tau (1-x)$ for odd $\tau$ and $b_\tau (x) = b_\tau (1-x)$ for even $\tau$, we have

$$\begin{aligned} \mathrm {sym}[b_\tau (x)b_\tau (y)] = {\left\{ \begin{array}{ll} 0 &{} \text {for odd }\tau ,\\ b_\tau (x)b_\tau (y) &{} \text {for even }\tau . \end{array}\right. } \end{aligned}$$

and the first term of (9) equals 0. Considering the Fourier series of $b_\alpha$ for even $\alpha$:

$$\begin{aligned} b_\alpha (x) = \frac{(-1)^{1+\alpha /2}}{(2\pi )^\alpha } \sum _{k\in \mathbb {Z}{\setminus } \{0\}}\frac{e^{2\pi \mathrm {i}kx}}{k^\alpha } = \frac{2 \cdot (-1)^{1+\alpha /2}}{(2\pi )^\alpha } \sum _{k=1}^{\infty }\frac{\cos (2\pi kx)}{k^\alpha }, \end{aligned}$$

the Fourier series of the second term of (9) is given by

$$\begin{aligned} \mathrm {sym}\left[ b_{\alpha }(x)b_{\alpha }(y)\right] = b_{\alpha }(x)b_{\alpha }(y) = \frac{4}{(2\pi )^{2\alpha }}\sum _{k,\ell =1}^{\infty }\frac{1}{k^\alpha \ell ^\alpha }\cos (2\pi kx)\cos (2\pi \ell y). \end{aligned}$$

Finally, using the Fourier series of $\tilde{b}_{2\alpha }$, the Fourier series of the third term of (9) is given by

$$\begin{aligned} \mathrm {sym}\left[ \tilde{b}_{2\alpha }(x-y) \right] = \frac{-2}{(2\pi )^{2\alpha }} \sum _{k=1}^{\infty }\frac{\mathrm {sym}\left[ \cos (2\pi k(x-y))\right] }{k^{2\alpha }} \\ = \frac{-2}{(2\pi )^{2\alpha }} \sum _{k=1}^{\infty }\frac{\cos (2\pi kx)\cos (2\pi ky)}{k^{2\alpha }}. \end{aligned}$$

By substituting these results on the Fourier series into (9), we have

$$\begin{aligned} -1+\mathrm {sym}\left[ K_{\alpha ,1,1}^{\mathrm {sob}(\mathrm {odd}+\alpha )}(x,y)\right] = \frac{1}{(2\pi )^{2\alpha }}\sum _{k,\ell =1}^{\infty }c^{\mathrm {sym}}(k,\ell )\cos (2\pi kx)\cos (2\pi \ell y), \end{aligned}$$

where

$$\begin{aligned} c^{\mathrm {sym}}(k,\ell ) = {\left\{ \begin{array}{ll} \displaystyle \frac{6}{k^{2\alpha }} &{} k= \ell , \\ \displaystyle \frac{4}{k^\alpha \ell ^\alpha } &{} k\ne \ell . \end{array}\right. } \end{aligned}$$

The rest of the proof follows exactly in the same manner as the proof of Theorem 2.