Sparse grid cubature and mixed regularity — Examples

Fix an interior $y\in (0,1)$ and write $g(h)=f(y+h,y)=\mathrm{\mid }h{\mathrm{\mid }}^3$ for sufficiently small positive and negative $h$. By [F4]–[F5], on the two sides $g^{\mathrm{\prime }}(h)=3h\mathrm{\mid }h\mathrm{\mid }$ and $g^{\mathrm{\prime }\mathrm{\prime }}(h)=6\mathrm{\mid }h\mathrm{\mid }$ for $h\mathrm{\ne }0$. The difference quotients at zero give $g^{\mathrm{\prime }}(0)=0$ and $g^{\mathrm{\prime }\mathrm{\prime }}(0)={\lim }_{h\to 0}3\mathrm{\mid }h\mathrm{\mid }=0$. The derivative of $g^{\mathrm{\prime }\mathrm{\prime }}$ at zero would require ${\lim }_{h\to 0}(g^{\mathrm{\prime }\mathrm{\prime }}(h)-g^{\mathrm{\prime }\mathrm{\prime }}(0))\mathrm{/}h={\lim }_{h\to 0}6\mathrm{\mid }h\mathrm{\mid }\mathrm{/}h$, whose right-hand value is $6$ and left-hand value is $-6$. Thus $g^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(0)$ does not exist, so $g^{(4)}(0)$ cannot exist. By [F3], the required derivative therefore fails at the interior diagonal point $(y,y)$.

Since [F1] requires $D_{\{1\}}f={\mathrm{\partial }}_x^{4}f$ to exist continuously everywhere, the failure in step 1.1 violates a stated hypothesis of the rate theorem [F2]. The theorem makes no such rate claim for this ridge.

For every fixed $y$, the same formulas give ${\mathrm{\partial }}_x^{2}f(x,y)=6\mathrm{\mid }x-y\mathrm{\mid }$, including its value zero on the diagonal; the same one-sided calculations apply at $x=0,1$. By symmetry ${\mathrm{\partial }}_y^{2}f(x,y)=6\mathrm{\mid }x-y\mathrm{\mid }$ as well. These partials are continuous and bounded by $6$ on the square since $\mathrm{\mid }x-y\mathrm{\mid }\le 1$, illustrating that lower-order smoothness does not imply the fourth-order condition used here.

The Given rules are the $n=1$ and $n=2$ rules. By [F3], their normalized nodes are $(1+\cos (j\pi \mathrm{/}n))\mathrm{/}2$; [F5] and [F8] give $X_0=\{0,1\}$ and $X_1=\{0,1\mathrm{/}2,1\}$. By [F4], these rules are exact through degrees $1$ and $2$, respectively. Writing the level-zero weights as $w_0,w_1$, exactness on $1$ and $x$ gives $w_0+w_1=1$ and $w_1=1\mathrm{/}2$, hence $Q_0(x)=Q_0(x^2)=1\mathrm{/}2$. Also $Q_1(x)=1\mathrm{/}2$ and $Q_1(x^2)=1\mathrm{/}3$ by exactness and [F6]. Thus ${\mathrm{\Delta }}_1(x)=0$ and ${\mathrm{\Delta }}_1(x^2)=-1\mathrm{/}6$.

Since $f$ is a product, its tensor details factor into one-dimensional values. The zero-index term is $Q_0(x^2)Q_0(x{)}^3=1\mathrm{/}16$. The $e_1$ term is ${\mathrm{\Delta }}_1(x^2)Q_0(x{)}^3=-1\mathrm{/}48$, and each $e_j$ term for $j=2,3,4$ vanishes because ${\mathrm{\Delta }}_1(x)=0$. Therefore $Q_{{\mathrm{\Lambda }}_1}(f)=1\mathrm{/}16-1\mathrm{/}48=1\mathrm{/}24$. By [F6]–[F7], $I_4(f)=(1\mathrm{/}3)(1\mathrm{/}2{)}^3=1\mathrm{/}24$, so the cubature is exact.

The node union is $X_0^4$ together with, for each coordinate $j$, the $8$ points whose $j$th coordinate is $1\mathrm{/}2$ and whose other coordinates are endpoints. Those four midpoint sets are disjoint from one another and from the $16$ corners, so $W({\mathrm{\Lambda }}_1)=16+4\cdot 8=48$. The full tensor level-one grid has $\mathrm{\mid }{X}_1{\mathrm{\mid }}^4=3^4=81$ nodes.

Every real square is nonnegative by [F13], so each factor in $\varphi$ is nonnegative and [F12] gives $\varphi (x)\ge 0$ on $[0,1]$. The polynomial $\varphi$ is continuous by [F4] and satisfies $\varphi (0)=1$. Continuity at zero gives $\eta >0$ such that $\varphi (x)>1\mathrm{/}2$ for $0\le x<\eta$; choose $\delta =\min (\eta \mathrm{/}2,1\mathrm{/}2)$, so $0<\delta <\eta$ and $\delta <1$. By [F5]–[F6], ${\int }_0^{\delta }\varphi \ge \delta \mathrm{/}2$ and ${\int }_{\delta }^1\varphi \ge 0$; [F7] then gives ${\int }_0^1\varphi \ge \delta \mathrm{/}2>0$. Applying [F8] to the continuous coordinate factors establishes integrability of $u$ and gives $c_{00}=({\int }_0^1\varphi {)}^2>0$.

For each $m=0,\dots ,8$, $\varphi (m\mathrm{/}8)=0$ when $m>0$, because the factor indexed by $k=m$ vanishes; $\varphi (0)=1$. The nested Simpson grids through level two are $X_0=\{0,1\mathrm{/}2,1\}$, $X_1=\{m\mathrm{/}4:0\le m\le 4\}$, and $X_2=\{m\mathrm{/}8:0\le m\le 8\}$. Thus every sampled value of $u$ on any tensor rule in ${\mathrm{\Lambda }}_2$ vanishes except at $(0,0)$, where it is one; all sampled values are nonnegative by [F13] and [F12].

Let $w_{\mathrm{\ell }}$ be the level-$\mathrm{\ell }$ endpoint weight. By [F2], $w_0=1\mathrm{/}6$, $w_1=1\mathrm{/}12$, and $w_2=1\mathrm{/}24$. Expanding the six details from [F3] and using finite distributivity [F12] gives $Q_{{\mathrm{\Lambda }}_2}=Q_2\otimes Q_0+Q_0\otimes Q_2-Q_1\otimes Q_0-Q_0\otimes Q_1+Q_1\otimes Q_1$. By step 2.1, its value on $u$ is the combined coefficient at $(0,0)$: $2w_2{w}_0-2w_1{w}_0+w_1^2=2\mathrm{/}(24\cdot 6)-2\mathrm{/}(12\cdot 6)+1\mathrm{/}{12}^2=-1\mathrm{/}144<0$. This differs in sign from $c_{00}>0$ in step 1.1.

For tensor products of Gaussian rules with valid positive-measure moment data, [F10] makes every coordinate weight positive and [F11] makes each tensor weight a product of positive numbers. Since all samples of $u$ are nonnegative by [F12]–[F13], [F9] and [F12] make the resulting finite sum nonnegative. Together with the negative Simpson estimate in step 3.1, this is a sign comparison; it does not claim that such a Gaussian rule exactly integrates $u$.