Composite periodic trapezoidal error in a uniform tensor grid

In one coordinate, on a cell $[a,a+h]$, put $K(x)=(x-a)(a+h-x)\mathrm{/}2$. Two integrations by parts give $\frac{h}{2}(f(a)+f(a+h))-{\int }_a^{a+h}f(x)dx={\int }_a^{a+h}K(x)f^{\mathrm{\prime }\mathrm{\prime }}(x)dx\mathrm{.}$ Since $K\ge 0$ and ${\int }_a^{a+h}K(x)dx=h^3\mathrm{/}12$, the absolute cell error is at most $h^3\mathrm{\parallel }{f}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}12$. Sum over the $N$ periodic cells, whose endpoint contributions assemble into the periodic trapezoid sum $h{\sum }_{j}f(jh)$. This gives the one-dimensional error bound $L{h}^2\mathrm{\parallel }{f}^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}12$.

For each coordinate $a$, let $Q_a$ denote the periodic trapezoid operator in that coordinate and $I_a$ integration over that coordinate. Both have sup-norm operator bound at most $L$, since their weights are nonnegative and sum to $L$. The telescoping identity $Q_1\cdots Q_d-I_1\cdots I_d={\sum }_{a=1}^d{Q}_1\cdots Q_{a-1}(Q_a-I_a)I_{a+1}\cdots I_d$ holds by cancellation. In the $a$th summand, the one-dimensional bound from step 1.1 applies after the later-coordinate integrations; differentiation in coordinate $a$ commutes with those integrations. The other $d-1$ operators contribute at most $L^{d-1}$ in sup norm. Thus that summand is bounded by $L^d{h}^2\mathrm{\parallel }{\mathrm{\partial }}_{aa}f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}12$. Summing these bounds gives the displayed estimate, because $Q_1\cdots Q_{d}f=h^d{\sum }_{j}f(x_j)$ and $I_1\cdots I_{d}f={\int }_{[0,L{]}^d}f$.