Tensor multilinear interpolation has a second-derivative error bound

Fix one coordinate interval $[a,b]$ of length $h_j$ and hold the other coordinates fixed. If $x_j$ is an endpoint, the linear interpolant is exact by [F2]. Otherwise [F2] gives $\mathrm{\mid }u(x)-I_{j}u(x)\mathrm{\mid }\le \frac{1}{2}\mathrm{\parallel }{\mathrm{\partial }}_{jj}u{\mathrm{\parallel }}_{\mathrm{\infty }}(x_j-a)(b-x_j)$. Writing $\theta =(x_j-a)\mathrm{/}{h}_j\in (0,1)$, the product equals $h_j^2\theta (1-\theta )\le h_j^2\mathrm{/}4$, hence the one-coordinate error is at most $h_j^2\mathrm{\parallel }{\mathrm{\partial }}_{jj}u{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}8$.

Write the tensor interpolant as $I_1\cdots I_d$, where $I_j$ interpolates only coordinate $j$. The telescoping identity $u-I_1\cdots I_{d}u=(u-I_{1}u)+I_1(u-I_{2}u)+\cdots +I_1\cdots I_{d-1}(u-I_{d}u)$ follows by cancellation of adjacent terms.

Apply the one-coordinate estimate of step 1.1 to each term before its preceding-coordinate operators. Every preceding product $I_1\cdots I_{j-1}$ is sup-norm nonexpansive by [F3], so the norm of the $j$th term is at most $h_j^2\mathrm{\parallel }{\mathrm{\partial }}_{jj}u{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}8$. Summing the $d$ bounds proves the result. Periodic endpoint cells use the periodic lift of $u$, which has the same derivative suprema, and at every grid node the interpolation is exact by the definition.