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Let $P\subset\mathbb{R}^{n}$ be a lattice polytope (vertices are in $\mathbb{Z}^{n}$). Set $P_{k}=P\cap k^{-1}\mathbb{Z}^{n}$ for $k\geq1$. Given a concave function $\phi:P\rightarrow\mathbb{R}$ (continuous on $P$). My question is: whether we have two-terms Euler-Maclaurin asymptotics \begin{equation} \sum_{u\in P_{k}}\phi(u)=\int_{P}\phi\ \mathrm{d}x\cdot k^{n}+\frac{1}{2}\int_{\partial P}\phi\ \mathrm{d}\sigma\cdot k^{n-1}+\mathrm{o}(k^{n-1})\ ?\label{eq:expansion} \end{equation} Where $\mathrm{d}x$ is Lebesgue measure and the boundary measure $\mathrm{d}\sigma$ is defined as follows.

For each facet $F\subset\partial P$, let $A_{F}\subset\mathbb{R}^{d}$ be the affine subspace generated by $F$, then $\mathrm{d}\sigma|_{F}$ is the Lebesgue measure normalized by the lattice $A_{F}\cap\mathbb{Z}^{d}$. For example, $P$ is the convex hull of $(0,0),(1,0),(0,2)$, then the length (measured by $\mathrm{d}\sigma$) of the hypotenuse is $1$ (not $\sqrt{5}$).

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    $\begingroup$ It does not seem to hold for $n=1$, $P=[0,1]$, $\phi(x)=\sqrt{x}$ (a remainder is worse). $\endgroup$ Commented Nov 24 at 5:21
  • $\begingroup$ Thanks! Sorry the remainder term should be little "o". I have modified the question. Actually, when n=1, it holds for all absolutely continuous functions. $\endgroup$
    – Yee Yao
    Commented Nov 25 at 2:04
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    $\begingroup$ What you write now is even strongest than it was $\endgroup$ Commented Nov 25 at 10:33
  • $\begingroup$ Oops. There are more refined asymptotic formulas for smooth functions, such as arXiv:0908.3073. For rational piecewise linear concave function, we can take the sum as counting lattice points under the graph (need to deal with the error of floor function), then Ehrhart polynomial theory can give the leading two-terms. Then a natural ideal is approximating a general concave funtion by smooth ones or rational PL ones (index by j). To let j goes to infinity, we need an uniform (for j) estimate on the remainder term. This is difficult. Is it possible to prove it without using approximation? $\endgroup$
    – Yee Yao
    Commented Nov 26 at 9:04

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