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Let $P$ be a polytope with $M$ vertices. (The polytope $P$ is the intersection of the hypercube $0≤x _j ≤1$ with the hyperplane $\sum_{j=1}^nx_j=t$, $0\leq t\leq n$). Suppose that the volume of $P$ is $Vol(P)=A$. We subdivide $P$ into $M$ pieces each of the volume $A/M$.

Let $f$ be a function, such that the integral below exist. Then,

$$ \int_{P}f(x)ds\geq \sum_{i=1}^{M}\min_{P_i} f(x)|P_i|=\frac{A}{M}\sum_{i=1}^{M}\min_{P_i} f(x). $$

Could you please help me to understand the notion of the $\min_{P_i} f(x)$, i.e. what is this and how one can find this minimum?

Thank you for your help.

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1 Answer 1

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It's no big deal: The $M$ pieces $P_i$ are essentially disjoint and make up $P$. Therefore $$\int_P f(x)\ {\rm d}(x)=\sum_{i=1}^n \int_{P_i} f(x)\ {\rm d}(x)\ .$$ If $f$ is, e.g., continuous on $P$ then for each $i$ there is a point $\xi_i\in P_i$ such that $\mu_i:=f(\xi_i)\leq f(x)$ for all $x\in P_i$. The minimal value $\mu_i=\min\{f(x)\ |\ x\in P_i\}$ is well defined, but there might be several points $\xi\in P_i$, where the minimum is attained.

From basic properties of the integral it follows that $$\int_{P_i} f(x)\ {\rm d}(x)\geq\int_{P_i} \mu_i\ {\rm d}(x)=\mu_i\ |P_i|\ ,$$ and putting it all together we arrive at $$\int_P f(x)\ {\rm d}(x)\geq \sum_{i=1}^n \mu_i\ |P_i|={A\over M}\sum_{i=1}^n \mu_i\ .$$ Remains the problem of determining the $\mu_i$, resp., finding the $\xi_i$. Since the combinatorial description of $P_i$ can be quite complicated the use of calculus methods is very cumbersome (one would have to look at all lower-dimensional faces of $P_i$). When $f$ is in fact linear you can be sure that the $\min$ is taken at a vertex of $P_i$; in this case it suffices comparing the vertex values and selecting the lowest one.

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