Another refinement of the right-hand side of the Hermite–Hadamard inequality for simplices

We establish a new refinement of the right-hand side of the Hermite–Hadamard inequality for convex functions of several variables defined on simplices.

The classical Hermite-Hadamard inequality states that if f : I → R is a convex function then for all a < b ∈ I the inequality is valid. This powerful tool has found numerous applications and has been generalized in many directions (see e.g. [1,2]). One of those directions is its multivariate version: Theorem 1. ( [1]) Let f : U → R be a convex function defined on a convex set U ⊂ R n and Δ ⊂ U be an n-dimensional simplex with vertices x 0 , x 1 , . . . , x n . Then is the barycenter of Δ and the integration is with respect to the n-dimensional Lebesgue measure.
The aim of this note is to prove a refinement of the right-hand side of (1) stated in Theorem 2.
Let us start with a set of definitions. A function f : I → R defined on an interval I is called convex if for any x, y ∈ I and t ∈ (0, 1) the inequality AEM If U is a convex subset of R n , then a function f : U → R is convex if its restriction to every line segment in U is convex.
For n + 1 points x 0 , . . . , x n ∈ R n in general positions the set Δ = conv{x 0 , . . . , x n } is called an n-dimensional simplex. If K is a nonempty subset of the set N = {0, . . . , n} of cardinality k, the set The barycenter of Δ will be denoted by b. By card K we shall denote the cardinality of the set K.
For each k − 1-face Δ K we calculate the average value of f over Δ K using the formula where the integration is with respect to the k−1-dimensional Lebesgue measure (in case k = 1 this is the counting measure). For k = 1, 2, . . . , n + 1 we define Note that the right-hand side of the inequality (1) can be rewritten as A(n + 1) ≤ A (1). It turns out, that Theorem 2. The following chain of inequalities holds: In the proof we shall use the following Proof of Theorem 2. We shall prove first the inequality A(n + 1) ≤ A(n). Let us use the notation from Lemma 1. For i = 0, 1, . . . , n we have Summing (2) we obtain Vol. 92 (2018) Hermite-Hadamard inequality for simplices 1165 Now using Lemma 1 and the convexity of f applied to (3) we get thus, by the left-hand side of (1)

This shows the inequality A(n + 1) ≤ A(n).
Let K ⊂ N be a set of cardinality k > 1. Applying the above reasoning to Δ K we obtain Summing the above for all k-element subsets we get The equality follows from the fact that every k − 2-face belongs to n − k + 2 distinct k − 1-faces so every term Avg(f, Δ K ) appears in the sum exactly n − k + 2 times. Dividing both sides by n+1 k we get A(k) ≤ A(k − 1), which completes the proof.
Just for completeness note that a similar refinement of the left-hand side of (1) can be found in [4,Corollary 2.6]. It reads as follows: Theorem 3. For a nonempty subset K of N define the simplex Σ K as follows: let A K be the affine span of Δ K and A K be the affine space of the same dimension, parallel to A K and passing through the barycenter of Δ. Then Σ K = Δ ∩ A K .
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