A "reversed" Jensen inequality

Question

Let $S= \langle v_0, \ldots, v_n \rangle$ be a non-degenerate simplex in some affine space, Consider $(x_i)_{i\in I}$ a (finite) family of points in $S$ and $\lambda_i\ge 0$, $\sum_{i\in I}\lambda_i =1$ a family of weights. Write $\sum_{i\in I} \lambda_i x_i = \sum_{k=0}^n \mu_k v_k$ ( $\mu_0$, $\ldots$, $\mu_n$ uniquely determined).

Conclusion: for every $f\colon S \to \mathbb{R}$ a convex function we have

$$\sum_{i\in I} \lambda_i f(x_i) \le \sum_{k=0}^n \mu_k f(v_k)$$

Notes:

1. A bit surprising to me, and also interesting. I have found a variant of it in the paper by Andreas Maurer, indicated to me by @FinleyMarsh.

2. By Jensen's inequality we have $f(\sum \mu_i x_i) \le \sum \lambda_i f(x_i)$, What is new is an upper estimate for the RHS. Perhaps a stretch to call it a reversed Jensen, suggestions are welcome

The proof is not difficult, so I will leave it as a reference. If somebody could provide another reference, that would be great. I will post a solution later on.

Any feedback would be appreciated!

Answer 1

The proof is straightforward. Every element $x_i$ admits a unique representation $$x_i=\sum_{k=0}^n\mu_{ik}v_k$$ where $\mu_{ik}\ge 0$ and $\sum_{k=0}^n\mu_{ik}=1.$ We have $$\sum_{i\in I}\lambda_ix_i=\sum_{k=0}^n\mu_kv_k.$$ Hence $\mu_k=\sum_{i\in I}\lambda_i\mu_{ik}.$ Therefore by the convexity of $f$ we get $$\sum_{i\in I}\lambda_if(x_i)\le \sum_{i\in I}\lambda_i\sum_{k=0}^n\mu_{ik}f(v_k)\\ = \sum_{k=0}^nf(v_k)\sum_{i\in I}\lambda_i\mu_{ik}=\sum_{k=0}^n\mu_kf(v_k)$$

Answer 2

Let $\ell$ be an affine function (uniquely determined) such that $$\ell(v_k) = f(v_k)$$ for all $0\le k\le n$.

For every $x\in S$ we have $f(x) \le \ell(x)$ ( by Jensen). Therefore

$$\sum_{i\in I} \lambda_i f(x_i) \le \sum_i\lambda_i \ell(x_i) = \ell( \sum_i \lambda_i x_i) = \ell ( \sum_k \mu_k v_k) = \sum \mu_k \ell( v_k) = \sum_k \mu_k f(v_k)$$

The proof follows the intuition: the graph of the function is below some hyperplane through the vertices of a simplex.