Pome

Lemma 21.11.6281D915E5313D

Lemma 21.11.62.81D915E5313D Let $\beta _1,\ldots ,\beta _ m\in \mathbb {N}^ n$ and $\beta \in \mathbb {R}^ n$. Then the following are equivalent:

  1. EE49D9E1DF50

    $\log \sum _{i=1}^ m \mathrm{e}^{x\cdot \beta _ i}-x\cdot \beta $ is bounded from below.

  2. BD8E8347B6CF

    $\beta $ is in the convex hull of the $\beta _ i$'s.

Proof. If $\beta =\sum _ i t_ i\beta _ i$ with $t_ i\in [0,1]$ and $\sum _ i t_ i=1$, then by the weighted arithmetic–geometric mean inequality,

\[ x\cdot \beta =\sum _ i t_ i\, x\cdot \beta _ i \leq \log \sum _ i t_ i\mathrm{e}^{x\cdot \beta _ i} \leq \log \sum _ i \mathrm{e}^{x\cdot \beta _ i}. \]

This proves that the function in 1 is bounded from below.

Conversely, if $\beta $ does not lie in the convex hull of the $\beta _ i$'s, choose $a\in \mathbb {R}^ n$ such that

\[ a\cdot \beta >\max _ i a\cdot \beta _ i. \]

Evaluating at $x=ta$ and letting $t\to \infty $, we find

\[ \log \sum _ i \mathrm{e}^{t a\cdot \beta _ i}-t\, a\cdot \beta \to -\infty , \]

contradicting 1. $\square$

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