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21.11.59 Several lemmata6F0D797F0564

Lemma 21.11.60.B1B557E6282F Let $\alpha ,\beta _1,\ldots ,\beta _ m\in \mathbb {Z}^ n$. Let $\Delta $ be the polytope generated by $\beta _1,\ldots ,\beta _ m$. Then the following are equivalent:

  1. 45E8CA3F6710
    EF4C4E285442 21.11.61
    \begin{equation} |z^{\alpha }|^2\left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1} \end{equation}

    is a bounded function on $\mathbb {C}^{*n}$.

  2. B62429A6EB39

    $\alpha \in \Delta $.

Proof. 2 $\implies $ 1. Write $\alpha =\sum _ i t_ i\beta _ i$, where $t_ i\in [0,1]$, $\sum _ i t_ i=1$. Then

\[ \begin{aligned} |z^{\alpha }|^2\left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1} & =\prod _ i |z^{\beta _ i}|^{2t_ i}\left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1}\\ & \leq \prod _ i\left(\sum _{j=1}^ m |z^{\beta _ j}|^2\right)^{t_ i} \left(\sum _{i=1}^ m |z^{\beta _ i}|^2 \right)^{-1}=1. \end{aligned} \]

1 $\implies $ 2. Assume that $\alpha \not\in \Delta $. By the strict separation theorem, there is $a=(a_1,\ldots ,a_ n)\in \mathbb {R}^ n$ such that

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

Set

\[ z(t)\coloneqq (t^{a_1},\ldots ,t^{a_ n}). \]

Then (21.11.61) evaluated at $z(t)$ grows like

\[ \frac{t^{2a\cdot \alpha }}{\sum _ i t^{2a\cdot \beta _ i}} \]

as $t\to \infty $, and hence is not bounded. $\square$

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. $\square$

Lemma 21.11.63.66B643EAC79A Let $I\subseteq \mathbb {R}$ be an open interval containing $0$. Let $(f_ k\colon I\rightarrow \mathbb {R})_{k\geq 1}$ be a sequence of convex functions, and let $g\colon I\to \mathbb {R}$ be a function. Assume that

\[ \varlimsup _{k\to \infty } f_ k(t)\leq g(t)\quad \text{for every } t\in I, \]

and that

\[ \lim _{k\to \infty } f_ k(0)=g(0). \]

If each $f_ k$ and $g$ is differentiable at $0$, then

\[ \lim _{k\to \infty } f_ k'(0)=g'(0). \]

Proof. Since $f_ k$ is convex and differentiable at $0$, for every $t\in I$,

\[ f_ k(t)\geq f_ k(0)+t f_ k'(0). \]

Taking the upper limit as $k\to \infty $ and using the assumptions gives

\[ \varlimsup _{k\to \infty } t f_ k'(0) \leq g(t)-g(0). \]

For $t>0$, this yields

\[ \varlimsup _{k\to \infty } f_ k'(0) \leq \frac{g(t)-g(0)}{t}. \]

Letting $t\to 0+$ and using the differentiability of $g$ at $0$, we get

\[ \varlimsup _{k\to \infty } f_ k'(0)\leq g'(0). \]

For $t<0$, the same inequality gives

\[ \varliminf _{k\to \infty } f_ k'(0) \geq \frac{g(t)-g(0)}{t}. \]

Letting $t\to 0-$ and again using the differentiability of $g$ at $0$, we obtain

\[ \varliminf _{k\to \infty } f_ k'(0)\geq g'(0). \]

Therefore

\[ \lim _{k\to \infty } f_ k'(0)=g'(0), \]

as claimed. $\square$

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