Gelfond-Schneider Theorem/Lemma 3

Lemma

Let $r$ and $R$ be two real numbers such that $1 \le r \le R$.

Let $f_1 \left({z}\right), f_2 \left({z}\right), \ldots, f_L \left({z}\right)$ be:

analytic in $D \subseteq \C: D = \left\{{z : \left|{z}\right| < R}\right\}$

continuous on the closure $D$, that is, $D^- = \left\{{z : \left \vert{z}\right \vert \le R}\right\}$.

Let $\zeta_1, \ldots, \zeta_L$ be complex numbers such that:

$\forall j \in \left\{{1, 2, \ldots, L}\right\}: \left|{\zeta_j}\right| \le r$

Then the determinant:

$\Delta = \det \begin{bmatrix} f_1 \left({\zeta_1}\right) & \cdots & f_L \left({\zeta_1}\right) \\ \vdots & \ddots & \vdots \\ f_1 \left({\zeta_L}\right) & \cdots & f_L \left({\zeta_L}\right) \end{bmatrix}$

satisfies:

$\displaystyle \left|{\Delta}\right| \le \left({\frac R r}\right)^{−L \left({L−1}\right) / 2} L! \prod_{\lambda=1}^L \left|{f_λ}\right|_R$

Proof

Let $h \left({z}\right)$ be the determinant of the $L \times L$ matrix $\left[{f_j \left({\zeta_i z}\right)}\right]$.

Then $h \left({z}\right)$ is:

analytic in $D' = \left\{{z : \left \vert {z}\right \vert < R/r}\right\}$

continuous on $D'^- = \left\{{z : \left \vert{z}\right \vert \le R/r}\right\}$.

Let $M = L \left({L − 1}\right) / 2$, and write:

$\displaystyle f_j \left({\zeta_i z}\right) = \sum_{k=0}^{M−1} b_k \left({j}\right) \zeta_i^k z^k + z^M g_{i,j} \left({z}\right)$

where:

$b_k \left({j}\right) \in \C$ for each $k$

$g_{i,j} \left({z}\right)$ is analytic in $D'$

$g_{i,j} \left({z}\right)$ is continuous on $D'^-$

By evaluating along the columns it is seen that the determinant is linear in its columns.

So we can view $h \left({z}\right)$ as $z^M$ times an analytic function plus terms involving the factor:

$z^{n_1 + \cdots + n_L} \det \left[{\zeta_i^{n_j} }\right]$

where the $n_j$ denote non-negative integers.

Observe that the determinant in this last expression is zero if the $n_j$ are not distinct.

Therefore, the non-zero terms of this form satisfy:

$n_1 + n_2 + \cdots + n_L \ge 0 + 1 + \cdots + \left({L − 1}\right) = \dfrac {L \left({L − 1}\right)} 2$

Hence, we deduce that $h \left({z}\right)$ is divisible by $z^M$.

More precisely, $\dfrac {h \left({z}\right)} {z^M}$ is analytic in $D'$ and continuous on $D'^-$.

Therefore, by Gelfond-Schneider Theorem: Lemma 2, for any $w \in D'$:

$\displaystyle \left|{\frac {h \left({w}\right)} {w^M}}\right| \le \left|{\frac {h \left({z}\right)} {z^M}}\right|_{R/r} = \left({\frac r R}\right)^M \left|{h \left({z}\right)}\right|_{R/r}$

For $\left|{z}\right| = R/r$, we get that $\left|{\zeta_i z}\right| \le R$.

We bound $\left|{h \left({z}\right)}\right|_{R/r}$ by multiplying the number of terms in $\det \left[{f_j \left({\zeta_i z}\right)}\right]$ by an obvious upper bound on the maximum such term.

Thus:

$\displaystyle \left|{h \left({z}\right)}\right|_{R/r} \le L! \prod_{\lambda=1}^L \left|{f_λ}\right|_R$

Observe that $\left|{\Delta}\right| = \left|{h \left({1}\right)}\right|$ and $1 \le R/r \le R$.

We deduce that

$\displaystyle \left|{\Delta}\right| \le \left({\frac r R}\right)^M \left|{h \left({z}\right)}\right|_{R/r} \le \left({\frac r R}\right)^M L! \prod_{\lambda=1}^L \left|{f_λ}\right|_R$

giving the desired conclusion.

$\Box$