Gelfond-Schneider Theorem/Lemma 3
== Lemma ==  Let $r$ and $R$ be two real numbers such that $0 < r \le R$. 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: :$\displaystyle \left|{\Delta}\right| \le \left({\frac R r}\right)^{−L \left({L−1}\right) / 2} L! \prod_{\lambda \mathop = 1}^L \left|{f_λ}\right|_R$  == 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: :$\displaystyle f_j \left({\zeta_i z}\right) = \sum_{k \mathop = 0}^{M−1} b_k \left({j}\right) \zeta_i^k z^k + z^M g_{i,j} \left({z}\right)$ 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 create a bound for $\left|{h \left({z}\right)}\right|_{R/r}$ 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: :$\displaystyle \left|{h \left({z}\right)}\right|_{R/r} \le L! \prod_{\lambda \mathop = 1}^L \left|{f_λ}\right|_R$ Observe that $\left|{\Delta}\right| = \left|{h \left({1}\right)}\right|$ and $1 \le R/r$. We deduce that :$\displaystyle \left|{\Delta}\right| \le \left({\frac r R}\right)^M \left|{h \left({z}\right)}\right|_{R/r} \le \left({\frac r R}\right)^M L! \prod_{\lambda \mathop = 1}^L \left|{f_λ}\right|_R$ giving the desired conclusion. {{qed|lemma}}