高维Montel定理证明中等度连续性步骤的积分项估计方法问询

Hey there, I totally get your frustration with those "trivial generalization" handwaves in several complex variables textbooks—nothing kills a deep dive like a proof that skips the messy, important details. Let's work through the equicontinuity step carefully, since that's where you're stuck, using the multi-variable Cauchy integral formula and some algebraic tricks to generalize the one-variable argument.

First, let's set up the context properly:
Given a compact set ( K \subset D ) (your region in ( \mathbb{C}^n )), we can first pick a ( \delta > 0 ) such that for every ( z \in K ), the closed polydisc ( \overline{P(z, 2\delta)} = { \zeta \in \mathbb{C}^n \mid |\zeta_k - z_k| \leq 2\delta \text{ for all } k } ) is entirely contained in ( D ). Since ( K ) is compact, we can cover it with finitely many open polydiscs ( P(z_1, \delta), \dots, P(z_m, \delta) ), and use the Lebesgue number lemma to get an ( r > 0 ): any two points ( z, w \in K ) with ( |z - w| < r ) lie inside one of these ( P(z_i, \delta) ).

Now, fix such a ( z \in K ), and consider the integral path ( \partial P(z, \delta) ) (the distinguished boundary of the polydisc, where each ( \zeta_k ) lies on the circle ( |\zeta_k - z_k| = \delta )). By the multi-variable Cauchy integral formula:
[
f_j(z) = \frac{1}{(2\pi i)^n} \int_{\partial P(z, \delta)} \frac{f_j(\zeta)}{\prod_{k=1}^n (\zeta_k - z_k)} d\zeta_1 \wedge \dots \wedge d\zeta_n
]
[
f_j(w) = \frac{1}{(2\pi i)^n} \int_{\partial P(z, \delta)} \frac{f_j(\zeta)}{\prod_{k=1}^n (\zeta_k - w_k)} d\zeta_1 \wedge \dots \wedge d\zeta_n
]

Taking the modulus of their difference, we get:
[
|f_j(z) - f_j(w)| \leq \frac{1}{(2\pi)^n} \int_{\partial P(z, \delta)} |f_j(\zeta)| \cdot \left| \frac{1}{\prod_{k=1}^n (\zeta_k - z_k)} - \frac{1}{\prod_{k=1}^n (\zeta_k - w_k)} \right| |d\zeta_1| \dots |d\zeta_n|
]

Since ( {f_j} ) is uniformly bounded on compact sets, let ( M = \sup_{j, \zeta \in \partial P(z, \delta)} |f_j(\zeta)| < \infty ). Our goal is to bound the fractional difference term—this is where the multi-variable twist comes in.

Estimating the Fractional Difference #

Let ( A = \prod_{k=1}^n (\zeta_k - z_k) ) and ( B = \prod_{k=1}^n (\zeta_k - w_k) ). We start with the standard identity:
[
\left| \frac{1}{A} - \frac{1}{B} \right| = \frac{|B - A|}{|A||B|}
]

First, expand ( B - A ) using a telescoping sum trick (this is the key generalization of the one-variable case):
[
B - A = \prod_{k=1}^n (\zeta_k - w_k) - \prod_{k=1}^n (\zeta_k - z_k) = \sum_{m=1}^n (z_m - w_m) \cdot \left( \prod_{k=1}^{m-1} (\zeta_k - w_k) \right) \cdot \left( \prod_{k=m+1}^n (\zeta_k - z_k) \right)
]

Now, use the triangle inequality on this sum:
[
|B - A| \leq \sum_{m=1}^n |z_m - w_m| \cdot \prod_{k=1}^{m-1} |\zeta_k - w_k| \cdot \prod_{k=m+1}^n |\zeta_k - z_k|
]

Since ( \zeta \in \partial P(z, \delta) ), we know ( |\zeta_k - z_k| = \delta ) for all ( k ). For ( |z - w| < \delta/2 ), we also have ( |\zeta_k - w_k| = |(\zeta_k - z_k) + (z_k - w_k)| \geq |\zeta_k - z_k| - |z_k - w_k| \geq \delta - |z - w| > \delta/2 ).

Substitute these bounds into the sum:

• ( \prod_{k=m+1}^n |\zeta_k - z_k| = \delta^{n - m} )
• ( \prod_{k=1}^{m-1} |\zeta_k - w_k| \geq (\delta/2)^{m - 1} )
• ( |z_m - w_m| \leq |z - w| )

So each term in the sum is bounded by ( |z - w| \cdot (\delta/2)^{m-1} \cdot \delta^{n - m} = |z - w| \cdot \delta^{n-1} / 2^{m-1} ). Summing over all ( m ):
[
|B - A| \leq |z - w| \cdot \delta^{n-1} \sum_{m=1}^n \frac{1}{2^{m-1}} \leq 2 |z - w| \cdot \delta^{n-1}
]
(The sum is a geometric series with sum ( 2 - 1/2^{n-1} < 2 ).)

Next, bound the denominator ( |A||B| ):

• ( |A| = \prod_{k=1}^n |\zeta_k - z_k| = \delta^n )
• ( |B| = \prod_{k=1}^n |\zeta_k - w_k| \geq (\delta/2)^n )

So ( |A||B| \geq \delta^n \cdot (\delta/2)^n = \delta^{2n}/2n ).

Putting it all together, the fractional difference becomes:
[
\left| \frac{1}{A} - \frac{1}{B} \right| \leq \frac{2 |z - w| \cdot \delta^{{n-1}}{\delta}{2n}/2^n} = \frac{2^{n+1} |z - w|}{\delta^{n+1}}
]

Finalizing the Equicontinuity Estimate #

Now plug this back into the original difference bound. The integral over ( \partial P(z, \delta) ) has a total "length" (product of circumferences) of ( (2\pi \delta)^n ). Substituting all constants:
[
|f_j(z) - f_j(w)| \leq \frac{1}{(2\pi)^n} \cdot M \cdot \frac{2^{n+1} |z - w|}{\delta^{n+1}} \cdot (2\pi \delta)^n
]

Simplify the constants:
[
\frac{1}{(2\pi)^n} \cdot (2\pi \delta)^n = \delta^n, \quad \text{so we get} \quad |f_j(z) - f_j(w)| \leq M \cdot \frac{2^{n+1} |z - w|}{\delta}
]

Let ( C = M \cdot 2^{n+1}/\delta )—this constant is independent of ( j ), ( z ), and ( w ) (as long as ( |z - w| < \delta/2 )). This means for any ( \epsilon > 0 ), we can choose ( r = \min(\delta/2, \epsilon/C) ), so ( |z - w| < r ) implies ( |f_j(z) - f_j(w)| < \epsilon ) for all ( j ).

Since we can do this for every compact subset ( K ) (using finite covering to extend the estimate across all of ( K )), we've proven equicontinuity on compact sets. From here, Arzelà-Ascoli gives pointwise convergence on ( K ), and a diagonalization argument extends this to compact convergence on all of ( D ), completing the proof of Montel's theorem for several complex variables.

备注：内容来源于stack exchange，提问作者Maths Matador