全纯函数与拉普拉斯算子：证明Δln(|f|+|g|)≥0并求简便证法

Great question! We can tackle this neatly using properties of subharmonic functions instead of grinding through partial derivative expansions—let's break it down step by step:

步骤1：核心背景回顾 #

A continuous function ( u: D \to \mathbb{R} ) is subharmonic if it satisfies either:

• The integral mean inequality: For any closed disk ( \overline{B}(z_0, r) \subset D ), ( u(z_0) \leq \frac{1}{2\pi} \int_0^{2\pi} u(z_0 + re^{i\theta}) d\theta );
• Or, ( u = \sup_{\alpha} u_\alpha ) where each ( u_\alpha ) is harmonic, and ( u ) is upper-semicontinuous.

The key property we care about here: Subharmonic functions satisfy ( \Delta u \geq 0 ) (in either the classical or distributional sense).

步骤2：Rewrite ( |f| + |g| ) as a supremum of holomorphic function moduli #

For any complex numbers ( a, b ), we have the identity:
[ |a| + |b| = \max_{|\lambda|=1} |\lambda a + b| ]
Quick verification: If ( a,b \neq 0 ), take ( \lambda = \frac{\overline{a}|b| + \overline{b}|a|}{|a||b|} )—this gives ( |\lambda a + b| = |a| + |b| ). If ( a=0 ), the max becomes ( |b| = |a| + |b| ), and vice versa for ( b=0 ).

Applying this to our holomorphic functions ( f,g ) on ( D ):
[ |f(z)| + |g(z)| = \sup_{|\lambda|=1} |\lambda f(z) + g(z)| ]

步骤3：Transform the target function #

Since the natural logarithm is strictly increasing, we can swap it with the supremum:
[ \ln(|f(z)| + |g(z)|) = \sup_{|\lambda|=1} \ln|\lambda f(z) + g(z)| ]

Now, for each fixed ( \lambda \in \mathbb{S}^1 ) (the unit circle):

• ( \lambda f + g ) is holomorphic on ( D ) (linear combinations of holomorphic functions are holomorphic);
• Because ( f ) and ( g ) have no common zeros, if ( \lambda f(z_0) + g(z_0) = 0 ), then ( f(z_0) \neq 0 ) and ( g(z_0) = -\lambda f(z_0) \neq 0 )—but other values of ( \lambda' ) will give ( \lambda' f(z_0) + g(z_0) \neq 0 ), so ( \ln|\lambda' f + g| ) is finite at ( z_0 );
• Where ( \lambda f + g \neq 0 ), ( \ln|\lambda f + g| ) is harmonic (the logarithm of the modulus of a non-vanishing holomorphic function is always harmonic).

步骤4：Final Conclusion #

( \ln(|f| + |g|) ) is continuous (since ( |f|+|g| ) is continuous and the logarithm is continuous), and it's the supremum of a family of harmonic functions. By the subharmonic function characterization, this makes it a subharmonic function.

And since subharmonic functions satisfy ( \Delta u \geq 0 ), we get exactly what we need:
[ \Delta \ln(|f| + |g|) \geq 0 ]

Quick check for edge cases #

• If ( f(z_0) = 0 ), then ( g(z_0) \neq 0 ), so ( \ln(|f|+|g|) = \ln|g| )—a harmonic function, so ( \Delta \ln|g| = 0 \geq 0 );
• If ( g(z_0) = 0 ), the same logic applies: ( \Delta \ln|f| = 0 \geq 0 ).

内容的提问来源于stack exchange，提问作者Mathworld