复函数连续性证明：求$f(z)=\frac{2|z|}{i+\text{Re}z}$的δ选取方法

Great question—let's walk through how to construct the required $\delta$ for continuity step by step. First, quick sanity check: this function is well-defined everywhere on $\mathbb{C}$ because the denominator $i+\text{Re}(z)$ has an imaginary part of 1, so its modulus $\sqrt{(\text{Re}(z))^2 +1}$ is always at least 1 (never zero). No gaps in the domain to worry about.

To prove continuity at a fixed point $w \in \mathbb{C}$, we need to show that for every $\epsilon > 0$, there's a $\delta > 0$ such that if $|z - w| < \delta$, then $|f(z) - f(w)| < \epsilon$. Here's how to build that $\delta$:

Step 1: Rewrite the difference $f(z) - f(w)$ #

Start by expanding the function difference and combining the fractions:
$$
f(z) - f(w) = 2\left( \frac{|z|}{\text{Re}(z) + i} - \frac{|w|}{\text{Re}(w) + i} \right) = 2 \cdot \frac{|z|(\text{Re}(w) + i) - |w|(\text{Re}(z) + i)}{(\text{Re}(z) + i)(\text{Re}(w) + i)}
$$
Split the numerator into real and imaginary components to make it easier to bound:
$$
\text{Numerator} = \left(|z|\text{Re}(w) - |w|\text{Re}(z)\right) + i\left(|z| - |w|\right)
$$

Step 2: Bound the numerator's modulus #

Use the triangle inequality to split the numerator's modulus into two manageable parts:
$$
|\text{Numerator}| \leq \left||z|\text{Re}(w) - |w|\text{Re}(z)\right| + \left||z| - |w|\right|
$$
Now rewrite the first term to apply standard inequalities:
$$
|z|\text{Re}(w) - |w|\text{Re}(z) = \text{Re}(w)\left(|z| - |w|\right) + |w|\left(\text{Re}(w) - \text{Re}(z)\right)
$$
Applying the triangle inequality again to this expression gives:
$$
\left||z|\text{Re}(w) - |w|\text{Re}(z)\right| \leq |\text{Re}(w)| \cdot \left||z| - |w|\right| + |w| \cdot \left|\text{Re}(w) - \text{Re}(z)\right|
$$
We can leverage two key inequalities here:

• Reverse triangle inequality: $\left||z| - |w|\right| \leq |z - w|$ (this holds for all complex numbers, just like the regular triangle inequality)
• Real part bound: $\left|\text{Re}(w) - \text{Re}(z)\right| = \left|\text{Re}(z - w)\right| \leq |z - w|$ (the real part of a complex number can never have a larger modulus than the number itself)

Substituting these into our numerator bound:
$$
|\text{Numerator}| \leq \left( |\text{Re}(w)| + |w| + 1 \right) \cdot |z - w|
$$

Step 3: Bound the denominator's modulus #

The denominator is the product of two complex numbers, so its modulus is the product of their moduli:
$$
|\text{Denominator}| = |\text{Re}(z) + i| \cdot |\text{Re}(w) + i|
$$
Notice that $|\text{Re}(z) + i| = \sqrt{(\text{Re}(z))^2 + 1} \geq 1$ for any real $\text{Re}(z)$—since squaring a real number gives a non-negative result, adding 1 and taking the square root can't be less than 1. This means:
$$
|\text{Denominator}| \geq 1 \cdot |\text{Re}(w) + i| = \sqrt{(\text{Re}(w))^2 + 1}
$$
This is a positive constant that only depends on our fixed point $w$.

Step 4: Choose your $\delta$ #

Combine the bounds for the numerator and denominator to get an overall bound on $|f(z) - f(w)|$:
$$
|f(z) - f(w)| \leq 2 \cdot \frac{\left( |\text{Re}(w)| + |w| + 1 \right) \cdot |z - w|}{\sqrt{(\text{Re}(w))^2 + 1}}
$$
Let’s define a constant $C$ (dependent only on $w$) as:
$$
C = \frac{2\left( |\text{Re}(w)| + |w| + 1 \right)}{\sqrt{(\text{Re}(w))^2 + 1}}
$$
To make $|f(z) - f(w)| < \epsilon$, we just need $C \cdot |z - w| < \epsilon$. Solving for $|z - w|$ gives:
$$
|z - w| < \frac{\epsilon}{C}
$$
So we can choose $\delta = \frac{\epsilon}{C}$. (If you want to be thorough, you can set $\delta = \min\left(1, \frac{\epsilon}{C}\right)$—this ensures any implicit bounds we used (like the behavior of $\text{Re}(z)$ near $\text{Re}(w)$) hold, though in this case $C$ already accounts for the worst-case denominator behavior.)

Since this works for any arbitrary $w \in \mathbb{C}$, the function is continuous at every point in $\mathbb{C}$.

内容的提问来源于stack exchange，提问作者TomSmith