【第八章 习题24】
令γ\gammaγ是习题23中那样的曲线，又假定γ\gammaγ的值域与负实轴不交。试证Ind(γ)=0Ind(\gamma) = 0Ind(γ)=0。提示：Ind(γ+c)Ind(\gamma + c)Ind(γ+c)在0≤c<+∞0 \leq c < + \infty0≤c<+∞上是ccc的连续整数值函数。又当c→+∞c \rightarrow + \inftyc→+∞时，Ind(γ+c)→0Ind(\gamma + c) \rightarrow 0Ind(γ+c)→0。
【证明】
由于γ\gammaγ是定义在紧集上的连续函数，所以γ\gammaγ的值域是紧集。而(−∞,0]( - \infty,0\rbrack(−∞,0]是闭集，根据习题4.21 的结论，对于任意的c≥0c \geq 0c≥0存在δ>0\delta > 0δ>0，使得
∣γ(t)−(−c)∣=∣γ(t)+c∣>δ\left| \gamma(t) - ( - c) \right| = \left| \gamma(t) + c \right| > \delta∣γ(t)−(−c)∣=∣γ(t)+c∣>δ
设∣γ′(t)∣≤M\left| \gamma'(t) \right| \leq M∣γ′(t)∣≤M，对于任意的c1、c2≥0c_{1}、c_{2} \geq 0c1、c2≥0
∣Ind(γ+c2)−Ind(γ+c1)∣=∣12πi∫abγ′(t)γ(t)+c2−12πi∫abγ′(t)γ(t)+c1∣=∣12πi∣∣∫abγ′(t)(c1−c2)(γ(t)+c1)(γ(t)+c2)∣≤∣12πi∣∫ab∣γ′(t)(c1−c2)(γ(t)+c1)(γ(t)+c2)∣<(b−a)M2πδ2∣c1−c2∣{\left| Ind\left( \gamma + c_{2} \right) - Ind\left( \gamma + c_{1} \right) \right| }{= \left| \frac{1}{2\pi i}\int_{a}^{b}\frac{\gamma'(t)}{\gamma(t) + c_{2}} - \frac{1}{2\pi i}\int_{a}^{b}\frac{\gamma'(t)}{\gamma(t) + c_{1}} \right| }{= \left| \frac{1}{2\pi i} \right|\left| \int_{a}^{b}\frac{\gamma'(t)\left( c_{1} - c_{2} \right)}{\left( \gamma(t) + c_{1} \right)\left( \gamma(t) + c_{2} \right)} \right| \leq \left| \frac{1}{2\pi i} \right|\int_{a}^{b}\left| \frac{\gamma'(t)\left( c_{1} - c_{2} \right)}{\left( \gamma(t) + c_{1} \right)\left( \gamma(t) + c_{2} \right)} \right| < \frac{(b - a)M}{2\pi\delta^{2}}\left| c_{1} - c_{2} \right|}∣Ind(γ+c2)−Ind(γ+c1)∣= 2πi1∫abγ(t)+c2γ′(t)−2πi1∫abγ(t)+c1γ′(t) = 2πi1 ∫ab(γ(t)+c1)(γ(t)+c2)γ′(t)(c1−c2) ≤ 2πi1 ∫ab (γ(t)+c1)(γ(t)+c2)γ′(t)(c1−c2) <2πδ2(b−a)M∣c1−c2∣
所以Ind(γ+c)Ind(\gamma + c)Ind(γ+c)在0≤c<+∞0 \leq c < + \infty0≤c<+∞上是ccc的一致连续函数
而当ccc足够大时
∣γ(t)+c∣>∣c∣−∣γ(t)∣≥12∣c∣\left| \gamma(t) + c \right| > |c| - \left| \gamma(t) \right| \geq \frac{1}{2}|c|∣γ(t)+c∣>∣c∣−∣γ(t)∣≥21∣c∣
∣Ind(γ+c)∣=∣12πi∫abγ′(t)γ(t)+c∣=∣12πi∣∣∫abγ′(t)γ(t)+c∣≤∣12πi∣∫ab∣γ′(t)γ(t)+c∣<(b−a)Mπ∣c∣\left| Ind(\gamma + c) \right| = \left| \frac{1}{2\pi i}\int_{a}^{b}\frac{\gamma'(t)}{\gamma(t) + c} \right| = \left| \frac{1}{2\pi i} \right|\left| \int_{a}^{b}\frac{\gamma'(t)}{\gamma(t) + c} \right| \leq \left| \frac{1}{2\pi i} \right|\int_{a}^{b}\left| \frac{\gamma'(t)}{\gamma(t) + c} \right| < \frac{(b - a)M}{\pi|c|}∣Ind(γ+c)∣= 2πi1∫abγ(t)+cγ′(t) = 2πi1 ∫abγ(t)+cγ′(t) ≤ 2πi1 ∫ab γ(t)+cγ′(t) <π∣c∣(b−a)M
此时总会有c0c_{0}c0使得
Ind(γ+c0)=0Ind\left( \gamma + c_{0} \right) = 0Ind(γ+c0)=0
又因为Ind(γ)Ind(\gamma)Ind(γ)的值是连续整数函数，所以
Ind(γ)=0Ind(\gamma) = 0Ind(γ)=0