4)设曲线C的长度为L,函数∫(2)在C上满 足|f(2)≤M,那么 f()d1s/Uf()ds≤ML 段的长度
4) C ÝǑ L, ¼ê f(z) 3 C þ÷ v |f(z)| 6 M, o Z C f(z)dz 6 Z C |f(z)|ds 6 ML. y² |△zk| = |zk − zk−1|, △sk Ǒùü:m lãÝ. w, |△zk| 6 △sk. ¤± Xn k=1 f(ζk)△zk 6 Xn k=1 |f(ζk)△zk| 6 Xn k=1 |f(ζk)|△sk - δ = max 16k6n {△sk} → 0, Z C f(z)dz 6 Z C |f(z)|ds. ÏǑ3 C þ |f(z)| 6 M, Z C |f(z)|ds 6 M Z C ds = ML. 13/127
4)设曲线C的存度为L,函数∫(z)在C上满 足|f(2)≤M,那么 f(2)d2|≤/f(2)lds≤ 证明设|△k=|2k-xk-1,且△Sk为这两点之间的 从段的存度.显然
4) C ÝǑ L, ¼ê f(z) 3 C þ÷ v |f(z)| 6 M, o Z C f(z)dz 6 Z C |f(z)|ds 6 ML. y² |△zk| = |zk − zk−1|, △sk Ǒùü:m lãÝ. w, |△zk| 6 △sk. ¤± Xn k=1 f(ζk)△zk 6 Xn k=1 |f(ζk)△zk| 6 Xn k=1 |f(ζk)|△sk - δ = max 16k6n {△sk} → 0, Z C f(z)dz 6 Z C |f(z)|ds. ÏǑ3 C þ |f(z)| 6 M, Z C |f(z)|ds 6 M Z C ds = ML. 13/127
4)设曲线C的长度为L,函数∫(2)在C上满 足|f(x)≤M,那么 f(2)d≤/f()ds≤ML 证明设|△xk=|2k-2k-1|,且△Sk为这两点之间的 弧段的长度.显然|△k≤△Sk.所以
4) C ÝǑ L, ¼ê f(z) 3 C þ÷ v |f(z)| 6 M, o Z C f(z)dz 6 Z C |f(z)|ds 6 ML. y² |△zk| = |zk − zk−1|, △sk Ǒùü:m lãÝ. w, |△zk| 6 △sk. ¤± Xn k=1 f(ζk)△zk 6 Xn k=1 |f(ζk)△zk| 6 Xn k=1 |f(ζk)|△sk - δ = max 16k6n {△sk} → 0, Z C f(z)dz 6 Z C |f(z)|ds. ÏǑ3 C þ |f(z)| 6 M, Z C |f(z)|ds 6 M Z C ds = ML. 13/127
4)设曲线C的长度为L,函数∫(z)在C上满 足‖f(z)≤M,那么 f(2)d2|≤/f(2)lds≤ 证明设|△k=|2k-xk-1,且△Sk为这两点之间的 弧段的长度.显然|△2≤△Sk.所以 f()△≤∑(A≤∑|)△s k=1 令6=max{△s}0.得
4) C ÝǑ L, ¼ê f(z) 3 C þ÷ v |f(z)| 6 M, o Z C f(z)dz 6 Z C |f(z)|ds 6 ML. y² |△zk| = |zk − zk−1|, △sk Ǒùü:m lãÝ. w, |△zk| 6 △sk. ¤± Xn k=1 f(ζk)△zk 6 Xn k=1 |f(ζk)△zk| 6 Xn k=1 |f(ζk)|△sk - δ = max 16k6n {△sk} → 0, Z C f(z)dz 6 Z C |f(z)|ds. ÏǑ3 C þ |f(z)| 6 M, Z C |f(z)|ds 6 M Z C ds = ML. 13/127
4)设曲线C的长度为L,函数f(2)在C上满 足|f(2)≤M,那么 f(2)d2|≤/f(2)lds≤ML 证明设|△xk=|2k-2k-1|,且△Sk为这两点之间的 弧段的长度.显然|△2≤△sk.所以 f()△≤∑(A|≤∑)△s k=1 令6=max{△sk}→0,得 f()d2|≤/|f()ds 因为在C上f(2)≤M,得到 f(2)s≤M 口+2
4) C ÝǑ L, ¼ê f(z) 3 C þ÷ v |f(z)| 6 M, o Z C f(z)dz 6 Z C |f(z)|ds 6 ML. y² |△zk| = |zk − zk−1|, △sk Ǒùü:m lãÝ. w, |△zk| 6 △sk. ¤± Xn k=1 f(ζk)△zk 6 Xn k=1 |f(ζk)△zk| 6 Xn k=1 |f(ζk)|△sk - δ = max 16k6n {△sk} → 0, Z C f(z)dz 6 Z C |f(z)|ds. ÏǑ3 C þ |f(z)| 6 M, Z C |f(z)|ds 6 M Z C ds = ML. 13/127