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:ƒm 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