极限函数及和函数的性质.doc

十九. 极限函数及和函数的性质交换极限号定理 fn (x) = (f (x) = )fn (x).条件: 在x0R*的某去心邻域内fn I f, nfn (x)存在. 证 (先用Cauchy准则证明右端的极限存在) 设x0R,fn (x) = an, 在去心邻域U内fn I f , 则e 0 $N m, nN xU : | fn (x) - fm (x)| 0 $N1 n N1: | an - a| N2 xU : | f (x) - fN (x) | 0当xU, 0 |x - x0 |d n时| fn (x) - an | N1 + N2 , d = dn , 则当xU, 0 |x - x0 |d 时| f (x) - a| f (x)- fn (x)| + | fn (x) - an | + | an - a| 3e . 故f (x) = a.叙述二. 由fn I f, $NN1 xU : | f (x) - fn (x)| 0当xU, 0 |x - x0 |d 时| fN (x) - aN |e . 故当xU, 0 |x - x0 |d 时| f (x) - a | f (x) - fN (x)| + | fN (x) - aN | + |aN -a|0 $N xa, b: | fN (x) - f (x) | e . 由fN 可积, 存在a, b的分割T = x0 = a, x1 , , xn = b, 使e (wk (f )表示f在Dk = xk-1, xk 上的振幅). (下证wk (f )wk (fN ) +2e .) 因为xa, b, fN (x) -e f (x) fN (x)+e, f (x)fN (x)+e,f(x)fN(x)-e ,wk (f)wk (fN )+2e, 2e (b-a) + 0, m , n充分大时 | fm (x0) - fn (x0)| e , xa, b: | fm (x) - fn (x)| e , 故m, n充分大时 xa, b $ x, x 0间的x , 使| fm (x) - fn (x)| ( fm (x) - fn (x) - ( fm (x0) - fn (x0)| + | fm (x0) - fn (x0)| = | fm (x ) - fn (x )| | x - x0| + | fm (x0) - fn (x0)| 0)上的一致收敛性, 与极限函数的连续、可积、可微性。解 （1) fn (x)f (x) =故在0,)上不一致收敛, f不连续, 不可积, 不可微. (2) fn (x)f (x) = 1 (xa 0), 显然, f 在a, )上连续, 不可积, 可微. | fn (x) - f (x)| =0, 故在a,)上一致收敛.例3(p.41例3). 证明n -3 ln (1 + n2 x2 )在0, 1上一致收敛, 并讨论其和函数的连续、可积、可微性.解 设 fn (x) = n -3 ln (1 + n 2 x2 ), 则 fn (x) =0, 即fn 是x的增函数, 故| fn (x)|fn (1) = n -3 ln (1 + n2 ) = o (n -2) , 由M法, fn 一致收敛. 设其和为f . 因为在0,1上每个 fn 连续, 可积, 故f 连续, 可积. 又, | fn (x)|n -2 , 由M法, fn 一致收敛. 因为 fn C1, 故 f C 1.例4. 证明: 1) 若在集D上 fn If , gn I g, 则 fn gn I f g .2) 若在D上fn = f 一致, gn = g一致, 则( fn gn) = f g一致.3) 若在D上fn If , gn I g, 且fn , gn 一致有界, 则fn gn I f g.证 1) |( fn gn) - (f g)| fn - f | + | gn - g| sup|( fn gn) - (f g)|sup| fn - f | + sup | gn - g|0.3) f 有界. | fn gn - fg| | gn | fn - f | + | f | gn - g|.注. 对3), 若引用p.42.2(1) ( fn I f 且f 有界 除有限项外 fn 一致有界), 条件可改为f , g有界.例5. 证明: 在D上| fn |一致收敛fn 一致收敛. 反之, fn 一致收敛 + | fn |收敛| fn |一致收敛.证 | fn |一致收敛 e | fn+1 (x)| + + | fn+p (x)| e | fn+1 (x) + + fn+p (x)| e fn 一致收敛. (也可用余项I0证明.) 反例见p.36.9: 设fn (x) = (-1) n x n (1 - x), D = 0, 1, 则 = 1 - x n+1故| fn | 收敛于不连续函数, 不可能一致收敛. 但fn 一致收敛: | Rn (x)|x n - x n-1| = x n - x n-1(x n - x n-1) 0. (也可用D法证明.)例6(p.42.10). 证明z (x) = n -x 在(1,)内连续且有连续的各阶导数(zC ).证 设a, b (1,), 则在a, b上| n - x |n - a , 由M法, n - x在a, b上一致收敛, 即在(1,)内闭一致收敛, 故z 在(1,)内连续. 又, 在a, b上| (n - x ) | = | - n - x ln n|n - a ln n. 由M法(n - x )