max n³1 | xn - xn(k) | = || xk - x || < e.Р| å n ³1 xn/2n | = | å n ³1 xn/2n - å n ³1 xn(k)/2n | = | å n ³1 (xn - xn(k))/2n | Р£ å n ³1 |xn - xn(k)|/2n £ å n ³1 e/2n = e.Р所以,å n ³1 xn/2n = 0,即x = (x1, x2, ..., xn, ...)ÎM.Р所以M是C0的闭线性子空间.Р(2) x0 = (2, 0, 0, ...),"z = (x1, x2, ..., xn, ...)ÎM,Р|| x0 – z || = max{| 2 - x1 |, | x2 |, | x3 |, ... }.Р如果| 2 - x1 | > 1,则|| x0 – z || > 1.Р如果| 2 - x1 | £ 1,则| x1 | ³ 1,我们断言{| x2 |, | x3 |, ... }中至少有一个大于1者.Р否则,假若它们都不超1,因为xn ® 0 (n ® ¥),故它们不能全为1.Р由å n ³1 xn/2n = 0知| x1 |/2 = | å n ³2 xn/2n | £ å n ³2 | xn | /2n < å n ³2 1/2n = 1/2,Р这样得到| x1 | < 1,矛盾.Р故{| x2 |, | x3 |, ... }中至少有一个大于1者.Р因此也有|| x0 – z || > 1.Р综上所述,但"yÎ M,有|| x0 – y || > 1.Р由此,立即知道inf z Î M || x0 – z || ³ 1.Р下面证明inf z Î M || x0 – z || £ 1.Р"nÎN+,令zn = (1 - 1/2n, -1, -1, ..., -1, 0, 0, ...).
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