※ 引述《wyob (Go Dolphins)》之铭言： : 1.Let f be a continuous real-valued funtion defind on [a,b],and let : b : M= max│f(x)│.Show that lim (∫│f(x)│^n dx)^(1/n)=M : x属於[a.b] n→∞ a : 2.Let f:(-1,2)→R be a real analytic function, : If f(1/k)=0 for all k 属於N,show that f is identically zero : 第二题我自己解的怪怪的，因为好像没有用到f定义在(-1,2)上 : 第一题该如何下手呢? 1.(1) If M = 0 => f(x) = 0 for all x b => ∫ |f(x)|^n dx = 0 for all n a b => (∫ |f(x)|^n dx)^(1/n) = 0 for all n a b => lim (∫ |f(x)|^n dx)^(1/n) = 0 = M n→∞ a then we are done (2) If M > 0 ∵ f is continuous on [a,b] ∴ M exists and |f| is continuous on [a,b] Let c belonging to [a,b] such that f(c) = M => |f(c)| = M By definition of continuity of |f| at c , Given 0 < ε < M/2 , there exists δ > 0 such that |x - c| < δ => ||f(x)| - |f(c)|| < ε => c - δ < x < c + δ => M - ε < |f(x)| < M + ε c+δ c+δ b b ∴ ∫ (M-ε)^n dx ≦ ∫ |f(x)|^n dx ≦ ∫|f(x)|^n dx ≦ ∫ M^n dx c-δ c-δ a a n b n n => (M-ε) (2δ) ≦ ∫ |f(x)| dx ≦ M (b-a) a 1/n b n 1/n 1/n => (M-ε)(2δ) ≦ (∫|f(x)| dx) ≦ M(b-a) a 1/n b n 1/n 1/n ∵ limsup(M-ε)(2δ) ≦ limsup(∫|f(x)| dx) ≦ limsupM(b-a) n→∞ n→∞ a n→∞ b n 1/n b n 1/n => M-ε ≦ limsup(∫|f(x)| dx) ≦ M => limsup(∫|f(x)| dx) = M n→∞ a n→∞ a 1/n b n 1/n 1/n and liminf(M-ε)(2δ) ≦ liminf(∫|f(x)| dx) ≦ liminfM(b-a) n→∞ n→∞ a n→∞ b n 1/n b n 1/n => M-ε ≦ liminf(∫|f(x)| dx) ≦ M => liminf(∫ |f(x)| dx) = M n→∞ a n→∞ a b n 1/n ∴ lim (∫ |f(x)| dx) = M n→∞ a -- 本周抽中:安 心 亚 本周最心碎:吴 怡 霈 本周最亮眼:王 薇 欣 动园木万社万医辛 麟六犁科大大忠复南东中国松机大剑路西港文内大公葫东南软园南展 物 栅芳区芳院亥 光张 技楼安孝兴京路山中山场直南 湖墘德湖湖园洲湖港体区港览 ○ ○○ ○ ○ ○◎ ○ ◎◎ ◎ ○ ○ ○◎ ○○○○○ ○◎○ ◎馆 王桦邵艾丝小桦张甯莎王欣李慧啾豆妹安亚吴霈廖娴小徐翊舒虎瑶可蜜儿蔓小刘萍 林玲 彩 庭莉 欣 钧 拉薇 怡 啾花 心 怡 书 娴裴 舒牙瑶乐雪 蔓蔓秀 志 --

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