※ 引述《wyob (Go Dolphins)》之銘言： : 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屬於[a.b] n→∞ a : 2.Let f:(-1,2)→R be a real analytic function, : If f(1/k)=0 for all k 屬於N,show that f is identically zero : 第二題我自己解的怪怪的，因為好像沒有用到f定義在(-1,2)上 : 第一題該如何下手呢? 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 -- 本週抽中:安 心 亞 本週最心碎:吳 怡 霈 本週最亮眼:王 薇 欣 動園木萬社萬醫辛 麟六犁科大大忠復南東中國松機大劍路西港文內大公葫東南軟園南展 物 柵芳區芳院亥 光張 技樓安孝興京路山中山場直南 湖墘德湖湖園洲湖港體區港覽 ○ ○○ ○ ○ ○◎ ○ ◎◎ ◎ ○ ○ ○◎ ○○○○○ ○◎○ ◎館 王樺邵艾絲小樺張甯莎王欣李慧啾豆妹安亞吳霈廖嫻小徐翊舒虎瑤可蜜兒蔓小劉萍 林玲 彩 庭莉 欣 鈞 拉薇 怡 啾花 心 怡 書 嫻裴 舒牙瑤樂雪 蔓蔓秀 志 --

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