Suppose $g:[0,1]\to\mathbb R$ such that $g\in C^1$ (i.e., $g'$ exists and is continuous). I want to show $$\lim_{n\to+\infty }\int_0^1x^ndg(x)=0.$$ Thanks.
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This is trivial by Lebesgue dominated convergence theorem.
And pretty straightforward as follows. Let $M$ be the max of $|g'|$ on $[0,1]$. Then $$ |\int_0^1x^ng'(x)dx|\leq M\int_0^1 x^ndx=\frac{M}{n+1}. $$ Conclude by the squeeze theorem.
Julien
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:thanks your answer is nice too – M.H Feb 05 '13 at 19:39
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@julien: +1 for your fast way. – user 1591719 Feb 05 '13 at 19:43
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@julien:why g' is bounded ? – M.H Feb 05 '13 at 19:50
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1@MaisamHedyelloo You said that $g'$ was continuous, and $[0,1]$ is compact. – Julien Feb 05 '13 at 19:53
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@julien:thanks i understand it . – M.H Feb 05 '13 at 19:54
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We use this that yields $$\lim_{n\to+\infty }\left(\frac{1}{n}\times n\int_0^1x^ndg(x)\right)=\lim_{n\to+\infty } \frac{g'(1)}{n}=0$$
user 1591719
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Nice, +1. I think the OP means $dg(x)=g'(x)dx$, so your numerator should simply be $g'(1)$. – Julien Feb 05 '13 at 19:32
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@MaisamHedyelloo: there was $"x"$ instead of $"n"$ in your post. Now it's corrected. – user 1591719 Feb 05 '13 at 19:45
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@chriss sister:i check it my question is true .but i think your answer is true according to refered question – M.H Feb 05 '13 at 19:48
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