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Suppose $f$ is a Riemann integrable function on $[0,1]$. Prove that $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$

This is what I am thinking: Fix $n$. Then by Jensen's Inequality we have $$0\leq\left(\int_0^1x^nf(x)dx\right)^2 \leq \left(\int_0^1x^{2n}dx\right)\left(\int_0^1f^2(x)dx\right)=\left(\frac{1}{2n+1}\right)\left(\int_0^1f^2(x)dx\right).$$Thus, if $n\to\infty$ then $$0\leq \lim_{n\to \infty}\left(\int_0^1x^nf(x)dx\right)^2 \leq 0$$ and hence we get what we want. How correct (or incorrect) is this?

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We didn't say that $f^2$ was integrable but that $f$ was. This is not very useful. And I don't think this is Jensen's inequality ; it looks more like Cauchy-Schwarz to me. –  Patrick Da Silva Feb 21 '12 at 4:04
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Just because $f$ is Riemann integral over $[0,1]$ doesn't mean that $f^2$ is. Consider $f(x)=x^{-1/2}$. (And if needed, define $f$ at $0$.) –  alex.jordan Feb 21 '12 at 4:10
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@alex Riemann integrable functions are bounded. Your function is "improperly integrable" over $[0,1]$. –  David Mitra Feb 21 '12 at 4:13
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@alex.jordan Your example is only improperly Riemann integrable. –  Ragib Zaman Feb 21 '12 at 4:14
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@alex Yes, it does. Consider your function with $[0,\epsilon]$ as the first subinterval in a partition. Select the "test point" to be $1/ n^2$. The contribution to the Riemann sum from this subinterval is $n\epsilon$, which can be made as large as you wish no matter how small $\epsilon$ is. –  David Mitra Feb 21 '12 at 4:20
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That looks great. If someone doesn't know Jensen's inequality, this is still seen just with Cauchy-Schwarz. Another quick method is the dominated convergence theorem. Gerry's and Peters answers are both far simpler though.

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Sorry but I do not understand what you mean by Jensen's inequality (...) is (...) seen just with Cauchy-Schwarz. –  Did Feb 21 '12 at 7:07
    
@DidierPiau I meant that if someone does not know Jensen's inequality, then the manipulations Galois did in his question post can just be seen as an application of Cauchy-Schwarz rather than one of Jensen's inequality. –  Ragib Zaman Feb 21 '12 at 7:15
    
Right. $ $ $ $ $ $ –  Did Feb 21 '12 at 12:13
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If $f$ is Riemann integrable on $I$ then it is bounded in $I$, i.e. $m \leq f \leq M$. Thus one has

$$\eqalign{ & m\int\limits_0^1 {{x^n}dx} < \int\limits_0^1 {{x^n}f\left( x \right)dx} < M\int\limits_0^1 {{x^n}dx} \cr & \frac{m}{{n + 1}} < \int\limits_0^1 {{x^n}f\left( x \right)dx} < \frac{M}{{n + 1}} \cr} $$

This, with the squeeze theorem proves the assertion.

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Given $\epsilon$, choose $\delta$ and $n$ such that $x^n$ is small for $0\le x\le1-\delta$ and $\int_{1-\delta}^1f$ is small.

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Just so people can agree : Wikipedia states that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere (just type Riemann integral on wiki). Since the function "squaring" is continuous and that composition of continuous function at a point preserves continuity, $f^2$ is continuous almost everywhere as well, and an obvious bound for $f^2$ is the bound for $f$, squared. The rest is taken care of by $OP$'s proof.

Hope that helps,

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What I think is more helpful than the fact that someone wrote it on Wikipedia is that those statements are theorems that can be proven from the usual definition of the Riemann integral on a bounded interval. David Mitra's comment on the question makes clear why boundedness is a requirement. –  Jonas Meyer Feb 21 '12 at 4:43
    
I precised "Just so people can agree : Wikipedia states...", because indeed this can be proven. I just wanted to end discussion about whether OP's proof was legitimate or not. But yes, boundedness is indeed a requirement, otherwise alex.jordan had found a counter-example. –  Patrick Da Silva Feb 21 '12 at 5:59
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