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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.2012-02-21
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    @PatrickDaSilva But the product of integrable functions on an interval yields an integrable function. I agree this is more usually called Cauchy-Schwarz or Holder inequality. But, for instance, in papa Rudin both Holder and Minkowski's inequalities are derived from Jensen's inequality.2012-02-21
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    @PatrickDaSilva My guess is that Galois is viewing it from a very advanced standpoint. Jensen's inequality does imply Cauchy-Schwarz after all.2012-02-21
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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 $0$.)2012-02-21
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    @dls : My memory is a little tired, as always. But it still looks like Cauchy-Schwarz to me ; $\langle x^n, f \rangle^2 \le \langle x^n,x^n \rangle \langle f,f \rangle$ with the scalar product meaning integrating from $0$ to $1$ the product of both functions. If $f^2$ is integrable then I agree with this answer. But my memory serves me wrong at the moment so I can't confirm.2012-02-21
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    @alex Riemann integrable functions are bounded. Your function is "improperly integrable" over $[0,1]$.2012-02-21
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    @alex.jordan Your example is only improperly Riemann integrable.2012-02-21
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    @David I consider "Riemann integrable" to mean that the limit of Riemann sums as the largest $\Delta x$ approaches $0$ will exist. This does not require $f$ to be bounded.2012-02-21
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    @alex.jordan : OP is in a more general context, where Riemann integration is always done "properly". Your consideration is not the one we have here, although it is correct in other contexts.2012-02-21
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    @ David,Ragib,Patrick If $f$ is supposed to be bounded, then this question is a whole lot easier to answer than the OP's argument. Just bound $|f|$ by $M$ and pull $M$ to the outside of the integral. That's what makes me suspect something else is going on here.2012-02-21
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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.2012-02-21
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    @David OK you won me over. It still seems like a much simpler argument is apparent if $f$ is bounded.2012-02-21

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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.2012-02-21
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    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.2012-02-21
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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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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*.2012-02-21
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    @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.2012-02-21
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    Right. $ $ $ $ $ $2012-02-21