TITLE: For $f$ Riemann integrable prove $\lim_{n\to\infty} \int_0^1x^nf(x)dx=0.$
QUESTION [5 upvotes]: 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?

REPLY [6 votes]: 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.

REPLY [5 votes]: 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.

REPLY [2 votes]: 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.