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I need to prove a limit on a closed interval $[0,1]$.

$f$ is integrable.

$$\lim_{n\to \infty}\int_0^1 x^n f(x) \, dx=0$$

Where should I start from?

Michael Hardy
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loradd
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3 Answers3

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If by "integrable" you mean Riemann integrable, then an integrable function must be bounded, and so $$\left|\int\limits_{0}^{1}{x^nf(x)\text{ d}x}\right|\le\left(\max\limits_{x\in[0,1]}{|f(x)|}\right)\int\limits_{0}^{1}{x^n\text{ d}x}. $$ Note that $\int\limits_{0}^{1}{x^n\text{ d}x} = \frac{1}{n+1}$.

If by "integrable" you mean Lebesgue integrable, then the functions $f_n(x) = x^nf(x)$ converges pointwise to zero almost everywhere, and $|f_n(x)|\le|f(x)|$ everywhere, so we can apply the Dominated Convergence Theorem.

Joey Zou
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I assume you mean $f\in L^1(0,1)$. Note $C[0,1]$ is dense in $L^1(0,1)$ under $L^1$ norm, so it suffices to prove it with $f\in C[0,1]$ and then pass to limit (Please see the comments for details).

Given $f\in C[0,1]$, by uniformly continuous, for any $\epsilon>0$, you can find a step function $g$, $0\le f(x)-g(x)\le \epsilon$ for all $x\in[0,1]$, so it suffices to prove it for step functions and then pass to limit.

Then by linearity of integral, it suffices to prove it for $f=1_{[a,b]}$, which is easy by simple calculation.

John
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  • More explicitly: you are approximating the given function $f$ with a sequence $f_m$ such that $f_m\to f$ in some sense. You have established that for any fixed $m$ you have $\lim_{n\to \infty} \int x^n f_m(x), dx=0$. Now you have to let $m$ go to infinity. How can you justify that $\lim_{n\to \infty} \lim_{m\to \infty} \int x^nf_m(x), dx=0$? – Giuseppe Negro Dec 18 '15 at 10:57
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    @GiuseppeNegro Given $\epsilon>0$, we find $g$ such that $|g-f|_{L^1}<\frac{\epsilon}{2}$ . Then we prove $\exists N$ such that $\forall n\ge N$, we have $|\int x^n g|<\frac{\epsilon}{2}$. Hence $\forall n\ge N$, $|\int x^n f|\le |\int x^n f-\int x^n g|+|\int x^n g|\le \int |f-g|+|\int x^n g|<\epsilon$, which gives the desired convergence. – John Dec 18 '15 at 11:17
  • @GiuseppeNegro Thank you. – John Dec 18 '15 at 11:41
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Since $f$ is integrable, it is finite almost everywhere. Then $x^n\,f(x)$ converges to $0$ almost everywhere in$[0,1]$. Moreover $|x^n\,f(x)|\le|f(x)|$ for all $x\in[0,1]$. The result follows now from the Dominated Convergence Theorem.

Julián Aguirre
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