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Assume $f(x)$ is continuous on $[0,1]$ , and $f(1)=0$. Prove $$\lim_{n\to \infty} n\int_0^1x^nf(x)\,\text dx=0 $$

I already know that $\lim_{n\to \infty}\int_0^1x^nf(x) \, \text dx=0$. Is this helpful in the question above?

Michael Hardy
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LOIS
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  • Where did you use f(1)=0 in the limit you already known? – Cloud JR K Aug 23 '18 at 17:06
  • Actually we don't need $f(1)=0$ to proof $\lim_{n\to \infty}\int_0^1x^nf(x)\text dx=0$ – LOIS Aug 23 '18 at 17:13
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    Sorry , This already have a answer here . https://math.stackexchange.com/questions/455193/for-f-continuous-show-lim-n-to-infty-n-int-01-fxxn-dx-f1 – LOIS Aug 23 '18 at 17:15
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    You may also use Weierstrass approximation theorem in the following form: for any $\varepsilon>0$, there is a polynomial $p(x)=\sum_{k=1}^{\deg p}a_k (1-x)^k$ such that $\left|p(x)-f(x)\right|\leq\varepsilon$ for any $x\in[0,1]$. The behaviour of $\int_{0}^{1}nx^n(1-x)^k,dx$ is fairly simple to study. – Jack D'Aurizio Aug 23 '18 at 17:30

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Fix some $\epsilon>0$. Then there is a $\delta>0$ (smaller one) so that on the interval $[1-\delta,1]$ we have $|f|<\epsilon$. Now we can easily estimate: $$ \begin{aligned} 0 &\le \left|n\int_0^1 x^n\; f(x)\; dx\right| \\ &\le \int_0^1 (n+1)x^n\; |f(x)|\; dx \\ &= \int_0^{1-\delta}(n+1)x^n\; |f(x)|\; dx + \int_{1-\delta}^1(n+1)x^n\; |f(x)|\; dx \\ &\le \int_0^{1-\delta}(n+1)x^n\; \|f\|\; dx + \int_{1-\delta}^1(n+1)x^n\; \epsilon\; dx \\ & \le (1-\delta)^{n+1} \|f\| + \epsilon \ . \end{aligned} $$ We pass to the limit (superior) w.r.t. $n$ now in the obtained inequality, getting $$ \limsup_n\left|n\int_0^1 x^n\; f(x)\; dx\right|\le \epsilon\ . $$ Now we let $\epsilon$ go to zero. So the limit exists, and is zero.

dan_fulea
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