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Let $f$ be a continuous function. I wish to show $$\lim_{n \rightarrow \infty} \int_0^1 n x^n f(x) \; dx = f(1)$$ I can try to split up the integral over intervals $[0, 1-\delta]$, $[1-\delta, 1]$. The integral over $[0, 1-\delta]$ vanishes as $n \rightarrow \infty$, but I am having difficulty estimating the integral over $[1-\delta, 1]$. Thanks!

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Your idea is great. One can do the following. First, note that for any integrable $f$ $$\int_0^1 t^nf(t)dt\to 0$$ so we may consider $n+1$ instead of $n$ inside the integrand. This helps, because now $f(1)=\displaystyle\int_0^1(n+1)t^nf(1)dt$. Pick $1>\delta>0$, let $g(t)=f(t)-f(1)$ and note $g(t)\to 0$ as $t\to 1$, $g(1)=0$, and write as you say $$\begin{align}\left|\int_0^1 (n+1)t^n(f(t)-f(1))dt\right|&=\left|\int_0^{1-\delta} (n+1)t^n g(t)dt+\int_{1-\delta}^1 (n+1)t^n g(t)dt\right|\\ &\leqslant \left|\int_0^{1-\delta} (n+1)t^n g(t) dt\right|+\left|\int_{1-\delta}^1 (n+1)t^n g(t) dt\right|\end{align} $$

The first integrand goes to zero since $(n+1)t^n (f(t)-f(1))\to 0$ uniformly over that interval for any $\delta >0$. Given $\varepsilon>0$; can you make the second integral $<\varepsilon$ knowing that $f(t)-f(1)\to 0$ as $t\to 1$?

SPOILERS

Given $\varepsilon>0$; choose $\delta>0$ so that $|f(t)-f(1)|<\varepsilon$ if $t\in[1-\delta,1]$. Then $$\begin{align}\left|\int_{1-\delta}^1 (n+1)t^n(f(t)-f(1))dt\right|&\leqslant \int_{1-\delta}^1 (n+1)t^n|f(t)-f(1)|dt\\ &\leqslant \varepsilon\int_{1-\delta}^1 (n+1)t^n dt \\ &\leqslant \varepsilon\int_0^1 (n+1)t^n dt =\varepsilon\end{align}$$

Note no $n$ appeared! This is why $n+1$ is much more convenient that $n$, namely, because it normalizes the integral.

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  • $\begingroup$ @user8123123 It should read $f(1)$ in the integrand. Fixing. $\endgroup$ – Pedro Tamaroff Feb 16 '14 at 2:56
  • $\begingroup$ Ok. Using your hint, I think I have the idea for the second integral. For $\delta$ sufficiently small, the integral over $[1-\delta, 1]$ is approximately $f(1) \int_{1-\delta}^{1} n x^n = f(1)\frac{n}{n+1} x^{n+1} \Big|^1_{1-\delta} \rightarrow f(1)$ as $n \rightarrow \infty$. $\endgroup$ – user8123123 Feb 16 '14 at 3:06
  • $\begingroup$ @user8123123 You can make that more rigorous. See what I added. $\endgroup$ – Pedro Tamaroff Feb 16 '14 at 3:36
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Alternatively, let me give you an idea which works remarkably well for problems like this:

1) Prove the result for $f(x)$ a polynomial (easy)

2) Use the fact that polynomials are dense in $C[0,1]$ to deduce the general result.

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Let $r\in (0,1).$

Let $M=\sup_{x\in [0,1]}|f(x)|.$

For $r\in (0,1)$ let $U(r)=\sup_{x\in [r,1]}f(x)$ and $L(r)=\inf_{x\in [r,1]}f(x).$

For all $n\in \Bbb N$ we have $$(I). \quad\frac {-n}{n+1}r^{n+1}M\le \int_0^rnx^nf(x)fdx\le \frac {n}{n+1}r^{n+1}M.$$ $$(II). \quad \frac {n}{n+1}(1-r^{n+1})L(r)\le \int_r^1nx^nf(x)dx\le \frac {n}{n+1}(1-r^{n+1})U(r).$$

For brevity let $B(r,m)= r^{m+1}M+(1-r^{m+1})U(r).$

Then $\lim_{m\to \infty}B(r,m)=U(r).$ So $$\lim_{m\to \infty}\sup_{n\ge m}B(r,m)=U(r).$$

For brevity let $I(n)=\int_0^1nx^nf(x)dx.$

By $(I)$ and $(II) $ we have $$\lim_{m\to \infty}\sup_{n\ge m}\frac {n+1}{n}I(n)\le$$ $$\le \lim_{m\to \infty}\sup_{n\ge m}r^{n+1}M+(1-r^{n+1})U(r)=$$ $$=\lim_{m\to \infty}\sup_{n\ge m}B(r,m)=$$ $$=\lim_{m\to \infty}B(r,m)=U(r).$$

This holds for every $r\in (0,1),$ so $$(III).\quad \lim_{m\to \infty}\sup_{n\ge m}\frac {n}{n+1}I(n)\le \lim_{r'\to 1^-} \sup_{r\in (r',1)}U(r)=f(1).$$

By similar techniques we obtain $$(IV).\quad \lim_{m\to \infty}\inf_{n\ge m}\frac {n}{n+1}I(n)\ge \lim_{r'\to 1^-}\inf_{r\in (r',1)}L(r)= f(1).$$

By $(III)$ and $(IV)$ we have $\lim_{n\to \infty}\frac {n}{n+1}I(n)=f(1).$ So $\lim_{n\to \infty} I(n)=f(1).$

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