For $f$ continuous, show $\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).$

Suppose $f:[0,1]\to \mathbb{R}$ is continuous. Show that

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

My answer so far: First I want to assume that $f\in C^1$. Then

$$n\int_0^1f(x)x^n\,dx = \left[\frac{n}{n+1}x^{n+1}f(x)\right]_0^1 - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\\ \frac{n}{n+1}f(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx,$$

which goes to $f(1)$ because the last integral goes to zero.

But approximating $f$ by $\phi\in C^1$ won't necessarily work, because $\phi(1)$ may not equal $f(1)$... how can we finish the argument?

• I don't think you could argue by proving it in the case when $f$ is continuously differentiable and then somehow using that to prove it when $f$ is just continuous - you'd have to argue via other methods. Jul 29, 2013 at 22:06
• Hint: $f(1)=(n+1)\int_{0}^{1} f(1)x^n dx$. Jul 29, 2013 at 22:09
• For a different approach, see the first answer here. Jul 29, 2013 at 22:09
• You can simplify your part of the argument: Only polynomials, rather than arbitrary $C^1$ functions, need to be considered. This case is trivial, as we only need to argue for monomials $x^k$, and everything is explicit here. The uniform convergence argument as in the answer below then goes unchanged. Jul 29, 2013 at 22:25
• math.stackexchange.com/q/128823/321264 Sep 6, 2020 at 18:26

Here is a more elementary method than you proposed:

First, note that if $f$ is continuous on $[0,1]$, then it is necessarily bounded on $[0,1]$; say $\lvert f(x)\rvert\leq M$ for all $x\in[0,1]$. If we define $\delta_n:=\frac{1}{\sqrt{n}}$, then $$\left\lvert n\int_0^{1-\delta_n}f(x)x^n\,dx\right\rvert\leq Mn\int_0^{1-\delta_n}x^n\,dx=\frac{n}{n+1}\left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\rightarrow0\text{ as }n\rightarrow\infty.$$ Now, let $\epsilon>0$ be given. Continuity of $f$ at $1$ implies that there exists $\delta>0$ such that $\lvert 1-x\rvert<\delta$ implies $\lvert f(x)-f(1)\rvert<\epsilon$. Choose $N\in\mathbb{N}$ such that $0<\delta_n<\delta$ for all $n\geq N$. Then for $n\geq N$, $$n\int_{1-\delta_n}^1(f(1)-\epsilon)x^n\,dx\leq n\int_{1-\delta_n}^1 f(x)x^n\,dx\leq n\int_{1-\delta_n}^{1}(f(1)+\epsilon)x^n\,dx.$$ Computing the left integral $$\frac{n}{n+1}\left(1-\left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right)\left(f(1)-\epsilon\right)\rightarrow f(1)-\epsilon\text{ as }n\rightarrow\infty;$$ the right integral yields the same, except with $f(1)+\epsilon$. Then $$f(1)-\epsilon\leq\liminf_{n\rightarrow\infty}\ n\int_0^1f(x)x^n\,dx\leq\limsup_{n\rightarrow\infty}\ n\int_0^1 f(x)x^n\,dx\leq f(1)+\epsilon.$$ But, this holds for any $\epsilon>0$; so, letting $\epsilon\rightarrow0$, we get the desired result.

• This is a great answer, Nicholas. Thanks for your contribution! (I would upvote it except that I've exhausted my daily upvote quota (sorry!). I'll return here in a couple of hours and upvote!) Jul 29, 2013 at 22:36

We can finish the argument as follows. (Note: We'll assume that the limit in question exists for $f$ and establish that it's equal to $f(1)$. Technically, we should prove that this limit exists as Peter Tamaroff notes below (thanks!). A minor modification of the following argument simultaneously establishes the existence of the limit and its value but we'll leave that as an exercise to the reader.) Let $\epsilon>0$. Choose $\phi\in C^1$ such that $\left|f(x)-\phi(x)\right|<\epsilon$ for all $x\in [0,1]$. You've proven that $$\lim_{n\to\infty} n\int_{0}^{1} \phi(x)x^n=\phi(1).$$ Therefore,

\begin{align}\left|\lim_{n\to\infty} n\int_{0}^{1} f(x)x^n dx -\lim_{n\to\infty} n\int_{0}^{1} \phi(x)x^n dx\right|&=\left|\lim_{n\to\infty} n\int_{0}^{1} (f(x)-\phi(x))x^n dx\right|\\ &\leq \lim_{n\to\infty} n\int_{0}^{1} \left|(f(x)-\phi(x))x^n\right| dx\\ &< \lim_{n\to\infty} n\int_{0}^{1} \epsilon x^n dx\\ &=\lim_{n\to\infty} \epsilon \frac{n}{n+1}\\ &=\epsilon\end{align}

Therefore,

\begin{align}\left|\lim_{n\to\infty} n\int_{0}^{1} f(x)x^n dx - f(1)\right|\leq \left|\lim_{n\to\infty} n\int_{0}^{1} f(x)x^n dx - \phi(1)\right| + \left|\phi(1)-f(1)\right|&<\epsilon + \epsilon\\&=2\epsilon\end{align}

Since $\epsilon>0$ was arbitrary, we conclude that $$\lim\limits_{n\to\infty} n\int_{0}^{1} f(x)x^n dx=f(1)$$ for all continuous functions $f:[0,1]\to \mathbb{R}$.

• Why'd there exist such $\varphi \in C^1$? Jul 29, 2013 at 22:22
• @Kunnysan I'm not sure which tools you'd like to use but this follows, e.g., from the Stone-Weierstrass theorem. Jul 29, 2013 at 22:27
• Yeah of course. You could have chosen even a polynomial. Jul 29, 2013 at 22:32
• I was writing exactly same solution. Discarded it seeing yours, so +1 instead, :) Jul 29, 2013 at 22:36
• Thanks @Kunnysan and I'm really sorry that you had to discard your answer! (The same thing happens to me sometimes. In a few years when there are more users on this website, it will hardly be possible to read the question, I suspect, before someone posts an answer!) Jul 29, 2013 at 22:39

First, note that $$\int_0^1 x^n f(x)dx\to 0$$

since $f$ is bounded, so we can prove that $$(n+1)\int_0^1 x^n f(x)dx\to f(1)$$

But note $$\left( {n + 1} \right)\int_0^1 {x^n}f (1)dx = f(1).$$ so it suffices to consider the case $f(1)=0$.

THM Suppose that $f:[0,1]\to \Bbb R$ is continuous and $f(1)=0$. Then $$\mathop {\lim }\limits_{n \to \infty } \left( {n + 1} \right)\int_0^1 f (x){x^n}dx = 0$$

P Let $\epsilon >0$ be given. By continuity, there exists a neighborhood $[1-\delta,1]$ such that $$|f(x)|<\frac\varepsilon2$$ whenever $x\in[1-\delta,1]$. Write $$\left( {n + 1} \right)\left| {\int_0^1 f (x){x^n} dx} \right| \leqslant \left( {n + 1} \right)\left| {\int_0^{1 - \delta } f (x){x^n} dx} \right| + \left( {n + 1} \right)\left| {\int_{1 - \delta }^1 f (x){x^n} dx} \right|$$ so that $$\left( {n + 1} \right)\left| {\int_{1 - \delta }^1 {f\left( x \right){x^n} dx} } \right| \leqslant \left( {n + 1} \right)\frac{\varepsilon }{2}\int_{1 - \delta }^1 {{x^n} dx} \leqslant \left( {n + 1} \right)\frac{\varepsilon }{2}\int_0^1 {{x^n} dx} = \frac{\varepsilon }{2}$$

On the other hand, $|f|$ attains a maximum on $[0,1-\delta]$ and we have $$\left( {n + 1} \right)\left| {\int_0^{1 - \delta } {f\left( x \right){x^n}{\mkern 1mu} dx} } \right| \leqslant \left( {n + 1} \right)\int_0^{1 - \delta } {\left| {f\left( x \right)} \right|{x^n}{\mkern 1mu} dx} \leqslant M\left( {n + 1} \right)\int_0^{1 - \delta } {{x^n}{\mkern 1mu} dx} \leqslant M{\left( {1 - \delta } \right)^{n + 1}}$$

Since $1-\delta <1$, this goes to $0$; so the claim follows. Note we could have also used that $(n+1)x^n$ converges to zero uniformly on $[0,1-\delta]$ for any $0<\delta <1$ $\blacktriangle$

OBS Note how the proof works: $x^n$ crunches everything away from $1$, and continuity of $f$ plus $f(1)=0$ crunches everything near $1$.

• Thanks @Peter for this excellent contribution! (In general, your fantastic contributions to this website have had an enormous positive impact! So, thanks for all of your contributions!) It's great to see that so many different approaches to the question are being posted. (I would certainly upvote this except that I've exhausted my daily quota of upvotes. I'm sorry but I'll definitely return here in a couple of hours to upvote!) Jul 29, 2013 at 22:44
• @AmiteshDatta Thank you for those kind words! =)
– Pedro
Jul 29, 2013 at 22:45
• I did it :-) ${}$
– leo
Jul 29, 2013 at 22:46
• @leo Ah? ${}{}{}{}{}$
– Pedro
Jul 29, 2013 at 22:46
• I did upvote this
– leo
Jul 29, 2013 at 23:07

First, note that $$(n+1)\color{#C00000}{\int_0^ax^n\,\mathrm{d}x}=a^{n+1}\tag{1}$$ and $$(n+1)\color{#00A000}{\int_0^1x^n\,\mathrm{d}x}=1\tag{2}$$ Pick an $\epsilon>0$. Since $f$ is continuous, there is a $\delta>0$ so that for all $x\in[1-\delta,1]$, we have $|f(x)-f(1)|\le\epsilon$. Since $f$ is continuous on $[0,1]$, there is an $M$ so that $|f(x)|\le M$ for $x\in[0,1]$. Furthermore, there is an $N$ so that for $n\ge N$, we have $2M(1-\delta)^{n+1}\le\epsilon$.

Thus, for $n\ge N$ \begin{align} &\left|f(1)-(n+1)\int_0^1x^nf(x)\,\mathrm{d}x\right|\\ &=(n+1)\left|\int_0^1x^n(f(1)-f(x))\,\mathrm{d}x\right|\\ &=(n+1)\left|\color{#C00000}{\int_0^{1-\delta}x^n(f(1)-f(x))\,\mathrm{d}x} +\color{#00A000}{\int_{1-\delta}^1x^n(f(1)-f(x))\,\mathrm{d}x}\right|\\ &\le\color{#C00000}{2M(1-\delta)^{n+1}}+\color{#00A000}{\epsilon}\\ &\le2\epsilon\tag{3} \end{align} Thus, $$\lim_{n\to\infty}(n+1)\int_0^1x^nf(x)\,\mathrm{d}x=f(1)\tag{4}$$ Since $\lim\limits_{n\to\infty}\dfrac n{n+1}=1$, we get $$\lim_{n\to\infty}n\int_0^1x^nf(x)\,\mathrm{d}x=f(1)\tag{5}$$

• Nice, but it can be written a bit easier: $$\lim\limits_{n\to+\infty}n\int_0^1x^nf(x)dx =\lim\limits_{\delta\to 0^{+}}\lim\limits_{n\to+\infty}n\int_0^{1-\delta}x^nf(x)dx +\lim\limits_{\delta\to 0^{+}}\lim\limits_{n\to+\infty}n\int_{1-\delta}^1x^nf(x)dx$$ For any $\delta\in (0,1] \lim\limits_{n\to+\infty}n\int_0^{1-\delta}x^nf(x)dx=0$ while $$\inf\{f(x):x\in[1-\delta,1]\}\leqslant\lim\limits_{n\to+\infty}n\int_{1-\delta}^1x^nf(x)dx\leqslant\sup\{f(x):x\in[1-\delta,1]\}$$ so, when $\delta\to 0^+$ then $$\lim\limits_{\delta\to 0^+}\lim\limits_{n\to+\infty}n\int_{1-\delta}^1x^nf(x)dx=f(1)$$ Jun 23, 2014 at 6:53

By changing the variable, let $x=t^{\frac{1}{n}}$ and we have $$n\int_0^1 x^n f(x)dx=\int_0^1 f\left(t^{\frac{1}{n}}\right)t^{\frac{1}{n}}dt,$$

and by dominated convergence theorem we conclude: $$\lim_n n\int_0^1 x^n f(x)dx=f(1).$$

• $dx= \frac 1n t^{\frac 1n -1} dt$. Dec 4, 2022 at 11:02

Thought I'd write out the polynomial method @Kunnysan mentioned in the comments.

Consider an arbitrary polynomial $$p(x) = a_0 + a_1x + \dots + a_kx^k$$. We can calculate \begin{align} \lim_{n \rightarrow \infty} n \int_0^1 p(x)x^n \, dx &= \lim_{n \rightarrow \infty} n \int_0^1 a_0x^n + a_1x^{n+1} + \dots + a_kx^{n+k} \, dx \\ &= \lim_{n \rightarrow \infty} \left( \frac{n}{n+1} a_0 + \frac{n}{n+2} a_1 + \dots + \frac{n}{n+k+1} a_k \right) \\ &= a_0 + a_1 + \dots + a_k \\ &= p(1) \end{align}

By the Weierstrass approximation theorem there exists a sequence of polynomials $$\{p_m\}$$ such that $$p_m(x) \rightarrow f(x)$$ uniformly.

We then write \begin{align} \lim_{n \rightarrow \infty} n\int_0^1 f(x)x^n \, dx &= \lim_{n \rightarrow \infty} n \int_0^1 \lim_{m \rightarrow \infty} p_m(x)) x^n \, dx \\ &= \lim_{n\rightarrow n} n \int_0^1 \lim_{m \rightarrow \infty} p_m(x)x^n \, dx \\ &= \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \int_0^1 p_m(x)x^n \, dx \end{align} where the interchanging of the limit and the integral is valid because the sequence $$p_m(x)x^n$$ converges uniformly to $$f(x)x^n$$.

Since $$p_m(x)$$ is a polynomial, we can use our preliminary work to write $$\lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \int_0^1 p_m(x)x^n \, dx = \lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \left( \frac{n}{n+1} a_0 + \frac{n}{n+2} a_1 + \dots + \frac{n}{n+k+1} a_k \right)$$ where $$a_i$$ are the coefficients of the polynomial $$p_m(x)$$.

Lastly we interchange limits again, first taking the limit as $$n \rightarrow \infty$$:

$$\lim_{n \rightarrow \infty} \lim_{m \rightarrow \infty} \left( \frac{n}{n+1} a_0 + \frac{n}{n+2} a_1 + \dots + \frac{n}{n+k+1} a_k \right) = \lim_{m \rightarrow \infty} p_m(1) = f(1)$$

• The last term should be $\frac{n}{n+k+1}a_k.$ Don't forget the $+1$ in the denominator. Jun 16, 2021 at 17:30
• The last switching of limits is not justified. Since the values $a_i$ and even $k$ depend on $m,$ it's not clear you can do this switch. Jun 16, 2021 at 17:36

Since $f$ is continuous, it is bounded on the compact interval $[0,1]$, say $|f(x)|<M$ for all $x\in[0,1]$. Also, for any $\epsilon>0$, we find delta such that $|f(x)-f(1)|<\epsilon$ for all $x>1-\delta$. Then $$\int_0^1 x^nf(x)\,dx = \int_0^{1-\delta} x^n f(x)\,dx+\int_{1-\delta}^1 x^n f(1)\,dx+\int_{1-\delta}^1 x^n (f(x)-f(1))\,dx$$ The first summand can be estimated by $$\left|\int_0^{1-\delta} x^n f(x)\,dx\right|\le \int_0^{1-\delta}\left| x^n f(x)\right|\,dx\le M\int_0^{1-\delta}x^n\,dx=\frac1{n+1} M(1-\delta)^{n+1}.$$ The second is just $$\int_{1-\delta}^1 x^n f(1)\,dx=\frac{f(1)}{n+1}\cdot(1-(1-\delta)^{n+1}).$$ The last can be estimated as $$\left|\int_{1-\delta}^1 x^n (f(x)-f(1))\,dx\right|\le \int_{1-\delta}^1 \left|x^n (f(x)-f(1))\right|\,dx\\\le\epsilon\int_{1-\delta}^1x^n=\frac\epsilon{n+1}\cdot(1-(1-\delta)^{n+1}).$$ As $n\to\infty$, we have $(1-\delta)^{n+1}\to 0$. If you stick these results together, you'll find that $$\lim_{n\to\infty}n\int_0^1x^nf(x)\,dx=f(1).$$

Hint: Try $f(x)=x^k$, then a polynomial, and then a general continuous function.

This answer is almost identical to the answer of user @Daniel Watkins. The difference is that it will be shorter by using his proof for functions $$C^{1}$$ on the interval $$[0,1]$$. Fixed any continuous function $$f:[0,1]\to \mathbb{R}$$, by Weierstrass' approximation theorem there exists a sequence of polynomials $$p_{k}:[0,1]\to \mathbb{R}$$ that converges uniformally to $$f:[0,1]\to \mathbb{R}$$. In particular $$\lim_{k\to\infty}p_{k}(1)=f(1)$$. From what has already been worked out for $$C^{1}$$ functions in this question ( polynomials are $$C^{1}$$ functions ) we have \begin{align} \lim_{n\to \infty}\left(n\int_0^1f(x)x^n\,dx\right) =& \lim_{n\to \infty}\left(\frac{n}{n+1}f(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right), \\ =& \lim_{n\to \infty}\left(\frac{n}{n+1}\left(\lim_{k\to \infty}p_{k}(1)\right) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right), \\ =& \lim_{n\to \infty}\;\lim_{k\to \infty}\;\frac{n}{n+1}\left(p_{k}(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right), \\ =& \lim_{k\to \infty}\;\lim_{n\to \infty}\;\frac{n}{n+1}\left(p_{k}(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\right), \\ =& \lim_{k\to \infty}p_{k}(1) \\ =& f(1) \end{align}

In this answer we apply the useful inequality

$$\frac{t}{1+t}\leq \ln( 1+t ) \leq t \qquad \mbox{ for all } \qquad t\geq -1$$

In particular we have for $$t=(-1/n)$$ and $$n>1$$ $$\color{\red}{\frac{-1}{n-1}=}\frac{(-1/n)}{1+(-1/n)}\leq \ln( 1+(-1/n) ) \leq (-1/n)$$ This inequality implies $$e^{\frac{-1}{n-1}}\leq e^{\ln(1-1/n)}\leq e^{-1/n}$$ and $$e^{\frac{-\sqrt{n}}{n-1}}\leq e^{\sqrt{n}\cdot \ln(1-1/n)}\leq e^{-\sqrt{n}/n} \color{\red}{=e^{-1/\sqrt{n}}}$$ Now, note that $$n\int_{0}^{1}f(x)\cdot x^{n} \, \mathrm{d} x = n\int_{0}^{\frac{1}{n}}f(x)\cdot x^{n} \, \mathrm{d} x + n\int_{1/n}^{\left(1-\frac{1}{n}\right)}f(x)\cdot x^{n} \, \mathrm{d} x + n\int_{\left(1-\frac{1}{n}\right)}^{1}f(x)\cdot x^{n} \, \mathrm{d} x$$ There is $$M>0$$ such that $$|f(x)| for all $$x\in [0,1]$$ and we have \begin{align} \left| n\int_{1/n}^{\left(1-\frac{1}{n}\right)}f(x)\cdot x^{n} \, \mathrm{d} x \right| \leq & n\int_{1/n}^{\left(1-\frac{1}{n}\right)}|f(x)|\cdot x^{n} \, \mathrm{d} x \\ \leq & n\int_{1/n}^{\left(1-\frac{1}{n}\right)}|f(x)|\cdot x^{\sqrt{n}} \, \mathrm{d} x \\ \leq & n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot x^{\sqrt{n}} \, \mathrm{d} x \\ = & n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{\sqrt{n}\cdot ln x} \, \mathrm{d} x \\ \leq & n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{\sqrt{n}\cdot ln \left(1-\frac{1}{n}\right)} \, \mathrm{d} x \\ \leq & n\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{\sqrt{n}\cdot ln \left(1-\frac{1}{n}\right)} \, \mathrm{d} x \\ \leq & (\sqrt{n})^{2}\int_{1/n}^{\left(1-\frac{1}{n}\right)}M \cdot e^{-1/\sqrt{n}} \, \mathrm{d} x \\ =& (\sqrt{n})^{2}M \cdot e^{-1/\sqrt{n}}\left[\left(1-\frac{1}{n}\right) -\frac{1}{n} \right] \overset{n\to \infty}{\longrightarrow}0 \end{align}

It is easy to verify that $$\lim_{n\to\infty} n\int_{0}^{\frac{1}{n}}f(x)\cdot x^{n} \, \mathrm{d} x=0$$ Finally, by Dominated Convergence Theorem and Lebesgue's Differentiation Theorem $$\lim_{n\to \infty} \left( n\int_{\left(1-\frac{1}{n}\right)}^{1}f(x)\cdot x^{n} \, \mathrm{d} x \right) = \lim_{n\to \infty} \left( \cfrac{\int_{\left(1-\frac{1}{n}\right)}^{1}f(x)\cdot x^{n} \, \mathrm{d} x}{\frac{1}{n}} \right) = f(1)$$