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

Solution 1:

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.

Solution 2:

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 $f:[0,1]\to \mathbb{R}$.

Solution 3:

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 $1$.

Solution 4:

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 $$ \lim_{n\to\infty}n\int_0^1x^nf(x)\,\mathrm{d}x=f(1)\tag{5} $$

Solution 5:

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: $$\lim_n n\int_0^1 x^n f(x)dx=f(1).$$