Prove that $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx$ is equal to $f(1)$.

$\mathbf{Question}:$ Let $f$ be a continuous function on $[0,1]$. Then prove that the limit $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx$ is equal to $f(1)$.

$\mathbf{Attempt}$: First, we try to show that the sequence of functions $\{nx^nf(x)\}_{x\in [0,1]}$ is uniformly convergent to $0$ on the restricted domain $[0,1-\epsilon]$, $0<\epsilon<1$.

Let $\sup_{x\in[0,1]}f(x)= \mathcal{M}$. Then $|{nx^nf(x)}|\leq n(1-\epsilon)^n|\mathcal{M}|$ for any $x$ and $n(1-\epsilon)^n \to 0$ as $n\to \infty$.

Thereby, $\lim_{n \to \infty}\int_0^{1-\epsilon}nx^nf(x)dx=\int_0^{1-\epsilon}{\lim_{n \to \infty} }nx^nf(x)dx=0$

Now, denote $\sup_{x\in [1-\epsilon,1]} f(x)=M(\epsilon)$ and $\inf_{x\in[1-\epsilon,1]}f(x)=m(\epsilon)$. $\int_{1-\epsilon}^1nx^n\ m(\epsilon)dx\leq\int_{1-\epsilon}^{1}nx^nf(x)dx\leq \int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx$.

Now,

$\lim_{n \to \infty}\int_{1-\epsilon}^1 nx^n\ M(\epsilon)dx =\displaystyle \lim_{n\to\infty}\bigg[ \frac{nx^{n+1}}{n+1}M(\epsilon)\bigg]_{1-\epsilon}^1=M(\epsilon)$ [Similarly the other one is $m(\epsilon)$]

Clearly, as $\epsilon \to 0 , \ \ M(\epsilon), m(\epsilon) \to f(1)$, so $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx=f(1)$.

Is the procedure correct? Kindly verify.

Solution 1:

I will prove a more general statement : Let $f:[0,1]\to \mathbb{R}$ be a Riemann integrable function. If $f$ is continuous at $x=1$, then prove that $\lim\limits_{n \to \infty} \int\limits_0^1{nx^nf(x)}dx=f(1)$.
Firstly, we will prove that $\lim\limits_{n\to \infty}\int\limits_0^1x^nf(x) dx=0$(*).
Since $f$ is Riemann integrable, it is bounded, so $\exists M \ge 0$ such that $|f(x)|\le M$,$\forall x \in [0,1]$.
We have that $\bigg|\int\limits_0^1x^nf(x) dx\bigg|\le \int\limits_0^1|x^nf(x)dx|\le M \int\limits_0^1x^ndx=\frac{M}{n+1}\to0$, so $\lim\limits_{n\to \infty}\int\limits_0^1x^nf(x) dx=0$.
Now, let $\epsilon >0$. Since $f$ is continuous at $x=1$, we deduce that $\exists a\in (0,1)$ such that $|f(x)-f(1)|<\frac{\epsilon}{2},\forall x\in(a,1)$.
We have that $$\bigg|(n+1)\int\limits_0^1 x^nf(x)dx -f(1)\bigg|=(n+1)\bigg|\int\limits_0^1(x^nf(x)-x^nf(1))dx\bigg|\le (n+1)\int\limits_0^1x^n|f(x)--f(1)|dx=(n+1)\int\limits_0^ax^n|f(x)-f(1)|dx+(n+1)\int\limits_a^1x^n|f(x)-f(1)|dx\le (n+1)\cdot 2M \cdot $$ $$\cdot \int\limits_0^1 x^n dx+(n+1)\frac{\epsilon}{2}\cdot\int\limits_a^1 x^n dx=2M\cdot a^{n+1}+\frac{\epsilon}{2}(1-a^{n+1})<2M\cdot a^n+\frac{\epsilon}{2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$ because for $a\in (0,1)$ we have that $\lim\limits_{n\to \infty}2M\cdot a^n=0$, so $\exists n_0 \in \mathbb{N}$ such that $2M\cdot a^n <\frac{\epsilon}{2},\forall n\ge n_0$.
Thus,$\bigg|(n+1)\int\limits_0^1 x^nf(x)dx -f(1)\bigg|<\epsilon$,$\forall n\ge n_0$, so $\lim\limits_{n\to\infty}(n+1)\int\limits_{0}^{1} x^n f(x)dx=f(1)$.
By using $(*)$ we get that $\lim\limits_{n \to \infty} \int\limits_0^1{nx^nf(x)}dx=f(1)$ as desired.

Solution 2:

Another way is to change the variable $u=x^n$: $$\int_0^1nx^nf(x)dx=\int_0^1 u^{\frac 1 n}f(u^{\frac 1 n})du$$ The integrand converges pointwise to $f(1)$ for $u\in (0,1]$ and is obviously bounded by a constant since $f$ is continuous.

The result is obtained by applying the dominated convergence theorem.