limit of integral $n\int_{0}^{1} x^n f(x) \text{d}x$ as $n\rightarrow \infty$

Question

I am trying to solve the following problem at the level of a senior undergrad analysis level. So, the problem is as follows: We are given a function $f$ which is continuous on the interval $\left [ 0,1 \right ]$, and the question is to find the limit: $$\lim_{n\rightarrow \infty}\int_{0}^{1}x^{n}f(x)dx\;.$$ The second part of the problem is to deduce the following limit: $$\lim_{n\rightarrow \infty}n\int_{0}^{1}x^{n}f(x)dx\;.$$

For the first part: I just did the following: For every $0\leq x< 1$: $x\leq M$, where $0< M< 1$. Then: $$\int_{0}^{1}x^{n}f(x)dx\leq M^{n}\int_{0}^{1}f(x)dx\;.$$ Then: $$\lim_{n\rightarrow \infty }\int_{0}^{1}x^{n}f(x)dx\leq \lim_{n\rightarrow \infty }M^{n}\int_{0}^{1}f(x)dx= 0.\int_{0}^{1}f(x)dx=0\;,$$ so $$\lim_{n\rightarrow \infty }\int_{0}^{1}x^{n}f(x)dx=\lim_{n\rightarrow \infty }f(1)\int_{0}^{1}1dx=f(1)\;.$$ Does that make sense? If not, please show me the correct one.

As for the second part, I have no idea what to do. Any help?

Answer 1

For the first part, use the fact that a continuous function on $[0,1]$ is bounded, say $|f(x)|\le M$. Then $$ \Bigg|\int_0^1 x^nf(x)\,dx\Bigg|\le\int_0^1 Mx^n\,dx =\frac{M}{n+1}\to 0 \quad\text{as}\quad n\to\infty. $$

For the second part $\dots$ $$ n\int_0^1 x^nf(x)\,dx = n\int_0^1 x^n\Big(f(x)-f(1)\Big)\,dx + \frac{n}{n+1}f(1). $$ The last term tends to $f(1)$ as $n\to\infty$. To show that the first term on the right tends to zero, suppose $\varepsilon>0$. Choose $a<1$ so that $|f(x)-f(1)|<\varepsilon$ for $a<x<1$. Then $$ \Bigg|\,n\int_0^1 x^n\Big(f(x)-f(1)\Big)\,dx\Bigg| \le \,n\int_0^a x^n\Big|f(x)-f(1)\Big|\,dx+ \,n\int_a^1 x^n\Big|f(x)-f(1)\Big|\,dx\\ \le 2M\frac{n}{n+1}a^n + \frac{n}{n+1}\varepsilon $$ and this tends to $\varepsilon$ as $n\to\infty$.

Answer 2

As to the second question: The integral can easily be computed explicitly when $f$ is a polynomial, and the limit then calculated. Because the integral $n\int_0^1 x^n dx$ is bounded above by $1$, the Weierstrass approximation theorem can be applied to deal with general continuous $f$.

Answer 3

A different approach: First let’s show the equality for polynomials; Let $p=a_0+\cdots+a_mx^m$ then $$\int_0^1nx^np(x)dx=n(\frac{a_0}{n+1}+\cdot+\frac{a_m}{n+m+1})$$ So as $n\to\infty$ the RHS tends to $a_0+\cdots+a_m=p(1).$ If $f$ is a continuous function in the unit interval for the given $\epsilon>0$ Find a polynomial p so that $\|p-f\|<\epsilon$ Then $$\Big|\int_0^1nx^nf(x)dx-f(1)\Big|=\Big|\int_0^1nx^nf(x)dx-f(1)+p(1)-p(1)-\int_0^1nx^np(x)dx+\int_0^1nx^np(x)dx\Big|\leq \Big|\int_0^1nx^n[f(x)-p(x)]dx\Big|+\Big|\int_0^1nx^np(x)dx-p(1)\Big|+|p(1)-f(1)|\leq \int_0^1n\epsilon x^ndx+\epsilon\leq \frac{n}{n+1}\epsilon+\epsilon<2\epsilon$$

Answer 4

For second part, we can find limit using little functional analysis. Let denote with $A_n:C[0,1] \to \mathbb{C}$ linear operator $A_n(f)=n\displaystyle\int_0^1 x^n f(x)\, dx$. We see that

$$\| A_n f \|_{\mathbb{C}} = |A_n (f)|=\left\lvert n\int_0^1 x^n f(x)\, dx\right\rvert \leqslant n \|f\|_{\infty} \int_0^1 x^n \, dx=\frac{n}{n+1}\| f\|_{\infty} < \|f\|_{\infty},$$

so $\sup\limits_{n \in \mathbb{N}} \|A_n \| < +\infty$.

Now, we are going to check our operator on fundamental set $\{x \mapsto x^k \mid k \in \mathbb{N}_0\}$ (because $\overline{\operatorname{span}\{x \mapsto x^k \mid k \in \mathbb{N}_0\}}=C[0,1]$):

$$\lim_{n\to \infty}A_n(x^k)=\lim_{n\to \infty} n \int_0^1 x^n x^k\, dx=\lim_{n \to \infty} \frac{n}{n+k+1}=1=x^k(1).$$

Banach–Steinhaus theorem gives us that limit exist and

$$\lim_{n \to \infty} A_n(f)=f(1).$$

Answer 5

This solution is not mine

I asked one of my friends about this problem 2 months ago and this was his proof.

$$\int_0^1 nx^nf(x)dx=\int_0^1x^{\frac{1}{n}} f(x^\frac{1}{n})dx$$

Since $f$ is continuous at $[0,1]$ then $\exists M,m \in \mathbb{R}$ such that $$m\le f(x^\frac{1}{n})\le M \ ,\forall x \in [0,1]$$ $$mx^{\frac{1}{n}}\le x^{\frac{1}{n}}f(x^\frac{1}{n})\le Mx^{\frac{1}{n}}$$ $$m\le \left(1+\frac{1}{n} \right)\int_0^1x^{\frac{1}{n}}f(x^\frac{1}{n})dx\le M$$ By IVT $\exists t \in (0,1)$ such that $ \left(1+\frac{1}{n} \right)\int_0^1x^{\frac{1}{n}}f(x^\frac{1}{n})dx = f(t^{\frac{1}{n}})$

$$\int_0^1x^{\frac{1}{n}}f(x^\frac{1}{n})dx=\left(\frac{n}{n+1}\right) f(t^{\frac{1}{n}})$$

$$\lim_{n \to \infty}\int_0^1 nx^nf(x)dx=\lim_{n \to \infty}\int_0^1x^{\frac{1}{n}} f(x^\frac{1}{n})dx$$

$$=\lim_{n \to \infty} \bigg( \frac{n}{n+1}\bigg) f(t^{ \frac{1}{n} })=f\bigg( \lim_{n \to \infty} t^{\frac{1}{n}} \bigg)= f(1)$$

Answer 6

In his answer, John Dawkins approximated $f$ by polynomials, but in fact it is sufficient to approximated $f$ by $C^1$-functions (for that, use uniform continuity to approximate $f$ by piecewise constant functions, and then use bump functions). Indeed, if we suppose without loss of generality that $f$ is $C^1$, we have thanks to an integration by parts:

$$n \int_0^1 t^nf(t)dt= \int_0^1 (n+1)t^nf(t)dt - \int_0^1 t^nf(t)dt = f(1) - \int_0^1 t^{n+1}f'(t)dt - \int_0^1 t^nf(t)dt.$$

Now apply the first part to $f$ and $f'$ to show that the last two integrals vanish as $n \to + \infty$.

Answer 7

For the second part, fix $M = 1 - 1/n^2$, and look at the integrals $n\int_0^M x^n f(x)dx + n\int_M^1 x^nf(x)dx$. The left one is bounded above by

$$ n\int_0^M x^nf(x)dx \leq nM^n\int_0^M f(x)dx \leq n(1 - \frac{1}{n^2})^n \int_0^M f(x)dx = Cn(1 - \frac{1}{n^2})^n $$

The second one is bounded by

$$ n\int_M^1 x^n f(x)dx \leq n\int_1^M f(x)dx = n(1-M) = \frac{1}{n} \rightarrow 0. $$

What does the top integral tend to as $n \rightarrow \infty$? It looks like it should converge to 0, but I'm not sure.