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

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?

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As for your attempted proof of the first part, no such (constant) $M < 1$ exists. – Martin Wanvik Apr 6 '12 at 21:38

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 $n\to\infty$.

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To make it more precise: and so $0 \le \liminf |blah_n| \le \limsup |blah_n| \le \varepsilon$ for all $\varepsilon \gt 0$, and thus $\lim blah_n = 0$. – Aryabhata Apr 7 '12 at 16:39

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

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

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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 $f$.

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Please elaborate... – Aryabhata Apr 7 '12 at 16:43
If $f(x) = \sum_{k=0}^m a_kx^k$ then $n\int_0^1 x^nf(x)dx =\sum_{k=0}^m a_k\cdot(n/n+k+1)$ which converges to $\sum_{k=1}^m a_k = f(1)$ as $n\to\infty$. – John Dawkins Apr 7 '12 at 17:23
If $f(x) = \sum_{k=0}^m a_kx^k$ then $n\int_0^1 x^nf(x)dx =\sum_{k=0}^m a_k\cdot(n/(n+k+1))$ which converges to $\sum_{k=1}^m a_k = f(1)$ as $n\to\infty$. For general continuous $f$ and a given $\epsilon>0$, there is a polynomial $g$ such that $|f(x)-g(x)|<\epsilon$. The integral $n\int_0^1 x^nf(x)dx$ differs from $n\int_0^1 x^n g(x)dx$ by no more than $\epsilon$ and the latter integral congerges to $g(1)$ (which differs from $f(1)$ by at most $\epsilon$) by the preceding discussion. This shows that $\lim_n n\int_0^1 x^n f(x) dx$ exists and is equal to $f(1)$. – John Dawkins Apr 7 '12 at 17:32
Thanks! Looks right (and my comment to kiwi's answer applies here too I think). I actually meant to say: can you please edit your answer? – Aryabhata Apr 7 '12 at 18:46

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.

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$\lim\limits_{n\to\infty}\left(1-\frac1{n^2}\right)^n=1$, so the first bound blows up. You have some errors in the second calculation. – Brian M. Scott Apr 6 '12 at 22:29

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 $n \to + \infty$.

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A different approach: First lets show that the equality holds 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 $a_0+\cdots+a_m=p(1).$

Now 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 $$\Big|\int_0^1nx^nf(x)dx-f(1)\Big|=\Big|\int_0^1nx^nf(x)dx-f(1)+p(1)-p(1)-\int_0^1nx^np(x)dx+\int_0^1nx^np(x)dx\Big|\leq \Big|\int_0^1nx^n[f(x)-p(x)]dx\Big|+\Big|\int_0^1nx^np(x)dx-p(1)\Big|+|p(1)-f(1)|\leq \int_0^1n\epsilon x^ndx+\epsilon\leq \frac{n}{n+1}\epsilon+\epsilon<2\epsilon$$

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